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

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

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

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

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## Application of Ampere’s circuit law-infinite sheet of current

consider an infinite sheet in the z=0 plane, which has uniform current density  $\overrightarrow{k}&space;=&space;k_{y}\overrightarrow{a_{y}}$    A/m .

Let us suppose the current is flowing in the positive y direction.

the sheet of current is assumed to be in rectangular co-ordinate system

$\overrightarrow{H}&space;=&space;H_{x}\overrightarrow{a_{x}}+H_{y}\overrightarrow{a_{y}}+H_{z}\overrightarrow{a_{z}}$

Let us suppose the conductor is carrying a current I , by right hand thumb rule magnetic field is produced around the conductor is right angles to the direction of I.

In this case of infinite sheet , the current is in the y-direction there is no component of H along the direction of y  and also the z components cancel each other because of opposite direction of the fields produced so only x components of H exists.

$\overrightarrow{H}&space;=&space;\left\{\begin{matrix}&space;H_{o}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z>0\\&space;-H_{o}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z<0&space;\end{matrix}\right.$

from Ampere’s Circuit law  $\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;\int&space;\left&space;[\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}&space;+\overrightarrow{H}_{2-3}.&space;\overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}.&space;\overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}.&space;\overrightarrow{dl}_{4-1}&space;\right&space;]$ .

the component  $\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;\int&space;\left&space;[\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}&space;+\overrightarrow{H}_{2-3}.&space;\overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}.&space;\overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}.&space;\overrightarrow{dl}_{4-1}&space;\right&space;]$

$\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}=&space;\overrightarrow{H}_{1-0}.&space;\overrightarrow{dl}_{1-0}+\overrightarrow{H}_{0-2}.&space;\overrightarrow{dl}_{0-2}$

$\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}=&space;H_{o}&space;\overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})-&space;H_{o}&space;\overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})$ .

$\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}=0$ .

similarly $\overrightarrow{H}_{3-4}.&space;\overrightarrow{dl}_{3-4}&space;=0$ .

$\therefore&space;\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;\int&space;\left&space;[\overrightarrow{H}_{2-3}.&space;\overrightarrow{dl}_{2-3}+\overrightarrow{H}_{4-1}.&space;\overrightarrow{dl}_{4-1}&space;\right&space;]$ .

$\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;-(H_{o}\&space;\overrightarrow{a_{x}}).(b&space;\&space;-\overrightarrow{a_{x}})+(H_{o}\&space;\overrightarrow{a_{x}}).(b&space;\&space;\overrightarrow{a_{x}})$ .

$\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;2H_{o}\&space;b$ .

$I=&space;2H_{o}&space;\&space;b$

$k_{y}\&space;b&space;=&space;2H_{o}&space;\&space;b$ .

$H_{o}&space;=\frac{1}{2}&space;k_{y}$ .

as   $\overrightarrow{H}&space;=&space;\left\{\begin{matrix}&space;H_{o}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z>0&space;\\&space;-H_{o}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z<0&space;\end{matrix}\right.$

this will be changed to $\overrightarrow{H}&space;=&space;\left\{\begin{matrix}&space;\frac{1}{2}&space;k_{y}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z>0&space;\\&space;-\frac{1}{2}&space;k_{y}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z<0&space;\end{matrix}\right.$

In general for a finite sheet of current density  $\overrightarrow{k}$  A/m  Magnetic field is generalized as     $\overrightarrow{H}&space;=&space;\frac{1}{2}&space;(\overrightarrow{k}&space;X&space;\overrightarrow{a_{n}})$ .

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## Magnetic Forces

Magnetic forces are required to study the force , a magnetic field exerts on charged particles, current elements and loops which is used in electrical devices in ammeters, volt meters, Galvano meters.

There are 3 ways in which force due to magnetic fields can be experienced.

1. The force can be due to a  moving charged particle in a  Magnetic field.
2. on a current element in an external B  field.
3. between two current elements.

Force on a charged particle:-

we know that $\overrightarrow{E}&space;=&space;\frac{\overrightarrow{F}}{Q}$  .

$\overrightarrow{F_{e}}&space;=&space;Q&space;\overrightarrow{E}-----EQN(1)$ .

where  $\overrightarrow{F_{e}}$  is the electric force on a stationary (or) moving electric charge in an electric field and is related to  $\overrightarrow{E}$ .  where  $\overrightarrow{F_{e}}$  and   $\overrightarrow{E}$ are in the same direction.

a magnetic field can exert force only on a moving charge , suppose a charge Q is moving with velocity u  (or) v in a magnetic field (B) is

$\overrightarrow{F_{m}}&space;=&space;(Q\&space;\overrightarrow{u}&space;\&space;X&space;\overrightarrow{B})-----EQN(2)$ .

from the  equations    $\overrightarrow{F_{e}}$   is independent of velocity of the charge and performs work on the charge which changes its kinetic energy but  $\overrightarrow{F_{m}}$  depends on the charge velocity and is normal to it so work done $\overrightarrow{F_{m}}.\overrightarrow{dl}&space;=0$ it does not cause increase in the kinetic energy of the charge.

$\overrightarrow{F_{m}}$  is small compared to    $\overrightarrow{F_{e}}$  except at high velocities.

so a charge which is in movement has both electric and magnetic fields.

## H due to finite long straight conuctor

Consider a conductor of finite length placed along z-axis as shown in the figure

The conductor has a finite length  AB , where A and B are located at distances Z1 and Z2 above the origin with it’s upper and lower ends respectively subtending angles $\alpha&space;_{2}$ and $\alpha&space;_{1}$ at P.

P is the point at which  $\overrightarrow{H}$   is to be determined.

Consider a differential element $\overrightarrow{dl}$  along the Z-axis at a distance Z from the origin.

where $\overrightarrow{dl}&space;=dl&space;\&space;\overrightarrow{a_{z}}$ .

$\overrightarrow{R}&space;=&space;-Z&space;\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}$ .

$\widehat{a_{R}}&space;=&space;\frac{\overrightarrow{R}&space;=&space;-Z&space;\overrightarrow{a_{z}}+\rho&space;\overrightarrow{a_{\rho&space;}}}{\sqrt{(Z^{2}+\rho&space;^{2})}}$ .

$\overrightarrow{dl}&space;X&space;\&space;\widehat{a_{R}}&space;=&space;\frac{\rho&space;dz&space;\&space;\overrightarrow{a_{\phi&space;}}}{\sqrt{(Z^{2}+\rho&space;^{2})}}$ .

$\overrightarrow{H}&space;=&space;\oint&space;\frac{&space;I&space;\&space;\rho&space;\&space;dz&space;\&space;\overrightarrow{a_{\phi&space;}}}{4\pi&space;(Z^{2}+\rho&space;^{2})^{\frac{3}{2}}}$ .

$\overrightarrow{H}&space;=&space;\oint_{Z_{1}}^{Z_{2}}&space;\frac{&space;I&space;\&space;\rho&space;\&space;dz&space;\&space;\overrightarrow{a_{\phi&space;}}}{4\pi&space;(Z^{2}+\rho&space;^{2})^{\frac{3}{2}}}$ .

as $Z_{1}&space;=&space;\rho&space;\&space;\cot&space;\alpha&space;_{1}$   ,  $Z_{2}&space;=&space;\rho&space;\&space;\cot&space;\alpha&space;_{2}$  and  $dz&space;=&space;-&space;\rho&space;\&space;cosec^{2}&space;\alpha&space;\&space;d\alpha$  .

$\overrightarrow{H}&space;=\frac{-I}{4\pi&space;}&space;\int_{\alpha&space;_{1}}^{\alpha&space;_{2}}&space;\frac{\rho&space;^{2}&space;\&space;cosec^{2}\alpha&space;\&space;d\alpha&space;\&space;\overrightarrow{a_{\phi&space;}}}{&space;(\rho&space;^{2}+\rho&space;^{2}&space;\cot&space;^{2}\alpha&space;)^{\frac{3}{2}}}$ .

$\overrightarrow{H}&space;=\frac{-I}{4\pi\&space;\rho&space;}&space;\int_{\alpha&space;_{1}}^{\alpha&space;_{2}}&space;\rho&space;^{2}&space;\&space;\sin&space;\alpha&space;\&space;d\alpha&space;\&space;\overrightarrow{a_{\phi&space;}}$ .

$\overrightarrow{H}&space;=\frac{I}{4\pi\&space;\rho&space;}&space;\left&space;[&space;\cos&space;\alpha&space;_{2}&space;-\cos&space;\alpha&space;_{1}\right&space;]\overrightarrow{a_{\phi&space;}}$ .

Case 1 :-

when the conductor is semi-finite   that is A is located at origin and B at $\infty$ .

i.e,    $Z_{1}&space;=0&space;\&space;\Rightarrow&space;\&space;\alpha&space;_{1}&space;=&space;90^{o}$     and    $Z_{2}&space;=\infty&space;\&space;\Rightarrow&space;\&space;\alpha&space;_{2}&space;=&space;0^{o}$ .

then  $\overrightarrow{H}&space;=&space;\frac{I}{4\pi&space;\&space;\rho&space;}&space;\overrightarrow{a_{\phi&space;}}$ .

Case 2:-

when conductor is infinite in length   A is at  $-\infty$  and B at $\infty$  implies  $Z_{1}&space;=&space;-\infty&space;\Rightarrow&space;\alpha&space;_{1}&space;=&space;180&space;^{o}$  and  $Z_{2}&space;=&space;\infty&space;\Rightarrow&space;\alpha&space;_{2}&space;=&space;0&space;^{o}$ .

$\overrightarrow{H}&space;=&space;\frac{I}{2\pi&space;\&space;\rho&space;}&space;\overrightarrow{a_{\phi&space;}}$ .

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## application of Ampere’s circuit law to infinite line current element

Consider as infinitely long straight conductor placed along z-axis carrying a current I .

In order to determine $\overrightarrow{H}$ at some point P. we allow a closed path which passes through the point P and encloses the current  carrying conductor symmetrically such path is known as Amperian path.

To apply Ampere’s law the conditions t be satisfied are

1. The field $\overrightarrow{H}$ is either tangential (or) Normal to the path at each point of the closed path.
2. The magnitude of $\overrightarrow{H}$ must be same at all points of the path where  $\overrightarrow{H}$ is tangential.

Now,   $\overrightarrow{H}$  is given by  $\overrightarrow{H}&space;=H&space;_{\rho&space;}&space;\overrightarrow{a}_{\rho&space;}+H&space;_{\phi&space;}&space;\overrightarrow{a}_{\phi&space;}+H&space;_{z}&space;\overrightarrow{a}_{z}$ .

