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## Full Wave Rectifier

Full Wave Rectifier (FWR) contains two diodes $D_{1}$ and $D_{2}$.

FWR converts a.c voltage into pulsating DC in two-half cycles of the applied input signal.

Here  we use a  Transformer, whose secondary winding has been split equally into two half waves with a common center tapped connection ‘c’.

This configuration results in each diode conducting in turn when it’s anode terminal is positive with respect to Center point ‘c’ of  the Transformer.

Working of Full Wave Rectifier:-

During positive half cycle of applied i/p signal

• point ‘P’ is more positive w.r.to ‘c’.
• point ‘Q’ is more negative w.r.to ‘c’.

i.e, Diode $D_{1}$ is Forward Biased and $D_{2}$ is Reverse Biased , under this condition the equivalent circuit is as shown below

$\therefore&space;V_{o}&space;\approx&space;V_{i}&space;=i_{L}R_{L}$, when there is no diode resistance.

Similarly the conditions of diodes will be reversed for the negative half cycle of i/p signal.

• point ‘P’ is negative w.r.to ‘c’.
• point ‘Q’ is positive w.r.to ‘c’.

i.e, Diode $D_{1}$ is Reverse Biased and $D_{2}$ is Forward Biased , under this condition the equivalent circuit is  and output voltage is $V_{o}&space;\approx&space;i_{L}R_{L}$.

the i/p and o/p wave forms are as shown below

FWR is advantageous compared to HWR in terms of its efficiency and ripple factor.

Ripple Factor ($\Gamma$):-

$\Gamma&space;=&space;\frac{V_({ac})rms}{V_{dc}}&space;=&space;\sqrt{\frac{(V_{rms})^2}{V_{dc}}-1}$

to find out $V_{rms}$ and $V_{dc}$ of output signal

$\therefore&space;V_{rms}&space;=&space;\sqrt{\frac{1}{\pi&space;}\int_{0}^{\pi&space;}V_{m}^{2}&space;(\sin&space;^{2}\omega&space;t&space;)&space;d\omega&space;t}$                     ,     $\therefore&space;V_{dc}&space;=&space;\frac{1}{\pi&space;}&space;\int_{0}^{\pi&space;}V_{m}&space;(\sin&space;\omega&space;t&space;)&space;d\omega&space;t$

$V_{rms}&space;=&space;\sqrt{\frac{1}{\pi&space;}\int_{0}^{\pi&space;}V_{m}^{2}&space;(\frac{1-\cos&space;2\omega&space;t}{2}&space;)&space;d\omega&space;t}$          ,               $V_{dc}&space;=&space;\frac{-V_{m}}{\pi&space;}&space;[-2&space;]$

$V_{rms}&space;=&space;\sqrt{\frac{V_{m}^{2}}{\pi&space;}\int_{0}^{\pi&space;}&space;}(\frac{1}{2}\pi&space;)$                                           ,              $V_{dc}&space;=&space;\frac{2V_{m}}{\pi&space;}$

$V_{rms}&space;=&space;\frac{V_{m}}{\sqrt{2}}$.

$I_{rms}&space;=&space;\frac{V_{rms}}{R_{L}}=\frac{V_{m}}{\sqrt{2}R_{L}}&space;=&space;\frac{I_{m}}{\sqrt{2}}$      and  $I_{dc}&space;=&space;\frac{2I_{m}}{\pi&space;}$.

now the ripple factor results to be  $\Gamma&space;=&space;\sqrt{\frac{(\frac{V_{m}}{\sqrt{2}})^{2}}{(\frac{2V_{m}}{\pi&space;})^{2}}-1}$

$\Gamma&space;=&space;\sqrt{\frac{V_{m}^{2}\pi&space;^{2}}{8V_{m}^{2}}-1}$

$\Gamma&space;=&space;\sqrt{\frac{\pi&space;^{2}}{8}-1}$

$\Gamma&space;=&space;0.482$

(No Ratings Yet)

## Surface impedance

At high frequencies, the current is almost confined to a very thin sheet at the surface of the conductor which is used in many applications.

The  surface impedance may be defined as the ratio of the tangential component of the electric field $\overrightarrow{E_{tan}}$ at the surface of the conductor to the current density (linear) $\overrightarrow{J_{s}}$ which flows due to this electric field.

given as $Z_{s}$ (or) $\eta&space;_{s}&space;=&space;\frac{\overrightarrow{E_{tan}}}{\overrightarrow{J_{s}}}$.

$\overrightarrow{E_{tan}}$   is the Electric field strength parallel to and at the surface of the conductor.

and $\overrightarrow{J}$  is the total linear current density which flows due to $\overrightarrow{E_{tan}}$.

The $\overrightarrow{J_{s}}$ represents the total conduction per meter width flowing in this sheet.

Let us consider a conductor of the type plate, is placed at the surface y=0 and the current distribution in the y-direction is given by

Assume that the depth of penetration ($\delta$) is very much less compared with the thickness of the conductor.

$J_{s}=&space;\int_{0}^{\infty&space;}&space;\overrightarrow{J}.\overrightarrow{dy}$

$J_{s}=&space;\int_{0}^{\infty&space;}&space;J_{o}e^{-\gamma&space;y}dy$

$J_{s}=&space;J_{o}(e^{-\gamma&space;y})_{0}^{\infty&space;}$

$J_{s}=&space;\frac{J_{o}}{\gamma&space;}$

from ohm’s law $\overrightarrow{J_{o}}&space;=&space;\sigma&space;\overrightarrow{E_{tan}}$

$E&space;=&space;\frac{J_{o}}{\sigma&space;}$ .

then  $\eta&space;_{s}&space;=&space;\frac{\gamma&space;}{\sigma&space;}$ .

$Z_{s}$  (or)  $\eta&space;_{s}&space;=&space;\frac{\gamma&space;}{\sigma&space;}$ .

we know that $\gamma&space;=&space;\sqrt{j\omega&space;\mu&space;(\sigma&space;+j\omega&space;\epsilon&space;)}$

for good conductors $\sigma&space;>&space;>&space;\omega&space;\epsilon$ .

then $\gamma&space;\approx&space;\sqrt{j\omega&space;\mu&space;\sigma&space;}$

$\eta&space;_{s}&space;=&space;\frac{\gamma&space;}{\sigma&space;}&space;=&space;\sqrt{\frac{j\omega&space;\mu&space;}{\sigma&space;}}$ .

therefore the surface impedance of a plane good conductor which is very much thicker than the skin depth is equal to the characteristic impedance of the conductor.

This impedance is also known s input impedance of the conductor when viewed as transmission line conducting energy into the interior of metal.

when the thickness of the plane conductor is not greater compared to the depth of penetration , reflection of wave occurs at the back surface of the conductor.

(No Ratings Yet)

## oblique incidence

when a uniform plane wave  is incident obliquely (making an angle $\theta&space;_{i}$ other than $90^{o}$) to the boundary between the two media then it is known as oblique incidence.

Now consider the situation that is more general case  that is the oblique incidence.

In this case the EM wave (incident wave) not strikes normally the boundary. i.e,  the incident wave is not propagating  along any standard axes (like x,y and z).

Therefore EM wave is moving in a random direction then the general form is    $\overrightarrow{E}&space;=&space;E_{o}&space;\cos&space;(\omega&space;t-\overrightarrow{k}.\overrightarrow{r})$

it is also in the form $\overrightarrow{E}&space;=&space;E_{o}&space;\cos&space;(\overrightarrow{k}.\overrightarrow{r}-\omega&space;t)$.

then $\overrightarrow{k}&space;=&space;k_{x}\overrightarrow{a_{x}}+k_{y}\overrightarrow{a_{y}}+k_{z}\overrightarrow{a_{z}}$ is called the wave number vector (or) the propagation vector.

and $\overrightarrow{r}&space;=&space;x\overrightarrow{a_{x}}+y\overrightarrow{a_{y}}+z\overrightarrow{a_{z}}$ is called the position vector (from origin to any point on the plane of incidence) , then the magnitude of $\overrightarrow{k}$ is related to $\omega$ according to the dispersion.

$k^{2}&space;=&space;k_{x}^{2}+k_{y}^{2}+k_{z}^{2}&space;=&space;\omega&space;^{2}\mu&space;\epsilon$

$\overrightarrow{k}&space;X&space;\overrightarrow{E}&space;=&space;\omega&space;\mu&space;\overrightarrow{H}$.

$\overrightarrow{k}&space;X&space;\overrightarrow{H}&space;=&space;-\omega&space;\epsilon&space;\overrightarrow{E}$.

$\overrightarrow{k}&space;.&space;\overrightarrow{H}&space;=0$.

$\overrightarrow{k}&space;.&space;\overrightarrow{E}&space;=0$.

i.    $\overrightarrow{E},\overrightarrow{H}$ and $\overrightarrow{k}$ are mutually orthogonal.

ii.  $\overrightarrow{E}$  and  $\overrightarrow{H}$ lie on the plane $\overrightarrow{k}&space;.&space;\overrightarrow{r}&space;=&space;k_{x}x+k_{y}y+k_{z}z=constant$.

then the $\overrightarrow{H}$ field corresponding to $\overrightarrow{E}$ field is $\overrightarrow{H}&space;=&space;\frac{1}{\omega&space;\mu&space;}&space;(\overrightarrow{k}&space;X&space;\overrightarrow{E})&space;=&space;\frac{\overrightarrow{a_{k}}&space;X&space;\overrightarrow{E}}{\eta&space;}$.

Now choose oblique incidence of a uniform plane wave at a plane boundary.

the plane defined by the propagation vector $\overrightarrow{k}$ and a unit normal vector $\overrightarrow{a_{n}}$ to the boundary is called the plane of incidence.

the angle $\theta&space;_{i}$ between $\overrightarrow{k}$ and $\overrightarrow{a_{n}}$ is the angle of incidence.

both the incident and reflected waves are in medium 1 while the transmitted wave is in medium 2 .

Now,

$\overrightarrow{E_{i}}&space;=E_{i}\cos&space;(k_{ix}x+k_{iy}y+k_{iz}z-\omega&space;_{i}t)$

$\overrightarrow{E_{r}}&space;=E_{i}\cos&space;(k_{rx}x+k_{ry}y+k_{rz}z-\omega&space;_{r}t)$

$\overrightarrow{E_{t}}&space;=E_{t}\cos&space;(k_{tx}x+k_{ty}y+k_{tz}z-\omega&space;_{t}t)$.

the wave propagates

1.     $\omega&space;_{i}=\omega&space;_{r}=\omega&space;_{t}=\omega$.
2.     $k_{ix}&space;=&space;k_{rx}=k_{tx}=k_{x}$.
3.    $k_{iy}&space;=&space;k_{ry}=k_{ty}=k_{y}$.

(1) indicates that all waves are propagating with same frequency. (2) and (3) shows that the tangential components of propagation vectors be continuous.

$k_{i}&space;\sin&space;\theta&space;_{i}=k_{r}&space;\sin&space;\theta&space;_{r}$  implies  $k_{i}&space;=&space;k_{r}&space;=\beta&space;_{1}&space;=\omega&space;\sqrt{\mu&space;_{1}\epsilon&space;_{1}}$    since $\theta&space;_{r}=\theta&space;_{i}$.

$k_{i}&space;\sin&space;\theta&space;_{i}=k_{t}&space;\sin&space;\theta&space;_{t}$   implies $k_{t}&space;=\beta&space;_{2}&space;=\omega&space;\sqrt{\mu&space;_{2}\epsilon&space;_{2}}$.

$\frac{\sin&space;\theta&space;_{t}}{sin&space;\theta&space;_{i}}=\sqrt{\frac{\mu&space;_{1}\epsilon&space;_{1}}{\mu&space;_{2}\epsilon&space;_{2}}}$

now velocity $u=\frac{\omega&space;}{k}$.

then from Snell’s law       $r_{1}\sin&space;\theta&space;_{i}&space;=&space;r_{2}\sin&space;\theta&space;_{t}$,  where $r_{1}$ and $r_{2}$  are the refractive indices of the two media.

(No Ratings Yet)

## Poynting theorem

Introduction:-

Poynting theorem is used to get an expression for propagation of energy in a medium.

It gives the relation between energy stored in a time-varying magnetic field and the energy stored in time-varying electric field and the instantaneous power flow out of a given region.

EM waves propagates through space from source to destination. In order to find out power in a uniform plane wave it is necessary to develop a power theorem for the EM field known as poynting theorem.

The direction of power flow is perpendicular to $\vec{E}$ and $\overrightarrow{H}$ in the direction of plane containing $\overrightarrow{E}$ and $\overrightarrow{H}$.

i.e, it gives the direction of propagation .

$\overrightarrow{P}&space;=&space;\overrightarrow{E}X\overrightarrow{H}$  Watts/m2  (or)   VA/m2.

Proof:-

from Maxwell’s  equations $\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J}+\overrightarrow{J_{d}}$

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

$\overrightarrow{J}=\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}-\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

the above equation has units of the form current density Amp/m2. When it gets multiplied by $\overrightarrow{E}$ V/m. The total units  will  have of the form power per unit volume.

$\overrightarrow{J}\rightarrow$ Amp/m2$\overrightarrow{E}\rightarrow$ Volts/m.

$EJ\rightarrow$ Amp. Volt/m3 $\rightarrow$ Watts/m3 $\rightarrow$ Power/volume.

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

$\overrightarrow{E}.(\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

$\overrightarrow{E}.(\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial&space;(\epsilon&space;\overrightarrow{E})}{\partial&space;t}$

by using the vector identity

$\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{H}.(\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{E}&space;)-\overrightarrow{E}.(\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;)$

then

$\overrightarrow{H}.(\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{E}&space;)-\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial&space;(\epsilon&space;\overrightarrow{E})}{\partial&space;t}$.

from the equation of Maxwell’s $\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}=-\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}&space;=&space;-\frac{\partial&space;(\mu&space;\overrightarrow{&space;H})}{\partial&space;t}$

$\overrightarrow{H}.(-\frac{\partial&space;(\mu&space;\overrightarrow{&space;H})}{\partial&space;t}&space;)-\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial&space;(\epsilon&space;\overrightarrow{E})}{\partial&space;t}$

$\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})=-\overrightarrow{E}.\overrightarrow{J}-\overrightarrow{E}.\frac{\partial&space;(\epsilon&space;\overrightarrow{E})}{\partial&space;t}-\overrightarrow{H}.(\frac{\partial&space;(\mu&space;\overrightarrow{&space;H})}{\partial&space;t}&space;)------EQN(I)$

by using the vector identity  $\frac{\partial&space;(\overrightarrow{A}.\overrightarrow{B})}{\partial&space;t}=\overrightarrow{A}.\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}+\overrightarrow{B}.\frac{\partial&space;\overrightarrow{A}}{\partial&space;t}$

if $\overrightarrow{A}&space;=&space;\overrightarrow{B}$    $\Rightarrow&space;\frac{\partial&space;(\overrightarrow{A}.\overrightarrow{A})}{\partial&space;t}=2&space;\overrightarrow{A}.\frac{\partial&space;\overrightarrow{A}}{\partial&space;t}$

$\overrightarrow{A}.\frac{\partial&space;\overrightarrow{A}}{\partial&space;t}=\frac{1}{2}\frac{\partial&space;A^{2}}{\partial&space;t}$

from EQN(I)   ,  $\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})=-\overrightarrow{E}.\sigma\overrightarrow{E&space;}-\frac{1}{2}\epsilon\frac{\partial&space;E^{2}}{\partial&space;t}-\frac{1}{2}\mu&space;\frac{\partial&space;H^{2}}{\partial&space;t}$

$\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})=-\sigma&space;E^{2}-\frac{1}{2}\frac{\partial&space;}{\partial&space;t}(\epsilon&space;E^{2}+\mu&space;H^{2})$

by integrating the above equation by over  a volume

$\int_{v}\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})&space;dv=-\int_{v}\sigma&space;E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial&space;}{\partial&space;t}(\epsilon&space;E^{2}+\mu&space;H^{2})dv$

by converting the volume integral to surface integral

$\oint_{s}(\overrightarrow{E}&space;X&space;\overrightarrow{H}).&space;\overrightarrow{ds}=-\int_{v}\sigma&space;E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial&space;}{\partial&space;t}(\epsilon&space;E^{2}+\mu&space;H^{2})dv$.

the above equation gives the statement of Poynting theorem.

Poynting theorem:-

It states that the net power flowing out of a given volume is equal to the time rate of decrease in the energy stored with in that volume V and the ohmic losses.

## Lag compensator

Lag compensator:-

A Lag compensator has a Transfer function of the form $G(s)&space;=&space;\frac{s+z_{c}}{s+p_{c}}------EQN(I)$

$G(s)&space;=&space;\frac{s+\frac{1}{\tau&space;}}{s+\frac{1}{\beta&space;\tau&space;}}$,    where $\beta&space;=\frac{z_{c}}{p_{c}}>&space;1$     and $\tau&space;>&space;0$

Pole-Zero Plot of Lag compensator:-

i.e, the pole is located to the right of the zero.

Realization of Lag compensator as Electrical Network:-

The lag compensator can be realized by an electrical Network.

Assume impedance of source is zero  $[Z_{s}&space;=0]$ and output load impedance to be infinite  .

The transfer function is $\frac{E_{o}(s)}{E_{i}(s)}&space;=&space;\frac{(R_{2}+\frac{1}{Cs})}{R_{1}+(R_{2}+&space;\frac{1}{Cs})}$

$\frac{E_{o}(s)}{E_{i}(s)}&space;=&space;\frac{R_{2}Cs+1}{(R_{1}+R_{2})Cs+1}$

after simplification

$\frac{E_{o}(s)}{E_{i}(s)}&space;=\frac{R_{2}}{(R_{1}+R_{2})}(\frac{s+\frac{1}{R_{2}C}}{s+\frac{1}{(R_{1}+R_{2})C}})$

after comparing the above equation with the transfer function of lag compensator has a zero at  $Z_{c}&space;=\frac{1}{R_{2}C}$  and has a pole at $p_{c}=\frac{1}{(R_{1}+R_{2})C}=\frac{1}{\beta&space;\tau&space;}$ .

from the pole $\beta&space;=\frac{(R_{1}+R_{2})}{R_{2}}$ and  $\tau&space;=R_{2}C$.

therefore  the transfer function  has a zero at $-\frac{1}{\tau&space;}$   and a pole at $-\frac{1}{\beta&space;\tau&space;}$.

$\frac{E_{o}(s)}{E_{i}(s)}&space;=\frac{1}{\beta&space;}\frac{s+\frac{1}{\tau&space;}}{s+\frac{1}{\beta&space;\tau&space;}}&space;=&space;(\frac{\tau&space;s+1}{\beta&space;\tau&space;s+1})--------EQN(II)$.

the values of the three parameters $R_{1}$ , $R_{2}$  and C are determined from the two compensator parameters $\tau$  and $\beta$.

using the EQN(II)

$\tau&space;=R_{1}C>&space;0$,    $\beta&space;=\frac{(R_{1}+R_{2})}{R_{2}}>&space;1$.

there is an additional degree of freedom in the choice of the values of the network components which is used to set the impedance level of the N/w.

the gain is $\left&space;|&space;G(j\omega&space;)&space;\right&space;|=\left&space;|&space;\frac{E_{o}(j\omega&space;)}{E_{i}(j\omega&space;)}&space;\right&space;|&space;=&space;\left&space;|&space;\beta(\frac{1+j\omega&space;\tau&space;}{1+\beta&space;\tau&space;j\omega&space;})&space;\right&space;|$

D.C gain at $\omega&space;=0$  is $\beta$  which is greater than 1.

Let the zero-frequency gain as unity, then the Transfer function is $G(j\omega&space;)&space;=&space;(\frac{1+j\omega&space;\tau&space;}{1+\beta&space;\tau&space;j\omega&space;})$.

Frequency-response of Lag compensator:-

Note:-“lag” refers to the property that the compensator adds positive phase to the system over some appropriate frequency range.

$G(j\omega&space;)&space;=&space;(\frac{1+j\omega&space;\tau&space;}{1+\beta&space;\tau&space;j\omega&space;})$,   let  $\beta&space;=1$.

the frequency response of lag compensator is $\left&space;|&space;G(j\omega&space;)&space;\right&space;|=&space;\sqrt{\frac{1+\omega&space;^{2}\tau&space;^{2}}{1+\omega&space;^{2}\beta&space;^{2}\tau&space;^{2}}}$

at $\omega&space;=\frac{1}{\tau&space;}$ $\Rightarrow\left&space;|&space;G(j\omega&space;)&space;\right&space;|=&space;\sqrt{\frac{2}{1+\beta&space;^{2}}}$.

