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Energy Density in Electrostatic Fields

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

EQN (I)+EQN(II) gives

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

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

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

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

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

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

simillarly,

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

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

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

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Switches-Circuit Switching

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

Circuit Switches (or) Structure of Circuit Switch:-

The Switches used in Circuit Switching are called Circuit-Switches

Space-Division Switch:-

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

Cross-bar Switch:-

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

The number of cross points required = n X m.

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

Multi stage Switch:-

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

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

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

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

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

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

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

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

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GATE problems in EMT

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Solved Example problems in Electro Magnetic Theory

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

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

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

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

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

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

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

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

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

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

$\fn_cm&space;r=5.91$

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

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

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

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

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

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

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

Phase Locked Loop (PLL)

Demodulation of an FM signal using PLL:-

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

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

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

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

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

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

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

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

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

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

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

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

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

the over all output of a loop filter is

Frequency domain representation of a Wide Band FM

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Continuous Fourier Series  expansion of

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

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

Now substituting this in the Equation (I)

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

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

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

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

The  Frequency spectrum  can be obtained by taking Fourier Transform

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

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

From the above Equation it is clear that

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

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Mutual Information I(X ; Y) Properties

Property 1:- Mutual Information is Non-Negative

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

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

Substitute Equation (II) in Equation (I)

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

The above Equation can be written as

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

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

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

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

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

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

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Example Problems in Electro Magnetic Theory Wave propagation

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

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

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

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

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

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

So given medium is a Conductor (Copper)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(1 votes, average: 5.00 out of 5)

Delta modulation and Demodulation

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

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

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

DM transmits one bit per sample.

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

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

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

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

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

This process can be illustrated in the following figure

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

Delta Modulator:-

Mathematical equations involved in DM Transmitter are

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

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

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

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

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

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

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

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

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

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

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

(1 votes, average: 5.00 out of 5)

Fourier Series and it’s applications

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(No Ratings Yet)

Block Diagram of Digital Communication System/Elements of DCS

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

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

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

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

this involves the following steps

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

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

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

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

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

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

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

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

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

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

There are two types of spreading techniques available

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

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

SS techniques are used in Military applications.

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

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

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

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Why Digital Communication is preferred over Analog Communication?

Introduction:-

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

Examples of different types of communications are

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

Why Digital?

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

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

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

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

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

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

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

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

3.Digital Circuits Vs Analog Circuits:-

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

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

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

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

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

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

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

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

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

(1 votes, average: 3.00 out of 5)

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

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

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

word statements of the field Equations:-

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

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

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

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

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

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

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

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

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

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

the first two Maxwell’s Equations can be stated as

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

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

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

(2 votes, average: 5.00 out of 5)

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

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

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

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

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

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

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

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

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

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

then by taking the Divergence of the above equation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(1 votes, average: 5.00 out of 5)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(2 votes, average: 4.00 out of 5)

Relation between E and V

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

i.e, $V_{AB}=-V_{BA}$

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

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

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

(No Ratings Yet)

Electric Potential (V)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

similarly, $V_{BA}=V_{A}-V_{B}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$ACF\leftrightarrow&space;ESD$

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

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

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

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

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

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

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

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

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

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

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

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

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

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Capture effect in Frequency Modulation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Two more advantages of FM over AM are:

1. Non-linearity in the Channel ,FM cancels it very nicely due to it’s inherent modulation and demodulation technique.
2. Capture effect( a near by interference) FM overrides this by $A_{c}$.
3. Noise cancellation.

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Propagation of plane EM wave in conducting medium (or) lossy dielectrics

A lossy dielectric medium is one which an EM wave as it propagates losses power owing to imperfect dielectric,that is a lossy dielectric is an imperfect conductor that is a partially conducting medium $(\sigma&space;\neq&space;0)$ .

where as a lossless dielectric is a  $(\sigma&space;=0)$ perfect dielectric,then wave equations for conductors are also holds good here

i.e, $\bigtriangledown&space;^{2}\overrightarrow{E}=\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}&space;+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}$

$\frac{\partial&space;}{\partial&space;t}=&space;jw$

then $\bigtriangledown&space;^{2}\overrightarrow{E}=&space;j\omega&space;\mu&space;\sigma&space;\overrightarrow{E}+\mu&space;\epsilon(j\omega&space;)&space;^{2}\overrightarrow{E}$

$\bigtriangledown&space;^{2}\overrightarrow{E}=&space;(\sigma&space;+&space;j&space;\omega&space;\epsilon&space;)j\omega&space;\mu&space;\overrightarrow{E}$

$\bigtriangledown&space;^{2}\overrightarrow{E}=&space;\gamma&space;^{2}&space;\overrightarrow{E}$

$\bigtriangledown&space;^{2}\overrightarrow{E}-&space;\gamma&space;^{2}&space;\overrightarrow{E}=0---------Equation&space;(1)$

Equation (1) is called helm holtz equation and $\gamma$ is  called propagation constant.