The path we are assuming is in the direction of $\phi$  so  $\overrightarrow{dl}&space;=&space;dl&space;\overrightarrow{a}_{\phi&space;}$ .

$\overrightarrow{dl}&space;=&space;\rho&space;\&space;d\phi&space;\overrightarrow{a}_{\phi&space;}$ .

Ampere’s law is used to find out $\overrightarrow{H}$  at P

i.e, from Ampere’s circuit law  $\oint&space;\overrightarrow{H}&space;.&space;\overrightarrow{dl}&space;=&space;I_{enc}$  .

$\oint&space;\overrightarrow{H}&space;.&space;\overrightarrow{dl}&space;=&space;I$ .

$\oint&space;\overrightarrow{H}&space;.&space;\overrightarrow{dl}&space;=\oint(H&space;_{\rho&space;}&space;\overrightarrow{a}_{\rho&space;}+H&space;_{\phi&space;}&space;\overrightarrow{a}_{\phi&space;}+H&space;_{z}&space;\overrightarrow{a}_{z})&space;.&space;\rho&space;\&space;d\phi&space;\overrightarrow{a}_{\phi&space;}$ .

$=\oint&space;(H&space;_{\phi&space;}&space;\overrightarrow{a}_{\phi&space;})&space;.&space;\rho&space;\&space;d\phi&space;\overrightarrow{a}_{\phi&space;}$

$=\int_{\phi&space;=&space;0}^{2\pi&space;}&space;H&space;_{\phi&space;}&space;\rho&space;\&space;d\phi$

$=&space;H&space;_{\phi&space;}&space;\&space;\rho&space;\&space;2\pi$ .

from Ampere’s law      $H&space;_{\phi&space;}&space;\&space;\rho&space;\&space;2\pi&space;=&space;I$ .

$H&space;_{\phi&space;}&space;=\frac{&space;I}{\&space;2\pi\&space;\rho&space;}$ .

$\therefore&space;\overrightarrow{H&space;_{\phi&space;}&space;}&space;=&space;\frac{&space;I}{\&space;2\pi\&space;\rho&space;}&space;\overrightarrow{a_{\phi&space;}}$ .

Ampere’s law is applied to find the value of $\overrightarrow{H}$  at any point P in it’s field.

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## Ampere’s Circuit law

Ampere’s Circuit law states that the line integral of the tangential component of $\overrightarrow{H}$ around a closed path is the same as the net current (Ienc) enclosed by the path.

i.e, $\oint&space;\overrightarrow{H}&space;.\overrightarrow{dl}&space;=&space;I_{enclosed}$ .

This is similar to Gauss’s law and can be applied to determine $\overrightarrow{H}$ when the current distribution is symmetrical it’s a special case of Biot-savart’s law.

Proof:-

Consider a circular loop which encloses a current element . Let the current be in upward direction then the field is in anti- clock wise .

The current which is enclosed by the circular loop is of infinite length then  $\overrightarrow{H}$at  any point A is given by

$\overrightarrow{H}&space;=&space;\frac{I_{enc}}{2\pi&space;R}&space;\overrightarrow{a}_{\phi&space;}$ .

$\overrightarrow{H}.\overrightarrow{dl}&space;=&space;\frac{I_{enc}}{2\pi&space;R}&space;\overrightarrow{a}_{\phi&space;}.dl&space;\overrightarrow{a}_{\phi&space;}$ .

$\overrightarrow{H}.\overrightarrow{dl}&space;=&space;\frac{I_{enc}}{2\pi&space;R}.dl$.

$\overrightarrow{H}.\overrightarrow{dl}&space;=&space;\frac{I_{enc}}{2\pi&space;R}.R&space;d\phi$ .

$\overrightarrow{H}.\overrightarrow{dl}&space;=&space;\frac{I_{enc}}{2\pi&space;}d\phi$ .

$\oint&space;\overrightarrow{H}&space;.&space;\overrightarrow{dl}&space;=&space;\int_{\phi&space;=0}^{2\pi&space;}&space;\frac{I_{enc}}{2\pi&space;}d\phi$ .

$\oint&space;\overrightarrow{H}&space;.\overrightarrow{dl}&space;=&space;I_{enc}$ .

which is known as the integral form of Ampere’s circuit law.

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## Optimum filter

The function of a receiver in a binary Communication system is to distinguish between two transmitted signals $x_{1}(t)\&space;and&space;\&space;x_{2}(t)$  (or) ($s_{1}(t)\&space;and&space;\&space;s_{2}(t)$) in the presence of noise.

The performance of Receiver is usually measured in terms of the probability of error Pe an the receiver is said to be optimum if it yields the minimum probability of error.

i.e, optimum receiver is the one with minimum probability of error Pe .

optimum receiver takes the form of Matched filter when the noise at the receiver input is white noise.

The block diagram of optimum receiver is as shown in the figure below

the decision boundary is set to $\frac{x_{o1}(T)+x_{o2}(T)}{2}$ .

Probability of error of optimum filter:-

The probability of error can be obtained as similar to Integrate and dump receiver. Here we will consider noise as Gaussian Noise.

The output of optimum filter is  $y(t)&space;=&space;x_{o1}(t)+n_{o}(t)$ .

The output of sampler is  $y(T)&space;=&space;\left\{\begin{matrix}&space;x_{o1}(T)+n_{o}(T)&space;\&space;for&space;\&space;binary&space;\&space;i/p&space;\&space;'1'\\&space;x_{o2}(T)+n_{o}(T)&space;\&space;for&space;\&space;binary&space;\&space;i/p&space;\&space;'0'&space;\end{matrix}\right.$

suppose if Binary ‘1’ is transmitted then the input is $x(t)&space;=&space;x_{1}(t)$ , to find the probability of error this transmitted ‘1’ should be received as ‘0’.

this is possible  when the condition  $\left&space;|&space;y(T)&space;\right&space;|&space;<\frac{x_{o1}(T)+x_{o2}(T)}{2}$ is true.

1 will be received as 0    $\Rightarrow&space;x_{o1}(T)+n_{o}(T)&space;<\frac{x_{o1}(T)+x_{o2}(T)}{2}$ .

$n_{o}(T)&space;<\frac{x_{o2}(T)-x_{o1}(T)}{2}$ .

similarly a Binary ‘0’ will be  received as ‘1’ if and only if

$\left&space;|&space;y(T)&space;\right&space;|&space;>\frac{x_{o1}(T)+x_{o2}(T)}{2}$ .

$\Rightarrow&space;x_{o2}(T)+n_{o}(T)&space;>\frac{x_{o1}(T)+x_{o2}(T)}{2}$ .

$n_{o}(T)&space;>\frac{x_{o1}(T)-x_{o2}(T)}{2}$ .

the conditions are  summarized in the table

Noe the Probability Distribution Function of Gaussian noise with zero mean and standard deviation $\sigma$  is given by

$f(n_{o}(T))&space;=&space;\frac{1}{\sigma&space;\sqrt{2\pi&space;}}&space;e^{-\frac{n_{o}^{2}(T)}{2}}$ .

Probability of error= probability ‘1’ will be received as ‘0’ =probability ‘0’ will be received as ‘1’.

$\therefore&space;P_{e}&space;=$  area under the curve $n_{o}(T)&space;>\frac{x_{o1}(T)-x_{o2}(T)}{2}$   (or) area under the curve $n_{o}(T)&space;<\frac{x_{o2}(T)-x_{o1}(T)}{2}$ .

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## aliasing effect in Sampling

Effect of under sampling (aliasing effect):-

When a CT band limited signal is sampled at  $f_{s}&space;<&space;2f_{m}$ , then the successive cycles of the spectrum of the sampled signal overlap with each other as shown below

Some aliasing is produced in the signal this is due to under sampling.

aliasing is the phenomenon in which a high frequency component in the frequency spectrum of the signal takes as a low frequency component in the spectrum of the sampled signal.

Because of aliasing it is not possible to reconstruct x(t) from g(t) by low pass filtering.

The spectral components in the overlapping regions and hence the signal is distorted.

Since any information signal contains a large no.of frequencies so the decision of sampling frequency is always become a problem.

A signal is first passed through LPF  before sampling.

i.e, it is band limited by this LPF which is known as pre-alias filter.

To avoid aliasing

1. Pre-alias filter must be used to limit the band width of the signal to $f_{m}$  Hz.
2. Sampling frequency must be  $f_{s}>2f_{m}$ .

Pre-alias filter means before sampling is passed through a LPF to make a perfect band limited signal.

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## Reconstruction filter(Low Pass Filter)

Reconstruction filter (Low Pass Filter) Procedure to reconstruct actual signal from sampled signal:-

Low Pass Filter is used to recover original signal from it’s samples. This is also known as interpolation filter.

An LPF is that type of filter which passes only low frequencies up to cut-off frequency and rejects all other frequencies above cut-off frequency.

For an ideal LPF, there is a sharp change in the response at cut-off frequency as shown in the figure.

i.e, Amplitude response becomes suddenly zero at cut-off frequency which is not possible practically that means an ideal LPF is not physically realizable.

i.e, in place of an  ideal LPF a practical filter is used.

In case of a practical filter, the amplitude response decreases slowly to zero (this is one of the reason why we choose  $f_{s}>2f_{m}$)

This means that there exists a transition band in case of practical Low Pass Filter in the reconstruction of original signal from its samples.

Signal Reconstruction (Interpolation function):-

The process of reconstructing a Continuous Time signal x(t) from it’s samples is known as interpolation.

Interpolation gives either approximate (or) exact reconstruction (or) recovery of CT signal.

One of the simplest interpolation procedures is known as zero-order hold.

Another procedure is linear interpolation. In linear interpolation the adjacent samples (or) sample points are connected by straight lines.

We may also use higher order interpolation formula for reconstructing the CT signal from its sample values.

If we use the above process (Higher order interpolation) the sample points are connected by higher order polynomials (or) other mathematical functions.

For a Band limited signal, if the sampling instants are sufficiently large then the signal may be reconstructed exactly by using a LPF.

In this case an exact interpolation can be carried out between sample points.

Mathematical analysis:-

A Band limited signal x(t) can be reconstructed completely from its samples, which has higher frequency component fm Hz.

If we pass the sampled signal through a LPF having cut-off frequency of  fm  Hz.

From sampling theorem

$g(t)&space;=&space;x(t).\delta&space;_{T_{s}}(t)$.