$(1+j\omega&space;\tau&space;)\rightarrow$ has a slope +20 dB/decade with corner frequency $\frac{1}{\tau&space;}$.

$(1+\beta&space;\tau&space;j\omega)\rightarrow$ slope is -20 dB/decade with corner frequency $\frac{1}{\beta&space;\tau&space;}$.

$\Phi&space;=&space;\angle&space;G(j\omega&space;)=tan^{-1}\omega&space;\tau&space;-tan^{-1}\beta&space;\omega&space;\tau$

to find at which frequency the phase is minimum , differentiate $\Phi$ w.r to $\omega$ and equate it to zero.

$\Phi&space;=&space;tan^{-1}(\frac{\omega&space;\tau-\beta\omega&space;\tau}{1+\beta&space;\omega^{2}&space;\tau^{2}})$

$\frac{d\Phi&space;}{d\omega&space;}=0$

$\frac{1}{1+(\frac{\omega&space;\tau-\beta&space;\omega&space;\tau}{1+\alpha&space;\omega^{2}&space;\tau^{2}})^{2}}(\frac{((1+\beta&space;\omega^{2}&space;\tau^{2})\tau&space;(1-\beta&space;))-(\omega&space;\tau&space;(1-\beta&space;)2\omega&space;\beta&space;\tau&space;^{2})}{(1+\beta&space;\omega^{2}&space;\tau^{2})^{2}})=0$

${((1+\beta&space;\omega^{2}&space;\tau^{2})\tau&space;(1-\beta&space;))-(\omega&space;\tau&space;(1-\beta&space;)2\omega&space;\beta&space;\tau&space;^{2})}=0$

$\tau&space;(1-\beta&space;)(1+\beta&space;\omega^{2}&space;\tau^{2}-2\omega^{2}&space;\beta\tau&space;^{2})=0$

$\tau&space;(1-\beta)(1-\omega^{2}&space;\beta&space;\tau&space;^{2})=0$

$\because&space;\tau&space;\neq&space;0$    implies $(1-\beta)=0\Rightarrow&space;\beta&space;=1$   , which is invalid because $\beta&space;>&space;1$.

$(1-\omega^{2}&space;\beta&space;\tau&space;^{2})=0\Rightarrow&space;\omega^{2}&space;=\frac{1}{\beta&space;\tau&space;^{2}}$.

$\omega&space;=\frac{1}{\sqrt{\beta}&space;\tau&space;}$  , at this $\omega$  lead compensator has minimum phase given by

$\Phi&space;_{m}&space;=&space;tan^{-1}(\frac{1-\beta&space;}{2\sqrt{\beta}})$

$tan&space;\Phi&space;_{m}&space;=&space;\frac{1-\beta&space;}{2\sqrt{\beta}}$ implies $sin&space;\Phi&space;_{m}&space;=&space;\frac{1-\beta}{1+\beta&space;}$.

$\beta&space;=\frac{1-sin&space;\Phi&space;_{m}}{1+sin&space;\Phi&space;_{m}}$.

at $\omega&space;=\omega&space;_{m}$ ,    $\left&space;|&space;G(j\omega&space;)&space;\right&space;|&space;=&space;\frac{1}{\sqrt{\beta&space;}}$.

Choice of $\beta$ :-

Any phase lag at the gain cross over frequency of the compensated system is undesirable.

To prevent the effects of lag compensator , the corner frequency of the lag compensator must be located substantially lower than the $\omega&space;_{gc}$ of compensated system.

In the high frequency range , the lag compensator has an attenuation of $20&space;log(\beta&space;)$ dB, which is used to obtain required phase margin.

The addition of a lag compensator results in an improvement in the ratio of control signal to noise in the loop.

high frequency noise signals are attenuated by a factor $\beta&space;>&space;1$, while low-frequency control signals under go unit amplification (0 dB gain).

atypical value of $\beta&space;=10$.

Procedure for bode-plot of a lead compensator:-

Step 1:- Sketch the Bode-plot of the uncompensated system with the gain k. Set the value of k according to the steady-state error requirement.

Measure the gain cross over frequency and the phase margin of uncompensated system.

Step 2:-  find $\omega&space;_{gc}^{'}$ at which phase angle of uncompensated system is

$-180^{o}$ + given Phase Margin+ $\epsilon$.

$\epsilon&space;=5^{o}(or)15^{o}$   is a good assumption for phase-lag contribution.

Step 3:- find gain of the uncompensated system at $\omega&space;_{gc}^{'}$ and equate it to 20 log ($\beta$)  and then find $\beta$.

Step 4:- choose the upper corner frequency of the compensator to one octave to one decade  below $\omega&space;_{gc}^{'}$ and find $\tau$ value.

Step 5:- Calculate phase lag of compensator  at $\omega&space;_{gc}^{'}$, if it is less than $\epsilon$ go to next step.

Step 6:- Draw the Bode plot of compensated system  to meet the desired specifications.

(No Ratings Yet)

A Lead compensator has a Transfer function of the form $G(s)&space;=&space;\frac{s+z_{c}}{s+p_{c}}------EQN(I)$

$G(s)&space;=&space;\frac{s+\frac{1}{\tau&space;}}{s+\frac{1}{\alpha&space;\tau&space;}}$,    where $\alpha&space;=\frac{z_{c}}{p_{c}}<&space;1$     and $\tau&space;>&space;0$

i.e, the pole is located to the left of the zero.

• A lead compensator speeds up the transient response and increases margin of stability of a system.
• It also helps  to increase the system error constant through a limited range.

Realization of Lead compensator as an Electrical Network:-

The lead compensator can be realized by an electrical Network.

Assume impedance of source is zero  $[Z_{s}&space;=0]$ and output load impedance to be infinite  .

The transfer function is $\frac{E_{o}(s)}{E_{i}(s)}&space;=&space;\frac{R_{2}}{R_{2}+(R_{1}||&space;\frac{1}{Cs})}$

$\frac{E_{o}(s)}{E_{i}(s)}&space;=&space;\frac{R_{2}}{R_{2}+\frac{(R_{1}\frac{1}{Cs})}{(R_{1}+\frac{1}{Cs})}}$

$\frac{E_{o}(s)}{E_{i}(s)}&space;=&space;\frac{R_{2}(R_{1}+\frac{1}{Cs})}{R_{2}(R_{1}+\frac{1}{Cs})+R_{1}\frac{1}{Cs}}$

after simplification

$\frac{E_{o}(s)}{E_{i}(s)}&space;=\frac{s+\frac{1}{R_{1}C}}{s+\frac{1}{R_{1}C}+\frac{1}{R_{2}C}}$

$\frac{E_{o}(s)}{E_{i}(s)}&space;=\frac{s+\frac{1}{R_{1}C}}{s+\frac{1}{(\frac{R_{2}}{R_{1}+R_{2}})R_{1}C}}$  by comparing this equation with the transfer function of lead compensator has a zero at  $Z_{c}&space;=\frac{1}{R_{1}C}$  and the pole is $p_{c}=\frac{1}{(\frac{R_{2}}{R_{1}+R_{2}})R_{1}C}$ .

from the pole $\alpha&space;=\frac{R_{2}}{R_{1}+R_{2}}$ and  $\tau&space;=R_{1}C$.

therefore  the transfer function  has a zero at $-\frac{1}{\tau&space;}$   and a pole at $-\frac{1}{\alpha&space;\tau&space;}$.

$\frac{E_{o}(s)}{E_{i}(s)}&space;=\frac{s+\frac{1}{\tau&space;}}{s+\frac{1}{\alpha&space;\tau&space;}}&space;=&space;\alpha&space;(\frac{\tau&space;s+1}{\alpha&space;\tau&space;s+1})--------EQN(II)$.

the values of the three parameters $R_{1}$ , $R_{2}$  and C are determined from the two compensator parameters $\tau$  and $\alpha$.

using the EQN(II)

$\tau&space;=R_{1}C>&space;0$,    $\alpha&space;=\frac{R_{2}}{R_{1}+R_{2}}<&space;1$.

there is an additional degree of freedom in the choice of the values of the network components which is used to set the impedance level of the N/w.

the gain is $\left&space;|&space;G(j\omega&space;)&space;\right&space;|=\left&space;|&space;\frac{E_{o}(j\omega&space;)}{E_{i}(j\omega&space;)}&space;\right&space;|&space;=&space;\left&space;|&space;\alpha&space;(\frac{1+j\omega&space;\tau&space;}{1+\alpha&space;\tau&space;j\omega&space;})&space;\right&space;|$

D.C gain at $\omega&space;=0$  is $\alpha$  which is less than 1.

attenuation $\frac{1}{\alpha&space;}$ is used to determine the steady state performance.

while using a lead N/w , it is important to increase the loop gain by an amount of $\frac{1}{\alpha&space;}$.

A lead compensator is visualized as a combination of a N/w and an amplifier.

Note:-“lead” refers to the property that the compensator adds positive phase to the system over some appropriate frequency range.

$G(j\omega&space;)&space;=&space;\alpha&space;(\frac{1+j\omega&space;\tau&space;}{1+\alpha&space;\tau&space;j\omega&space;})$,   let  $\alpha&space;=1$.

the frequency response of lead compensator is

$\Phi&space;=&space;\angle&space;G(j\omega&space;)=tan^{-1}\omega&space;\tau&space;-tan^{-1}\alpha&space;\omega&space;\tau$

to find at which frequency the phase is maximum , differentiate $\Phi$ w.r to $\omega$ and equate it to zero.

$\Phi&space;=&space;tan^{-1}(\frac{\omega&space;\tau-\alpha&space;\omega&space;\tau}{1+\alpha&space;\omega^{2}&space;\tau^{2}})$

$\frac{d\Phi&space;}{d\omega&space;}=0$

$\frac{1}{1+(\frac{\omega&space;\tau-\alpha&space;\omega&space;\tau}{1+\alpha&space;\omega^{2}&space;\tau^{2}})^{2}}(\frac{((1+\alpha&space;\omega^{2}&space;\tau^{2})\tau&space;(1-\alpha&space;))-(\omega&space;\tau&space;(1-\alpha&space;)2\omega&space;\alpha&space;\tau&space;^{2})}{(1+\alpha&space;\omega^{2}&space;\tau^{2})^{2}})=0$

${((1+\alpha&space;\omega^{2}&space;\tau^{2})\tau&space;(1-\alpha&space;))-(\omega&space;\tau&space;(1-\alpha&space;)2\omega&space;\alpha&space;\tau&space;^{2})}=0$

$\tau&space;(1-\alpha&space;)(1+\alpha&space;\omega^{2}&space;\tau^{2}-2\omega^{2}&space;\alpha&space;\tau&space;^{2})=0$

$\tau&space;(1-\alpha&space;)(1-\omega^{2}&space;\alpha&space;\tau&space;^{2})=0$

$\because&space;\tau&space;\neq&space;0$    implies $(1-\alpha&space;)=0\Rightarrow&space;\alpha&space;=1$   , which is invalid because $\alpha&space;<&space;1$.

$(1-\omega^{2}&space;\alpha&space;\tau&space;^{2})=0\Rightarrow&space;\omega^{2}&space;=\frac{1}{\alpha&space;\tau&space;^{2}}$.

$\omega&space;=\frac{1}{\sqrt{\alpha}&space;\tau&space;}$   , at this $\omega$  lead compensator has maximum phase given by

$\Phi&space;_{m}&space;=&space;tan^{-1}(\frac{1-\alpha&space;}{2\sqrt{\alpha&space;}})$

$tan&space;\Phi&space;_{m}&space;=&space;\frac{1-\alpha&space;}{2\sqrt{\alpha&space;}}$  implies $sin&space;\Phi&space;_{m}&space;=&space;\frac{1-\alpha&space;}{1+\alpha&space;}$.

$({1+\alpha&space;})sin&space;\Phi&space;_{m}&space;=&space;{1-\alpha&space;}$

$sin&space;\Phi&space;_{m}+\alpha&space;sin&space;\Phi&space;_{m}&space;=&space;{1-\alpha&space;}$

$\alpha&space;=\frac{1-sin&space;\Phi&space;_{m}}{1+sin&space;\Phi&space;_{m}}$.

at $\omega&space;=\omega&space;_{m}$ ,    $\left&space;|&space;G(j\omega&space;)&space;\right&space;|&space;=&space;\frac{1}{\sqrt{\alpha&space;}}$.

when there is a need for phase leads of more than $60^{o}$, two cascaded lead networks are used where each N/w provides half of the required phase.

for phase leads more than $60^{o}$$\alpha$ decreases sharply and if single N/w is used $\alpha$ will be too low.

Choice of $\alpha$ :-

In choosing parameters of compensator $\tau$ depends on $R_{1}$ and C . The $\tau$ value may be anything but for $\alpha$ there is a constraint. It depends on inherent noise in Control systems.

from the lead N/w , it’s been observed that the high frequency noise is amplified by $\frac{1}{\alpha&space;}$ while low frequencies by unity.

more (or) less $\alpha$ should not be less than 0.07.

Procedure for bode-plot of a lead compensator:-

Step 1:- Sketch the Bode-plot of the uncompensated system with the gain k. Set the value of k according to the steady-state error requirement.

Measure the gain cross over frequency and the phase margin of uncompensated system.

Step 2:- using the relation

Additional phase lead required = specified phase margin- Phase Margin of uncompensated system.

$\epsilon$  is a margin of safety required by the fact that the gain cross over frequency will increase due to compensation.

for example :-  $\epsilon&space;=5^{o}$ is a good assumption for -40 dB/decade.

$\epsilon&space;=15^{o}$  (or) $20^{o}$ $\Rightarrow$ -60 dB/decade.

Step 3:- Set the maximum phase of the lead compensator    $\Phi&space;_{m}&space;=$ Additional phase lead required  and compute $\alpha&space;=\frac{1-sin&space;\Phi&space;_{m}}{1+sin&space;\Phi&space;_{m}}$.

Step 4:- Find the frequency at which the uncompensated system has a gain of $-20&space;log(\frac{1}{\sqrt{\alpha&space;}})$ dB, which gives the new gain cross over frequency.

with $\omega&space;_{gc}$ as the gain cross over frequency the system has a phase margin of $\gamma&space;_{1}$

where as with $\omega&space;_{gc}^{'}$ as the gain cross over frequency the system has a phase margin of $\gamma&space;_{2}$

Step 5:- Now $\omega&space;_{gc}^{'}&space;=&space;\omega&space;_{m}&space;=&space;\frac{1}{\tau&space;\sqrt{\alpha&space;}}$.    find the value of $\tau$ and the transfer function of lead compensator  $\frac{1+j\omega&space;\tau&space;}{1+\alpha&space;j\omega&space;\tau&space;}$.

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## Introduction to Root Locus

The introduction of a feedback to a system causes some instability , therefore an unstable system can not perform the control task requires of it.

while in the analysis of  a given system, the very first investigation that needs to be made is whether the system is stable or not?

However, the determination of stability of a system is necessary but not sufficient.

A stable system with low damping is also unwanted.

a design problem in which the designer is required to achieve the desired performance for a system by adjusting the location of its close loop poles in the S-plane by varying one (or) more system parameters.

The Routh’s criterion obviously does not help much in such problems.

for determining the location of closed-loop poles one may resort to the classical techniques of factoring the characteristic equation and determining it’s roots.

when the degree is higher (or) repeated calculations are required as a system parameter is varied for adjustments.

a simple technique, known as the root locus technique, for finding the roots of the ch.eqn introduced by W.R.Evans.

This technique provides a graphical method of plotting the locus of the roots in the S-plane as a given system parameter is varied from complete range of values (may be from zero to infinity).

The roots corresponding to a particular value of the system parameter can then be located on the locus (or) value of the parameter for a desired root location can be determined from the locus.

Root Locus:-

• In the analysis and design for stable systems and gives information about transient response of control systems.
• It gives information about absolute stability and relative stability of a system.
• It clearly shows the ranges of stability and instability.
• used for higher order differential equations.
• value of k for a particular root location can be determined.
• and the roots for a particular k can be determined using Root Locus.

$\frac{C(s)}{R(s)}=\frac{G(s)}{1+G(s)H(s)}$

ch. equation is $1+G(s)H(s)=0$

let $G(s)H(s)=D(s)$

$1+D(s)=0$

$D(s)=-1$

To find the whether the roots are on the Root locus (or) not

They have to satisfy ‘2’ criteria known as

1. Magnitude Criterion.
2. Angle Criterion.

Magnitude criterion:-

$\left&space;|&space;D(s)&space;\right&space;|=1$

$\left&space;|&space;G(s)H(s)&space;\right&space;|=1$

the magnitude criterion states that  $s=s_{a}$  will be a point on root locus, if for that value of s

i.e, $\left&space;|&space;G(s)H(s)&space;\right&space;|=1$

Angle criterion:-

$\angle&space;D(s)&space;=&space;\angle&space;G(s)H(s)=\pm&space;180^{o}(2q+1)$

where q=0,1,2……….

if $\angle&space;D(s)&space;=&space;\pm&space;180^{o}(2q+1)$ is odd multiple of $180^{o}$, a point s on the root locus, if $\angle&space;D(s)$ is odd multiple of at $s=s_{a}$ of $180^{o}$, then that point is on the root locus.

Root Locus definition:-

The locus of roots of the Ch. eqn in the S-plane by the variation of system parameters (generally gain k) from $0$ to $\infty$ is known as Root locus.

It is a graphical method

$-\infty$ to $0$       $\rightarrow$     Inverse Root Locus

$0$  to $\infty$    $\rightarrow$  Direct Root Locus

generally Root Locus means Direct Root Locus.

$D(s)&space;=&space;G(s)H(s)&space;=&space;k&space;\frac{(s+z_{1})(s+z_{2})(s+z_{3})....}{(s+p_{1})(s+p_{2})(s+p_{3})....}$

$\left&space;|&space;D(s)&space;\right&space;|&space;=&space;k&space;\frac{\left&space;|&space;s+z_{1}&space;\right&space;|\left&space;|&space;s+z_{2}&space;\right&space;|\left&space;|&space;s+z_{3}&space;\right&space;|....}{\left&space;|&space;s+p_{1}&space;\right&space;|\left&space;|&space;s+p_{2}&space;\right&space;|\left&space;|&space;s+p_{3}&space;\right&space;|....}$

$\left&space;|&space;D(s)&space;\right&space;|&space;=&space;k&space;\frac{\prod_{i=1}^{m}\left&space;|&space;s+z_{i}&space;\right&space;|}{\prod_{i=1}^{n}\left&space;|&space;s+p_{i}&space;\right&space;|}$

m= no .of zeros

n= no.of poles

from magnitude criterion $\left&space;|&space;D(s)&space;\right&space;|&space;=&space;1$

$k&space;=\frac{\prod_{i=1}^{n}\left&space;|&space;s+p_{i}&space;\right&space;|}{\prod_{i=1}^{m}\left&space;|&space;s+z_{i}&space;\right&space;|}$

The open loop gain k corresponding to a point $s=s_{a}$ on Root Locus can be calculated

$k=$ product of length of vectors from open loop poles to the point $s=s_{a}$/product of length of vectors from open loop zeros to the point $s=s_{a}$.

from the Angle criterion,

$\angle&space;D(s)&space;=&space;\angle&space;(s+z_{1})+\angle&space;(s+z_{2})+\angle&space;(s+z_{3}).....&space;-\angle&space;(s+p_{1})+\angle&space;(s+p_{2})+\angle&space;(s+p_{3}).....$

$\angle&space;D(s)&space;=&space;\sum_{i=1}^{m}\angle&space;(s+z_{i})&space;-\sum_{i=1}^{n}\angle&space;(s+p_{i})$

$\sum_{i=1}^{m}\angle&space;(s+z_{i})&space;-\sum_{i=1}^{n}\angle&space;(s+p_{i})=\pm&space;180^{o}(2q+1)$

i.e,( sum of angles of vectors from Open Loop zeros to point $s=s_{a}$)-(sum of angles of vectors from Open Loop poles to point$s=s_{a}$$=\pm&space;180^{o}(2q+1)$

where q=0,1,2………

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## Inductance of a Co-axial cable, Solenoid & Toroid

Inductance of a Co-axial cable:-

Consider a coaxial cable with inner conductor having radius a and outer conductor of radius b and the current is flowing in the cable along z-axis  in which the cable is placed such that the axis of rotation of the cable co-incides with z-axis.

the length of the co-axial cable be d meters.

we know that the $\overrightarrow{H_{\phi&space;}}$ between the region $a<&space;\rho&space;<&space;b$  is $\overrightarrow{H_{\phi&space;}}=\frac{I}{2\pi&space;\rho&space;}\overrightarrow{a_{\phi&space;}}$

and $\overrightarrow{B_{\phi&space;}}=\frac{\mu&space;_{o}I}{2\pi&space;\rho&space;}\overrightarrow{a_{\phi&space;}}$   since $\overrightarrow{B_{\phi&space;}}=\mu&space;_{o}\overrightarrow{H_{\phi&space;}}$

as $L&space;=&space;\frac{\lambda&space;}{I}$, here the flux linkage $\lambda&space;=1\phi$

where $\phi$-Total flux coming out of the surface

$\phi&space;=&space;\oint_{s}&space;\overrightarrow{B}.\overrightarrow{ds}$

Since the magnetic flux will be radial plane extending from $\rho&space;=a$   to   $\rho&space;=b$ and $z=0$ to $z=d$.