$\gamma&space;^{2}&space;=j\omega&space;\mu&space;(\sigma&space;+j\omega&space;\epsilon&space;)$

$\gamma&space;^{2}=&space;j\omega&space;\mu&space;\sigma&space;-\omega&space;^{2}\mu&space;\epsilon$

Since $\gamma$ is a complex quantity it can be expressed as $\gamma&space;=&space;\alpha&space;+j\beta$

$\alpha$– is attenuation constant measured in Nepers/meter.

$\beta$-is phase constant measured in radians/meter.

$(\alpha&space;+j\beta&space;)&space;^{2}=&space;j\omega&space;\mu&space;\sigma&space;-\omega&space;^{2}\mu&space;\epsilon$

$\alpha^{2}&space;+2j\alpha&space;\beta-\beta&space;^{2}&space;=&space;j\omega&space;\mu&space;\sigma&space;-\omega&space;^{2}\mu&space;\epsilon$

by equating real and imaginary parts separately $\alpha&space;^{2}-\beta&space;^{2}=&space;-\omega&space;^{2}\mu&space;\epsilon------Equation(2)$

and $2\alpha&space;\beta&space;=\omega&space;\mu&space;\sigma$

$\alpha&space;=\frac{\omega&space;\mu&space;\sigma}{2\beta&space;}$

by substituting  $\alpha$ value in the equation (2)   $\frac{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}{4\beta&space;^{2}}-\beta&space;^{2}=-\omega&space;^{2}\mu&space;\epsilon$

${\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}-4\beta&space;^{4}=-4\omega&space;^{2}\beta&space;^{2}\mu&space;\epsilon$

$4\beta&space;^{4}-4\omega&space;^{2}\beta&space;^{2}\mu&space;\epsilon&space;-{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}=0$

let $\beta&space;^{2}=t$

$4t^{2}-4\omega&space;^{2}t\mu&space;\epsilon&space;-{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}=0$

$t^{2}-\omega&space;^{2}t\mu&space;\epsilon&space;-\frac{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}{4}=0$

the roots of the above quadratic expression are

$t=\frac{\omega&space;^{2}\mu&space;\epsilon&space;\pm&space;\sqrt{\omega&space;^{4}\mu&space;^{2}\epsilon&space;^{2}-4(-\frac{\omega&space;^{2}\mu&space;^{2}\sigma&space;^{2}}{4})}}{2}$

$t=\frac{\omega&space;^{2}\mu&space;\epsilon&space;\pm&space;\sqrt{\omega&space;^{4}\mu&space;^{2}\epsilon&space;^{2}(1+\frac{\sigma&space;}{\omega&space;\epsilon&space;})^{2}}}{2}$

$\beta&space;^{2}=\frac{\omega&space;^{2}\mu&space;\epsilon&space;\pm&space;\sqrt{\omega&space;^{4}\mu&space;^{2}\epsilon&space;^{2}(1+\frac{\sigma&space;}{\omega&space;\epsilon&space;})^{2}}}{2}$

$\beta&space;=\sqrt{\frac{\omega&space;^{2}\mu&space;\epsilon&space;\pm&space;\sqrt{\omega&space;^{4}\mu&space;^{2}\epsilon&space;^{2}(1+\frac{\sigma&space;}{\omega&space;\epsilon&space;})^{2}}}{2}}$

$\beta&space;=\sqrt{\frac{\omega&space;^{2}\mu&space;\epsilon&space;(1+&space;\sqrt{(1+\frac{\sigma&space;}{\omega&space;\epsilon&space;})^{2}})}{2}}$

similarly,

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Drift and Diffusion currents

The flow of charge (or) current through a semi conductor material is of two types. Similarly the net current that flows through a PN diode is also of two types (i) Drift current and  (ii) Diffusion current.

Drift current:-

When an Electric field is applied across the semi conductor, the charge carriers attains certain velocity known as drift velocity $v_{d}=&space;\mu&space;E$ with this velocity electrons move towards positive terminal and holes move towards negative terminal of the battery. This movement of charge carriers constitutes a current known as ‘Drift current’.

Drift current is defined as the flow of electric current due to the motion of the charge carriers under the influence of an external field.

Drift current density due to free electrons $J_{n}&space;=&space;qn\mu&space;_{n}E$ atoms/Cm2  and the Drift current density due to free holes $J_{p}&space;=&space;qp\mu&space;_{p}E$ atoms/Cm2.