$g(t)=\frac{1}{T_{s}}\left&space;\{&space;1+2\cos&space;\omega&space;_{s}t+2\cos&space;2\omega&space;_{s}t+2\cos&space;3\omega&space;_{s}t+.....&space;\right&space;\}$.

g(t)     has a multiplication factor  $\frac{1}{T_{s}}$. To reconstruct  x(t)  (or)  X(f) , the sampled signal must be passed through an ideal LPF of Band Width of  $f_{m}$   Hz and gain  $T_{s}$ .

$\left&space;|&space;H(\omega&space;)&space;\right&space;|=T_{s}&space;\&space;for&space;\&space;-\omega&space;_{m}\leq&space;\omega&space;\leq&space;\omega&space;_{m}$.

$h(t)&space;=&space;\frac{1}{2\pi&space;}&space;\int_{-\omega&space;_{m}}^{\omega&space;_{m}}T_{s}e^{j\omega&space;t}\&space;d\omega$.

$h(t)&space;=&space;2f_{m}T_{s}&space;\&space;sinc(2\pi&space;f_{m}t)$.

If sampling is done at Nyquist rate , then Nyquist interval is  $T_{s}&space;=&space;\frac{1}{2f_{m}}$ .

therefore  $h(t)&space;=&space;\&space;sinc(2\pi&space;f_{m}t)$ .

h(t) = 0.      at all Nyquist instants  $t=&space;\pm&space;\frac{n}{2f_{m}}$  , when    g(t)    is applied at the input to this filter the output will be  x(t)  .

Each sample in g(t)  results a sinc pulse having amplitude equal to the strength of sample. If we add all these sinc pulses that gives the original signal  x(t) .

$g(t)&space;=&space;x(kT_{s})\delta&space;(t-kT_{s})$ .

$x(t)&space;=\sum_{k}&space;x(kT_{s})\&space;h&space;(t-kT_{s})$ .

$x(t)&space;=\sum_{k}&space;x(kT_{s})\&space;sinc(2\pi&space;f_{m}&space;(t-kT_{s}))$.

$x(t)&space;=\sum_{k}&space;x(kT_{s})\&space;sinc(2\pi&space;f_{m}t-k\pi&space;)$ .

This is known as interpolation formula

It is assumed that the signal  x(t) is strictly band limited but in general an information signal may contain a wide range of frequencies and can not be strictly band limited this means that the maximum frequency in the signal can not be predictable.

then it is not possible to select suitable sampling frequency  fs  .

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## Region of Convergence (ROC)

The range of values of the complex variable s for which Laplace Transform $X(S)=\int_{-\infty&space;}^{\infty&space;}x(t)e^{-st}\&space;dt$ converges is called the Region of Convergence (ROC).

i.e, The region of Convergence (or) existence of signal’s Laplace transform X(S) is the set of values of s for which the integral defining the direct L T/F X(S) converges.

The ROC is required for evaluating the inverse L T/F of x(t) from X(S).

i.e, the operation of finding the inverse T/F requires an integration in the complex plane.

i.e, $x(t)=&space;\frac{1}{2\pi&space;j}\int_{\sigma&space;-j\infty&space;}^{\sigma&space;+j\infty&space;}X(S)&space;e^{St}&space;\&space;ds$ .

The path of integration is along S-plane $S&space;=&space;\sigma&space;+j\omega$ that is along $\sigma&space;+j\omega$ with $\omega$ varying from $-\infty&space;\&space;to&space;\&space;\infty$  and moreover , the path of integration must lie in the ROC for X(S).

for example the signal $e^{-at}u(t)$ , this is possible if $\sigma&space;>-a$  so the path of integration is shown in the figure

Thus to obtain $x(t)&space;=&space;e^{-at}u(t)$   from $X(S)&space;=&space;\frac{1}{s+a}$   , the integration is performed through this path for the function  $\frac{1}{s+a}$ . such integration in the complex plane requires a back ground in the theory of functions of complex variables.

so we can avoid this integration by compiling a Table of L T/F’s . so for inverse L T/F’s we use this table instead of performing complex integration.

specific constraints on the ROC are closely associated with time-domain properties of x(t).

Properties of ROC/ constraints (or) Limitations:-

1.The ROC of  X(S) consists of strips parallel to the $j\omega$ axis in the S-plane.

i.e, The ROC of X(S) consists of the values of s for which Fourier T/F of $x(t)e^{-\sigma&space;t}$ converges this is possible if $x(t)e^{-\sigma&space;t}$ is fully integrable thus the condition depends only on $\sigma$ . Hence ROC is the strips (bands) which is only in terms of values of $\sigma$.

2.

3. For Rational Laplace T/F’s , the ROC does not contain any poles. This is because X(S) is finite at poles and the integral can not be converge at this point.

4. If x(t) is of finite duration and absolutely integrable, then the ROC is the entire S-plane.

5. If x(t) is right-sided and if the line $Re\left&space;\{&space;s&space;\right&space;\}&space;=\sigma&space;_{o}$ is in the ROC, then all values of s for which $Re\left&space;\{&space;s&space;\right&space;\}&space;>&space;\sigma&space;_{o}$ will also be in the ROC.

i.e, if the signal is $x(t)&space;=&space;e^{-at}u(t)$  right-sided $[0&space;\&space;to&space;\&space;\infty&space;]$  then $X(S)&space;=&space;\frac{1}{s+a}$  for ROC : $Re\left&space;\{&space;s&space;\right&space;\}&space;>&space;-a$ .

6. If x(t) is left-sided and if the line $Re\left&space;\{&space;s&space;\right&space;\}&space;=\sigma&space;_{o}$ is in the ROC, then all values of s for which $Re\left&space;\{&space;s&space;\right&space;\}&space;<&space;\sigma&space;_{o}$ will also be in the ROC.

7. If x(t) is two-sided and if the line $Re\left&space;\{&space;s&space;\right&space;\}&space;=\sigma&space;_{o}$ is in the ROC, then the ROC consists of a strip  in the s-plane that includes the line $Re\left&space;\{&space;s&space;\right&space;\}&space;=&space;\sigma&space;_{o}$ .

for the both sided signal , the ROC lies in the region $\sigma&space;_{1}&space;<&space;Re\left&space;\{&space;s&space;\right&space;\}<\sigma&space;_{2}$ . This ROC is the strip parallel to $j\omega$  axis in the s-plane.

8. If the L T/F X(S) of x(t) is rational, then it’s ROC is bounded by poles (or) extends to infinity in addition no poles of X(s) are contained in the ROC.

If the function has two poles , then ROC will be area  between these two poles for two sided signal, if for single sided signal the area extends from one pole to infinity.

But is does not include any pole.

9. If the L T/F X(S) of x(t) is rational, then if x(t) is right-sided. The ROC is the region in the s-plane to the right of the right most pole and if x(t) is left-sided, the ROC is the region in the s-plane to the left of the left most pole.

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## Analogy between Vectors and Signals

we have already defined the signal as any ordinary function of time. To understand more about signal we consider it as a problem. A problem is better understood (or) better remembered if it can be associated with some familiar phenomenon.

we always search for analogies while studying a new problem.

i.e, In the study of abstract problems analogies are very helpful. Particularly if the problem can be shown to be analogous to some concrete phenomenon.

It is then easier to gain some insight into the new problem from the knowledge of the analogous phenomenon.

There is a perfect analogy that exists between vectors and signals which leads to a better understanding of signal analysis. we shall now briefly review the properties of vectors.

Vectors:-

A vector is specified by magnitude and direction $\overrightarrow{A}$.

Let us consider two vectors $\overrightarrow{V_{1}}$  and $\overrightarrow{V_{2}}$ . It is possible to find out the component of one vector along the other vector.

In order to find out the component of vector $\overrightarrow{V_{1}}$ along  $\overrightarrow{V_{2}}$ . Let us assume it as $C_{12}V_{2}$ ,  which is only the magnitude.

how do we represent physically the component of one vector $\overrightarrow{V_{1}}$ along  $\overrightarrow{V_{2}}$ ? This is possible by finding the projection of one vector on to the other.

i.e, by drawing a perpendicular from $\overrightarrow{V_{1}}$   to   $\overrightarrow{V_{2}}$

$\overrightarrow{V_{1}}&space;=&space;C_{12}\overrightarrow{V_{2}}&space;+&space;\overrightarrow{V_{e}}$ .

There exists two other possibilities.

but these are not suitable. $\because$  the error vectors are more in these cases.

$\overrightarrow{V_{1}}.\overrightarrow{V_{2}}=V_{1}V_{2}\cos&space;\theta$ .

If $\theta$ is the angle between two vectors $\overrightarrow{V_{1}}$  and $\overrightarrow{V_{2}}$ , the component of $\overrightarrow{V_{1}}$  along $\overrightarrow{V_{2}}$ is

$\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{\left&space;|&space;V_{2}&space;\right&space;|}=V_{1}\cos&space;\theta-----EQN(1)$.

The component  of $\overrightarrow{V_{1}}$  along $\overrightarrow{V_{2}}$ is $C_{12}V_{2}----EQN(2)$.

$\therefore&space;(1)&space;=&space;(2)$ .

$\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{\left&space;|&space;V_{2}&space;\right&space;|}&space;=&space;C_{12}V_{2}$ .

$C_{12}=\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{V_{2}^{2}}$ .

If two vectors are orthogonal  $\overrightarrow{V_{1}}.\overrightarrow{V_{2}}&space;=0$ .

i.e, $C_{12}&space;=0$.

Signals:-

The concept of vector comparison & orthogonality can be extended to signals.

i.e, a signal is nothing but a single-valued function of independent variable. Assume two signals  $f_{1}(t)$   and $f_{2}(t)$, now to approximate  $f_{1}(t)$  in terms of $f_{2}(t)$  over  $t_{1}<&space;t .

$f_{1}(t)&space;\approx&space;C_{12}f_{2}(t)$ .

$\therefore&space;f_{1}(t)&space;\approx&space;C_{12}f_{2}(t)+f_{e}(t)$ .

$f_{e}(t)&space;=&space;f_{1}(t)-C_{12}f_{2}(t)$ .

Now, we choose in order to achieve the best approximation.

i.e, which keeps the error as minimum as possible.

One possible way for minimizing error  $f_{e}(t)$ is to choose minimize the average value of $f_{e}(t)$ .

i.e, as    $\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}[f_{1}(t)-C_{12}f_{2}(t)]dt$.