$\overrightarrow{ds_{\phi&space;}}&space;=&space;d\rho&space;dz&space;\overrightarrow{a_{\phi&space;}}$

$\phi&space;=&space;\oint_{s}&space;\overrightarrow{B}.\overrightarrow{ds}$

$\phi&space;=&space;\oint_{s}\frac{\mu&space;_{o}I}{2\pi&space;\rho&space;}\overrightarrow{a_{\phi&space;}}&space;.d\rho&space;dz&space;\overrightarrow{a_{\phi&space;}}$

$\phi&space;=&space;\int_{\rho&space;=a}^{b}\int_{z&space;=0}^{d}\frac{\mu&space;_{o}I}{2\pi&space;\rho&space;}&space;d\rho&space;dz$

$\phi&space;=&space;\frac{\mu&space;_{o}I}{2\pi&space;}&space;ln&space;d$

$L=\frac{\phi&space;}{I}&space;=\frac{\mu&space;_{o}d}{2\pi&space;}&space;ln$  Henries.

Inductance of a Solenoid:-

Consider a Solenoid of N turns and  let the current flowing inside it is  ‘I’ Amperes. The length of the solenoid is ‘l’ meters and ‘A’ is its cross sectional area.

$\phi$ – Total flux coming out of solenoid.

flux linkage $\lambda&space;=&space;N\phi$      $\Rightarrow&space;\lambda&space;=&space;NBA----------EQN(I)$        $\because&space;\frac{\phi&space;}{A}&space;=&space;B$

$L&space;=&space;\frac{\lambda&space;}{I}$, from the definition

As B is the Magnetic flux density given $B=&space;\frac{flux}{unit&space;area}&space;=\frac{\phi&space;}{A}$

from EQN(I) ,

$\lambda&space;=&space;N\mu&space;_{o}HA-------EQN(II)$  because $B&space;=&space;\mu&space;_{o}H$

The field strength H of a solenoid is $H&space;=&space;\frac{NI}{l}&space;A/m$

EQN (II) becomes  $\lambda&space;=&space;N&space;\mu&space;_{o}\frac{NI}{l}A$

$\lambda&space;=&space;\frac{N^{2}&space;\mu&space;_{o}IA}{l}$

from the inductance definition $L&space;=&space;\frac{\lambda&space;}{I}$

$L=&space;\frac{N^{2}&space;\mu&space;_{o}A}{l}$   Henries.

Inductance of a Toroid:-

Consider a toroidal ring with N-turns and carrying current I.

let the radius of the toroid be ‘R’ and the total flux emerging be $\phi$

then flux linkage  $\lambda&space;=&space;N\phi$

the magnetic flux density inside a toroid is given by $B&space;=&space;\frac{\mu&space;_{o}NI}{2\pi&space;R}$

$\lambda&space;=&space;NBA$

where A is the cross sectional area of the toroid then $\lambda&space;=&space;N&space;\frac{\mu_{o}&space;NI}{2\pi&space;R}A$

$\lambda&space;=&space;\frac{\mu_{o}&space;N^{2}I}{2\pi&space;R}A$

$L=\frac{\lambda&space;}{I}$

$L=\frac{\mu_{o}&space;N^{2}A}{2\pi&space;R}$  Henries.

if the toroid has a height ‘h’ , inner radius $\rightarrow&space;r_{1}$ and outer radius $\rightarrow&space;r_{2}$ then its Inductance is $L=\frac{\mu_{o}&space;N^{2}h}{2\pi&space;}ln$   Henries.

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## Classification (or) topologies of feedback Amplifiers

There are 4 different combinations possible with negative feedback in Amplifiers as given below

1. Voltage-Series.
2. Current-Series.
3. Voltage-Shunt.
4. Current-Shunt.

The first part represents the type of sampling at the output .

• i.e ,  Voltage- Shunt connection.
• Current-Series connection.

and the second part represents the type of Mixing at the input

• Series- Voltage is applied at the input.
• Shunt-Current is applied at the input.

For any Amplifier circuit we require

• High Gain
• High Band Width
• High Input Impedance
• and Low Output Impedance.

Classification of feedback Amplifiers is also known as feedback Topologies.

Voltage-Series feedback Connection:-

at i/p side connection is Series and at o/p side connection used is Shunt  since o/p is collected is voltage.

Series connection increases i/p impedance and Voltage at the o/p indicates a decrease in o/p impedance.

i.e, $Z_{if}&space;=&space;Z_{i}(1+A\beta&space;)$    and    $Z_{of}&space;=&space;\frac{Z_{o}}{(1+A\beta&space;)}$.

Current-Series feedback Connection:-

Series connection increases i/p impedance and Current at the o/p indicates an increase in o/p impedance.

i.e, $Z_{if}&space;=&space;Z_{i}(1+A\beta&space;)$    and    $Z_{of}&space;=&space;Z_{o}(1+A\beta&space;)$.

Voltage-Shunt feedback Connection:-

In this connection, both i/p and o/p impedance decreases .

i.e, $Z_{if}&space;=&space;\frac{Z_{i}}{(1+A\beta&space;)}$    and    $Z_{of}&space;=&space;\frac{Z_{o}}{(1+A\beta&space;)}$.

Current-Shunt feedback Connection:-

Shunt connection decreases i/p impedance and Current at the o/p indicates an increase in o/p impedance.

i.e, $Z_{if}&space;=&space;\frac{Z_{i}}{(1+A\beta&space;)}$    and    $Z_{of}&space;=&space;Z_{o}(1+A\beta&space;)$.

Effect of negative feedback on different topologies:-

 Type of f/b Voltage gain Band Width with f/b i/p impedance o/p impedance Voltage-Series decreases increases increases decreases Current-Series decreases increases increases increases Voltage-Shunt decreases increases decreases decreases Current-Shunt decreases increases decreases increases

Similarly negative feedback decreases noise and harmonic distortion for all the four topologies.

Note:-  for any of the characteristics in the above table, increase ‘s shown by multiplying the original value with $(1+A\beta&space;)$ and decrease ‘s shown by dividing with $(1+A\beta&space;)$.

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## continuity equation

The continuity equation of the current is based on the principle of conservation of charge that is  the charge can neither be created not destroyed.

consider a closed surface S with a current density $\overrightarrow{J}$, then the total current I crossing the surface S is given by the volume V

The current coming out of the closed surface is $I_{out}&space;=&space;\oint_{s}&space;\overrightarrow{J}.\overrightarrow{ds}$

since the direction of current is in the direction of positive charges, positive charges also move out of the surface because of the current I.

According to principle of conversation of charge, there must be decrease of an equal amount of positive charge inside the closed surface.

therefore the time rate of decrease of charge with in a given volume must be equal to the net outward current flow through the closed surface of the volume.

$I_{out}&space;=&space;\frac{-dQ_{in}}{dt}=&space;\oint_{s}&space;\overrightarrow{J}.\overrightarrow{ds}$

By Divergence theorem $\oint_{s}&space;\overrightarrow{J}.\overrightarrow{ds}&space;=&space;\oint_{v}\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;dv$

$\frac{-dQ_{in}}{dt}=&space;\oint_{v}\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;dv$

$Q_{in}=\int_{v}\rho&space;_{v}dv$    implies  $-\frac{dQ_{in}}{dt}=-\frac{d}{dt}\int_{v}\rho&space;_{v}dv$

for a constant surface the derivative becomes the partial derivative

$\oint_{v}\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;dv&space;=-\int_{v}\frac{\partial&space;\rho&space;_{v}}{\partial&space;t}dv$   -this is for the whole volume.

for a differential volume $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;dv&space;=-\frac{\partial&space;\rho&space;_{v}}{\partial&space;t}dv$

$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;=-\frac{\partial&space;\rho&space;_{v}}{\partial&space;t}$ , which is called as continuity of current equation (or) Point form (or) differential form of the continuity equation.

This equation is derived based on the principle of conservation charge states that there can be no accumulation of charge at any point.

for steady  (dc) currents     $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;=0$ $\Rightarrow&space;-\frac{\partial&space;\rho&space;_{v}}{\partial&space;t}=0$

from $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;=0$. The total charge leaving a volume is the same as total charge entering it. Kirchhoff’s law follows this equation.

This continuity equation states that the current (or) the charge per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point.

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## Effect of negative feedback on Band width of an Amplifier

Let the Band Width of an amplifier without feedback is = BW. Band width of an amplifier with negative feed back is $BW_{f}=&space;BW&space;(1+A\beta&space;)$. Negative feedback increases Band width.

Proof:-  Consider an amplifier with gain ‘A’

Now the frequency response of the amplifier is as shown in the figure. Frequency response curve means gain (dB) Vs frequency (Hz)

the frequency response of an amplifier consists of three regions

1. Low frequency region ($<&space;f_{1}$ -lower cut off frequency).
2. Mid frequency region ( between $f_{1}$ and $f_{2}$).
3. High frequency region ( the region $>&space;f_{2}$ -upper cutoff frequency)

Gain in low- frequency region is given as$A_{vl}&space;=&space;\frac{A_{v}}{1-j\frac{f_{1}}{f}}---------EQN(I)$,

$A_{v}$ -open loop gain,

$f$– frequency,

$f_{1}$– lower cut off frequency, where Gain in constant region is $A_{v}$.

Gain in High-frequency region is $A_{vh}&space;=&space;\frac{A_{v}}{1+j\frac{f}{f_{2}}}$  .

In low-frequency region:-

since open loop gain  in low-frequency region is $A_{vl}$ and gain with feedback is $A_{vlf}&space;=&space;\frac{A_{vl}}{1+A_{vl}\beta&space;}$

From EQN(I) $A_{vl}&space;=&space;\frac{A_{v}}{1-j\frac{f_{1}}{f}}$   after substituting $A_{vl}$ in the above equation

$A_{vlf}&space;=&space;\frac{\frac{A_{v}}{1-j\frac{f_{1}}{f}}}{1+\frac{A_{v}}{1-j\frac{f_{1}}{f}}\beta&space;}$

$=\frac{A_{v}}{1-j\frac{f_{1}}{f}+A_{v}\beta&space;}$

$=&space;\frac{A_{v}}{1+A_{v}\beta-j\frac{f_{1}}{f}&space;}$

Now by dividing the whole expression with $(1+A_{v}\beta&space;)$

$A_{vlf}=&space;\frac{\frac{A_{v}}{(1+A_{v}\beta&space;)}}{\frac{1+A_{v}\beta-j\frac{f_{1}}{f}}{(1+A_{v}\beta&space;)}&space;}$

$A_{vlf}=&space;\frac{\frac{A_{v}}{(1+A_{v}\beta&space;)}}{1-j\frac{f_{1}}{f}\frac{1}{(1+A_{v}\beta&space;)}&space;}$

$A_{vlf}&space;=&space;\frac{A_{vf}}{1-j\frac{f_{1}^{'}}{f}}$, where $A_{vf}&space;=&space;\frac{A_{v}}{1+A_{v}\beta&space;}$  and $f_{1}^{'}&space;=&space;\frac{f_{1}}{1+A_{v}\beta&space;}$

for example lower cut-off frequency $f_{1}&space;=20&space;Hz$  implies $f_{1}^{'}&space;=&space;\frac{20}{1+A_{v}\beta&space;}$  is decreasing with negative feedback.

In High-frequency region:-

Gain with out feed back in High frequency region is $A_{vh}&space;=&space;\frac{A_{v}}{1+j\frac{f}{f_{2}}}$

Now Gain with negative feed back is $A_{vhf}&space;=&space;\frac{A_{vh}}{1+A_{vh}\beta&space;}$

Substituting $A_{vh}$ in the above equation

$A_{vhf}&space;=&space;\frac{\frac{A_{v}}{1+j\frac{f}{f_{2}}}}{1+\frac{A_{v}}{1+j\frac{f}{f_{2}}}\beta&space;}$

$A_{vhf}=\frac{A_{v}}{1+j\frac{f}{f_{2}}+A_{v}\beta&space;}$

$A_{vhf}=&space;\frac{A_{v}}{1+A_{v}\beta+j\frac{f}{f_{2}}&space;}$

Now by dividing the whole expression with $(1+A_{v}\beta&space;)$

$A_{vhf}=&space;\frac{\frac{A_{v}}{(1+A_{v}\beta&space;)}}{\frac{1+A_{v}\beta+j\frac{f}{f_{2}}}{(1+A_{v}\beta&space;)}&space;}$

$A_{vhf}=&space;\frac{\frac{A_{v}}{(1+A_{v}\beta&space;)}}{1+j\frac{f}{f_{2}}\frac{1}{(1+A_{v}\beta&space;)}&space;}$

$A_{vhf}&space;=&space;\frac{A_{vf}}{1+j\frac{f}{f_{2}^{'}}}$, where $A_{vf}&space;=&space;\frac{A_{v}}{1+A_{v}\beta&space;}$  and $f_{2}^{'}&space;=&space;f_{2}(1+A_{v}\beta&space;)$

for example lower cut-off frequency $f_{2}&space;=20K&space;Hz$  implies $f_{1}^{'}&space;=&space;20&space;K&space;(1+A_{v}\beta&space;)$  is increasing with negative feedback.

In Mid-frequency region:-

Gain with out feed back is $A_{v}$

and the gain with negative feed back is $A_{vf}&space;=&space;\frac{A_{v}}{1+A_{v}\beta&space;}$

 With out feedback With feedback lower cut-off frequency  is $f_{1}$ lower cut-off frequency $f_{1}^{'}=&space;\frac{f_{1}}{1+A_{v}\beta&space;}$, increases upper cut-off frequency is $f_{2}$ upper cut-off frequency is $f_{2}^{'}&space;=&space;f_{2}(1+A_{v}\beta&space;)$ BW = $f_{2}-f_{1}$ $BW_{f}&space;=&space;f_{2}^{'}-f_{1}^{'}$ increases

Thus negative feedback decreases lower cut-off frequency and increases upper cut-off frequency.

$\therefore$ Over all gain decreases with  negative feedback and Band Width increases.

(No Ratings Yet)

## Time-domain representation of SSB-SC signal

Let the signal produced by SSB-SC modulator is S(t), a Band pass signal

$S_{USB}(t)=S_{I}(t)&space;cos&space;\omega&space;_{c}t-S_{Q}(t)&space;sin&space;\omega&space;_{c}t$, where $S_{I}(t)$ is  In-Phase component of S(t) obtained by

i. Multiplying S(t) with $\cos&space;\omega&space;_{c}t$.

ii. and passing the product through a LPF with suitable cut-off frequency.

$S_{I}(t)&space;=&space;S(t)&space;\cos&space;\omega&space;_{c}t$

By finding the Fourier Transform of in-phase component

$S_{I}(f)&space;=&space;\frac{1}{2}(S(f-f_{c})+S(f+f_{c}))$

after restricting the signal $S_{I}(f)$ between   $-B\leq&space;f&space;\leq&space;B$

$S_{I}(f)&space;=\left\{\begin{matrix}&space;\frac{1}{2}(S(f-f_{c})+S(f+f_{c})),-B\leq&space;f&space;\leq&space;B&space;&&space;\\0&space;,&space;otherwise&space;&&space;\end{matrix}\right.$

Similarly $S_{Q}(t)$ is the quadrature phase component of s(t), obtained by multiplying S(t) with $\sin&space;\omega&space;_{c}t$  and by passing the resultant signal through a LPF .

$S_{Q}(t)&space;=&space;S(t)&space;\sin&space;\omega&space;_{c}t$

By finding the Fourier Transform of Q-phase component

$S_{Q}(f)&space;=&space;\frac{1}{2j}(S(f-f_{c})-S(f+f_{c}))$

after restricting the signal $S_{Q}(f)$ between   $-B\leq&space;f&space;\leq&space;B$

$S_{Q}(f)&space;=\left\{\begin{matrix}&space;\frac{1}{2j}(S(f-f_{c})-S(f+f_{c})),-B\leq&space;f&space;\leq&space;B&space;&&space;\\0&space;,&space;otherwise&space;&&space;\end{matrix}\right.$

Now Let’s assume S(f) is the required frequency spectrum of SSB-SC signal when only USB has been transmitted.

i.e,

from the above figure,

one can obtain $S(f-f_{c})$ ,  $S(f+f_{c})$ by shifting the signal S(f) towards right by $f_{c}$  and  left by  $f_{c}$

Now by adding  $S(f-f_{c})$  and  $S(f+f_{c})$

from the above figure, $S_{I}(f)$ results to be

from the frequency spectrum of $S_{I}(f)$ , the time-domain representation turns out to be $S_{I}(t)=\frac{1}{2}A_{c}m(t)-----EQN(I)$

Similarly,

The resultant signals $S(f-f_{c})-S(f+f_{c})$ and $S_{Q}(f)$

from the frequency spectrum of $S_{Q}(f)$ turns out to be

$S_{Q}(f)&space;=\left\{\begin{matrix}&space;\frac{1}{2j}A_{c}M(f),-B\leq&space;f&space;\leq&space;0&space;&&space;\\&space;-\frac{1}{2j}A_{c}M(f),0\leq&space;f&space;\leq&space;B&space;&&space;\end{matrix}\right.$

since Signum function is

$Sign(f)&space;=\left\{\begin{matrix}&space;+1,f>0&space;&&space;\\&space;-1&space;f<0&space;&&space;\end{matrix}\right.$

$S_{Q}(f)$ when expressed in terms of Signum function $s_{Q}(f)&space;=&space;\frac{j}{2}A_{c}M(f)(-sign(f))$

$s_{Q}(f)&space;=&space;(-jsign(f)M(f))\frac{A_{c}}{2}$

By using Hilbert transform of m(t) , the time-domain representation turns out to be $S_{Q}(t)=\frac{1}{2}A_{c}\widehat{m(t)}-----EQN(II)$

From EQN’s (I) and (II) , the time-domain representation of SSB-SC signal results

$S_{USB}(t)&space;=&space;\frac{A_{c}}{2}m(t)\cos&space;\omega&space;_{c}t-\frac{A_{c}}{2}\widehat{m(t)}\sin&space;\omega&space;_{c}t$.

similarly, SSB signal when only LSB has been transmitted

$S_{LSB}(t)&space;=&space;\frac{A_{c}}{2}m(t)\cos&space;\omega&space;_{c}t+\frac{A_{c}}{2}\widehat{m(t)}\sin&space;\omega&space;_{c}t$

(No Ratings Yet)

## Input and Output characteristics of transistor in Common Base Configuration

Input Characteristics:-

Input characteristics in Common Base configuration means input voltage Vs input current by keeping output voltage  as constant.

i.e, $V_{EB}$ Vs $I_{E}$ by keeping $V_{CB}$ constant.

Therefore the curve between Emitter current $I_{E}$ and Emitter to Base voltage $V_{EB}$ for a given value of Collector to Base voltage $V_{CB}$ represents input characteristic.

for a given output voltage  $V_{CB}$, the input circuit acts as a PN-junction diode under Forward Bias.

from the curves there exists a cut-in (or) offset (or) threshold voltage $V_{EB}$ below which the emitter current is very small  and a  substantial amount of Emitter-current flows after cut-in voltage ( 0.7 V for Si and 0.3 V for Ge).

the emitter current $I_{E}$ increases rapidly with the small increase in $V_{EB}$. with the low dynamic input resistance of a transistor.

i.e, $r_{i}&space;=\frac{\Delta&space;V_{EB}}{\Delta&space;I_{E}}|_{V_{CB}\approx&space;Constant}$

$input&space;resistance&space;=\frac{change&space;in&space;input&space;voltage}{change&space;in&space;emitter&space;current}|V_{CB}{\approx&space;Constant}$

This is calculated by measuring the slope of the input characteristic.

i.e, input characteristic determines the input resistance $r_{i}$.

The value of $r_{i}$ varies from point to point on the Non-linear portion of the characteristic and is about $100\Omega$ in the linear region.