The current densities are perpendicular to the direction of current flow.

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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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Electro Magnetic Wave Equation

Assume a Uniform, Homogeneous,linear,isotropic and Stationary medium with Non-zero current $i.e,&space;\overrightarrow&space;J_{c}(\sigma&space;\overrightarrow{E})&space;\neq&space;0$.

When an EM wave is travelling in a conducting medium in which  $\overrightarrow&space;J\neq&space;0$. The wave is rapidly attenuated in a conducting medium and in a good conductors, the attenuation is so high at Radio frequencies. The wave penetrates the conductor only to a small depth.

choose the equation $\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J_{C}}+\overrightarrow{J_{D}}$ the time-domain representation of it is    $\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J_{C}}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

since  $\frac{\partial&space;}{\partial&space;t}&space;=j\omega$ in phasor-notation $\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;=&space;\sigma&space;\overrightarrow{E}+j\omega&space;\epsilon&space;\overrightarrow{E}$ , $\because&space;\overrightarrow{J_{C}}=\sigma&space;\overrightarrow{E}$ and $\overrightarrow{D}=\epsilon&space;\overrightarrow{E}$ .

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

By differentiating the above equation with respect to time$\frac{\partial(\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H})}{\partial&space;t}&space;=\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}&space;+&space;\epsilon&space;\frac{\partial^{2}&space;\overrightarrow{E}}{\partial&space;t^{2}}--------------Equation&space;(2)$

From the Maxwell Equation $\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}&space;=-\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$

By taking curl on both sides $\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}&space;=-&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}&space;=-&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\frac{\partial&space;\overrightarrow{(\mu&space;H)}}{\partial&space;t}$  since $\overrightarrow{B}&space;=&space;\mu&space;\overrightarrow{H}$

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

By using the vector identity

$\bigtriangledown&space;^{2}\overrightarrow{E}&space;=&space;\overrightarrow{\bigtriangledown&space;}(\overrightarrow{\bigtriangledown&space;}.&space;\overrightarrow{E})&space;-\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}$

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

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

From Equation (2)

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

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

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

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

$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{E}&space;=&space;\frac{\rho&space;_{v}}{\epsilon&space;}$

$\bigtriangledown&space;^{2}\overrightarrow{E}=\overrightarrow{\bigtriangledown&space;}(\frac{\rho&space;_{v}}{\epsilon&space;})+\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}---------EquationI$

This is called Wave equation (Electric field)for a general medium when ${\rho&space;_{v}}\neq&space;0$.

wave equation for Magnetic field of a general  medium is

$\bigtriangledown&space;^{2}\overrightarrow{H}=\overrightarrow{\bigtriangledown&space;}(\frac{\rho&space;_{v}}{\epsilon&space;})+\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{H}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{H}}{\partial&space;t^{2}}---------EquationII$.

Case 1:-  wave equation for a conducting medium

for a conductor the net charge inside an isolated conductor is $\rho&space;_{v}=0$, then the wave equation for a conducting medium ($\sigma&space;\neq&space;0$) is

$\bigtriangledown&space;^{2}\overrightarrow{E}=\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{E}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}---------EquationI$

$\bigtriangledown&space;^{2}\overrightarrow{H}=\mu&space;\sigma&space;\frac{\partial&space;\overrightarrow{H}}{\partial&space;t}+\mu&space;\epsilon&space;\frac{\partial&space;^{2}\overrightarrow{H}}{\partial&space;t^{2}}---------EquationII$

The above equations are known as wave equations for conducting medium and are involving first and second order time derivatives, which are well known equations for damped(or) attenuated waves in absorbing medium of homogeneous, isotropic such as metallic conductor.

Case 2:-  Wave equation for free space/ Non-conducting medium/loss-less medium/Perfect Di-electric medium

The conditions of free space are $\rho&space;=0,\sigma&space;=0,\overline{J}&space;=0,\mu&space;=\mu&space;_{o}$ and $\epsilon&space;=\epsilon&space;_{o}$

By substituting the above equations in general wave equations, the resulting wave equations for non-conducting medium are

$\bigtriangledown&space;^{2}\overrightarrow{E}=\mu_{o}&space;\epsilon_{o}&space;\frac{\partial&space;^{2}\overrightarrow{E}}{\partial&space;t^{2}}---------EquationI$

$\bigtriangledown&space;^{2}\overrightarrow{H}=\mu_{o}&space;\epsilon_{o}&space;\frac{\partial&space;^{2}\overrightarrow{H}}{\partial&space;t^{2}}---------EquationII$.