But the process of averaging gives a false indication.

i.e, for example while approximating a function  $\sin&space;t$  with a null function  $f(t)=0$  is

$f_{1}(t)&space;=C_{12}f_{2}(t)$.

$\sin&space;t&space;=0.&space;\&space;\&space;0\leq&space;t\leq&space;2\pi$.

indicates that  $\sin&space;t&space;=0$  during    $0$   to  $2\pi$   without any error

i.e,  $f_{e}(t)&space;=&space;f_{1}(t)-C_{12}f_{2}(t)$ .

$f_{e}(t)&space;=&space;\sin&space;t$ .

Average value of error is $=&space;\frac{1}{2&space;\pi&space;}&space;\int_{0}^{2\pi&space;}&space;\sin&space;t&space;\&space;dt&space;=0$ .

This seems to be error is zero but actually there exists some error.

To avoid this false indication, we choose to minimize the average of the square of the error

i.e, Mean Square Error $\epsilon&space;=&space;\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}f_{e}^{2}(t)&space;\&space;dt$ .

$\epsilon&space;=&space;\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}(f_{1}(t)-C_{12}f_{2}(t))^{2}&space;\&space;dt$.

To find value which keeps error minimum  $\frac{d\epsilon&space;}{dC_{12}}=0$ .

$C_{12}&space;=&space;\frac{\int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t)&space;\&space;dt}{\int_{t_{1}}^{t_{2}}f_{2}^{2}(t)&space;\&space;dt}$ .

$C_{12}$  Which is similar to $C_{12}=\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{V_{2}^{2}}$ where $\int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t)&space;\&space;dt$   denotes the inner product between two  Real signals

$\therefore$  For the orthogonality of two signals $C_{12}&space;=0$

$\Rightarrow&space;\&space;\int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t)&space;\&space;dt&space;=0$.

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## Relation between Laplace and Fourier Transform

The Fourier transform  of a signal x(t) is given as

$X(j\omega&space;)&space;=&space;\int_{-\infty&space;}^{\infty&space;}&space;x(t)&space;e^{-j\omega&space;t}dt----EQN(I)$

Fourier Transform exists only if $\int_{-\infty&space;}^{\infty&space;}&space;\left&space;|&space;x(t)&space;\right&space;|dt<&space;\infty$

we know that $s=\sigma&space;+&space;j\omega$

$X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}&space;x(t)&space;e^{-s&space;t}dt$

$X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}&space;\left&space;|&space;x(t)e^{-\sigma&space;t}&space;\right&space;|&space;e^{-j\omega&space;t}dt----EQN(II)$

if we compare Equations (I) and (II) both are equal when  $\sigma&space;=0$.

i.e, $X(S)&space;=X(j\omega)|&space;\right&space;|_{s=j\omega&space;}$ .

This means that Laplace Transform is same as Fourier transform when $s=j\omega$.

Fourier Transform is nothing but the special case of Laplace transform where  $s=j\omega$ indicates the imaginary axis in complex-s-plane.

Thus Laplace transform is basically Fourier Transform on imaginary axis in the s-plane.

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## System in Continuous Time-domain

A system is defined as an entity that acts on an input signal x(t) and Transforms it into an output signal y(t)

In a system, there exists Cause and effect relationship between two (or) more signals

ex:- Electrical systems, mechanical systems etc.

A system is represented by a block, which accepts signals as inputs and produces signals as outputs.

input signal:- is called as excitation , source 9or) driving function -x(t)

output signal:- is called as response-y(t)

systems may be single input single output systems (or) Multi input Multi output systems.

Here we discuss only about SISO systems only.

$y(t)=T(x(t))$.

T- Transformation of x(t) (Or) operation on x(t).

In general systems are either Continuous -Time (or) Discrete Time systems.

 continuous-Time systems (CT system) Discrete-Time systems (DT system) A system that operates/produces on Continuous Time signals is a CT system. A system that operates/produces on Discrete Time signals  and produces a DT signals is a  DT system.

Classification of systems:-

Systems are classified into

## Distortion in a Transmission line

Signals transmitted over Transmission lines are mostly complex and consists of high frequency components, while passing through the line distortion occurs in the signal , this is commonly known as line distortion.

In ideal Transmission line , it is intended that the waveform at the receiving end of the line must be identical to the wave form at the sending end then the line is said to be distortion less line (or) distortion free line.

It has been seen that distortion in the waveform exists if all frequencies in the complex wave form do not have the same attenuation and same delay during propagation.

when the received signal is not the exact replica of the transmitted signal then the signal is said to be distorted.

causes of distortion in a Transmission Line:-

distortion is caused due to the following three reasons mainly

1. variation of characteristic impedance with respect to frequency:-The characteristic impedance of the line varies with change in frequency and line should be terminated in an impedance that does not vary with frequency to avoid distortion.
2. variation of attenuation with respect to frequency:- the attenuation of the line varies with frequency. Hence waves of different frequencies are attenuated by different amounts known as frequency distortion.
3. variation of phase constant with respect to frequency:-  phase distortion due to the variation of phase constant with the frequency which in turn varies  the velocity of propagation  with frequency . Therefore waves of different frequencies arrive at different times at the end of the line.

Thus for a line to be distortion less the characteristic impedance $Z_{o}$ , attenuation $(\alpha&space;)$ and phase constant  $(\beta&space;)$ should be independent of frequency.

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## Circuit Switched Networks

A Circuit Switched N/w consists of a set of switches connected by physical links.

A connection b/w ‘2’ stations is dedicated path made of one (or) more links. Each connection uses only one dedicated channel on each link.

i.e, each link is divided into n channels either by using TDM (or) FDM.

This circuit consists of 4 switches I, II, III and IV and Multiplexers with n=’3′ channels and one link.

In some circuits Multiplexing can be implicitly included in the switch fabric it self. In this circuit the end systems are connected to a switch for simplicity consider ‘2’ end systems A and M, connected to the switches I and III.

when A needs to communicate with M . A needs to request to a connection to M, which must be accepted by all switches and by M it self- which is called setup phase.

a channel circuit is reserved on each link and the combination of circuits forms a dedicated path. After establishing path data transfer can take place. The next phase is tear down.

i.e, after all data have been transferred. Generally circuit-switching takes place at the physical layer.

Before Communication (starting), the stations must make reservation for the resources like channels, switch buffers switch i/o ports switch processing time and are dedicated during the entire duration of data transfer until the tear down phase.

Data transferred is not packatized that is data is send as a continuous flow b/w source and destination.

there is end-to-end addressing in setup phase.

The 3 phases involved are:-

Circuit switched N/w’s requires ‘3’  setup phases

1. Connection-setup.
2. Data transfer.
3. Tear down.

Setup Phase:-

A dedicated circuit is established before the ‘2’ communicating parties talk to each other.

i.e, creating  a dedicated channels b/w switches. To communicate A with M . initially a requesting process as follows

A to I, I to IV and IV to III, III to M and an acknowledgement in the reverse order after the reception of ‘ack’ a connection is established.

Data Transfer Phase:-

In this phase data transfer occurs b/w the ‘2’ devices.

Tear down phase:-

To disconnect , a signal is sent to each switch to release the resources by any one of station.

Efficiency of Circuit Switched Network:-

These are less efficient in terms of allocated resources. Since all the resources are allocated during the entire duration of the connection  and these resources are un available to other connections.

Delay in this type of N/w’s is due to establishment of connection , data transfer and disconnecting the circuit.

Switching at the physical layer in the traditional telephone N/w uses the circuit switching approach.

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## Design issues of Network Layer

Introduction:-

The N/w layer is concerned with getting packets from source and makes many hops at intermediate routers along the way to the destination.

i.e, N/w layer is the lowest layer that deals with end-to-end transmission.

To achieve goals of N/w Layer (NL), NL must know the topology of communication subnet and to choose appropriate paths through it.

Network Layer Design issues:-

The design issues of NL are

Services provided to the Transport Layer at the N/w -Transport layer interface while providing services we must keep the following factors in mind.

the services should be independent of the router technology .

the Transport layer should be shielded from the number, type and topology of the routers present.

The N/w addresses made available to the Transport layer should use a uniform numbering plan even across LAN’s and WAN’s.

depending on above goals , the designers of N/w layer have more freedom and has assuming that the N/w layer should provide ‘2’ types of services.

1. Connection-oriented Service(COS)
2. Connection-Less Service(CLS).

Implementation of Connection-Less N/w layer service:-

In this type of N/w layer service packets are injected into subnet individually and are routed independently.

Packets are called datagrams and subnet as datagram subnet. Here no advanced setup is required.

Suppose in Transport Layer, a Process   in Host wants to send a long message to process in Host it then adds a TL header to the message and handed over to N/w layer.

Now in the N/w layer, the packet size is small compared to this message so it breaks the message into 4 packets 1, 2, 3 and 4 and gives the packets to router A.

Each router in the datagram subnet will maintain a Table (Routing Table), which gives the information about where to send the packets.

So Router A has a table, each table consists of ‘2’ columns Destination (where to go) and line (for outgoing) for that Destination.

i.e. from A to reach Destination A, No line is used, from A to reach Destination D,  B Router is used.

Now packets 1, 2, 3 and 4  (from ) Host are given to Router A then Router A stores all these packets and forwards them depending on the Routing Table of ‘A’.

Initially packets 1, 2, and 3 are forwarded using C to later on it uses a new Router B to forward 4 to depending on traffic.

Implementation of Connection-Oriented N/w layer service:-

In this service the subnet is called Virtual Circuit subnet.

A path from Source Router to Destination Router is established before sending any packets. That Router is use for all traffic flowing over that connection.

Ex: – a telephone system.

When the connection is released, the Virtual Circuit is also terminated.

Let us see an example.

In this Host wants to send packets to Host 1,2,3 and 4.

Now will establishes a connection 1 with Host.

Whenever packet comes from Host , we use connection 1 with identifier as 1.

i.e, from Routing Table of A, a packet is coming from and its identifier is ‘1’ and uses the outgoing Router C with identifier ‘1’

Similarly for   use identifier as ‘2’. This is called label switching. The comparison between these Connection-Oriented and Connection Less Services (or) Virtual Circuit (or) Datagram sub nets is given in the table.

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## Categories of Networks

Another alternating criterion for classifying N/w’s is their scale

i.e, the classification of multiple processor systems by their physical size.

Personal Area Network (PAN):-

PAN – is meant for one person. A wireless N/w connecting a computer with it’s mouse, keyboard and printer is a PAN also a PDA that controls the user’s hearing aid.