Output Characteristics:-

Output Characteristics are in between output current Vs output voltage with input current as kept constant.

i.e, $f(I_{c},V_{CE})_{I_{E}&space;=&space;Constant}$

i.e, O/p characteristics are in between $V_{CB}$ Vs $I_{c}$ by keeping $I_{E}$ as constant.

basically it has 4 regions of operation Active region, saturation region,cut-off region and reach-through region.

active region:-

from the active region of operation $I_{c}$ is almost independent of $I_{E}$

i.e, $I_{c}\approx&space;I_{E}$

when $V_{CB}$ increases, there is very small increase in $I_{c}$ .

This is because the increase in $V_{CB}$ expands the collector-base depletion region and shortens the distance between the two depletion regions.

with $I_{E}$ kept constant the increase in $I_{c}$ is so small. transistor operates in it’s normal operation mode in this region.

saturation region:-

here both junctions are Forward Biased.

Collector current $I_{c}$ flows even when $V_{CB}=0$(left of origin)  and this current reaches to zero when $V_{CB}$ is increased negatively.

cut-off region:-

the region below the curve $I_{E}=0$ ,transistor operates in this region  when  the two junctions are Reverse Biased.

$I_{c}\neq&space;0$ even though $I_{E}=0$ mA.  this is because of collector leakage current (or) reverse-saturation current $I_{CO}$ (or) $I_{CBO}$.

punch through/reach through region:-

$I_{c}$ is practically independent of $V_{CB}$ over certain transistor operating region of the transistor.

• If $V_{CB}$ is increased beyond a certain value, $I_{c}$ eventually increases rapidly because of avalanche (or) zener effects (or) both this condition is known as punch through (or) reach through region.
• If transistor is operated beyond the specified output voltage ($V_{CB}$) transistor breakdown occurs.
• If $V_{CB}$ is increased beyond certain limit, the depletion region($J_{c}$) of o/p junction penetrates into the base until it makes contact with emitter-base depletion region. we call this condition as punch-through (or) reach-through effect.
• In this region , the large collector current destroys the transistor.
• To avoid this $V_{CB}$ should be kept in safe limits specified by the manufacturer

(No Ratings Yet)

## Early effect in Common Base Configuration

Early effect (or) Base-width modulation:-

In Common Base configuration in the Reverse Bias, As the voltage $V_{CC}$ increases, the space-charge width between collector and base tends to increase,  with the result that the effective width of the base decreases. This dependency of Base-width on the Collector to emitter voltage is known as the early effect.

The early effect has three consequences:-

1. There is less chance for recombination with in the base region. Hence $\alpha$ increases with increasing $\left&space;|&space;V_{CB}&space;\right&space;|$.
2. The charge gradient is increased with in the base and consequently, the current of minority carriers injected across the emitter junction increases.
3. For extremely large voltages, the effective Base-width may be reduced to zero, causing voltage break-down in the transistor. This phenomenon is called the Punch-through.

For higher values of $V_{CB}$, due to early effect the value of $\alpha$ increases, for example $\alpha$ changes say from 0.98 to 0.985. Hence there is a very small positive slope in the CB output characteristics and hence the output resistance is not zero.

(No Ratings Yet)

## Hall effect

When a transverse magnetic field ‘B’is applied to a specimen (of metal (or) Semi conductor) carrying a current Ian Electric field E is induced perpendicular to both I and B. This phenomenon is known as Hall effect.

The figure shows the  experimental arrangement to observe Hall effect  Now

I $\rightarrow$ Current flowing in the semi conductor (x-direction)

B$\rightarrow$ Applied Magnetic field (z-direction)

E$\rightarrow$ Induced Electric field is along y-direction perpendicular to both I and B.

Now charge carrier electron is moving under the influence of two fields both electric field(E) and Magnetic field(B).

i.e, electron is under the influence of both E and B, E applies some force on electron similarly B.

under equilibrium $F_{E}&space;=&space;F_{B}$

$qE&space;=&space;Bqv_{d}------EQN(I)$, where $v_{d}$ is the drift velocity

Electric field Intensity due to Hall effect is $E=\frac{V_{H}}{d}--------------EQN(II)$

$V_{H}$ is the Hall voltage between plates 1 and 2.

and d- is the distance between the two plates.

In an N-type Semi conductor, the current is due to electrons , plate 1 is negatively charged compared to plate 2.

The current density J related to charge density $\rho$ is $J&space;=&space;\rho&space;v_{d}------------EQN(III)$

$J&space;=&space;\frac{Current}{Area}=\frac{I}{A}=\frac{I}{Wd}$

W- width of the specimen, d- height of the specimen.

From EQN(I) $E=Bv_{d}$ and From EQN(II) $V_{H}=Ed$

up on multiplying with ‘d’ on both sides $E&space;d&space;=&space;Bd&space;v_{d}$

$V_{H}&space;=&space;Bd&space;v_{d}$

$V_{H}&space;=&space;B&space;d&space;\frac{J}{\rho&space;}$    from EQN(III)

$V_{H}&space;=&space;B\frac{I}{Wd\rho&space;}d$

$V_{H}&space;=&space;\frac{BI}{\rho&space;W}$

$V_{H}&space;=&space;\frac{1}{\rho&space;}&space;.&space;\frac{BI}{W}$, let Hall coefficient $R_{H}&space;=&space;\frac{1}{\rho&space;}$

$V_{H}&space;=&space;R_{H}.&space;\frac{BI}{W}$ .

Uses of Hall effect (or) Applications of Hall effect:-

• Hall effect specifies the type of semi conductor that is P-type (or) N-type.when $R_{H}$ is positive it’s a P-type semi conductor and  $R_{H}$  negative means  it’s  N-type semi conductor.
• It is used to find out carrier concentrations ‘n’ and ‘p’ , by using either $\rho&space;=&space;nq$  or $\rho&space;=pq$.
• To find out mobilities $\mu&space;_{n}$ and $\mu&space;_{p}$ using the equation $\mu&space;=\sigma&space;R_{H}$.
• Some other applications of Hall effect are measurement of velocity, sorting,limit sensing etc.
• used to measure a.c power and the strength of Magnetic field and also finds the angular position of static magnetic fields in a magnetic field meter.
• used in Hall effect multiplier, which gives the output proportional to product of two input signals.

(No Ratings Yet)

## Energy Density in Electrostatic Fields

To determine the Energy present in an assembly of charges (or) group of charges one must first determine the amount of work necessary to assemble them.

It is seen that , when a unit positive charge is moved from infinity to a point in a field, the work is done by the external source and energy is expended.

If the external source is removed then the unit positive charge will be subjected to a force exerted by the field and will be moved in the direction of force.

Thus to hold the charge at a point in an electrostatic field, an external source has to do work , this energy gets stored in the form of Potential Energy when the test charge is hold at a point in a field.

when external source is removed , the Potential Energy gets converted to a Kinetic Energy.

In order to derive the expression for energy stored in electrostatics (i.e, the expression of such a Potential Energy)

Consider an empty space where there is no electric field at all, the Charge $Q_{1}$ is moved from infinity to a point in the space ,let us say the point as $P_{1}$, this requires no work to be done to place a charge $Q_{1}$ from infinity to a point $P_{1}$ in empty space.

i.e, work done = 0 for placing a charge $Q_{1}$ from infinity to a point $P_{1}$ in empty space.

now another charge $Q_{2}$ has to be placed from infinity to another point $P_{2}$ . Now there has to do some work to place $Q_{2}$ at $P_{2}$ because there is an electric field , which is produced by the charge $Q_{1}$ and $Q_{2}$ is required to move against the field of $Q_{1}$.

Hence the work required to be done is  $Potential=\frac{work&space;done}{unit&space;charge}$

i.e, $V&space;=&space;\frac{W}{Q}$ $\Rightarrow&space;W&space;=&space;V&space;X&space;Q$ .

$\therefore$ Work done to position $Q_{2}$ at $P_{2}$ = $V_{21}&space;X&space;Q_{2}$.

Now the charge $Q_{3}$ to be moved from infinity to $P_{3}$ , there are electric fields due to $Q_{1}$ and $Q_{2}$, Hence total work done is due to potential at $P_{3}$ due to charge at $P_{1}$ and Potential at $P_{3}$ due to charge at $P_{2}$.

$\therefore$ Work done to position $Q_{3}$ at $P_{3}$ = $V_{31}Q_{3}+V_{32}Q_{3}$.

Similarly , to place a  charge $Q_{n}$ at $P_{n}$ in a field created by (n-1) charges is ,work done to position $Q_{n}$ at $P_{n}$$=V_{n1}Q_{n}+V_{n2}Q_{n}+V_{n3}Q_{n}+.......$

$\therefore$Total Work done $W_{E}&space;=Q_{2}V_{21}+Q_{3}V_{31}+Q_{3}V_{32}+Q_{4}V_{41}+Q_{4}V_{42}+Q_{4}V_{43}+....&space;EQN(I)$

The total work done is nothing but the Potential energy in the system of charges hence denoted as $W_{E}$,

if charges are placed in reverse order (i.e, first $Q_{4}$ and then $Q_{3}$ and then  $Q_{2}$  and finally $Q_{1}$ is placed)

work done to place $Q_{3}&space;\Rightarrow&space;V_{34}Q_{3}$

work done to place $Q_{2}&space;\Rightarrow&space;V_{24}Q_{2}+V_{23}Q_{2}$

work done to place $Q_{1}&space;\Rightarrow&space;V_{14}Q_{1}+V_{13}Q_{1}++V_{12}Q_{1}$

Total work done $W_{E}&space;=Q_{3}V_{34}+Q_{2}V_{24}+Q_{2}V_{23}+Q_{1}V_{14}+Q_{1}V_{13}+Q_{1}V_{12}+....&space;EQN(II)$

EQN (I)+EQN(II) gives

$2W_{E}&space;=Q_{1}(V_{12}+V_{13}+V_{14}+....+V_{1n})&space;+Q_{2}(V_{21}+V_{23}+V_{24}+....+V_{2n})+Q_{3}(V_{31}+V_{32}+V_{34}+....+V_{3n})+.....$

let $V_{1}=(V_{12}+V_{13}+V_{14}+....+V_{1n})$, $V_{2}=(V_{21}+V_{23}+V_{24}+....+V_{2n})$ and $V_{n}=(V_{n1}+V_{n2}+V_{n3}+....+V_{nn-1})$ are the resultant Potentials due to all the charges except that charge.

i.e, $V_{1}$ is the resultant potential due to all the charges except $Q_{1}$.

$2W_{E}&space;=&space;Q_{1}V_{1}+Q_{2}V_{2}+Q_{3}V_{3}+......+Q_{n}V_{n}$

$W_{E}&space;=\frac{1}{2}&space;\sum_{m=1}^{n}Q_{m}V_{m}$ Joules.

The above expression represents the Potential Energy stored in the system of n point charges.

simillarly,

$W_{E}&space;=&space;\frac{1}{2}\int_{l}\rho&space;_{l}dl.&space;V$  Joules

$W_{E}&space;=&space;\frac{1}{2}\int_{s}\rho&space;_{s}ds.&space;V$ Joules

$W_{E}&space;=&space;\frac{1}{2}\int_{v}\rho&space;_{v}dv.&space;V$ Joules  for different types of charge distributions.

(No Ratings Yet)

## Switches-Circuit Switching

Switches are used in Circuit-Switched and Packet-Switched Networks.  The switches are used are different depending up on the structure and usage.

Circuit Switches (or) Structure of Circuit Switch:-

The Switches used in Circuit Switching are called Circuit-Switches

Space-Division Switch:-

• The paths are separated spatially from one switch to other.
• These were originally designed for analog circuits but currently used for both analog and digital Networks.

Cross-bar Switch:-

In this type of Switch we connect n inputs and m outputs using micro switches (Transistors) at each cross point to form a cross-bar switch of size n X m.

The number of cross points required = n X m.

As n and m increases, cross points required also increases, for example n=1000 and m=1000  requires n X m= 1000 X 1000 cross points. A cross-bar with these many number of cross points is impractical and statics show that 25% of the cross points are in use at any given time.

Multi stage Switch:-

The solution to Cross-bar Switch is Multi stage switching. Multi stage switching is preferred over cross-bar switches to reduce the number of cross points.  Here number of cross-bar switches are combined in several stages.

Suppose an N X N cross-bar Switch can be made into 3 stage Multi bar switch as follows.

1.  N is divided into groups , that is N/n Cross-bars with n-input lines and k-output lines forms n X k cross points.
2. The second stage consists of k Cross-bar switches with each cross-bar switch size as (N/n) X (N/n).
3. The third stage consists of N/n cross-bar switches with each switch size as k X n.

The total number of cross points = $2kN&space;+&space;k&space;(\frac{N}{n})^{2}$, so the number of cross points required are less than single-stage cross-bar Switch = $N^{2}$.

for example k=2 and n=3 and N=9 then a Multi-stage switch looks like as follows.

The problem in Multi-stage switching is Blocking during periods of heavy traffic, the idea behind Multi stage switch is to share intermediate cross-bars. Blocking means times when one input line can not be connected to an output because there is no path available (all possible switches are occupied). Blocking generally occurs in tele phone systems and this blocking is due to intermediate switches.

Clos criteria gives a condition for a non-blocking Multi stage switch

$n&space;=&space;\sqrt{\frac{N}{2}}$$k\geq&space;(2n-1)$  and  Total no.of Cross points $\geq&space;4N(\sqrt{2N}-1)$.

(2 votes, average: 5.00 out of 5)

## GATE problems in EMT

pb. An Electro Magnetic Wave Propagates through a loss less insulator with a velocity $1.5&space;X&space;10^{10}$ Cm/sec . Calculate the electric magnetic properties of the insulator if its intrinsic impedance is $90\pi$ Ohms.

Ans:- Given loss less insulator  $\Rightarrow&space;\alpha&space;=0$

$v_{p}&space;=1.5&space;X&space;10^{10}$ Cm/Sec

$v_{p}&space;=&space;\frac{1}{\sqrt{\mu&space;_{o}\epsilon&space;_{o}\mu&space;_{r}\epsilon&space;_{r}}}$

$1.5&space;X&space;10^{10}=&space;\frac{1}{\sqrt{\mu&space;_{o}\epsilon&space;_{o}\mu&space;_{r}\epsilon&space;_{r}}}$

$1.5&space;X&space;10^{10}=&space;\frac{3&space;X&space;10^{10}}{\sqrt{\mu&space;_{r}\epsilon&space;_{r}}}$

${\sqrt{\mu&space;_{r}\epsilon&space;_{r}}}&space;=&space;2&space;------------EQNI$

from $\eta&space;=\sqrt{\frac{\mu&space;_{o}\mu&space;_{r&space;}}{\epsilon&space;_{o}\epsilon&space;_{r&space;}}}$

$90\pi&space;=&space;\sqrt{\frac{\mu&space;_{o}\mu&space;_{r&space;}}{\epsilon&space;_{o}\epsilon&space;_{r&space;}}}$

$90\pi&space;=&space;\sqrt{\frac{\mu&space;_{r&space;}}{\epsilon&space;_{r&space;}}}&space;.&space;120&space;\pi$

$\sqrt{\frac{\mu&space;_{r&space;}}{\epsilon&space;_{r&space;}}}&space;=\frac{3}{4}--------EQNII$

from Equations I and II $\mu&space;_{r}&space;=&space;\sqrt{\frac{\mu&space;_{r&space;}}{\epsilon&space;_{r&space;}}}&space;\sqrt{\mu&space;_{r&space;}\epsilon&space;_{r}}$

$\mu&space;_{r}&space;=&space;2&space;X&space;\frac{3}{4}$

$\mu&space;_{r}&space;=&space;1.5$

$\sqrt{\mu&space;_{r&space;}\epsilon&space;_{r&space;}}&space;=&space;2$

$\sqrt{1.5&space;\epsilon&space;_{r&space;}}&space;=&space;2$

$\epsilon&space;_{r}&space;=&space;2.66$

(1 votes, average: 5.00 out of 5)

## Solved Example problems in Electro Magnetic Theory

1. Convert Points P(1,3,5)  from Cartesian to Cylindrical and Spherical Co-ordinate system.

Ans. Given P(1,3,5) $\fn_cm&space;\Rightarrow&space;x=1,y=3,&space;z=5$

Cylindrical :- $\fn_cm&space;\phi&space;=tan^{-1}(\frac{y}{x})$

$\fn_cm&space;=tan^{-1}(\frac{3}{1})$

$\fn_cm&space;=75^{o}$

Similarly $\fn_cm&space;\rho&space;=&space;\sqrt{x^{2}+&space;y^{2}}&space;\Rightarrow&space;\sqrt{1^{2}+&space;3^{2}}&space;=&space;\sqrt{10}=3.16$

$\fn_cm&space;P(\rho&space;,\phi&space;,z)=&space;P(3.16,71.5^{o},5)$

Spherical :- $\fn_cm&space;r=&space;\sqrt{x^{2}+y^{2}+z^{2}}$

$\fn_cm&space;r=&space;\sqrt{1^{2}+3^{2}+5^{2}}$

$\fn_cm&space;r=&space;\sqrt{35}$

$\fn_cm&space;r=5.91$

$\fn_cm&space;\theta&space;=tan^{-1}(\frac{\sqrt{x^{2}+y^{2}}}{z})$

$\fn_cm&space;\theta&space;=tan^{-1}(\frac{\sqrt{1^{2}+3^{2}}}{5})$

$\fn_cm&space;\theta&space;=32.31^{o}$

$\fn_cm&space;\phi&space;=tan^{-1}(\frac{y}{x})$

$\fn_cm&space;\phi&space;=tan^{-1}(\frac{3}{1})$

$\fn_cm&space;\phi&space;=&space;75^{o}$

$\fn_cm&space;P(r,\theta&space;,\phi&space;)&space;=&space;P(5.91,32.31^{o},71.5^{o})$

## Phase Locked Loop (PLL)

Demodulation of an FM signal using PLL:-

Let the input to PLL is an FM signal $S(t)&space;=&space;A_{c}&space;\sin&space;(2&space;\pi&space;f_{c}t+2\pi&space;k_{f}&space;\int_{0}^{t}m(t)dt)$

let  $\Phi&space;_{1}&space;(t)&space;=&space;2\pi&space;k_{f}&space;\int_{0}^{t}m(t)dt&space;------Equation&space;(I)$

Now the signal at the output of VCO is FM signal (another FM signal, which is different from input FM signal) Since Voltage Controlled Oscillator is an FM generator.

$\therefore&space;b(t)&space;=&space;A_{v}&space;\cos&space;(2&space;\pi&space;f_{c}t+2\pi&space;k_{v}&space;\int_{0}^{t}v(t)dt)$

the corresponding phase    $\Phi&space;_{2}&space;(t)&space;=&space;2\pi&space;k_{v}&space;\int_{0}^{t}v(t)dt&space;------Equation&space;(II)$

It is observed that S(t) and b(t) are out of phase by $90^{o}$. Now these signals are applied to a phase detector , which is basically a multiplier

$\therefore$ the error signal $e(t)&space;=S(t)&space;.b(t)$

$e(t)&space;=A_{c}&space;\sin&space;(2&space;\pi&space;f_{c}t+2\pi&space;k_{f}&space;\int_{0}^{t}m(t)dt).&space;A_{v}&space;\cos&space;(2&space;\pi&space;f_{c}t+2\pi&space;k_{v}&space;\int_{0}^{t}v(t)dt)$

$e(t)&space;=A_{c}A_{v}&space;\sin&space;(2&space;\pi&space;f_{c}t+\phi&space;_{1}(t)).&space;\cos&space;(2&space;\pi&space;f_{c}t+\phi&space;_{2}(t))$

on further simplification , the product yields a higher frequency term (Sum) and a lower frequency term (difference)

$e(t)&space;=A_{c}A_{v}k_{m}&space;\sin&space;(4&space;\pi&space;f_{c}t+\phi&space;_{1}(t)+\phi&space;_{2}(t))-&space;A_{c}A_{v}k_{m}\sin&space;(\phi&space;_{1}(t)-\phi&space;_{2}(t))$

$e(t)&space;=A_{c}A_{v}k_{m}&space;\sin&space;(2&space;\omega&space;_{c}t+\phi&space;_{1}(t)+\phi&space;_{2}(t))-&space;A_{c}A_{v}k_{m}\sin&space;(\phi&space;_{1}(t)-\phi&space;_{2}(t))$

This product e(t) is given to a loop filter , Since the loop filter is a LPF it allows the difference and term and rejects the higher frequency term.

the over all output of a loop filter is

## Frequency domain representation of a Wide Band FM

To obtain the frequency-domain representation of Wide Band FM signal for the condition $\beta&space;>&space;>&space;1$ one must express the FM signal in complex representation (or) Phasor Notation (or) in the exponential form

i.e, Single-tone FM signal is $S_{FM}(t)=A_{c}cos(2\pi&space;f_{c}t+\beta&space;sin&space;2\pi&space;f_{m}t).$

Now by expressing the above signal in terms of  Phasor notation ($\because&space;\beta&space;>&space;>&space;1$ , None of the terms can be neglected)

$S_{FM}(t)&space;\simeq&space;Re(A_{c}e^{j(2\pi&space;f_{c}t+\beta&space;sin&space;2\pi&space;f_{m}t)})$

$S_{FM}(t)&space;\simeq&space;Re(A_{c}e^{j2\pi&space;f_{c}t}e^{j\beta&space;sin&space;2\pi&space;f_{m}t})$

$S_{FM}(t)&space;\simeq&space;Re(e^{j2\pi&space;f_{c}t}&space;A_{c}e^{j\beta&space;sin&space;2\pi&space;f_{m}t})-------Equation(I)$

Let    $\widetilde{s(t)}&space;=A_{c}e^{j\beta&space;sin&space;2\pi&space;f_{m}t}$      is the complex envelope of FM signal.