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Automatic Gain Control (AGC)

Let us discuss about the facts why we need AGC in a Radio Receiver , as we all know that the voltage gain available at the Receiver from antenna to demodulator in several stages of amplification is very high, so that it can amplify a very weak signal But what if the signal is much stronger at the front end of the receiver ?

If same gain (gain maintained for an incoming weak signal) is maintained by different stages of the Receiver for a stonger  incoming signal, the signal is further amplified by these stages and the received signal strength is far beyond the expectations which can be avoided. so we need to have a mechanism which will measure the stength of the input signal and accordingly adjust the gain. AGC does precisely this job and improves the dynamic range of the antenna to (60-100)dB by adjusting the gain of the Intermediate Frequency and sometimes the Radio Frequency stages.

It is generally observed that as a result of fading, the amplitude of the IF carrier signal at the detecor input may vary  as much as 30 (or) 40 dB this results in the corresponding variation in general level of reproduced signal at the receiver output.

At IF carrier minimum loud speaker output becomes inaudible and mixed up with noise.

At IF  carrier maximum loud speaker output becomes intolerably large.

Therefore a properly designed AGC reduces the amplitude variation due to fading from a high value of (30-40)dB to (3-4)dB.

Basic need of AGC or AVC:-

AGC is a sub system by means of which the overall gain of a receiver is varied automatically with the variations in the stregth of the received signal to keep the output substantially constant.

i.e, the overall requirement of an AGC circuit in a receiver is to maintain a constant output level.

Some of the factors that explain why AGC is needed:-

• When a Receiver without AGC/AVC is tuned to a strong station, the received signal may overload the subsequent IF and AF stages this overloading causes carrier distortion in the incoming signal this can be prevented by using manual gain control on first RF stage but now a days AGC circuits are used for this purpose.
• When the Receiver is tuned from one station to another, difference in signal strengths of the two stations causes an unpleasant loud output if signal is moving from a weak station to a strong station unless we initially keep the volume control very low before changing the tuning from one station to another . Changing the volume control every time before attempting to re-tunethe receiver is howeve cumbersome. Therefore AGC/AVC enables the user to listen to a station without constantly monitoring the volume control.
• AGC is particularly important for mobile Receivers.
• AGC helps to smooth out the rapid fading which may occur with long distance short-wave reception.

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Compensators-Introduction

Compensators are corrective sub systems to compensate the deficiency in the performance of the plant or system, so given a plant and a set of specifications suitable compensators are to be designed so that the overall system will meet given specifications. Proper selection of performance specifications is the most important step in the design of compensators.

i.e, All the control systems are designed to achieve specific objectives that is the requirements are defined for the control system. A good control system has less error, good accuracy, good speed of response, good relative stability, good damping which will not cause unusual Overshoots etc.

For stationary performance of the system, gain is adjusted first but gain adjustment alone can not provide satisfactory results. When gain increases, Steady-state behavior of system improves but results into poor Transient response (or) even instability.

The desired behavior of a system is specified in terms of

• Transient response.
• Steady-state error($e_{ss}$).

$e_{ss}$ →is usually specified in terms of constants $k_{a},k_{v}$ and $k_{p}$ for $u(t),r(t)$ and $p(t)$ as inputs.

Transient response → it measures relative stability and speed of response which are specified in time or frequency domain.

In time domain the measure of relative stability is in terms of $\xi$ or $M_{p}$, while the speed response is measured in terms of rise time $t_{r}$ , settling time $t_{s}$ or natural frequency $\omega&space;_{n}$. Where as in frequency-domain the measure of relative stability is given by Resonant peak $M_{r}$ or Phase margin $\varphi&space;_{pm}$ and the speed response is measured by Resonant frequency $\omega&space;_{r}$ or band width $\omega&space;_{b}$.

Once a set of performance specifications have been selected, the next step is to chose the appropriate compensator. There exists Electrical,hydraulic,pneumatic and mechanical compensators and in this context we prefer Electrical compenators.

An external device which is used to alter the behavior of the system so as to achieve given specifications is called as compensator .

Compensators can be added to the system in series or in parallel or in combination of both.

Series Compensation:-

The flow of signal in series scheme is from lower energy level towards higher energy level. This requires additional amplifiers to increase the gain and also provide necessary isolation. The number of components required in series scheme is more than in parallel scheme.

Parallel Compensation:-

In this compensation technique energy flow is from higher energy level to lower energy level.  As there is no need of any amplifiers additional components required are less.

Series-Parallel Compensation:-

This is a compensation technique which utilizes the advantages of both series and parallel compensation techniques.