The next category is longer-range N/w’s  that is LAN, WAN, MAN -finally Inter network.

Those are categorized based on physical size, owner ship, the distance it covers and it’s physical architecture.

Local Area Network (LAN):-

• LANs which are generally called as LANs are privately owned Networks within a single building (or) campus.
• These are up to a few Km in size (10 m to 1 Km).
• These are used to connect Personal computers (or) work stations in company office (or) factories to share resources (ex: Printers) and exchange information.
• LANs are distinguished from other kinds of Networks by ‘3’ characteristics

i. their size.    ii. Transmission technology.      iii. topology.

• LANs size is restricted that is worst case transmission time is bounded and is well known before in hand makes it possible to use certain kinds of designs that would not otherwise be possible. which also simplifies N/w management.
• LANs may use a transmission technology consisting of a cable to which all the machines are attached.

Ex:-Telephone lines in rural areas.

• LANs (traditional) may run at speeds of 10 Mbps to 100 Mbps and newer ones up to 10 Gbps.

(1 Mbps  $\rightarrow$  1000000 bits per second), (1 Gbps  $\rightarrow$  1000000000 bits per second).

• The general possible topologies for LANs are bus, ring and star.

Bus Topology:-

In the Bus N/w (linear) at any instant at most one machine is the master and is allowed to transmit all other are required to refrain from sending.

An arbitration mechanism is needed to resolve conflicts when ‘2’ (or) more machines want to transmit simultaneously.

The arbitration mechanism may be centralized (or) de-centralized.

ex:- IEEE 802.3 ETHERNET a Bus based broad cast N/w is operating at 10 Mbps to 10 Gbps.

In Ethernet computers can transmit whenever they want to , if ‘2’ (or) more packets collide each computer just waits a random time and tries again later.

Ring topology:-

a second type broad casting system is the ring. In a ring each bit propagates on it’s own not waiting for the rest of the packet to which it belongs.

It also requires some arbitrating mechanism is required to the ring. IEEE 802.5 is a ring based LAN, which operates at 4 and 16 Mbps.

Ex:- FDDI.

LANs can be as simple as 2 pc’s and a printer (or) as long as with in a building.

LAN is used for sharing H/w (or) S/w (or) data.

## Hierarchial Routing(dynamic)

As the size of the N/w increases the entries in the routers routing table increases, this increase causes ‘2’ things to increase

1. Memory consumed by Routing Tables.
2. CPU processing time required to scan the entries in Routing Tables also increases.

and also the Band width needed also increases.

at particular point it is not possible almost all for a router to maintain routing tables still the size increases.

So the possible solutions for this is to use Hierarchial routing.

In this Hierarchial routing there are regions. The regions consists of no.of routers and he routers in a region are aware of how to route packets in it’s own regional routers but nothing about internal structure of other regions.

Hierarchial routing may be ‘2’ level Hierarchy as shown in the given figure.

Initially assume we don’t have Hierarchial routing that there exists 17 routers 1A,  1B, 2A, 2B, 2C, 2D……………….5E and the routing table for all these 17 routers by choosing no.of hops and destination line as parameters is given in the figure.

from router 1A to reach router 1B, it uses line 1B itself and the no.of hops are ‘1’. Similarly, from route 1A to router 4B it uses line 1C and no.of hops are ‘4’  as 1A to 1C to 3B to 4A to 4B.

if we use Hierarchial routing the no.of entries previously 17 are reduced to ‘7’. only.

The 17 outers are divided into 5 regions with region having some no.of routers.

If we observe the table , the table consists of 7 entries , the destination as region 2,3,4,5 but not routers 2A, 2B,…..etc.

but for it’s own region it is aware of other routers 1A, 1B, 1C  from 1A to reach region 4 it uses line 1C and no.of hops are 3……..

even there is a problem with Hierarchial routing is choosing the best path based on path lengths.

the best from 1A to 5C is Via region 2 rather than Via region 3.  If N/w size increases we go for other levels of Hierarchy that is a 3 level Hierarchy.

Kamour and Kleinrock (1979) discovered that the no.of levels for a N router subnet is  $ln&space;N$ entries for a router $=e&space;.ln&space;(N)$.

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## Flooding(static)

This is another type of static algorithm.

the main concept of flooding is to sent every incoming packet on a line to every other outgoing line except the line it arrived on.

flooding generates a large no.of duplicate packets, sometimes infinite unless we may take certain measures.

the measures are as follows:-

• one measure is use of hop count in the header of each packet and decrement this count at each hop when count reaches to zero discard the packet.
• How to take this hop count is another problem. Generally it is set to the length of path from source to destination and in worst cases the full diameter of the subnet.

• another way is avoid sending a packet more than once through a router this is possible by using sequence no.
• i.e, a source router (which generates packets) can put a sequence no. to each packet and each router will maintain a list of sequence nos. and if sees a packet with same sequence no in the list that packet is discarded (not flooded).

another way of flooding is of use selective flooding.

i.e, with this the router wouldn’t send every incoming packet on every line instead the router will send packets in a particular direction only.

i.e, east bound packets are sent on east side routers and similarly on  west side by west side routers.

even flooding is cumbersome, it has some uses

i.e,

1. used in military applications.
2. used in distributive data base applications in which to update all data bases concurrently.

flooding is used rather than any other algorithm since flooding chooses shorter path between two nodes where other algorithms may not.

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## Multicast Routing(dynamic)

Some applications require widely-separated processes to work together as groups.

i.e, for example a distributed  data base system.

so there is a need to send a message to well defined groups normally large in size but small compared as a whole (system).

sending a message to such group is called Multi casting and the routing algorithm used is called Multicast Routing.

therefore some mechanism is required to create and destroy groups and allow processes to leave and join a group.

i.e, Routers learn about the hosts belong to which group, this is possible by ‘2’ ways

1.  Either hosts must inform their routers about changes in groups.
2. (or) routers must query their hosts periodically.

Now let us see how  to route messages in  Multicast routing

Consider a N/W with ‘2’ groups 1 and 2 and some are members of both 1,2. a spanning tree for the left most router A is given in the figure.

when a process sends a multicast packet to a group the first router examines it’s spanning tree and prunes(cuts) it without having the members of the other group.

Then using this pruned trees, the router can send messages to the specific group only either to group 1 using Fig(a) and to groups using Fig(b).

while pruning we use Link State Routing (or) Distance Vector Routing.

if a router B is not a member of either 1 (or) 2 and it receives a multicast message if it doesn’t want to receive messages . It sends a PRUNE message saying  don’t send any multicast messages to it.

disadvantage of this algorithm is it scales poorly to large N/w’s  hence another alternative design is core-based tree.

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## Design issues of Data Link Layer

The study of design principles of Data Link Layer deals with the algorithms for achieving reliable, efficient communication between ‘2’ adjacent machines at DLL.

adjacent means those two machines are connected by a wire— a co-axial cable, telephone line (or) point-to point wireless channel.

The essential property of a channel –wire like— means sending the bits in the same order as they are sent .

Let us suppose we have a system A and machine B and these two are connected by a wire and assume that there is no means of any software in any machine then the picture seems to be as follows

as A just puts the bits on the wire and B just takes off the bits.

i.e, they have only a finite data rate and there is a non-zero propagation delay between the time a bit is sent and the time it is received and the communication devices makes errors occasionally.

these are the limitations that have been taken care for the efficiency of the data transfer.

The protocols used for communications must take all these factors into consideration.

i.e design issues  and also the nature of errors , their causes and how they can be detected and corrected are all taken care in DLL.

Design issues of DLL:-

The DLL has specific functions to carryout those are

1. Providing a well-defined service interface to the N/w layer.
2. Dealing with the transmission errors.
3. Regulating the flow of data so that slow receivers are not swamped by fast senders.

i.e, DLL takes the packets coming from N/W layer and are encapsulated into frames for Transmission.

each frame contains a frame header and a trailer and a payload field for holding the packet.

the heart of DLL does the frame formation.

Services provided to the N/W Layer:-

The function of DLL is to provide services to N/W layer

## Optical Communication System

Optical Communication System:-

An Optical fiber transmission link comprises the elements shown in the given figure.

The key sections are

1. A transmitter consisting of a light source and its associated drive circuitry.
2. A cable offering mechanical and environmental protection to the optical fibers contained inside it.
3. A receiver consisting of a Photo detector plus amplification and signal restoring circuitry.

Additional components include optical connectors, splicers, couplers (or) beam splitters and repeaters.

Cable:-

optical fiber is one of the most important elements in an optical fiber link. The cable may contain copper wires for powering repeaters which are needed for periodically amplifying and reshaping the signal in long distance communication.

The cable generally contains several cylindrical hair-thin glass fibers, each of which is an independent communication channel.

Similar to copper cables, the installation of optical fiber cables can be either aerial, in ducts, under sea (or) buried directly in the ground.

As a result of installation and (or) manufacturing limitations, individual cable lengths will range from several hundred meters to several kilo meters for long distance applications.

The real size and cable weight determines the actual length of a single cable section.

Cable in ducts—- shorter length

Aerial/ buried applications—– longer lengths.

The complete long-distance transmission line is formed by splicing (or) connecting together these cable sections.

In optical fibers attenuation is a function of wave length .

In early stages of technology, optical fibers were used in

First window: (800nm-900nm) wave length.

Later on optical fibers are used in the long-wave length region.

Long-wave length region (1100-1600) nm

Second window-centered around 1300nm.

Third window- centered around 1550nm.

Transmitter:-

Once the fiber cable is installed a light source (which is dynamically compatible with the fiber cores) is used to launch power into the fiber.

The electric i/p signal is either analog (or) digital form the Transmitter circuit converts this electric signal to an optical signal.

Optical source is a square-law device. In (800-900) nm region the light source is made up of

Ga Al As and in long distance region (1100nm-1600nm) In Ga AsP is the alloy used.

after an optical signal (light) has been launched into the fiber, it will be attenuated and distorted with increasing distance because of scattering, absorption and dispersion mechanisms in the wave guide.

The attenuated and distorted , modulated optical power emerging from the fiber end will be detected by photo diode (or) photo detector.

Photo detector converts optical power into electrical signal (it also uses a square-law).

photo detectors are PIN diodes, Avalanche photo diodes and the type of material it is made up of is In Ga As.

further the electrical signal will be amplified and restored.

therefore the design of the receiver is more complex than that of transmitter.

(1 votes, average: 5.00 out of 5)

## Colpitt’s Oscillator

Colpitt’s  Oscillator is an excellent circuit and is widely used in commercial signal generators upto 100MHz.