$\widetilde{s(t)}$ is a periodic function with period $\frac{1}{f_{m}}$ . This $\widetilde{s(t)}$ can be expressed in it’s Complex Fourier Series expansion.

i.e, $\widetilde{S(t)}&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}C_{n}&space;e^{jn\omega&space;_{m}t}$  this approximation is valid over $[-\frac{1}{2f_{m}},\frac{1}{2f_{m}}]$ . Now the Fourier Coefficient  $C_{n}&space;=&space;\frac{1}{T}&space;\int_{\frac{-T}{2}}^{\frac{T}{2}}&space;\widetilde{S(t)}&space;e^{-jn2\pi&space;f_{m}t}dt$

$T=&space;\frac{1}{f_{m}}$

$C_{n}&space;=&space;\frac{1}{\frac{1}{f_{m}}}&space;\int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}}&space;\widetilde{S(t)}&space;e^{-jn2\pi&space;f_{m}t}dt$

$C_{n}&space;=&space;f_{m}&space;\int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}}&space;A_{c}e^{j\beta&space;sin&space;2\pi&space;f_{m}t}&space;e^{-jn2\pi&space;f_{m}t}dt$

$C_{n}&space;=&space;f_{m}&space;\int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}}&space;A_{c}e^{{j\beta&space;sin&space;2\pi&space;f_{m}t-jn2\pi&space;f_{m}t}}dt$

$C_{n}&space;=&space;f_{m}&space;\int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}}&space;A_{c}e^{j({\beta&space;sin&space;2\pi&space;f_{m}t-n2\pi&space;f_{m}t})}dt$

let $x=2\pi&space;f_{m}t$       implies   $dx=2\pi&space;f_{m}dt$

as $x\rightarrow&space;\frac{-1}{2f_{m}}&space;\Rightarrow&space;t\rightarrow&space;-\pi$     and    $x\rightarrow&space;\frac{1}{2f_{m}}&space;\Rightarrow&space;t\rightarrow&space;\pi$

$C_{n}&space;=&space;\frac{A_{c}}{2\pi&space;}&space;\int_{-\pi&space;}^{\pi&space;}&space;e^{j({\beta&space;sin&space;x-nx})}dx$

let $J_{n}(\beta&space;)&space;=&space;\frac{1}{2\pi&space;}&space;\int_{-\pi&space;}^{\pi&space;}&space;e^{j({\beta&space;sin&space;x-nx})}dx$   as    $n^{th}$  order Bessel Function of first kind then   $C_{n}&space;=&space;A_{c}&space;J_{n}(\beta&space;)$.

Continuous Fourier Series  expansion of

$\widetilde{S(t)}&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}C_{n}&space;e^{jn\omega&space;_{m}t}$

$\widetilde{S(t)}&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}A_{c}&space;J_{n}&space;(\beta&space;)e^{jn\omega&space;_{m}t}$

Now substituting this in the Equation (I)

$S_{WBFM}(t)&space;\simeq&space;Re(e^{j2\pi&space;f_{c}t}&space;\sum_{n=-\infty&space;}^{\infty&space;}A_{c}&space;J_{n}&space;(\beta&space;)e^{jn\omega&space;_{m}t})$

$S_{WBFM}(t)&space;\simeq&space;A_{c}&space;Re(&space;\sum_{n=-\infty&space;}^{\infty&space;}J_{n}&space;(\beta&space;)&space;e^{j2\pi&space;f_{c}t}&space;e^{jn\omega&space;_{m}t})$

$S_{WBFM}(t)&space;\simeq&space;A_{c}&space;Re(&space;\sum_{n=-\infty&space;}^{\infty&space;}J_{n}&space;(\beta&space;)&space;e^{j2\pi&space;(f_{c}+nf&space;_{m}t)})$

$\therefore&space;S_{WBFM}(t)&space;\simeq&space;A_{c}&space;\sum_{n=-\infty&space;}^{\infty&space;}J_{n}&space;(\beta&space;)&space;cos&space;2\pi&space;(f_{c}+nf&space;_{m}t)$

The  Frequency spectrum  can be obtained by taking Fourier Transform

$S_{WBFM}(f)&space;=&space;\frac{A_{c}}{2}\sum_{n=-\infty&space;}^{\infty&space;}J_{n}(\beta&space;)&space;$

 n value wide Band FM signal 0 $S_{WBFM}(f)&space;=&space;\frac{A_{c}}{2}\sum_{n=-\infty&space;}^{\infty&space;}J_{0}(\beta&space;)&space;$ 1 $S_{WBFM}(f)&space;=&space;\frac{A_{c}}{2}\sum_{n=-\infty&space;}^{\infty&space;}J_{1}(\beta&space;)&space;$ -1 $S_{WBFM}(f)&space;=&space;\frac{A_{c}}{2}\sum_{n=-\infty&space;}^{\infty&space;}J_{-1}(\beta&space;)&space;$ … ….

From the above Equation it is clear that

• FM signal has infinite number of side bands at frequencies $(f_{c}\pm&space;nf_{m})$for n values changing from $-\infty$ to  $\infty$.
• The relative amplitudes of all the side bands depends on the value of  $J_{n}(\beta&space;)$.
• The number of significant side bands depends on the modulation index $\beta$.
• The average power of FM wave is $P=\frac{A_{c}^{2}}{2}$ Watts.

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## Mutual Information I(X ; Y) Properties

Property 1:- Mutual Information is Non-Negative

Mutual Information is given by equation $I(X&space;;&space;Y)&space;=\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i},&space;y_{j})\log&space;_{2}&space;\frac{P(\frac{x_{i}}{y_{j}})}{P(x_{i})}---------Equation(I)$

we know that $P(\frac{x_{i}}{y_{j}})=\frac{P(x_{i},&space;y_{j})}{P(y_{j})}-------Equation(II)$

Substitute Equation (II) in Equation (I)

$I(X&space;;&space;Y)&space;=\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i},&space;y_{j})\log&space;_{2}\frac{P(x_{i},&space;y_{j})}{P(x_{i})P(y_{j})}$

The above Equation can be written as

$I(X&space;;&space;Y)&space;=-\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i},&space;y_{j})\log&space;_{2}\frac{P(x_{i})P(y_{j})}{P(x_{i},&space;y_{j})}$

$-I(X&space;;&space;Y)&space;=\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i},&space;y_{j})\log&space;_{2}\frac{P(x_{i})P(y_{j})}{P(x_{i},&space;y_{j})}------Equation(III)$

we knew that $\sum_{k=1}^{m}&space;p_{k}\log&space;_{2}(\frac{q_{k}}{p_{k}})\leq&space;0---Equation(IV)$

This result can be applied to Mutual Information $I(X&space;;&space;Y)$ , If $p_{k}&space;=&space;P(x_{i},&space;y_{j})$ and $q_{k}$ be $P(x_{i})&space;P(&space;y_{j})$, Both $p_{k}$ and $q_{k}$ are two probability distributions on same alphabet , then Equation (III) becomes

$-I(X&space;;&space;Y)&space;\leq&space;0$

i.e, $I(X&space;;&space;Y)&space;\geq&space;0$  , Which implies that Mutual Information is always Non-negative (Positive).

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## Example Problems in Electro Magnetic Theory Wave propagation

1. A medium like Copper conductor which is characterized by the parameters $\bg_black&space;\sigma&space;=&space;5.8&space;X&space;10^{7}&space;Mho's/meter$ and $\epsilon&space;_{r}=1,\mu&space;_{r}=1$ uniform plane wave of frequency 50 Hz. Find $\alpha&space;,\beta&space;,v,\eta$  and $\lambda$.

Ans.  Given $\bg_black&space;\bg_black&space;\sigma&space;=&space;5.8&space;X&space;10^{7}&space;Mho's/meter$  ,     $\bg_black&space;\epsilon&space;_{r}=1,\mu&space;_{r}=1$    and $\bg_white&space;f=&space;50&space;Hz$

$\bg_white&space;\alpha&space;=?&space;,\beta&space;=?&space;,v&space;=&space;?,\eta&space;=?$ and $\bg_white&space;\lambda&space;=?$

Find the Loss tangent $\bg_white&space;\frac{\sigma&space;}{\omega&space;\epsilon&space;}&space;=&space;\frac{5.8X&space;10^{7}}{2&space;\pi&space;X50X\epsilon&space;_{o}\epsilon&space;_{r}}$

$\bg_white&space;\bg_white&space;\frac{\sigma&space;}{\omega&space;\epsilon&space;}&space;=&space;\frac{5.8X&space;10^{7}}{100\pi&space;X\epsilon&space;_{o}}$

$\bg_white&space;\bg_white&space;\frac{\sigma&space;}{\omega&space;\epsilon&space;}&space;=&space;2.08&space;X&space;10&space;^{16}>&space;>&space;1$

So given medium is a Conductor (Copper)

then $\bg_white&space;\alpha&space;(or)&space;\beta&space;=\sqrt{\frac{\omega&space;\mu&space;\sigma&space;}{2}}$

$\bg_white&space;=\sqrt{\frac{5.8X10^{7}X2\pi&space;X&space;50X\mu&space;_{o}}{2}}$

$\bg_white&space;\alpha&space;=&space;106.99$  , $\bg_white&space;\beta&space;=106.99$.

$\bg_white&space;v_{p}=\frac{\omega&space;}{\beta&space;}$  $\bg_white&space;=\frac{2\pi&space;X50}{106.99}$$\bg_white&space;=2.936&space;meters/Sec$.

$\bg_white&space;\lambda&space;=\frac{2\pi&space;}{\beta&space;}=\frac{2\pi&space;}{106.99}=0.0587&space;meters.$

$\bg_white&space;\eta&space;=\sqrt{\frac{j\omega&space;\mu&space;}{(\sigma&space;+j\omega&space;\epsilon&space;)}}$

$\bg_white&space;=\sqrt{\frac{jX2\pi&space;X50X\mu&space;_{o}}{(5.8X10^{7}+j2\pi&space;X50X\epsilon&space;_{o})}}$

$\bg_white&space;=&space;\sqrt{\frac{j&space;3.947&space;X10^{-4}}{(5.8X10^{7}+j&space;2.78&space;X10^{-9})}}$

$\bg_white&space;=&space;\sqrt{\frac{&space;3.947&space;X10^{-4}\angle&space;90^{o}}{(5.8X10^{7}\angle&space;-2.74&space;X&space;10^{-15})}}$

$\bg_white&space;=&space;\sqrt{0.68&space;X10&space;^{-11}}\angle&space;\frac{90-(2.74&space;X&space;10^{-5})}{2}$

$\eta&space;=&space;2.6&space;X&space;10^{-6}\angle&space;45^{o}$.

2. If $\bg_white&space;\epsilon&space;_{r}=9,\mu&space;=\mu&space;_{o}$ for a medium in which a wave with a frequency of $\bg_white&space;f=&space;0.3&space;GHz$ is propagating . Determine the propagation constant and intrinsic impedance of the medium when $\bg_white&space;\sigma&space;=0.$

Ans: Given $\bg_white&space;\epsilon&space;_{r}=9$,  $\bg_white&space;\mu&space;=\mu&space;_{o}$ , $\bg_white&space;f=0.3GHz$ and $\bg_white&space;\sigma&space;=0$.

$\bg_white&space;\gamma&space;=?,\eta&space;=?$

Since $\bg_white&space;\sigma&space;=0$, the given medium is a lossless Di-electric.

which implies $\bg_white&space;\alpha&space;=&space;\frac{\sigma&space;}{2}\sqrt{\frac{\mu&space;}{\epsilon&space;}}&space;=0.$

$\bg_white&space;\beta&space;=&space;\omega&space;\sqrt{\mu&space;\epsilon&space;}$

$\bg_white&space;=2\pi&space;X&space;o.3X10^{9}\sqrt{\mu&space;_{o}X9\epsilon&space;_{o}}$

$\bg_white&space;=&space;18.86$.

$\bg_white&space;\eta&space;=&space;\sqrt{\frac{\mu&space;}{\epsilon&space;}}$

$\bg_white&space;\eta&space;=&space;\sqrt{\frac{\mu_{o}\mu&space;_{r}&space;}{\epsilon_{o}\epsilon&space;_{r}&space;}}$

$\bg_white&space;\eta&space;=&space;\sqrt{\frac{\mu_{o}&space;}{9\epsilon_{o}&space;}}$

$\bg_white&space;\eta&space;=&space;\frac{120\pi&space;}{3}$

$\bg_white&space;\eta&space;=&space;40$ Ω.

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## Delta modulation and Demodulation

DM is done on an over sampled message signal in its basic form, DM provides a stair case approximated signal to the over sampled version of message signal.

i.e, Delta Modulation (DM) is a Modulation scheme in which an incoming  message signal is over sampled (i.e, at a rate much higher than the Nyquist rate $f_{s}>&space;2f_{m}$) to purposely increase the correlation between adjacent samples of the signal. Over sampling is done to permit the use of a sample Quantizing strategy for constructing the encoded signal.

Signaling rate and Transmission Band Width are quite large in PCM. DM is used to overcome these problems in PCM .

DM transmits one bit per sample.

The process of approximation in Delta Modulation is as follows:-

The difference between the input ($x[nT_{s}]$) and the approximation ($x$) is quantized into only two levels $\pm&space;\Delta$ corresponding to Positive and negative differences.

i.e, If the approximation  ($x$) falls below the signal ($x[nT_{s}]$)at any sampling epoch(the beginning of a period)output signal level is increased by $\Delta$.

On the other hand the approximation   ($x$) lies above the signal ($x[nT_{s}]$) , output signal level is diminished by $\Delta$ provided that the input signal does not change too rapidly from sample to sample.

it is observed that the  change in stair case approximation lies with in  $\pm&space;\Delta$ .

This process can be illustrated in the following figure

Delta Modulated System:- The DM system consists of Delta Modulator and Delta Demodulator.

Delta Modulator:-

Mathematical equations involved in DM Transmitter are

error signal: $e[nT_{s}]=x[nT_{s}]-x_{q}$

Present sample of the (input) sampled signal: $x[nT_{s}]$

last sample approximation of stair case signal: $x_{q}$

Quantized  error signal( output of one-bit Quantizer): $e_{q}[nT_{s}]$

if         $x[nT_{s}]\geq&space;x_{q}&space;\Rightarrow&space;e_{q}[nT_{s}]&space;=&space;\Delta$.

and  $x[nT_{s}]<&space;x_{q}&space;\Rightarrow&space;e_{q}[nT_{s}]&space;=&space;-\Delta$.

encoding has to be done on the after Quantization that is when the output level is increased by $\Delta$ from its previous quantized level, bit ‘1’ is transmitted .

similarly when output is diminished by $\Delta$ from the previous level  a ‘0’ is transmitted.

from the accumulator $x_{q}[nT_{s}]=x_{q}+&space;e_{q}[nT_{s}]$

$e_{q}[nT_{s}]=e[nT_{s}]+&space;q[nT_{s}]$

where $q[nT_{s}]$ is the Quantization error.

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## Fourier Series and it’s applications

The starting point of Fourier Series is the development of representation of signals as linear combination (sum of) of a set of basic signals.

$f(t)\approx&space;C_{1}x_{1}(t)+C_{2}x_{2}(t)+.......+C_{n}x_{n}(t)+....$

The alternative representation if  a set of complex exponentials are used,

$f(t)\approx&space;C_{1}e^{j\omega&space;_{o}t}+C_{2}e^{2j\omega&space;_{o}t}+.......+C_{n}e^{jn\omega&space;_{o}t}+....$

The resulting representations are known as Fourier Series in Continuous-Time . Here we focus on representation of Continuous-Time and Discrete-Time periodic signals in terms of basic signals as Fourier Series and extend the analysis to the Fourier Transform representation of broad classes of aperiodic, finite energy signals.

These Fourier Series & Fourier Transform representations are most powerful tools used

1. In the analyzation of signals and LTI systems.
2. Designing of Signals & Systems.
3. Gives insight to S&S.

The development of Fourier series analysis has a long history involving a great many individuals and the investigation of many different physical phenomena.

The concept of using “Trigonometric Sums”, that is sum of harmonically related sines and cosines (or) periodic complex exponentials are used to predict astronomical events.

Similarly, if we consider the vertical deflection $f(t,x)$ of the string at time t and at a distance x along the string then for any fixed instant of time, the normal modes are harmonically related sinusoidal functions of x.

The scientist Fourier’s work, which motivated him physically was the phenomenon of heat propagation and diffusion. So he found that the temperature distribution through a body can be represented by using harmonically related sinusoidal signals.

In addition to this he said that any periodic signal could be represented by such a series.

Fourier obtained a representation for aperiodic (or) non-periodic signals not as weighted sum of harmonically related sinusoidals but as weighted integrals of Sinusoids that are not harmonically related, which is known as Fourier Integral (or) Fourier Transform.

In mathematics, we use the analysis of Fourier Series and Integrals in

1. The theory of Integration.
2. Point-set topology.
3. and in the eigen function expansions.

In addition to the original studies of vibration and heat diffusion, there are numerous other problems in science and Engineering in which sinusoidal signals arise naturally, and therefore Fourier Series and Fourier T/F’s plays an important role.

for example, Sine signals arise naturally in describing the motion of the planets and the periodic behavior of the earth’s climate.

A.C current sources generate sinusoidal signals as voltages and currents. As we will see the tools of Fourier analysis enable us to analyze the response of an LTI system such as a circuit to such Sine inputs.

Waves in the ocean consists of the linear combination of sine waves with different spatial periods (or) wave lengths.

Signals transmitted by radio and T.V stations are sinusoidal in nature as well.

The problems of mathematical physics focus on phenomena in Continuous Time, the tools of Fourier analysis for DT signals and systems have their own distinct historical roots and equally rich set of applications.

In particular, DT concepts and methods are fundamental to the discipline of numerical analysis , formulas for the processing of discrete sets of data points to produce numerical approximations for interpolation and differentiation were being investigated.

FFT known as Fast Fourier Transform algorithm was developed, which suited perfectly for efficient digital implementation and it reduced the time required to compute transform by orders of magnitude (which utilizes the DTFS and DTFT practically).

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## Block Diagram of Digital Communication System/Elements of DCS

A General Communication System can be viewed as a Transmitting unit and a Receiving Unit connected by a medium(Channel). Obviously Transmitter and Receiver consists of various sub systems (or) blocks.

Our basic aim is to understand the various modules and sub systems in the system. If we are trying to understand the design and various features of DCS , it is plus imperative that we have to understand how we should design a transmitter and we must understand how to design a very good quality Receiver. Therefore one must know the features of the channel to design a good Transmitter as well as receiver that is the channel and it’s contribution will come repeatedly in digital Communications.

Source:- the primary block (or) the starting point of a DCS is an information source, it may be an analog/digital source , for example the signal considered is analog in nature, then the signal generated by the source is some kind of electrical signal which is random in nature. if the signal is a speech signal (not an electrical signal) that has to be converted into electrical signal by means of a Transducer, which can be considered as a part of source itself.

Sampling & Quantization:- the secondary block involves the conversion of analog to discrete signal

this involves the following steps

Sampling:- it is the process that involves in the conversion of Continuous Amplitude Continuous Time (CACT) signal into Continuous Amplitude Discrete Time (CADT) signal.

Quantization:- it is the process that involves in the conversion of Continuous Amplitude Discrete Time (CADT) signal into Discrete Amplitude Discrete Time (DADT) signal.