Table of Z-Transforms for some standard signals

 Signal Z-Transform Region of Convergence (ROC) $\delta&space;(n)$ $1$ entire Z-plane $u[n]$ $\frac{1}{1-z^{-1}}&space;or\frac{z}{z-1}$ $\left&space;|&space;z&space;\right&space;|>&space;1$ $u[-n-1]$ $\frac{-1}{1-z^{-1}}&space;or\frac{-z}{z-1}$ $\left&space;|&space;z&space;\right&space;|<&space;1$ $a^{n}u[n]$ $\frac{1}{1-az^{-1}}&space;or\frac{z}{z-a}$ $\left&space;|&space;z&space;\right&space;|>&space;a$ $-a^{n}u[-n-1]$ $\frac{1}{1-az^{-1}}&space;or\frac{z}{z-a}$ $\left&space;|&space;z&space;\right&space;|<&space;a$ $na^{n}u[n]$ $\frac{az^{-1}}{(1-az^{-1})^{2}}&space;(or&space;)\frac{az}{(z-a)^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;a$ $cos&space;(\omega&space;_{o}n)u[n]$ $\frac{1-cos(\omega&space;_{o})z^{-1}}{1-2cos(\omega&space;_{o})z^{-1}+z^{-2}}$ $\left&space;|&space;z&space;\right&space;|>&space;1$ $sin&space;(\omega&space;_{o}n)u[n]$ $\frac{sin(\omega&space;_{o})z^{-1}}{1-2cos(\omega&space;_{o})z^{-1}+z^{-2}}$ $\left&space;|&space;z&space;\right&space;|>&space;1$ $r^{n}cos&space;(\omega&space;_{o}n)u[n]$ $\frac{1-rz^{-1}cos(\omega&space;_{o})}{1-2rz^{-1}cos(\omega&space;_{o})+z^{-2}r^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;\left&space;|&space;r&space;\right&space;|$ $r^{n}sin&space;(\omega&space;_{o}n)u[n]$ $\frac{rz^{-1}sin(\omega&space;_{o})}{1-2rz^{-1}cos(\omega&space;_{o})+z^{-2}r^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;\left&space;|&space;r&space;\right&space;|$ $\frac{1}{n},n>&space;0$ $-\ln&space;$ $\left&space;|&space;z&space;\right&space;|>&space;1$ $a^{\left&space;|&space;n&space;\right&space;|}&space;\forall&space;n$ $\frac{(1-a^{2})}{(1-az)(1-az^{-1})}$ $\left&space;|&space;a&space;\right&space;|<&space;\left&space;|&space;z&space;\right&space;|<&space;\frac{1}{\left&space;|&space;a&space;\right&space;|}$ $-na^{n}u[-n-1]$ $\frac{az^{-1}}{(1-az^{-1})^{2}}&space;(or)\frac{az}{(z-a)^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;\left&space;|&space;a&space;\right&space;|$ $(n+1)a^{n}u[n]$ $\frac{1}{(1-az^{-1})^{2}}&space;(or)\frac{z^{2}}{(z-a)^{2}}$ $\left&space;|&space;z&space;\right&space;|>&space;\left&space;|&space;a&space;\right&space;|$

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Choice of Intermediate Frequency of a receiving system is usually a compromise , since there are reasons why it is neither low nor high, nor in a certain range between the two.

The following are the major factors influencing the choice of the Intermediate Frequency in any particular system.

1. If the IF is too high poor selectivity and poor adjacent channel rejection results unless sharp cut-off filters(crystal/mechanical filters) are used in the IF stage.
2. A high value of Intermediate Frequency(IF) increases tracking difficulties.
3. If we chose IF as low frequency, image frequency rejection becomes poorer. i.e, if $\frac{f_{si}}{f_{s}}$ is more IFRR(image Frequency Rejection Ratio) has been improved, which requires a high Intermediate Frequency($f_{si}$). Similarly when $f_{s}$ is more IFRR becomes worst.
4. Average Intermediate Frequency(IF) can make the selectivity too sharp cutting of the side bands.This problem arises because the Q must be low when the IF is low, unless crystal or mechanical filters are used and hence gain per stage is low. Thus a designer is more likely to raise Q rather than increasing the number of IF amplifiers.
5. If IF is very low , the frequency stability of local oscillator must be made correspondingly high.
6. IF must not fall in the tuning range of the receiver or else instability occurs and hetero dyne whistles (noise) will be heard.