It consists of a single-stage inverting amplifier and an LC phase shift Network.

The two capacitors $C_{1}$ and $C_{2}$ provides potential divider used for providing $V_{f}$. $C_{1}$ is the feedback element and which provides positive feedback required for sustained Oscillations.

The amplifier circuit is a self-Bias Circuit with $R_{1}$ , $R_{2}$ and parallel combination of $R_{E}$ with $C_{E}$.

$V_{CC}$ is applied through a resistor  $R_{C}$ (or) RFC choke some times. This RFC choke offers very high impedance to high frequency currents.

$R_{C}$ value has chosen in such a way that it offers high impedance. Two coupling Capacitors $C_{C1}$ and $C_{C2}$ are used to block d.c currents, that means they do not permit d.c currents into tank circuit.

These capacitors $C_{C1}$ and $C_{C2}$ provides a path from Collector to Base through LC Network.

when $V_{CC}$ is switched on , a transient current is produced in the tank circuit an consequently damped oscillations are setup in the circuit.

The oscillatory current in the tank circuit produces a.c voltages across $C_{1}$ and $C_{2}$ . If terminal 1 is more positive w.r. to 2 , then voltages across $C_{1}$ and $C_{2}$ are opposite thus providing a phase shift of $180^{o}$ between 1 and 2.

as the transistor is operating in CE mode , it provides a phase shift of $180^{o}$.

Therefore the over all phase shift provided by the circuit results $360^{o}$ which is an essential condition for developing oscillations.

If the feedback is adjusted so that the loop gain $A\beta&space;=1$ then then the  circuit acts as an Oscillator.

The frequency of oscillation depends on the tank circuit and is varied by gang (or) group tuning of $C_{1}$ and $C_{2}$ means $C_{1}=C_{2}$.

working:-

The capacitors $C_{1}$ and $C_{2}$ are charged by $V_{CC}$ and are discharged through the coil $L$ setting up of oscillations with frequency

$f_{o}=\frac{1}{2\pi&space;}\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$.

these oscillations across $C_{1}$ are applied to the Base-Emitter junction  and the amplified version of output is collected across Collector (the frequency of amplifier output is same as that of input of the amplifier) .

This amplified energy is given back to tank circuit to compensate losses.

therefore un damped oscillations results in the circuit.

Derivation for frequency of oscillations:-

chose $\left&space;|&space;A\beta&space;\right&space;|\geq&space;1$ for sustained oscillations.

Analysis(Qualitative):-

if $Z_{1}$ , $Z_{2}$  and $Z_{3}$  are pure reactive elements  such that $Z_{1}=\frac{1}{j\omega&space;C_{1}}&space;=\frac{-j}{\omega&space;C_{1}}$$Z_{2}=\frac{1}{j\omega&space;C_{2}}&space;=\frac{-j}{\omega&space;C_{2}}$   and  $Z_{3}=j\omega&space;L$.

from the general condition for an Oscillator

$\left&space;|&space;A\beta&space;\right&space;|&space;=1$  $\Rightarrow&space;h_{ie}(Z_{1}+Z_{2}+Z_{3})+Z_{1}Z_{2}(1+h_{fe})+Z_{1}Z_{3}=0$.

$h_{ie}(-\frac{j}{\omega&space;C_{1}}-\frac{j}{\omega&space;C_{2}}+j\omega&space;L)+\frac{j^{2}}{\omega&space;^{2}C_{1}C_{2}}(1+h_{fe})-\frac{j}{\omega&space;C_{1}}.j\omega&space;L=0$

find the real and imaginary parts,

$-j(\frac{1}{\omega&space;C_{1}}+\frac{1}{\omega&space;C_{2}}-\omega&space;L)h_{ie}-\frac{1}{\omega&space;^{2}C_{1}C_{2}}(1+h_{fe})+\frac{L}{C_{1}}=0$

equating imaginary part to zero  $(\frac{1}{\omega&space;C_{1}}+\frac{1}{\omega&space;C_{2}}-\omega&space;L)=0$  ,  since $h_{ie}\neq&space;0$ .

$\frac{\omega&space;C_{1}+\omega&space;C_{2}}{\omega^{2}&space;C_{1}C_{2}}=\omega&space;L$.

after simplification

$\omega&space;^{2}=\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$.

by substituting $\omega&space;=2\pi&space;f$    results $f_{o}=\frac{1}{2\pi&space;}\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$.

substituting the value of $\omega&space;^{2}$  in the real part gives $h_{fe}=\frac{C_{2}}{C_{1}}$  . this is the condition for sustained oscillations.

(1 votes, average: 5.00 out of 5)

## OSI reference model

The OSI model is based on ISO and is introduced in the year 1983 and was revised in 1995 .

This is also known as ISO-OSI model(International Standards Organization-Open System Interconnection model)

and is used to connect open systems(open- they are ready for communication)

The OSI model has 7 layers. These layers are formed by considering the following things

1.  A layer should be created where a different abstraction is required.
2. Each layer should perform a well defined function.
3. A layer  boundaries should be chosen to minimize the information flow across the interface.
4. The function of each layer chosen by keeping an eye toward defining international standardized protocols.
5. The no.of layers chosen  that same function is not performed in the each layer and the function performed is not so small.

Now the model looks like this

Physical layer:-

Physical layer is connected with

• Transmitting raw bits over a communication channel.
• i.e, the design issue makes sure that sending ‘1’  must be received as ‘1’ itself  but not as ‘0’.
• How many volts is required to represent 1 and 0?
• How many Nano Seconds a bit lasts?
• whether transmission may proceed in both the directions (or) uni-directional and how the initial connection is established?
• whether to terminate the connection (or) not?
• i.e, most of the design issues deal with mechanical, electrical and timing interfaces & the physical transmission medium.

At physical layer  the data rate, synchronization of bits, line configuration(point-to-point,Broadcasting) and the topology used and Transmission mode( simplex/duplex) are specified.

It gets services from physical layer and offers services to the N/w layer.

The DLL makes the raw transmission as reliable and is responsible for node to node delivery . It makes physical layer appears error free to the upper layer.

the main functions of DLL are:

Framing:- The DLL divides the stream of bits received from N/w layer into manageable data units called frames.

Physical addressing:- If frames are distributed to different systems on the N/w , DLL adds header to the frame to define physical address of sender and receiver of the frame.

Flow control:- DLL also keep a fast Transmitter from drowning a slow receiver in data. therefore it requires a mechanism for controlling the flow to prevent overwhelming of the receiver.

Error control:-

DLL provides a mechanism to

• detect damaged (or) lost frames and to re transmit this damaged (or) lost frames.
• and needs a mechanism to prevent duplication of frames. error control is normally achieved through a trailer added at the end of the frame.
• it accepts data from N/w layer and break up that data into data frames and transmit the frames sequentially. If transmission is reliable it is observed by acknowledgement frame.
• MAC layer in DLL takes care of how to share channel and control the access in case of broad casting used for end-to-end (or) node-to-node delivery.

Network Layer:-

It controls the operation of subnet. The processes involved in N/w layer are

Distance Vector routing was used in ARPANET (1979) till it is replaced by Link State Routing.

The two problems in Distance Vector Routing are

1. DVR doesn’t take line Band width into account since the design metric is delay.(initially all lines are 56 Kbps so line Band width is not an issue but when lines are upgraded with 230 Kbps, 1.54 Kbps then the problem arises if we will not consider Band width).
2. one more problem that occurs in DVR is count-to infinity problem.

for these reasons it was replaced by new algorithm known as  Link State  Routing algorithm (LSR) algorithm.

the main idea of LSR is as follows(for a Router)

Step 1:- Discover it’s neighbors and learn their N/W addresses.

Step 2:-Measure the delay (or) cost to each of it’s neighbors.

Step 3:- Construct a packet with the information a Router has learned.

Step 4:- Send this packet to all the Routers.

Step 5:-Compute the shortest path to every other Router.

for example,

first of all, when a Router is booted to learn about it’s neighbors it will send a packet called ‘HELLO’ on each point-to- point line.

the Router on the other hand is expected to send back a reply saying who it is?

when two (or) more Routers are connected by a LAN, the situation is more complicated and the Routers are named uniquely to avoid any conflicts.

In the LAN A, C and F are connected to LAN, when a distance Router hears that 3 Routers are all connected to F, it is essential to know whether all 3 means same F (or) not?

To avoid this we can treat LAN as an additional node N as below

N in the above figure is an artificial node, the path from A to C is represented as ANC.

S2:- each Router in LSR requires to know an estimate of delay to each of it’s neighbors.

one way to measure delay send an ECHO packet and get reply immediately and then calculate the round-trip-delay t/2 .

for better results perform this same no.of times and use the average.

while measuring delay one question that arises is to consider the load (or) not? If load is considered, the round trip timer must be started when ECHO packet is queued.

when load is ignored the timer shouted be started when the ECHO packet reaches the front queue.

when a Router has a choice between 2 lines with the same Band width one of which is loaded all the time and the other one  is not loaded at all.

Then Router will chose the time with less load as the shortest path, this will result in better performance.

Consider a Sub net which is divided into 2 parts X and Y an is connected by 2 lines CF and EI.

Suppose the line CF is heavily loaded with long delays(including Queuing delay) after the new Routing tables have been installed most of the traffic will now go on EI.

Consequently in the nest update CF will appear as best path. Routing Tables may oscillate wildly causing some potential problems.

One solution to this is to divide the load equally among the lines but that may disturb the concept of best path.

Once the transformation needed for the exchange has been collected the next step is for each Router is to build a link state packet.

link state packet consists of information regarding to sender , sequence no, Age, neighbors delays.

for example consider the Sub net

These link state packets have to build periodically and also when a Router going down etc.

S4:- Distributing the Link State Packets

The net thing is to distribute the  Ls packets reliably. in order to distribute the packets we may use flooding , to make flooding more efficient we use sequence numbers to packets.

The main problem is with sequence no’s repetition of Seq.nos one solution is to use a 32 bit Seq.no. which may take 137 years to repeat the same no.

If a Router crashes the sequence no becomes a zero then there is a possibility a Router may discards it.

To avoid all the above problems we use a parameter called Age whenever Age=0 the Router discards a packet .

after distribution process we use refinements to this distribution process(flooding).

whenever a packet arrives it first placed in a holding area later on another packet arrives the 2 Seq.nos are compared , if they are equal the duplicate is discarded.