Source Encoder:-  An important problem in  Digital Communications is the efficient representation of data generated by a Discrete Source, this is accomplished by source encoder.

” The process of representation of incoming data  from a Discrete source into a more suitable form required for Transmission is known as source encoding”

Note:-The blocks Sampler, Quantizer followed by an Encoder constructs ADC (Analog to Digital Converter).

∴ the output of Source encoder is a Digital Signal, the advantages of Source coding are

• It reduces the Redundancy.
• Minimizes the average bit rate.

Channel encoder:-Channel coding is also known as error control coding. Channel coding is a technique which reduces the probability of error $P_{e}$ by reducing Signal to Noise Ratio at the expense of Transmission Band Width.The device that performs the channel coding is known as Channel encoder.

Channel encoding increases the redundancy of incoming data , this also involves error detection and error correction  along with the channel decoder at the receiver.

Spreading Techniques:- Spread Spectrum techniques are the methods by which a signal generated with a particular Band Width is deliberately spread in the frequency domain, resulting in a signal with a wider Band width.

There are two types of spreading techniques available

1. Direct Sequence Spread Spectrum Techniques.
2. Frequency Hopping Spread Spectrum Techniques.

The output of a spreaded signal is very much larger than incoming sequence. Spreading increases the BW required for transmission, which is a disadvantage even though spreading is done for high security of data.

SS techniques are used in Military applications.

Modulator:- Spreaded sequence is modulated by using digital modulation schemes like ASK, PSK, FSK etc depending up on the requirement, now the transmitting antenna transmits the modulated data into the channel.

Receiver:- Once you understood the process involved in transmitter Block. One should perform reverse operations in the receiver block.

i.e the input of the demodulator is demodulated after that de-spreaded and then the channel decoder removes the redundancy added by the channel encoder ,the output of channel decoder is then source decoded and is given to Destination.

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## Why Digital Communication is preferred over Analog Communication?

Introduction:-

Communication is the process of establishing Connection (or) link between two points (which are separated by some distance) and transporting information between those two points. The electronic equipment used for communication purpose is called Communication equipment. The equipment when assembled together forms a communication system.

Examples of different types of communications are

• Line Telephony & Telegraphy.
• Point-to-Point Communication.
• Mobile Communication.
• Radar and Satellite Communications etc.

Why Digital?

A General Communication system has two devices and a medium (channel) connecting those two devices. This can be understood that a Transmitter and Receiver are separated by a medium called as Communication channel. To transport an information-bearing signal from one point to another point over a communication channel either Analog or digital modulation techniques are used.

Now Coming to the point, Why Digital communication is preferred over analog Communication?

Why are communication systems, military and commercial alike, going digital?

1. There are many reasons; the primary advantage is the ease with which digital signals compared with analog signals are generated. That is the generation of digital signals is much easier compared to analog signals.
2. Propagation of Digital pulse through a Transmission line:-

When an ideal binary digital pulse propagating along a Transmission line. The shape of the waveform is affected by two mechanisms

• Distortion caused on the ideal pulse because all Transmission lines and Circuits have some Non-ideal frequency Transfer function.
• Unwanted electrical noise (or) other interference further distorts the pulse wave form.

Both of these mechanisms cause the pulse shape to degrade as a function of line length. During the time that the transmitted pulse can still be reliably identified (i.e. before it is degraded to an ambiguous state). The pulse is amplified by a digital amplifier that recovers its original ideal shape. The pulse is thus “re-born” (or) regenerated.

Circuits that perform this function at regular intervals along Transmission system are called “regenerative repeaters’. This is one of the reasons why Digital is preferred over Analog.

3.Digital Circuits Vs Analog Circuits:-

Digital Circuits are less subject to distortion and Interference than are analog circuits because binary digital circuits operate in one of two states FULLY ON (or) FULLY OFF to be meaningful, a disturbance must be large enough to change the circuit operating point from one state to another. Such two state operation facilitates signal representation and thus prevents noise and other disturbances from accumulating in transmission.

However, analog signals are not two-state signals, they can take an infinite variety of shapes with analog circuits and even a small disturbance can render the reproduced wave form unacceptably distorted. Once the analog signal is distorted, the distortion cannot be removed by amplification because accumulated noise is irrecoverably bound to analog signals, they cannot be perfectly generated.

4. With digital techniques, extremely low error rates and high signal fidelity is possible through error detection and correction but similar procedures are not available with analog techniques.

5.  Digital circuits are more reliable and can be produced at a lower cost than analog circuits also; digital hardware lends itself to more flexible implementation than analog hardware.

Ex: – Microprocessors, Digital switching and large scale Integrated circuits.

6. The combining of Digital signals using Time Division Multiplexing (TDM) is simpler than the combining of analog signals using Frequency Division Multiplexing (FDM).

7. Digital techniques lend themselves naturally to signal processing functions that protect against interference and jamming (or) that provide encryption and privacy and also much data communication is from computer to computer (or) from digital instruments (or) terminal to computer, such digital terminations are normally best served by Digital Communication links.

8. Digital systems tend to be very signal-processing intensive compared with analog systems.

Apart from pros there exists a con in Digital Communications that is non-graceful degradation when the SNR drops below a certain threshold, the quality of service can change suddenly from very good to very poor. In contrast most analog Communication Systems degrade more gracefully.

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## Maxwell’s Equations in Point (or Differential form) and Integral form

##### Maxwell’s Equations for time-varying fields in point and Integral form are:
1. $\overrightarrow{\bigtriangledown&space;}X\overrightarrow{H}=\overrightarrow{J}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$      $\Rightarrow&space;\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}+\int_{s}\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}.\overrightarrow{ds}$.
2. $\overrightarrow{\bigtriangledown&space;}X\overrightarrow{E}=-\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$       $\Rightarrow&space;\oint_{l}\overrightarrow{E}.\overrightarrow{dl}=-\int_{s}\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}.\overrightarrow{ds}$ .
3. $\overline{\bigtriangledown&space;}.\overrightarrow{D}=\rho&space;_{v}$            $\Rightarrow&space;\oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho&space;_{v}dv$.
4. $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{B}=0$      $\Rightarrow&space;\oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0$.

The 4 Equations above are known as Maxwell’s Equations. Since Maxwell contributed to their development and establishes them as a self-consistent set.  Each differential Equation has its integral part. One form may be derived from the other with the help of Stoke’s theorem (or) Divergence theorem.

##### word statements of the field Equations:-

A word statement of the field Equations is readily obtained from their mathematical statement in the integral form.

1.$\overrightarrow{\bigtriangledown&space;}X\overrightarrow{H}=\overrightarrow{J}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$ $\Rightarrow&space;\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}+\int_{s}\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}.\overrightarrow{ds}$.

i.e, The magneto motive force ($\because&space;\oint_{l}\overrightarrow{H}.\overrightarrow{dl}\rightarrow$ is m.m.f)around a closed path is equal to the conduction current plus the time derivative of the electric displacement through any surface bounded by the path.

2. $\overrightarrow{\bigtriangledown&space;}X\overrightarrow{E}=-\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$$\Rightarrow&space;\oint_{l}\overrightarrow{E}.\overrightarrow{dl}=-\int_{s}\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}.\overrightarrow{ds}$.

The electro motive force ($\because&space;\oint_{l}\overrightarrow{E}.\overrightarrow{dl}\rightarrow$ is e.m.f)around a closed path is equal to the time derivative of the magnetic displacement through any surface bounded by the path.

3.$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{D}=\rho&space;_{v}$  $\Rightarrow&space;\oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho&space;_{v}dv$.

The total electric displacement through the surface enclosing a volume is equal to the total charge within the volume.

4.    $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{B}=0$  $\Rightarrow&space;\oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0$.

The net magnetic flux emerging through any close surface is zero.

the time-derivative of electric displacement is called displacement current. The term electric current is then to include both conduction current and displacement current. If the time-derivative of electric displacement is called an electric current, similarly $\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$ is known as magnetic current, e.m.f as electric voltage and m.m.f as magnetic voltage.

the first two Maxwell’s Equations can be stated as

1. The magnetic voltage around a closed path is equal to the electric current through the path.
2. The electric voltage around a closed path is equal to the magnetic current through the path.
##### Maxwell’s Equations for static fields in point and Integral form are:

Maxwell’s Equations of static-fields in differential form and integral form are:

1. $\overrightarrow{\bigtriangledown&space;}&space;X\overrightarrow{H}=\overrightarrow{J}$        $\Rightarrow&space;\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}$.
2. $\overline{\bigtriangledown&space;}&space;X\overrightarrow{E}=0$           $\Rightarrow&space;\oint_{l}\overrightarrow{E}.\overrightarrow{dl}=0$.
3. $\overline{\bigtriangledown&space;}.\overrightarrow{D}&space;=&space;\rho&space;_{v}$            $\Rightarrow&space;\oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho&space;_{v}dv$.
4. $\overline{\bigtriangledown&space;}.\overrightarrow{B}&space;=&space;0$             $\Rightarrow&space;\oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0$.

(2 votes, average: 5.00 out of 5)

## Inconsistensy in Ampere’s law (or) Displacement Current density

Faraday’s experimental law has been used to obtain one of Maxwell’s equations in differential form $\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{E}=-\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$ , which shows that a time-varying Magnetic field produces an Electric field.

From Ampere’s  Circuital law which is applicable to Steady Magnetic fields

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}=\overrightarrow{J}$

By taking divergence of Ampere’s law the Ampere’s law is not consistent with time-varying fields

$\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H})=\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}$

$\overrightarrow{\bigtriangledown&space;}&space;.&space;\overrightarrow{J}=0$ ,since $\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H})=0---------Equation(1)$

the divergence of the curl is identically zero which implies $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}=0------Equation(2)$, but from the continuity equation $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;=&space;-\frac{\partial&space;\rho&space;_{v}}{\partial&space;t}-------Equation(3)$ which is not equal to zero, as $\frac{\partial&space;\rho&space;_{v}}{\partial&space;t}\neq&space;0$ is an unrealistic limitation(i.e we can not assume $\frac{\partial&space;\rho&space;_{v}}{\partial&space;t}$ as zero) .

$\therefore$ to make a compromise between the above two situations we must add an unknown term $\overrightarrow{G}$ to Ampere’s Circuital law

i.e, $\overrightarrow{\bigtriangledown&space;}X\overrightarrow{H}&space;=&space;\overrightarrow{J}+\overrightarrow{G}$

then by taking the Divergence of the above equation

$\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{\bigtriangledown&space;}X\overrightarrow{H})&space;=&space;\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}+\overrightarrow{\bigtriangledown&space;}.\overrightarrow{G}------Equation(4)$

from Equation(1),Equation(4) becomes     $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}+\overrightarrow{\bigtriangledown&space;}.\overrightarrow{G}=0$

$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{G}=-\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}$

thus $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{G}&space;=&space;\frac{\partial&space;\rho&space;_{v}}{\partial&space;t}---------Equation(5)$

from Maxwell’s first Equation $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{D}=\rho&space;_{v}$

then Equation (5) becomes $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}&space;=&space;\frac{\partial&space;}{\partial&space;t}&space;(\overrightarrow{\bigtriangledown&space;}.\overrightarrow{D})$

$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{G}&space;=\overrightarrow{\bigtriangledown&space;}.&space;\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

then   $\overrightarrow{G}&space;=&space;\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

$\overrightarrow{\bigtriangledown&space;}X\overrightarrow{H}&space;=&space;\overrightarrow{J}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

This is the equation obtained which does not disagree with the continuity equation. It is also consistent with all other results. This is a second Maxwell’s Equation is time-varying fields so the term $\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$ has the dimensions of current density Amperes/Square-meter. Since it results from a time-varying electric flux density ($\overrightarrow{D}$ ) , Maxwell termed it as displacement current density $\overrightarrow{J_{D}}$.

$\overrightarrow{\bigtriangledown&space;}X\overrightarrow{H}&space;=&space;\overrightarrow{J}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

$\overrightarrow{\bigtriangledown&space;}X\overrightarrow{H}&space;=&space;\overrightarrow{J}+\overrightarrow{J_{D}}$

$\overrightarrow{J_{D}}=\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

up to this point three current densities are there $\overrightarrow{J}=\sigma&space;\overrightarrow{E}$ , $\overrightarrow{J}=\rho&space;_{v}&space;\overrightarrow{v}$ and $\overrightarrow{J_{D}}=&space;\frac{\partial\overrightarrow{D}&space;}{\partial&space;t}$.

when the medium is Non-conducting medium $\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}=\frac{\partial\overrightarrow{D}&space;}{\partial&space;t}$

the total displacement current crossing any given surface is expressed by the surface integral $I_{d}&space;=&space;\oint_{s}&space;\overrightarrow{J_{D}}.\overrightarrow{ds}$

$I_{d}&space;=&space;\oint_{s}&space;\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}.\overrightarrow{ds}$

from Ampere’s law $\oint_{s}(\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}).\overrightarrow{ds}=\int_{s}&space;\overrightarrow{J}.\overrightarrow{ds}&space;+\oint_{s}&space;\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}.\overrightarrow{ds}$

$\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\int_{s}&space;\overrightarrow{J}.\overrightarrow{ds}&space;+\oint_{s}&space;\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}.\overrightarrow{ds}$

$\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=I&space;+\oint_{s}&space;\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}.\overrightarrow{ds}$

$\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=I&space;+I_{d}$

(1 votes, average: 5.00 out of 5)

we know that one form of Optimum filter is Matched filter, we will now derive another form of Optimum filter that is different from Matched filter Let the input to the Optimum filter is $v(t)$ which is a noisy input that is $v(t)=x(t)+n(t)$

from the figure output of the filter after sampling at $t=T_{b}$ seconds is $v_{o}(T_{b})$

$v_{o}(t)=v(t)*h(t)$

$v_{o}(t)&space;=&space;\int_{-\infty&space;}^{T_{b}}v(\tau&space;)&space;h(t-\tau&space;)d\tau$

at $t=&space;T_{b}$ output becomes  $v_{o}(T_{b})&space;=&space;\int_{-\infty&space;}^{T_{b}}v(\tau&space;)&space;h(T_{b}-\tau&space;)d\tau&space;-----Equation(1)$

Now by substituting $h(\tau&space;)=x_{2}(T_{b}-\tau&space;)-x_{1}(T_{b}-\tau&space;)$

$h(T_{b}-\tau&space;)=x_{2}(T_{b}-T_{b}+\tau&space;)-x_{1}(T_{b}-T_{b}+\tau&space;)$

$h(T_{b}-\tau&space;)=x_{2}(\tau&space;)-x_{1}(\tau&space;)------Equation(2)$

by substituting the  Equation(2)  in Equation(1) over the limits $[0,T_{b}]$

$v_{o}(T_{b})&space;=&space;\int_{0&space;}^{T_{b}}v(\tau&space;)&space;(x_{2}(\tau&space;)-x_{1}(\tau&space;))d\tau$

$v_{o}(T_{b})&space;=&space;\int_{0&space;}^{T_{b}}v(\tau&space;)&space;x_{2}(\tau&space;)&space;d\tau-\int_{0}^{T_{b}}v(\tau&space;)x_{1}(\tau&space;)d\tau$

Now by replacing $\tau$ with $t$ the above equation becomes

$v_{o}(T_{b})&space;=&space;\int_{0&space;}^{T_{b}}v(t&space;)&space;x_{2}(t&space;)&space;dt-\int_{0}^{T_{b}}v(t&space;)x_{1}(t&space;)dt-----Equation(3)$

The equation (3) suggests that the Optimum Receiver can be implemented as shown in the figure, this form of the Receiver is called  as correlation Receiver. This receiver requires the integration operation be ideal with zero initial conditions. Correlation Receiver performs coherent-detection.

in general Correlation Receiver can be approximated with Integrate and dump filter.

(2 votes, average: 4.00 out of 5)

## Relation between E and V

The potential difference between two points A and B is $V_{AB}=&space;V_{A}-V_{B}$ and $V_{BA}=&space;V_{B}-V_{A}$.

i.e, $V_{AB}=-V_{BA}$

$V_{AB}+V_{BA}=0$.

This equation implies that the total work done in moving a charge from A to B $(V_{AB})$and then from B to A $(V_{BA})$ is zero.

i.e, $\oint_{l}\overrightarrow{E}.\overrightarrow{dl}=0$

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## Electric Potential (V)

Electric field intensity $\overrightarrow{E}$ can be calculated by using either Coulomb’s law/Gauss’s law . when the charge distribution is symmetric another way of obtaining $\overrightarrow{E}$ is from the electric scalar potential V

Assume a test charge $Q_{t}$ at A in an Electric field, let points A and B are located at $r_{A}and&space;r_{B}$ units from the origin O,from Coulomb’s law the force acting on a test charge $Q_{t}$ is $\overrightarrow{F}=&space;Q_{t}\overrightarrow{E}$

The work done in moving a point charge $Q_{t}$ along a differential length $\overrightarrow{dl}$ is $dW$ is given by $dW&space;=&space;-\overrightarrow{F}.\overrightarrow{dl}$

$dW&space;=&space;-Q_{t}\overrightarrow{E}.\overrightarrow{dl}$

so the total work done in moving a point charge $Q_{t}$ from A to B is $W=-Q_{t}\int_{A}^{B}\overrightarrow{E}.\overrightarrow{dl}$

the direction of work done is always opposite to the direction of displacement.

where A is the initial point and B is the final point. Dividing the work done by the charge $Q_{t}$ gives the potential energy per unit charge denoted by $V_{AB}$,this is also known as potential difference between  the two points A and B.

Thus $V_{AB}&space;=&space;\frac{W}{Q_{t}}=&space;-\int_{A}^{B}\overrightarrow{E}.\overrightarrow{dl}$

if we take B as initial point and A as final point , then $V_{BA}&space;=&space;\frac{W}{Q_{t}}=&space;-\int_{B}^{A}\overrightarrow{E}.\overrightarrow{dl}----Equation(1)$

To derive the expression for V in terms of charge Q and distance r , we can use the concept of Electric field intensity $\overrightarrow{E}$ produced by a charge Q, which is placed at a distance r

i.e, $\overrightarrow{E}&space;=&space;\frac{Q}{4\pi&space;\epsilon&space;_{o}r^{2}}\overrightarrow{a_{r}}$

from Equation(1) $V_{BA}&space;=&space;-\int_{B}^{A}\overrightarrow{E}.\overrightarrow{dl}$

$V_{AB}=&space;-\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi&space;\epsilon&space;_{o}r^{2}}\overrightarrow{a_{r}}.dr&space;\overrightarrow{a_{r}}$  since $\overrightarrow{dl}=dr.\overrightarrow{a_{r}}$

$V_{AB}=&space;-\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi&space;\epsilon&space;_{o}r^{2}}dr$

$V_{AB}=&space;-\frac{Q}{4\pi&space;\epsilon&space;_{o}}\left&space;[&space;\frac{-1}{r}&space;\right&space;]_{r_{A}}^{r_{B}}$

$V_{AB}=&space;-\frac{Q}{4\pi&space;\epsilon&space;_{o}}$

$V_{AB}=V_{B}-V_{A}$

similarly, $V_{BA}=V_{A}-V_{B}$

where $V_{A}$ and $V_{B}$ are the scalar potentials at the points A and B respectively. If A is  located at $\infty$ with respect to origin ,with zero potential $V_{A}&space;=0$ and B is located at a distance r with respect to origin. then the work done in moving a charge from  A (infinity) to B is given by

$V_{AB}&space;=&space;\frac{Q}{4\pi&space;\epsilon&space;_{o}r_{B}}$  here $r_{B}&space;=&space;r$

$\therefore&space;V&space;=&space;\frac{Q}{4\pi&space;\epsilon&space;_{o}r}$ volts.

hence the potential at any point is the potential difference between that point and a chosen point at which the potential is zero. In other words assuming Zero potential at infinity .