Frequencies used:-

1. Standard AM broadcast receivers tuned to (540 KHz-1650 KHz) or(6 MHz-18 MHz) and European long wave band (150 KHZ- 350 KHz) uses IF in the range (438 KHz- 465 KHz). 455 KHz is the most popular value used.
2. FM receivers using the standard (88 MHz -108 MHz) band have an IF which is almost always 10.7 MHz.
3. TV Receivers in the  VHF band (54 MHz-223 MHz),UHF band (470 MHz-940 MHz) uses IF between (26 MHz-46 MHz) and the popular values are 36 MHz and 46 MHz.
4. AM-SSB Receviers employed for short-wave reception in the short wave band / VHF band uses IF in the range (1.6 MHz to 2.3 MHz).

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Working /Operation of NPN and PNP Transistor

NPN Transistor Working:-

For Normal operation of NPN Transistor Emitter junction JE is Forward Biased and Collector junction JC is Reverse Biased.

The applied Forward Biased at Emitter-Base junction injects a large number of electrons into the N-region and these electrons have enough energy to overcome the JE junction and enter into the very thin lightly doped Base region.

Since Base is very lightly doped very few electrons recombine with the holes in the P-type Base region and constitutes a small Base current IB in μA.

The electrons in the Emitter region are more when compared to electrons in the Collector. Only 5% (or) 1% of injected electrons combines with the holes in Base to produce Iand remaining 95% (or0 99% of electrons diffuse into Collector region due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused electrons which enters Collector junction and are collected by the Collector(Positive electrode).

Thus injected electrons from Emitter reaches Collector constituting a current known as $I_{E}=I_{B}+I_{C}$ Thus Emitter current is sum of Base current and Collector current. $I_{B}$ is very small in the Base region.

Current directions are  always from negative to positive and Majority carriers are electrons in NPN Transistor.

NPN Transistor is preferred over PNP since the mobility of electron is more than that of hole that is electron moves faster than holes.

PNP Transistor Working:-

For Normal operation of PNP Transistor Emitter junction JE is forward Biased and Collector junction JC is reverse biased.

The applied FB at Emitter-Base junction injects a large number of holes in the P-type emitter region and these holes have enough energy to enter into very thin lightly doped Base region. Base is very lightly doped N-type region. Therefore very few holes combines with the Base region and constitutes a small Base current IB (in Micro Amperes).

The holes in the Emitter region are more when compared to holes in the collector region.Only 5% or 1% of injected holes from Emitter combines with the electrons in the Base to produce IB and remaining 95% (or) 99% of holes diffuse into Collector region  due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused holes which enters Collector junction and are collected by the Collector(negative electrode).

Thus injected holes from Emitter reaches Collector constituting a current known as $I_{E}=I_{B}+I_{C}$$I_{B}$ is very small in the Base region.

Majority carriers are holes in PNP Transistor.

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Working of PN-junction Diode under Forward and Reverse Bias Conditions

In order to consider the working of a diode,we shall consider the effect of forward and Reverse Bias across PN-junction.

Forward Bias:-

Forward Bias means the Positive terminal of the Battery has been  connected to P-type and negative terminal to N-type in a PN-junction diode that is when an external voltage is applied to PN-junction in such a way that it cancels the barrier potential and permits the current flow such a bias  is called as Forward-Bais.

Under No Bias voltage condition, Near the junction the holes moves towards the junction and electrons as well forms a region known as Depletion region, the region depleted with immobile ions .

when the applied voltage V establishes an electric field opposite to the potential barrier , as a result the width of the potential barrier is reduced as it is very small

0.3 Volts in Ge diode and

0.7 Volts in Si diode.

∴ a small voltage (V) is sufficient to completely eliminate the barrier that is the barrier is completely eliminated and the resistance at the junction becomes zero and the current flow across the diode can be explained as follows.

Now holes move towards junction simultaneously electrons since holes and electrons were repelled by the opposite terminals of the Battery, As the Battery voltage is sufficiently greater than barrier voltage electrons and holes gets sufficient energy to cross the barrier easily.

The continuous current in external circuit is due to electrons, the current in N-type material is due to movement of free electrons, when these electrons reaches the junction they combine with the holes at the junction and releases a new electron.Similarly, in the P-type region current is due to holes.

i.e, when an electron-hole combination takes place near the junction ,   A co-valent bond near positive terminal of the battery breaks down and it liberates an electron which moves towards positive terminal of the Battery as electron movement is  towards positive terminal of the Battery this can be treated as hole movement in opposite direction.

therefore the constant movement of electrons and holes towards opposite terminals creates a high forward current in the external circuit.

PN-juction Diode in Reverse-Bias:-

When an External voltage V is applied to a PN-junction in such a way(direction) that it increases the Potential barrier is called as Reverse Bias that is Positive terminal of the Battery connected to N-type and negative terminal to P-type.