The figure shows the buffer space at Router B

Suppose a packet is coming from source A with Seq.no 21 and Age as 60

we may expect an Acknowledgement from C and F but not from A.

Computing the new Routes for a Router :-

after constructing the LS packets to all the Routers

we may use Dijkstra’s algorithm to construct the shortest path to all the destinations and this can be updated from time to time.

(1 votes, average: 5.00 out of 5)

## Distance Vector Routing(dynamic)

Modern Computer Networks uses dynamic algorithms rather than static algorithms.

dynamic algorithms may consider the current  traffic (or) load on the Network.

Two types of  dynamic routing algorithms are there

1. Distance Vector Routing (DVR).

Distance Vector Routing operates by the following way

each Router maintains a table (gives the information about distance to other routers) and updates these routing tables by exchanging information with it’s neighbors.

It is also known as Bellman-Ford (or) Ford Fulkerson algorithm.

DVR is used in ARPANET  and also as RIP.

In DVR each Router will maintain a Routing Table regarding to each Router in the subnet and the estimate of the time (or) distance to the destination.

one can use different design metrics like no.of hops, time delay in (milli Seconds), no.of packets Queued etc.

Here time delay is used as a metric.

Therefore a Router knows a delay to each of it’s neighbors and once every T milli Seconds these delays get updated by exchanging information with it’s neighboring Routers.

Consider a subnet with Routers A,B,…..L . Now choose a Router J with immediate neighbors (directly connected to J) are A, I, H and K.

Now the estimated delay of J to A, I, H & K are 8, 10, 12 & 6 milli Seconds respectively.

Suppose J wants to calculate a new route from J to G this is possible  by finding the delay from J to G using the neighbors to J.

i.e, J to G delay (through A) = J to A delay +A to G delay = 8+18=26 mSec.

J to G delay (through I) = J to I delay +I to G delay = 10+31=41 mSec.

J to G delay (through H) = J to H delay +H to G delay = 12+6=18 mSec.

J to G delay (through K) = J to K delay +K to G delay = 6+31=37 mSec.

The best among the 4 possibilities is through H with less delay 18 mSec and makes as entry in it’s Routing table.

In this way Router J computes all possible delays to each router and updates it in it’s Routing table.

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## Shortest Path Routing (static)

The idea of this shortest path routing is simple, which is used to build a graph with each node as router and arc represents a communication line (or) link.

This algorithm just finds the shortest path between them on the graph.

There exists many design metrics to choose to get the shortest path are no.of hops, queue length,transmission delay etc.

for example if we choose no.of hops as metric, the paths ABC, ABE have equal no of hops means that those are equally long but ABC is much larger than ABE.

The labels on the above graph (2,2,7) are computed as a function of the distance, Band width, average traffic, cost etc.

one of the algorithm used for computing the shortest path between 2 nodes is Dijkstra’s algorithm.

it is as follows, Initially all nodes are labeled with infinite distance.

Let us consider the figure as shown below

to find the shortest path from A to D.

step 1:- choose the source node as A and mark it as permanent node.

step 2:- find the adjacent nodes to A  those are B and G then choose the node with the smallest label as the permanent node.

Now this node B becomes the new working node.

step 3:- Now start at B and repeat the same procedure

by following above procedure two paths are available ABEGHD with a distance of 11 from A and ABEFHD with a distance of 10 from A.

so the second path is chosen as a shortest path.

therefore the final shortest path is ABEFHD with nodes A,B,E,F,H and D as permanent nodes.

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In some applications hosts need to send messages to many (or) all other hosts like weather reports, stock market updates (or) live radio programs.

i.e, sending a packet to all destinations simultaneously is called Broadcasting.

• first method is to send a packet to all destinations. This is a method wasteful of Band width and source needs to know the complete list of all destinations.

so this is least desirable one.

• flooding is another way to broadcast a packet, the problem with flooding is that it generates too many packets and also consumes too much of Band width.
• Third way is to use multi destination routing

In this technique each packet contains a list of destinations (or) a bit map for those destinations.

when a packet arrives at a router,  the router checks all the output lines it requires. The router generates a new copy of the packet for each output line after sufficient number of hops each packet will carry only one destination.

i.e, multi destination routing is like separately addressed packets (to B,C,D,E & D) must follow the same route one of them pays full fare and rest are free.

• The fourth type of method is to use sink tree (or) spanning tree.

A spanning tree is a subset of subnet that includes all the routers but contains no loops.

if each router knows which of it’s lines belong to spinning tree then it broadcasts packet to all the lines except the one it arrived on.

This is efficient method in terms of Band width usage but problem is to maintain the knowledge of all the nodes of spanning tree at a routes.

• Last method is to use Reverse path forwarding to approximate behavior of spanning tree.

Consider a subnet and it’s sink tree for router I as root node and how reverse path algorithm works in figure (C)

on the first hop I sends packets to F, H, J & N. on the second hop eight packets are generated among them 5 are given to preferred paths indicated as circles (A,D,G,O,M)

of the 6 packets generated in third hop only 3 are given to preferred paths (C,E & K) the others are duplicates.

in the fourth hop to B and L after this broadcasting terminates.

• it is easy to implement.
• it does not require routers to known about spanning trees.
• it does not require any special mechanism to stop the process (as like flooding).

The principle is : if a packet arrives on a line if it is preferred one to reach source it gets forwarded.

if it arrives on a line that is not preferred one that packet is discarded as a duplicate.

ex:-

when a packet arrives at ‘L’ the preferred paths are N and P so it forward the packets to both N and P and if a packet arrives at ‘K’ , there the preferred path is M and N is not preferred so it forwards packet to M and discards to N.

This is reverse path forwarding.

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## Asynchronous Transfer Model(ATM)

Introduction:-

Asynchronous Transfer Model is another important connection oriented Network.

Why we call it asynchronous is most of  the transmission in telephone systems is synchronous (closed tied to a clock) but ATM is not such type.

ATM was designed in 1990’s, it was the cell ray protocol designed by the ATM forum and was adopted by ITU-T.

Design goals:-

1.  A technology is required that provides large data rates with the high data rate transmission media available (Optical Fiber Communication) and this media requires less susceptible to noise.
2. The system must interface with existing systems to provide wide-area inter connectivity.
3. The cost for such a system should not be more.
4. The new system must be connection-oriented type.
5. The new system must be able to work with the existing tele-communication hierarchies like local loops, long distance carriers etc.

The problems associated with existing networks:-

The design goals come into picture for ATM, since there exists some problems that are associated with the existing systems.

Frame Networks:-

Before ATM we have data communications at DLL are based on frame switching and frame networks .

i.e, different protocols use frames of varying size (frame has data and header). If header size is more than that of actual data  there is a burden so some protocols have enlarged the size of data unit relative to the header.

if there is no data in such cases there is a wastage , so there is to provide variable frame sizes to the users.

Mixed N/w Traffic:-

If there exists variable frame sizes

1. The switches Multiplexers and routers must require an elaborate Software to manage variable size frames.
2. Internet working among different frame N/w ‘s become slow and expensive too.

suppose we have two networks generating frames of variable sizes that is N/W 1 is connected to line 1 and the frame is X. N/W 2is connected to line 2 and of having 3 frames of equal sizes A,B,C are connected to a TDM.

If X has arrived a bit earlier than A,B,C (having more priority than X) on the output line . The frames has to wait for a time to move on to the output line, this causes delay for line 2 N/W.

i.e, Audio and video frames are small so mixing them with conventional data traffic often creates unacceptable delays and makes shared frame links unusable for audio and video information.

but we need to send all kinds of traffic over the same links.

Cell Networks:-

so a solution to frame internet working is by adopting a concept called cell networking.

In a cell N/W we use a small data unit of fixed size called cell so all types of data are loaded into identical cells and are multiplexed with other cells and are routed through the cell N/W.

because each cell is small and of same size the problems associated with multiplexing different sized frames are avoided.

Asynchronous TDM:-

ATM  uses asynchronous TDM- hence the name Asynchronous Transfer Model.

i.e, it multiplexes data coming from different channels. it also uses fixed size slots called cells.

ATM Mux’rs fill a slot with a cell from any input channel that has a cell and slot is empty if there is no cell.

ATM architecture:-

ATM was going to solve all the world’s networking and tele-communications problems by merging voice, data, cable TV, telex,telegraph…… and everything else into a single integrated system that could do everything for everyone.

i.e, ATM was much successful than OSI and is now widely used in telephone system for moving IP packets.

ATM is a cell-switched N/W the user access devices are connected through a user-to- N/W interface (UNI) to the switches inside the N/W. The switches are connected through N/W-to-N/W interface (NNI) as shown in the following figure

Virtual Connection:-

two end points is accomplished through transmission paths (TP’s), Virtual Paths (VP’s) and Virtual Circuits (VC’s)

ATM Virtual Circuits:-

Since ATM N/w’s are connection-oriented, sending data requires a connection , first sending a packet to setup the connection.

• as setup packet travels though the sub net all the routers on the path make an entry in their internal tables noting for existence and reserving the resources.
• connections are often called virtual circuits and most ATM N/W’s support permanent virtual circuits.  i.e, for permanent connections b/w two hosts.
• after establishing a connection either side can transmit data.
• all information is in small, fixed size packets called cells.
• cell routing is done in Hard ware at high speed.
• fixed size cells makes the building of Hard ware routers easier with short, fixed length cells.
• variable length IP packets have to be routed by Software which is a slower process.
• ATM uses the Hardware that can setup to copy one incoming cell to multiple output lines (ex:-TV).
• All cells follow the same route to the destination.
• cell delivery is not guaranteed but their order is.
• if cells lost along the way it is up to higher protocol levels to recover from lost cells but this also not guarantee.
• ATM N/W’s are organized like traditional WAN’s with lines and switches.
• the most common speeds for ATm are
• 155 Mbps-used for high definition TV.
• 155.52 Mbps-used for AT & T’s SONET transmission system
• 622 Mbps-4 X 155 Mbps channels can be sent over it.

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## 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 $[Z_{R}=\infty&space;]$ .

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.

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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 $[Z_{R}=\infty&space;]$ .

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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## 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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## 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.

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## 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

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

(1 votes, average: 5.00 out of 5)

## 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

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## 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.

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## 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.

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## Switches-Circuit Switches

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 $\geq&space;4N(\sqrt{2N}-1)$.