The potential at a distance r from a point charge is the work done per unit charge by an external agent in transferring a test charge from infinity to that point.

i.e, $V&space;=&space;-\int_{\infty&space;}^{r}\overrightarrow{E}.\overrightarrow{dl}$

So a point charge $Q_{1}$ located at a point P with position vector $\overline{r_{1}}$ then the potential at another point Q with a position vector $\overline{r}$ is

$V_{at&space;\overline{r}}&space;=&space;\frac{Q_{1}}{4\pi&space;\epsilon&space;_{o}\left&space;|&space;\overline{r}&space;-\overline{r_{1}}\right&space;|}$

As like $\overrightarrow{E}$ superposition principle is applicable to V also that is for n point charges $Q_{1},Q_{2},Q_{3},Q_{4},........Q_{n}$ located at points with position vectors  $\overline{r_{1}},\overline{r_{2}},\overline{r_{3}},.......\overline{r_{n}}$

then the potential at $\overline{r}$ is

$V_{at&space;\overline{r}}&space;=&space;\frac{Q_{1}}{4\pi&space;\epsilon&space;_{o}\left&space;|&space;\overline{r}&space;-\overline{r_{1}}\right&space;|}+\frac{Q_{2}}{4\pi&space;\epsilon&space;_{o}\left&space;|&space;\overline{r}&space;-\overline{r_{2}}\right&space;|}+\frac{Q_{3}}{4\pi&space;\epsilon&space;_{o}\left&space;|&space;\overline{r}&space;-\overline{r_{3}}\right&space;|}+........+\frac{Q_{n}}{4\pi&space;\epsilon&space;_{o}\left&space;|&space;\overline{r}&space;-\overline{r_{n}}\right&space;|}$

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## few problems on Auto correlatioon Function(ACF) and Energy Spectral Density(ESD)

1. Find the Auto correlation function of $x(t)&space;=&space;\frac{1}{\sqrt{2\pi&space;}}\exp&space;^{\frac{-t^{2}}{2}}$.

Ans. We know that  Auto correlation function forms fourier transform pair with Energy Spectral Density function

$ACF\leftrightarrow&space;ESD$

$R_{xx}(t)\leftrightarrow&space;S(f)$

the Fourier Transform of  $e^{-ct^{2}}\leftrightarrow&space;\frac{\sqrt{\pi&space;}}{c}e^{-\pi&space;^{2}f^{2}}$

$\frac{1}{\sqrt{2\pi&space;}}e^{\frac{-t^{2}}{2}}\leftrightarrow&space;\frac{1}{\sqrt{2\pi&space;}}.\sqrt{\frac{\pi&space;}{(1/2)}}e^{\frac{-\pi&space;^{2}f^{2}}{(1/2)}}$ here $c&space;=&space;\frac{1}{2}$

$\frac{1}{\sqrt{2\pi&space;}}e^{\frac{-t^{2}}{2}}\leftrightarrow&space;\frac{1}{\sqrt{2\pi&space;}}.\sqrt{2\pi&space;}e^{-2&space;\pi&space;^{2}f^{2}}$

$\frac{1}{\sqrt{2\pi&space;}}e^{\frac{-t^{2}}{2}}\leftrightarrow&space;e^{-2&space;\pi&space;^{2}f^{2}}$

$x(t)\leftrightarrow&space;X(f)$

$\therefore$ the Fourier Transform of x(t) is X(f) and is $X(f)&space;=&space;e^{-2\pi&space;^{2}f^{2}}$ and the Energy Spectral Density $S(f)&space;=&space;\left&space;|&space;X(f)&space;\right&space;|^{2}$

$S(f)&space;=&space;e^{-4\pi&space;^{2}f^{2}}$

By finding the inverse Fourier Transform of S(f) gives the Auto Correlation Function

$S(f)&space;=&space;e^{\frac{-\pi&space;^{2}f^{2}}{(1/4)}}$

$e^{\frac{-\pi&space;^{2}f^{2}}{(1/4)}}\leftrightarrow&space;\frac{e^{\frac{-t^{2}}{4}}}{\sqrt{4\pi&space;}}$

$e^{\frac{-\pi&space;^{2}f^{2}}{(1/4)}}\leftrightarrow&space;\frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi&space;}}$

$\therefore$ the ACF of the given signal is inverse Fourier Transform of S(f) which is $R_{xx}(t)&space;=&space;\frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi&space;}}$.

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## Capture effect in Frequency Modulation

The Amplitude Modulation schemes like AM,DSB-SC and SSB-SC systems can not handle inherent Non-linearities in a really good manner where as FM can handle it very well.

Let us suppose un Modulated FM carrier $S(t)&space;=&space;A_{c}cos\omega&space;_{c}(t)$

$S(t)&space;=&space;A_{c}cos(\omega&space;_{c}(t)+\phi&space;(t))$

By considering un modulated FM carrier in terms of frequency(by neglecting phase) i.e $S(t)&space;=&space;A_{c}cos&space;(\omega&space;_{c}t)$ has been interfered by a near by interference located at a frequency $(\omega&space;_{c}+\omega&space;)$ where $\omega$ is a small deviation from $\omega&space;_{c}$.

the nearby inerference is $I&space;cos(\omega&space;_{c}&space;+&space;\omega&space;)t$

when the original signal got interfered by this near by interference , the received signal is $r(t)=&space;A_{c}cos&space;\omega&space;_{c}t&space;+&space;I&space;cos(\omega&space;_{c}+\omega&space;)t$

$r(t)=&space;(A+&space;I&space;cos&space;\omega&space;t)cos&space;\omega&space;_{c}t&space;-I&space;sin&space;\omega&space;t&space;sin\omega&space;_{c}t$   Let $A_{c}=A$

$r(t)&space;=&space;E_{r}(t)&space;cos&space;(\omega&space;_{c}t+\Psi&space;_{d}(t))$

now the phase of the signal is $\Psi&space;_{d}(t)&space;=&space;tan^{-1}&space;(\frac{I&space;sin&space;\omega&space;t}{A+I&space;cos&space;\omega&space;t&space;})$

as $A>&space;>&space;I$ implies $\frac{I}{A}<&space;<&space;1$

$\Psi&space;_{d}(t)&space;=&space;tan^{-1}&space;(\frac{I&space;sin&space;\omega&space;t}{A})$

since $\frac{I}{A}<&space;<&space;1$ , $\tan&space;^{-1}\theta&space;=&space;\theta$

$\Psi&space;_{d}(t)&space;\approx&space;\frac{I&space;sin&space;\omega&space;t}{A}$

As the demodulated signal is the output of a discriminator $y&space;_{d}(t)&space;=\frac{d}{dt}&space;(\frac{I&space;sin&space;\omega&space;t}{A})$

$y&space;_{d}(t)&space;=\frac{I\omega&space;}{A}&space;({cos&space;\omega&space;t})$ , which is the detected at the output of the demodulator.

the detected output at the demodulator is $y_{d}(t)$ in the absence of message signal  i.e, $m(t)=0$.

i.e, when message signal is not being transmitted at the transmitter but detected some output $y_{d}(t)$ which is nothing but the interference.

As ‘A’ is higher the interference is less at t=0 the interference is $\frac{I\omega&space;}{A}$ and is a linear function of $\omega$, when $\omega$ is small interference is less. That is $\omega$ is closer to $\omega&space;_{c}$ interference is less in FM.

Advantage of FM :- is Noise cancellation property , any interference that comes closer with the carrier signal (in the band of FM) more it will be cancelled. Not only that it overridden by the carrier strength $A_{c}$ but also exerts more power in the demodulated signal.

This is known as ‘Capture effect’ in FM which is a very good property of FM. Over years it has seen that a near by interference is 35 dB less in AM where as the near by interference in FM is 6 dB less this is a big advantage.

Two more advantages of FM over AM are:

1. Non-linearity in the Channel ,FM cancels it very nicely due to it’s inherent modulation and demodulation technique.
2. Capture effect( a near by interference) FM overrides this by $A_{c}$.
3. Noise cancellation.

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## Propagation of plane EM wave in conducting medium (or) lossy dielectrics

A lossy dielectric medium is one which an EM wave as it propagates losses power owing to imperfect dielectric,that is a lossy dielectric is an imperfect conductor that is a partially conducting medium $(\sigma&space;\neq&space;0)$ .

where as a lossless dielectric is a  $(\sigma&space;=0)$ perfect dielectric,then wave equations for conductors are also holds good here

i.e, $\bigtriangledown&space;^{2}\overrightarrow{E}=\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}&space;+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}$

$\frac{\partial&space;}{\partial&space;t}=&space;jw$

then $\bigtriangledown&space;^{2}\overrightarrow{E}=&space;j\omega&space;\mu&space;\sigma&space;\overrightarrow{E}+\mu&space;\epsilon(j\omega&space;)&space;^{2}\overrightarrow{E}$

$\bigtriangledown&space;^{2}\overrightarrow{E}=&space;(\sigma&space;+&space;j&space;\omega&space;\epsilon&space;)j\omega&space;\mu&space;\overrightarrow{E}$

$\bigtriangledown&space;^{2}\overrightarrow{E}=&space;\gamma&space;^{2}&space;\overrightarrow{E}$

$\bigtriangledown&space;^{2}\overrightarrow{E}-&space;\gamma&space;^{2}&space;\overrightarrow{E}=0---------Equation&space;(1)$

Equation (1) is called helm holtz equation and $\gamma$ is  called propagation constant.

$\gamma&space;^{2}&space;=j\omega&space;\mu&space;(\sigma&space;+j\omega&space;\epsilon&space;)$

$\gamma&space;^{2}=&space;j\omega&space;\mu&space;\sigma&space;-\omega&space;^{2}\mu&space;\epsilon$

Since $\gamma$ is a complex quantity it can be expressed as $\gamma&space;=&space;\alpha&space;+j\beta$

$\alpha$– is attenuation constant measured in Nepers/meter.

$\beta$-is phase constant measured in radians/meter.

$(\alpha&space;+j\beta&space;)&space;^{2}=&space;j\omega&space;\mu&space;\sigma&space;-\omega&space;^{2}\mu&space;\epsilon$

$\alpha^{2}&space;+2j\alpha&space;\beta-\beta&space;^{2}&space;=&space;j\omega&space;\mu&space;\sigma&space;-\omega&space;^{2}\mu&space;\epsilon$

by equating real and imaginary parts separately $\alpha&space;^{2}-\beta&space;^{2}=&space;-\omega&space;^{2}\mu&space;\epsilon------Equation(2)$

and $2\alpha&space;\beta&space;=\omega&space;\mu&space;\sigma$

$\alpha&space;=\frac{\omega&space;\mu&space;\sigma}{2\beta&space;}$

by substituting  $\alpha$ value in the equation (2)   $\frac{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}{4\beta&space;^{2}}-\beta&space;^{2}=-\omega&space;^{2}\mu&space;\epsilon$

${\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}-4\beta&space;^{4}=-4\omega&space;^{2}\beta&space;^{2}\mu&space;\epsilon$

$4\beta&space;^{4}-4\omega&space;^{2}\beta&space;^{2}\mu&space;\epsilon&space;-{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}=0$

let $\beta&space;^{2}=t$

$4t^{2}-4\omega&space;^{2}t\mu&space;\epsilon&space;-{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}=0$

$t^{2}-\omega&space;^{2}t\mu&space;\epsilon&space;-\frac{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}{4}=0$

the roots of the above quadratic expression are

$t=\frac{\omega&space;^{2}\mu&space;\epsilon&space;\pm&space;\sqrt{\omega&space;^{4}\mu&space;^{2}\epsilon&space;^{2}-4(-\frac{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}{4})}}{2}$

$t=\frac{\omega&space;^{2}\mu&space;\epsilon&space;\pm&space;\sqrt{\omega&space;^{4}\mu&space;^{2}\epsilon&space;^{2}(1+\frac{\sigma&space;}{\omega&space;\epsilon&space;})^{2}}}{2}$

$\beta&space;^{2}=\frac{\omega&space;^{2}\mu&space;\epsilon&space;\pm&space;\sqrt{\omega&space;^{4}\mu&space;^{2}\epsilon&space;^{2}(1+\frac{\sigma&space;}{\omega&space;\epsilon&space;})^{2}}}{2}$

$\beta&space;=\sqrt{\frac{\omega&space;^{2}\mu&space;\epsilon&space;\pm&space;\sqrt{\omega&space;^{4}\mu&space;^{2}\epsilon&space;^{2}(1+\frac{\sigma&space;}{\omega&space;\epsilon&space;})^{2}}}{2}}$

$\beta&space;=\sqrt{\frac{\omega&space;^{2}\mu&space;\epsilon&space;(1+&space;\sqrt{(1+\frac{\sigma&space;}{\omega&space;\epsilon&space;})^{2}})}{2}}$

similarly,

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## Drift and Diffusion currents

The flow of charge (or) current through a semi conductor material is of two types. Similarly the net current that flows through a PN diode is also of two types (i) Drift current and  (ii) Diffusion current.

Drift current:-

When an Electric field is applied across the semi conductor, the charge carriers attains certain velocity known as drift velocity $v_{d}=&space;\mu&space;E$ with this velocity electrons move towards positive terminal and holes move towards negative terminal of the battery. This movement of charge carriers constitutes a current known as ‘Drift current’.

Drift current is defined as the flow of electric current due to the motion of the charge carriers under the influence of an external field.

Drift current density due to free electrons $J_{n}&space;=&space;qn\mu&space;_{n}E$ atoms/Cm2  and the Drift current density due to free holes $J_{p}&space;=&space;qp\mu&space;_{p}E$ atoms/Cm2.

The current densities are perpendicular to the direction of current flow.

Diffusion Current:-

It is possible for an electric current to flow in a semi conductor even in the absence of the applied Electric field (or) voltage provided there exists a concentration gradient.

concentration gradient exists if the number of electrons (or) holes is greater in one region than other region in a semi conductors.

Now the charge carriers move from higher concentration to that lower concentration of same type charged regions.

The  resulting current is known as diffusion current.

Diffusion current density ($J_{P}$) due to holes is  $J_{p}&space;=&space;-q&space;D_{p}\frac{dp}{dx}$    A/Cm2 .

Diffusion current density ($J_{P}$) due to holes is $J_{n}&space;=&space;q&space;D_{n}\frac{dn}{dx}$    A/Cm2 .

$\therefore$ Total current in a semi-conductor is the sum of drift and diffusion currents

In P-type total current density is $J=&space;qp\mu&space;_{p}&space;E-qD_{p}\frac{dp}{dx}$ .

In N-type total current density is $J=&space;qn\mu&space;_{n}&space;E+qD_{n}\frac{dn}{dx}$.

(No Ratings Yet)

## Conductivity of a Semi conductor

In a pure Semi conductor number of electrons = number of holes. Thermal agitation (increase in temperature) produces new electron-hole pairs and these electron-hole pair combines produces new charge particles.

one particle is of negative charge which is known as free electron with mobility $\mu&space;_{n}$ another in with positive charge known as free hole with mobility $\mu&space;_{p}$.

two particles moves in opposite direction in an electric field $\overrightarrow{E}$ and constitutes a current.

The total current density (J) with in the semi conductor.

$\overrightarrow{J}&space;=&space;\overrightarrow{J_{n}}&space;+&space;\overrightarrow{J_{p}}$

Total conduction current density = conduction current density due to electrons + conduction current density due to holes.

$J_{n}=&space;nq\mu&space;_{n}E$.

$J_{p}=&space;pq\mu&space;_{p}E$.

n- number of electrons/Unit-Volume.

p-number of holes/Unit-Volume.

E- applied Electric field strength V/m.

q-charge of electron/hole $\approx&space;1.6X10^{-19}C.$

$J&space;=&space;nq\mu&space;_{n}E&space;+pq\mu&space;_{p}E$.

$J&space;=&space;(n\mu&space;_n&space;+p\mu&space;_{p})qE$.

$J=\sigma&space;E$.

where $\sigma&space;=&space;(n\mu&space;_{n}+p\mu&space;_{p})q$ is the conductivity of semi conductor.

Intrinsic Semi conductor:-

In an  intrinsic semi conductor $n=p=n_{i}$

$\therefore$ conductivity $\sigma&space;_{i}=&space;(n_{i}\mu&space;_{n}+&space;n_{i}\mu&space;_{p})q$

$\sigma&space;_{i}=&space;n_{i}(\mu&space;_{n}+&space;\mu&space;_{p})q$

where $J_{i}$ is the current density in an intrinsic semi conductor $J_{i}&space;=&space;\sigma&space;_{i}&space;E$

Conductivity in N-type semi conductor:-

In N-type $n>&space;>&space;p$

number of electrons $>&space;>$ number of holes

$\therefore&space;\sigma&space;_{N}\simeq&space;n\mu&space;_{n}q$

$J&space;_{N}=&space;n\mu&space;_{n}q$.

Conductivity in P-type semi conductor:-

In P-type $p>&space;>&space;n$

number of holes $>&space;>$ number of electrons

$\therefore&space;\sigma&space;_{p}&space;\approx&space;p\mu&space;_{p}q$.

$J_{P}=&space;p\mu&space;_{p}q&space;E$.

(No Ratings Yet)

## Current components of a PNP Transistor

The various Current components which flow across a PNP Transistor are as shown in the figure.

For Normal operation

• Emitter Junction $J_{E}$ is Forward Biased.
• collector Junction $J_{C}$ is Reverse Biased.

The current flows into Emitter is Emitter current $I_{E}$,  $I_{E}&space;=&space;I_{hE}+I_{eE}$.

This current consists of two components

• $I_{hE}$ or $I_{pE}$– Current due to majority carriers(holes).
• $I_{eE}$  or $I_{nE}$– Current due to minority carriers(electrons).

since $I_{eE}$ is very small $I_{E}&space;\simeq&space;I_{hE}-----------Equation(1)$

All the holes crossing the Emitter junction $J_{E}$ do not reach the Collector junction because some of them combine with the electrons in the N-type Base.

$I_{hC}$ – is the hole current in the Collector.

∴ Base current = Total hole current in Emitter – hole current in Collector.

i.e, $I_{B}&space;=&space;I_{hE}-I_{hC}----------------Equation(2)$.

If emitter were open circuited $I_{E}&space;=&space;0$ Amperes which implies  $I_{E}&space;=&space;I_{hE}$ from Equation(1) $I_{hE}\approx&space;0$ Amperes.

Under these conditions, Base-Collector junction acts as Reverse-Biased Diode and gives rise to a small reverse-Saturation current known as $I_{CO}$.

when $I_{E}&space;\neq&space;0$  , Total Collector current  $I_{C}$ is the sum of current due to holes in the Collector and Reverse Saturation current $I_{CO}$.

i.e, $I_{C}&space;=&space;I_{hC}+I_{CO}$.

i.e, In a PNP Transistor $I_{CO}$ consists of holes moving across $J_{C}$ (from Base to Collector) that is $I_{hCO}$ and electrons crossing the junction $J_{C}$ (from Collector to Base) constitutes $I_{eCO}$.

$I_{CO}&space;=&space;I_{hCO}+I_{eCO}$

i.e, $I_{E}&space;=&space;0$  $\Rightarrow&space;I_{C}&space;=&space;I_{CO}$ only

when $I_{E}&space;\neq&space;0$ $\Rightarrow&space;I_{C}&space;=&space;I_{hC}+I_{CO}$.

$\therefore$ Total current in the transistor is given by  $I_{E}&space;=&space;I_{B}+I_{C}$.

$\therefore$ The general expression for Collector current is $I_{C}&space;=&space;\alpha&space;I_{E}+I_{CO}$

$I_{C}&space;=\frac{\alpha&space;}{(1-\alpha&space;)}&space;I_{B}+\frac{1}{(1-\alpha&space;)}I_{CO}$.

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## Electro Magnetic Wave Equation

Assume a Uniform, Homogeneous,linear,isotropic and Stationary medium with Non-zero current $i.e,&space;\overrightarrow&space;J_{c}(\sigma&space;\overrightarrow{E})&space;\neq&space;0$.

When an EM wave is travelling in a conducting medium in which  $\overrightarrow&space;J\neq&space;0$. The wave is rapidly attenuated in a conducting medium and in a good conductors, the attenuation is so high at Radio frequencies. The wave penetrates the conductor only to a small depth.

choose the equation $\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J_{C}}+\overrightarrow{J_{D}}$ the time-domain representation of it is    $\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J_{C}}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

since  $\frac{\partial&space;}{\partial&space;t}&space;=j\omega$ in phasor-notation $\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;=&space;\sigma&space;\overrightarrow{E}+j\omega&space;\epsilon&space;\overrightarrow{E}$ , $\because&space;\overrightarrow{J_{C}}=\sigma&space;\overrightarrow{E}$ and $\overrightarrow{D}=\epsilon&space;\overrightarrow{E}$ .