The applied voltage V acts in the Same direction to that of Potential Barrier.

that is when the PN-junction is Reverse Biased

• The junction Potential Barrier width increases.
• The junction offers higher resistance.
• electrons and holes move away from the junction and a very small current flows through the junction because of  minority carriers known as Reverse saturation current.

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Comparision table HWR ,FWR and Bridge Rectifier

 Parameter Half-Wave Rectifier Full-wave Rectifier Bridge Rectifier No of diodes 1 2 4 Maximum Efficiency ( η ) 40.6% 81.2% 81.2% $V_{dc}$ $\frac{V_{m}}{\pi&space;}$ $\frac{2V_{m}}{\pi&space;}$ $\frac{2V_{m}}{\pi&space;}$ $I_{dc}$ $\frac{I_{m}}{\pi&space;}$ $\frac{2I_{m}}{\pi&space;}$ $\frac{2I_{m}}{\pi&space;}$ Output RMS voltage $\frac{V_{m}}{2}$ $\frac{V_{m}}{\sqrt{2}}$ $\frac{V_{m}}{\sqrt{2}}$ Average current Idc $I_{dc}$ $\frac{I_{dc}}{{2}}$ $\frac{I_{dc}}{{2}}$ Ripple Factor ($\Gamma$) 1.21 0.48 0.48 Peak Inverse Voltage (PIV) $V_{m}$ $2V_{m}$ $V_{m}$ Output Frequency f 2f 2f TUF(Transformer Utilization Factor) 0.287 0.693 0.812 Form Factor 1.57 1.11 1.11 Peak factor 2 $\sqrt{2}$ $\sqrt{2}$

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Generation of PWM and PPM using Wave forms

PWM Generator:-

The circuit that generates PWM wave is as follows, Here in this circuit Op-Amp works in comparator mode.It compares two voltages, modulating voltage with Saw-tooth Voltage. Saw-tooth voltage is taken as reference voltage.

 condition Output voltage Vo(t) $m(t)>&space;V_{r}(t)$ Low $V_{r}(t)>m(t)$ High

from the graphs whenever modulating voltage dominates saw-tooth voltage corresponding output is low.

Similarly, when saw-tooth voltage dominates modulating voltage corresponding output is High.

Then the resultant output voltage is a PWM signal.

PPM Generator:-

Now, a PPM signal has been generated by passing the PWM signal through a Mono-stable Multi vibrator . Here the resultant signal is a PPM signal with the pulse starting with respect to trailing edge of PWM signal.

The width and Amplitude of Pulse remains constant only the position of the pulse changes with respect to m(t) .

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Band Pass Sampling

However when the given signal is a Band Pass signal then a different criterion must be used to sample the signal , the Band Pass signal x(t) whose maximum BW is ‘$2f_{m}/W$‘ Hz can be completely represented and recovered from it’s samples if it is  sampled at the minimum rate of greater than or  equals to twice that of the BW.

then sampling rate $f_{s}\geq&space;2&space;X&space;BW$

i.e, $f_{s}\geq&space;4f_{m}&space;or&space;f_{s}\geq&space;2W$

Any band pass signal in time-domain can be represented in it’s in-phase $x_{I}(t)$ and quadrature phase $x_{Q}(t)$ components as

$x(t)&space;=&space;x_{I}(t)cos&space;2\pi&space;f_{c}t&space;\pm&space;x_{Q}(t)sin&space;2\pi&space;f_{c}t$

after sampling the band pass signal, the signal after reconstruction is

$x(t)&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}sinc(2f_{m}t-\frac{n}{2})cos(2\pi&space;f_{c}(t-\frac{n}{4f_{m}}))$

$T_{s}&space;=&space;\frac{1}{4f_{m}}$, where BW of band pass signal is $2f_{m}$  Hz

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Current equation of the diode

The diode under Forward bias is as follows

The current equation related to the voltage V and current I is given by

$I=I_{o}(e^{\frac{V}{\eta&space;v_{T}}}-1)&space;Amperes$

I- Diode current

Io– Reverse saturation current of the diode at room temperature.

V-applied External voltage

$\eta$– constant   = 1  For Ge

= 2 For Si.

$v_{T}=\frac{kT}{q}$ – volt equivalent temperature 26 mV at room temp.

where k-Boltzmann constant = 1.38 X 10-23 J/K.

T-Temperature of Diode in kelvin   oK = o C + 273.

q- charge of electron  = 1.6X10-19 C.

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Reverse Saturation Current (Io) of PN-Diode:-

Pulse Position Modulation(PPM)

Pulse Position modulation is another type of Pulse Time modulation technique that is in PPM the position of the pulse carrier is varied in accordance with the instantaneous values of the message signal, where as the amplitude and width of the pulse remains constant. here message lies in the position(OFF periods) of the PPM signal.