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## 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

$S_{WBFM}(f)&space;=&space;\frac{A_{c}}{2}\sum_{n=-\infty&space;}^{\infty&space;}J_{n}(\beta&space;)&space;[\delta&space;(f-(f_{c}+nf_{m}))+\delta&space;(f+(f_{c}-nf_{m}))]$

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

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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## 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 [Fourier Transform in the case of Non-Periodic signal]. 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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## 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 $R_{xx}(t)&space;=&space;\frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi&space;}}$.

(No Ratings Yet)

## Figure of merit of FM

The block diagram of FM Receiver in the presence of noise is as follows

The incoming signal at the front end of the receiver is an FM signal $S_{FM}(t)&space;=&space;A_{c}&space;cos(2\pi&space;f_{c}t+2\pi&space;k_{f}\int&space;m(t&space;)dt&space;)--------Equation(1)$ got interfered by Additive noise $n(t)$, since the FM signal has a transmission band width $B_{T}$,the Band Pass filter characteristics are also considered over the band of interest i.e from$\frac{-B_{T}}{2}$ to $\frac{B_{T}}{2}$.

The output of Band Pass Filter is $x(t)&space;=&space;S_{FM}(t)+n_{o}(t)------------------Equation(2)$ is passed through a Discriminator for simplicity simple slope detector (discriminator followed by envelope detector) is used, the output of discriminator is $v(t)$ this signal is considered over a band of $(-W,W)$ by using a LPF .

The input noise to the BPF is n(t),  the  resultant output noise is band pass noise $n_{o}(t)$

$n_{o}(t)&space;=&space;n_{I}(t)cos&space;\omega&space;_{c}t-n_{Q}(t)sin&space;\omega&space;_{c}t-----------Equation(3)$

phasor representation of Band pass noise is $n_{o}(t)=r(t)cos&space;(\omega&space;_{c}t&space;+\Psi&space;(t))$ where $r(t)=\sqrt{n_{I}^{2}(t)+n_{Q}^{2}(t)}$ and $\Psi&space;(t)&space;=&space;\tan&space;^{-1}(\frac{n_{Q}(t)}{n_{I}(t)})$.

$n_{I}(t),n_{Q}(t)$ are orthogonal, independent and are Gaussian.

$r(t)$– follows a Rayleigh’s distribution and $\Psi&space;(t)$ is uniformly distributed over $(0,2\pi&space;)$$r(t)&space;and&space;\Psi&space;(t)$ are separate random processes.

substituting Equations (1), (3) in (2)

$x(t)&space;=&space;S_{FM}(t)+n_{o}(t)$

$x(t)=&space;A_{c}&space;cos(2\pi&space;f_{c}t+2\pi&space;k_{f}\int&space;m(t&space;)dt&space;)+n_{I}(t)cos&space;\omega&space;_{c}t-n_{Q}(t)sin&space;\omega&space;_{c}t$

$x(t)=&space;A_{c}&space;cos(2\pi&space;f_{c}t+2\pi&space;k_{f}\int&space;m(t&space;)dt&space;)+r(t)cos&space;(\omega&space;_{c}t&space;+\Psi&space;(t))$

$x(t)=&space;A_{c}&space;cos(2\pi&space;f_{c}t+\Phi&space;(t)&space;)+r(t)cos&space;(\omega&space;_{c}t&space;+\Psi&space;(t))-----Equation(4)$ where $\Phi&space;(t)=2\pi&space;k_{f}\int&space;m(t&space;)dt$.

now the analysis is being done from it’s phasor diagram/Noise triangle as follows

$x(t)$ is the resultant of two phasors $A_{c}&space;cos(2\pi&space;f_{c}t+\Phi&space;(t)&space;)$ and $r(t)cos&space;(\omega&space;_{c}t&space;+\Psi&space;(t))$.

$\theta&space;(t)-\Phi&space;(t)&space;=&space;\tan&space;^{-1}(\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}+r(t)\cos&space;(\Psi&space;(t)-\Phi&space;(t))})$

$\theta&space;(t)-\Phi&space;(t)&space;=&space;\tan&space;^{-1}(\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}})$ since $r(t)<&space;

$\theta&space;(t)=\Phi&space;(t)&space;+\tan&space;^{-1}(\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}})$

$\theta&space;(t)=\Phi&space;(t)&space;+\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}}$ because $\frac{r(t)}{A_{c}}<&space;<&space;1\Rightarrow&space;\tan&space;^{-1}\theta&space;=\theta$.

$\theta&space;(t)$ is the phase of the resultant signal $x(t)$ and when this signal is given to a discriminator results  an output$v(t)$.

i.e, $v(t)=\frac{1}{2\pi&space;}\frac{d\theta&space;(t)}{dt}$

i.e, $v(t)=&space;\frac{1}{2\pi&space;}\frac{d}{dt}(\Phi&space;(t)&space;+\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}})---------Equation(5)$

As $\Phi&space;(t)&space;=2\pi&space;k_{f}\int&space;m(t)dt$

$\frac{d\Phi&space;(t)}{dt}=2\pi&space;k_{f}m(t)$

the second term in the Equation $n_{d}(t)=\frac{1}{2\pi&space;}\frac{d}{dt}(\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}})$ where $n_{d}(t)$ – denotes noise after demodulation.

this can be approximated to $n_{d}(t)=\frac{1}{2\pi&space;A_{c}&space;}\frac{d}{dt}(r(t)\sin&space;\Psi&space;(t))-------Equation(6)$, which is a valid approximation. In this approximation $r(t)\sin&space;\Psi&space;(t)$ is Quadrature-phase noise with power spectral density $S_{NQ}(f)$ over $(\frac{-B_{T}}{2},\frac{B_{T}}{2})$

the power spectral density of $n_{d}(t)$ will be obtained from Equation (6) using Fourier transform property $\frac{d}{dt}\leftrightarrow&space;j2\pi&space;f$

$S_{Nd}(f)=\frac{1}{(2\pi&space;A_{c})^{2}}(2\pi&space;f)^{2}S_{NQ}(f)$

$S_{Nd}(f)=(\frac{f}{A_{c}})^{2}S_{NQ}(f)$  , $\left&space;|&space;f&space;\right&space;|\leq&space;\frac{B_{T}}{2}$

$S_{Nd}(f)=0$    elsewhere.

the power spectral density functions are drawn in the following figure

$\therefore&space;v(t)&space;=&space;k_{f}m(t)+n_{d}(t)-------Equation(7))$ , from Carson’s rule $\frac{B_{T}}{2}\geq&space;W$

the band width of v(t) has been restricted by passing it through a LPF.

Now, $S_{Nd}(f)=(\frac{f}{A_{c}})^{2}S_{NQ}(f),\left&space;|&space;f&space;\right&space;|\leq&space;W$

$S_{Nd}(f)=0&space;elsewhere$.

To calculate Figure of Merit $FOM&space;=&space;\frac{(SNR)_{output}}{(SNR)_{input}}$

Calculation of $(SNR)_{output}$:-

output Noise power $P_{no}&space;=&space;\int_{-W}^{W}(\frac{f}{A_{c}})^{2}&space;N_{o}df$

$P_{no}&space;=&space;\frac{N_{o}}{A_{c}^{2}}&space;\left&space;(&space;\frac{f^{3}}{3}&space;\right&space;)^{W}_{-W}$

$P_{no}&space;=&space;\frac{N_{o}}{A_{c}^{2}}&space;\left&space;(&space;\frac{2W^{3}}{3}&space;\right&space;)------Equation(I)$

The output signal power is calculated from $k_{f}m(t)$ tha is  $P_{so}&space;=&space;k_{f}^{2}P--------Equation(II)$

$(SNR)_{output}&space;=&space;\frac{P_{so}}{P_{no}}$

From Equations(I) and (II)

$(SNR)_{output}&space;=\frac{\frac{N_{o}}{A_{c}^{2}}&space;\left&space;(&space;\frac{2W^{3}}{3}&space;\right&space;)}{k_{f}^{2}P}$

$(SNR)_{output}&space;=\frac{3}{2}\frac{k_{f}^{2}PA_{c}^{2}}{N_{o}W^{3}}-------Equation(8)$

Calculation of $(SNR)_{input}$:-

$(SNR)_{input}&space;=&space;\frac{P_{si}}{P_{ni}}$

input signal power $P_{si}=&space;\frac{A_{c}^{2}}{2}---------------Equation(III)$

noise signal power  $P_{ni}=N_{o}W--------------Equation(IV)$

from Equations (III) and (IV)

$(SNR)_{input}&space;=&space;\frac{A_{c}^{2}}{2WN_{o}}-------------------Equation(9)$

Now the Figure of Merit of FM is $FOM&space;=&space;\frac{(SNR)_{output}}{(SNR)_{input}}$

$FOM&space;=&space;\frac{\frac{3}{2}\frac{k_{f}^{2}PA_{c}^{2}}{N_{o}W^{3}}}{\frac{A_{c}^{2}}{2WN_{o}}}$

$FOM_{FM}&space;=&space;\frac{3k_{f}^{2}P}{W^{2}}--------Equation(10)$

to match this with AM tone(single-tone) modulation is used i.e, $m(t)&space;=&space;cos&space;\omega&space;_{m}t$ then the signal power $P&space;=&space;\frac{1}{2}$  and $W&space;=&space;f_{m}$

$FOM_{FM}&space;=&space;\frac{3k_{f}^{2}}{f_{m}^{2}}\frac{1}{2}$

$FOM_{FM}&space;=&space;\frac{3}{2}&space;(\frac{k_{f}}{f_{m}})^{2}$

$FOM_{FM}&space;=&space;\frac{3}{2}\beta&space;^{2}$

since for tone(single-tone) modulation $\beta&space;=&space;\frac{k_{f}}{f_{m}}$.

when you compare single-tone FM with AM $FOM_{FM&space;(single-tone)}&space;=FOM_{AM(single-tone)}$

$\frac{3}{2}\beta&space;^{2}&space;>&space;\frac{1}{3}$

$\beta&space;>&space;\frac{\sqrt{2}}{3}$

$\beta&space;>&space;0.471$.

the modulation index $\beta&space;>0.471.$ will be beneficial in terms of noise cancellation, this is one of the reasons why we prefer WBFM over NBFM.

(2 votes, average: 3.50 out of 5)

## 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.

(1 votes, average: 5.00 out of 5)

## 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 $J=&space;qn\mu&space;_{n}&space;E+qD_{n}\frac{dn}{dx}$.

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## 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

$\therefore&space;\sigma&space;_{p}&space;\approx&space;p\mu&space;_{p}q$.

$J_{P}=&space;p\mu&space;_{p}q&space;E$.

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## 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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## 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.

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