$\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;=&space;\sigma&space;\overrightarrow{E}+\frac{\epsilon\partial&space;\overrightarrow{E}}{\partial&space;t}--------------Equation&space;(1)$

By differentiating the above equation with respect to time$\frac{\partial(\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H})}{\partial&space;t}&space;=\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}&space;+&space;\epsilon&space;\frac{\partial^{2}&space;\overrightarrow{E}}{\partial&space;t^{2}}--------------Equation&space;(2)$

From the Maxwell Equation $\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}&space;=-\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$

By taking curl on both sides $\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}&space;=-&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}&space;=-&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\frac{\partial&space;\overrightarrow{(\mu&space;H)}}{\partial&space;t}$  since $\overrightarrow{B}&space;=&space;\mu&space;\overrightarrow{H}$

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}&space;=-\mu&space;(\overrightarrow{\bigtriangledown&space;}&space;X&space;\frac{\partial&space;\overrightarrow{&space;H}}{\partial&space;t})$

By using the vector identity

$\bigtriangledown&space;^{2}\overrightarrow{E}&space;=&space;\overrightarrow{\bigtriangledown&space;}(\overrightarrow{\bigtriangledown&space;}.&space;\overrightarrow{E})&space;-\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}$

$\bigtriangledown&space;^{2}\overrightarrow{E}&space;=&space;\overrightarrow{\bigtriangledown&space;}(\overrightarrow{\bigtriangledown&space;}.&space;\overrightarrow{E})&space;-(-\mu&space;(\overrightarrow{\bigtriangledown&space;}&space;X&space;\frac{\partial&space;\overrightarrow{&space;H}}{\partial&space;t}))$

$\bigtriangledown&space;^{2}\overrightarrow{E}&space;=&space;\overrightarrow{\bigtriangledown&space;}(\overrightarrow{\bigtriangledown&space;}.&space;\overrightarrow{E})&space;+\mu&space;(\overrightarrow{\bigtriangledown&space;}&space;X&space;\frac{\partial&space;\overrightarrow{&space;H}}{\partial&space;t})$

From Equation (2)

$\mu&space;(\overrightarrow{\bigtriangledown&space;}&space;X&space;\frac{\partial&space;\overrightarrow{&space;H}}{\partial&space;t})=\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}$

$\bigtriangledown&space;^{2}\overrightarrow{E}-\overrightarrow{\bigtriangledown&space;}(\overrightarrow{\bigtriangledown&space;}.\overrightarrow{E})=\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}$

$\bigtriangledown&space;^{2}\overrightarrow{E}=\overrightarrow{\bigtriangledown&space;}(\overrightarrow{\bigtriangledown&space;}.\overrightarrow{E})+\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}$

from Maxwell’s equation $\overrightarrow{\bigtriangledown&space;}.\overrightarrow{D}&space;=&space;\rho&space;_{v}$

$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{E}&space;=&space;\frac{\rho&space;_{v}}{\epsilon&space;}$

$\bigtriangledown&space;^{2}\overrightarrow{E}=\overrightarrow{\bigtriangledown&space;}(\frac{\rho&space;_{v}}{\epsilon&space;})+\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}---------EquationI$

This is called Wave equation (Electric field)for a general medium when ${\rho&space;_{v}}\neq&space;0$.

wave equation for Magnetic field of a general  medium is

$\bigtriangledown&space;^{2}\overrightarrow{H}=\overrightarrow{\bigtriangledown&space;}(\frac{\rho&space;_{v}}{\epsilon&space;})+\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{H}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{H}}{\partial&space;t^{2}}---------EquationII$.

Case 1:-  wave equation for a conducting medium

for a conductor the net charge inside an isolated conductor is $\rho&space;_{v}=0$, then the wave equation for a conducting medium ($\sigma&space;\neq&space;0$) is

$\bigtriangledown&space;^{2}\overrightarrow{E}=\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}---------EquationI$

$\bigtriangledown&space;^{2}\overrightarrow{H}=\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{H}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{H}}{\partial&space;t^{2}}---------EquationII$

The above equations are known as wave equations for conducting medium and are involving first and second order time derivatives, which are well known equations for damped(or) attenuated waves in absorbing medium of homogeneous, isotropic such as metallic conductor.

Case 2:-  Wave equation for free space/ Non-conducting medium/loss-less medium/Perfect Di-electric medium

The conditions of free space are $\rho&space;=0,\sigma&space;=0,\overline{J}&space;=0,\mu&space;=\mu&space;_{o}$ and $\epsilon&space;=\epsilon&space;_{o}$

By substituting the above equations in general wave equations, the resulting wave equations for non-conducting medium are

$\bigtriangledown&space;^{2}\overrightarrow{E}=\mu_{o}&space;\epsilon_{o}&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}---------EquationI$

$\bigtriangledown&space;^{2}\overrightarrow{H}=\mu_{o}&space;\epsilon_{o}&space;\frac{\partial&space;^{2}\overrightarrow{H}}{\partial&space;t^{2}}---------EquationII$.

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## Automatic Gain Control (AGC)

Let us discuss about the facts why we need AGC in a Radio Receiver , as we all know that the voltage gain available at the Receiver from antenna to demodulator in several stages of amplification is very high, so that it can amplify a very weak signal But what if the signal is much stronger at the front end of the receiver ?

If same gain (gain maintained for an incoming weak signal) is maintained by different stages of the Receiver for a stonger  incoming signal, the signal is further amplified by these stages and the received signal strength is far beyond the expectations which can be avoided. so we need to have a mechanism which will measure the stength of the input signal and accordingly adjust the gain. AGC does precisely this job and improves the dynamic range of the antenna to (60-100)dB by adjusting the gain of the Intermediate Frequency and sometimes the Radio Frequency stages.

It is generally observed that as a result of fading, the amplitude of the IF carrier signal at the detecor input may vary  as much as 30 (or) 40 dB this results in the corresponding variation in general level of reproduced signal at the receiver output.

At IF carrier minimum loud speaker output becomes inaudible and mixed up with noise.

At IF  carrier maximum loud speaker output becomes intolerably large.

Therefore a properly designed AGC reduces the amplitude variation due to fading from a high value of (30-40)dB to (3-4)dB.

Basic need of AGC or AVC:-

AGC is a sub system by means of which the overall gain of a receiver is varied automatically with the variations in the stregth of the received signal to keep the output substantially constant.

i.e, the overall requirement of an AGC circuit in a receiver is to maintain a constant output level.

Some of the factors that explain why AGC is needed:-

• When a Receiver without AGC/AVC is tuned to a strong station, the received signal may overload the subsequent IF and AF stages this overloading causes carrier distortion in the incoming signal this can be prevented by using manual gain control on first RF stage but now a days AGC circuits are used for this purpose.
• When the Receiver is tuned from one station to another, difference in signal strengths of the two stations causes an unpleasant loud output if signal is moving from a weak station to a strong station unless we initially keep the volume control very low before changing the tuning from one station to another . Changing the volume control every time before attempting to re-tunethe receiver is howeve cumbersome. Therefore AGC/AVC enables the user to listen to a station without constantly monitoring the volume control.
• AGC is particularly important for mobile Receivers.
• AGC helps to smooth out the rapid fading which may occur with long distance short-wave reception.

(No Ratings Yet)

## Electric field due to infinite line charge distribution

Consider an infinitely long straight line carrying uniform line charge with density $\rho&space;_{L}&space;C/m$ and lies on Z-axis from $-\infty$ to $+\infty$.

Consider a point P at which Electric field intensity has to be determined which is produced by the line charge distribution.

from the figure let the co-ordinates of P are $(0,\rho&space;,0)$ ( a point on y-axis) and assume $dQ$ is a small differential charge confirmed to a point  M $(0,0,Z)$ as co-ordinates.

$\therefore&space;dQ$ produces a differential field $\overrightarrow{dE}$

$\overrightarrow{dE}=\frac{dQ}{4\pi&space;\epsilon&space;_{o}R^{2}}\widehat{a_{r}}$

the position vector $\overrightarrow{R}=-Z\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}$ and the corresponding unit vector $\widehat{a_{r}}&space;=\frac{-Z\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}}{\sqrt{\rho&space;^{2}+Z^{2}}}$

$\therefore&space;\overrightarrow{dE}&space;=\frac{dQ}{4\pi&space;\epsilon&space;_{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}}{(\rho&space;^{2}+Z^{2})^{\frac{3}{2}}}})$

$therefore&space;\overrightarrow{dE}&space;=\frac{\rho&space;_{L}dZ}{4\pi&space;\epsilon&space;_{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}}{(\rho&space;^{2}+Z^{2})^{\frac{3}{2}}}})$

then the Electric field strength $\overrightarrow{E}$ produced by the infinite line charge distribution $\rho&space;_{L}$ is

$\overrightarrow{E}&space;=&space;\int&space;\overrightarrow{dE}$

$\overrightarrow{E}&space;=&space;\int_{z=-\infty&space;}^{\infty&space;}\frac{\rho&space;_{L}dZ}{4\pi&space;\epsilon&space;_{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}}{(\rho&space;^{2}+Z^{2})^{\frac{3}{2}}}})$

to solve this integral  let $Z=&space;\rho&space;\tan&space;\theta&space;\Rightarrow&space;dZ=\rho&space;\sec&space;^{2}\theta&space;d\theta$

as $Z\rightarrow&space;-\infty&space;\Rightarrow&space;\theta&space;\rightarrow&space;\frac{-\pi&space;}{2}$

$Z\rightarrow&space;\infty&space;\Rightarrow&space;\theta&space;\rightarrow&space;\frac{\pi&space;}{2}$

$\therefore&space;\overrightarrow{E}&space;=&space;\int_{\theta&space;=&space;\frac{-\pi&space;}{2}}^{&space;\frac{\pi&space;}{2}}&space;\frac{\rho&space;_{L}}{4\pi\epsilon&space;_{o}}(\frac{-\rho&space;^{2}\\sec&space;^{2}\theta&space;\tan&space;\theta&space;d\theta&space;\overrightarrow{a_{z}}+\rho&space;^{2}\sec&space;^{2}\theta&space;d\theta&space;\overrightarrow{a_{\rho&space;}}}{(\rho&space;^{2}+\rho&space;^{2}\tan&space;^{2}\theta&space;)^{\frac{3}{2}}})$

$\overrightarrow{E}&space;=&space;\int_{\theta&space;=&space;\frac{-\pi&space;}{2}}^{&space;\frac{\pi&space;}{2}}&space;\frac{\rho&space;_{L}}{4\pi\epsilon&space;_{o}}(\frac{-\rho&space;^{2}\\sec&space;^{2}\theta&space;\tan&space;\theta&space;d\theta&space;\overrightarrow{a_{z}}+\rho&space;^{2}\sec&space;^{2}\theta&space;d\theta&space;\overrightarrow{a_{\rho&space;}}}{\rho&space;^{3}\sec&space;^{3}\theta&space;})$

$\overrightarrow{E}&space;=&space;\frac{\rho&space;_{L}}{4\pi\epsilon&space;_{o}\rho&space;}(\int_{\theta&space;=&space;\frac{-\pi&space;}{2}}^{&space;\frac{\pi&space;}{2}}&space;-\sin&space;\theta&space;\overrightarrow{a_{z}&space;}d\theta&space;+\int_{\theta&space;=&space;\frac{-\pi&space;}{2}}^{&space;\frac{\pi&space;}{2}}&space;\cos\theta&space;d\theta&space;\overrightarrow{a_{\rho&space;}})$

$\overrightarrow{E}=&space;\frac{\rho&space;_{L}}{4\pi\epsilon&space;_{o}\rho&space;}(0+2\overrightarrow{a_{\rho&space;}})$

$\therefore&space;\overrightarrow{E}=&space;\frac{\rho&space;_{L}}{2\pi\epsilon&space;_{o}\rho&space;}\overrightarrow{a_{\rho&space;}}&space;Newtons/Coulomb$.

$\overrightarrow{E}$ is a function of $\rho$ only, there is no $\overrightarrow{a_{z}}$ component and $\rho$ is the perpendicular distance from the point P to line charge distribution $\rho&space;_{L}$.

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## Compensators-Introduction

Compensators are corrective sub systems to compensate the deficiency in the performance of the plant or system, so given a plant and a set of specifications suitable compensators are to be designed so that the overall system will meet given specifications. Proper selection of performance specifications is the most important step in the design of compensators.

i.e, All the control systems are designed to achieve specific objectives that is the requirements are defined for the control system. A good control system has less error, good accuracy, good speed of response, good relative stability, good damping which will not cause unusual Overshoots etc.

For stationary performance of the system, gain is adjusted first but gain adjustment alone can not provide satisfactory results. When gain increases, Steady-state behavior of system improves but results into poor Transient response (or) even instability.

The desired behavior of a system is specified in terms of

• Transient response.
• Steady-state error($e_{ss}$).

$e_{ss}$ →is usually specified in terms of constants $k_{a},k_{v}$ and $k_{p}$ for $u(t),r(t)$ and $p(t)$ as inputs.

Transient response → it measures relative stability and speed of response which are specified in time or frequency domain.

In time domain the measure of relative stability is in terms of $\xi$ or $M_{p}$, while the speed response is measured in terms of rise time $t_{r}$ , settling time $t_{s}$ or natural frequency $\omega&space;_{n}$. Where as in frequency-domain the measure of relative stability is given by Resonant peak $M_{r}$ or Phase margin $\varphi&space;_{pm}$ and the speed response is measured by Resonant frequency $\omega&space;_{r}$ or band width $\omega&space;_{b}$.

Once a set of performance specifications have been selected, the next step is to chose the appropriate compensator. There exists Electrical,hydraulic,pneumatic and mechanical compensators and in this context we prefer Electrical compenators.

An external device which is used to alter the behavior of the system so as to achieve given specifications is called as compensator .

Compensators can be added to the system in series or in parallel or in combination of both.

Series Compensation:-

The flow of signal in series scheme is from lower energy level towards higher energy level. This requires additional amplifiers to increase the gain and also provide necessary isolation. The number of components required in series scheme is more than in parallel scheme.

Parallel Compensation:-

In this compensation technique energy flow is from higher energy level to lower energy level.  As there is no need of any amplifiers additional components required are less.

Series-Parallel Compensation:-

This is a compensation technique which utilizes the advantages of both series and parallel compensation techniques.

## Table of Z-Transforms for some standard signals

 Signal Z-Transform Region of Convergence (ROC) $\delta&space;(n)$ $1$ entire Z-plane $u[n]$ $\frac{1}{1-z^{-1}}&space;or\frac{z}{z-1}$ $\left&space;|&space;z&space;\right&space;|>&space;1$ $u[-n-1]$ $\frac{-1}{1-z^{-1}}&space;or\frac{-z}{z-1}$ $\left&space;|&space;z&space;\right&space;|<&space;1$ $a^{n}u[n]$ $\frac{1}{1-az^{-1}}&space;or\frac{z}{z-a}$ $\left&space;|&space;z&space;\right&space;|>&space;a$ $-a^{n}u[-n-1]$ $\frac{1}{1-az^{-1}}&space;or\frac{z}{z-a}$ $\left&space;|&space;z&space;\right&space;|<&space;a$ $na^{n}u[n]$ $\frac{az^{-1}}{(1-az^{-1})^{2}}&space;(or&space;)\frac{az}{(z-a)^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;a$ $cos&space;(\omega&space;_{o}n)u[n]$ $\frac{1-cos(\omega&space;_{o})z^{-1}}{1-2cos(\omega&space;_{o})z^{-1}+z^{-2}}$ $\left&space;|&space;z&space;\right&space;|>&space;1$ $sin&space;(\omega&space;_{o}n)u[n]$ $\frac{sin(\omega&space;_{o})z^{-1}}{1-2cos(\omega&space;_{o})z^{-1}+z^{-2}}$ $\left&space;|&space;z&space;\right&space;|>&space;1$ $r^{n}cos&space;(\omega&space;_{o}n)u[n]$ $\frac{1-rz^{-1}cos(\omega&space;_{o})}{1-2rz^{-1}cos(\omega&space;_{o})+z^{-2}r^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;\left&space;|&space;r&space;\right&space;|$ $r^{n}sin&space;(\omega&space;_{o}n)u[n]$ $\frac{rz^{-1}sin(\omega&space;_{o})}{1-2rz^{-1}cos(\omega&space;_{o})+z^{-2}r^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;\left&space;|&space;r&space;\right&space;|$ $\frac{1}{n},n>&space;0$ $-\ln&space;$ $\left&space;|&space;z&space;\right&space;|>&space;1$ $a^{\left&space;|&space;n&space;\right&space;|}&space;\forall&space;n$ $\frac{(1-a^{2})}{(1-az)(1-az^{-1})}$ $\left&space;|&space;a&space;\right&space;|<&space;\left&space;|&space;z&space;\right&space;|<&space;\frac{1}{\left&space;|&space;a&space;\right&space;|}$ $-na^{n}u[-n-1]$ $\frac{az^{-1}}{(1-az^{-1})^{2}}&space;(or)\frac{az}{(z-a)^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;\left&space;|&space;a&space;\right&space;|$ $(n+1)a^{n}u[n]$ $\frac{1}{(1-az^{-1})^{2}}&space;(or)\frac{z^{2}}{(z-a)^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;\left&space;|&space;a&space;\right&space;|$

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## Choice Of Intermediate Frequency (or) IF Amplifier in a Radio Receiver

Choice of Intermediate Frequency of a receiving system is usually a compromise , since there are reasons why it is neither low nor high, nor in a certain range between the two.

The following are the major factors influencing the choice of the Intermediate Frequency in any particular system.

1. If the IF is too high poor selectivity and poor adjacent channel rejection results unless sharp cut-off filters(crystal/mechanical filters) are used in the IF stage.
2. A high value of Intermediate Frequency(IF) increases tracking difficulties.
3. If we chose IF as low frequency, image frequency rejection becomes poorer. i.e, if $\frac{f_{si}}{f_{s}}$ is more IFRR(image Frequency Rejection Ratio) has been improved, which requires a high Intermediate Frequency($f_{si}$). Similarly when $f_{s}$ is more IFRR becomes worst.
4. Average Intermediate Frequency(IF) can make the selectivity too sharp cutting of the side bands.This problem arises because the Q must be low when the IF is low, unless crystal or mechanical filters are used and hence gain per stage is low. Thus a designer is more likely to raise Q rather than increasing the number of IF amplifiers.
5. If IF is very low , the frequency stability of local oscillator must be made correspondingly high.
6. IF must not fall in the tuning range of the receiver or else instability occurs and hetero dyne whistles (noise) will be heard.

Frequencies used:-

1. Standard AM broadcast receivers tuned to (540 KHz-1650 KHz) or(6 MHz-18 MHz) and European long wave band (150 KHZ- 350 KHz) uses IF in the range (438 KHz- 465 KHz). 455 KHz is the most popular value used.
2. FM receivers using the standard (88 MHz -108 MHz) band have an IF which is almost always 10.7 MHz.
3. TV Receivers in the  VHF band (54 MHz-223 MHz),UHF band (470 MHz-940 MHz) uses IF between (26 MHz-46 MHz) and the popular values are 36 MHz and 46 MHz.
4. AM-SSB Receviers employed for short-wave reception in the short wave band / VHF band uses IF in the range (1.6 MHz to 2.3 MHz).

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## Working /Operation of NPN and PNP Transistor

NPN Transistor Working:-

For Normal operation of NPN Transistor Emitter junction JE is Forward Biased and Collector junction JC is Reverse Biased.

The applied Forward Biased at Emitter-Base junction injects a large number of electrons into the N-region and these electrons have enough energy to overcome the JE junction and enter into the very thin lightly doped Base region.

Since Base is very lightly doped very few electrons recombine with the holes in the P-type Base region and constitutes a small Base current IB in μA.

The electrons in the Emitter region are more when compared to electrons in the Collector. Only 5% (or) 1% of injected electrons combines with the holes in Base to produce Iand remaining 95% (or0 99% of electrons diffuse into Collector region due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused electrons which enters Collector junction and are collected by the Collector(Positive electrode).

Thus injected electrons from Emitter reaches Collector constituting a current known as $I_{E}=I_{B}+I_{C}$ Thus Emitter current is sum of Base current and Collector current. $I_{B}$ is very small in the Base region.

Current directions are  always from negative to positive and Majority carriers are electrons in NPN Transistor.

NPN Transistor is preferred over PNP since the mobility of electron is more than that of hole that is electron moves faster than holes.

PNP Transistor Working:-

For Normal operation of PNP Transistor Emitter junction JE is forward Biased and Collector junction JC is reverse biased.

The applied FB at Emitter-Base junction injects a large number of holes in the P-type emitter region and these holes have enough energy to enter into very thin lightly doped Base region. Base is very lightly doped N-type region. Therefore very few holes combines with the Base region and constitutes a small Base current IB (in Micro Amperes).

The holes in the Emitter region are more when compared to holes in the collector region.Only 5% or 1% of injected holes from Emitter combines with the electrons in the Base to produce IB and remaining 95% (or) 99% of holes diffuse into Collector region  due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused holes which enters Collector junction and are collected by the Collector(negative electrode).

Thus injected holes from Emitter reaches Collector constituting a current known as $I_{E}=I_{B}+I_{C}$$I_{B}$ is very small in the Base region.

Majority carriers are holes in PNP Transistor.

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