PPM demodulator:-

The PPM Demodulator consists of a Transistor T1  which acts as a switch followed by a second order Low pass filter circuit( using OP-AMP).

As the  input to the demodulator is a PPM signal, the gaps between pulses contains the information in PPM signal. Let us consider a PPM signal with OFF and ON periods marked from A to F.

Here Transistor T1 acts as a switch  as follows

• input to the base of T1 is low  —–> Transistor T1 is in cut-off region.
• input to the base of T1 is high  —–> Transistor T1 is in Saturation region.

during  the time inerval AB, the input to the base of T1 is low and transistor T1 moves into cut-off region in this condition capacitor C charges to a vlotage proportional to length of time duaration AB that is the height of the ramp is equals to duration AB.

During the time interval BC, the input to T1 is high and T1 moves into Saturation region in this case Capacitor ‘C’ discharges through T1 , this discharge is rapid and the collector voltage remains low over the duartion BC.

This process continues and results a saw-tooth wave form at the output of transistor T1 , by applying this signal to a second order LPFn Demodulated signal has been obtained as the final output.

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This is  the most commonly used Receiver and it uses “hetero dyning” principle which is used almost in all types of receivers like TR Receiver and Radar Receiver etc. The word hetero(≈different) dyne(≈mixing) means mixing  different frequencies using a Mixer. Hence the name given as super hetero dyne Receiver.

The block diagram consists of  a receiving antenna followed by an RF stage as the primary block , the receiving signal has been fed to RF stage through the antenna.

In a Super hetero dyne Receiver the incoming RF signal frequency ($f_{s}$) is combined with local oscillator frequency($f_{o}$) through a mixer and converts a signal of a lower fixed frequency (IF) this lower fixed frequency is called as Intermediate Frequency ($f_{i}$ or $f_{IF}$). A constant frequency difference is maintained between the Local Oscillator and incoming RF signal. This is provided through Capacitance tuning that is all capacitors are ganged together and operated by a common control knob.

$\therefore$ incoming RF is  down translated to IF using a mixer now this IF is given as input to the secondary stage of the block diagram that is IF amplifier. IF amplifier consists of number of transformers each consisting of a pair of mutually tuned circuits thus with a large number of double tuned circuits, operating at a specially chosen frequency the IF amplifier provides most of the gain.

Thus IF stage full fills most of the gain (sensitivity) and Band width(selectivity) requirements of the Receiver. For a Super hetero dyne receiver Sensitivity and selectivity are quite uniform throughout it’s tuning range this is one of the advantage over TRF Receiver.

The amplified IF signal is given as an input to the Detector. The Detector or the demodulator demodulates the signal and down translates the IF signal to AF(Audio Frequency) signal.

The AF signal is amplified by Audio amplifier and further by power amplifier. The last stage of the receiver is a Loud speaker , which receives AF signal. Loud speaker is in general a transducer which converts electrical signal into a voice (or) Audio.

• It provides high gain through IF amplifier that is more sensitivity is being provided by it.
• Improved selectivity over TRF receiver.
• BW remains constant over the entire operating range.
• Selectivity and Sensitivity are uniform throughout it’s tuning range.

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The block diagram consists of  a receiving antenna followed by an RF stage as the primary block , the receiving signal has been fed to RF stage through the antenna. This RF stage consists of two (or) three RF Amplifiers, these amplifiers are tuned RF Amplifiers.i.e they have variable tuned circuits at input and output sides.

The received signal has been amplified by the RF amplifiers and the amplified signal is being given as an input to the Detector. The Detector or the demodulator demodulates the signal and down converts the RF signal to AF(Audio Frequency) signal.

The AF signal is amplified by Audio amplifier and further by power amplifier. The last stage of the receiver is a Loud speaker , which receives AF signal. Loud speaker is in general a transducer which converts electrical signal into a voice (or) Audio.

1. Selectivity of TRF Receiver is poor. This is because achieving sufficient selectivity at high frequencies is difficult due to enforced use of single-tuned Circuits.
2. Instability:-(RF Stage)  The TRF Receiver suffers from a tendency to oscillate at a higher frequencies (i.e, instability), this is because multi-stage RF amplifiers has to provide high gain at high frequencies. RF amplifiers provides high gain which results in positive feed back leads to oscillations and then causes instability of the circuit. This positive feedback (caused by the leakage of output of RF stage back to it’s input) could result from power supply coupling through any other element common to input and output stages.
3. Variation of band width over tuning range:- One more draw back in TRF receiver is the BW variation over the tuning range i.e the BW of TRF receiver varies with the incoming frequency.

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Sampling Theorem

Sampling of signals is the fundamental operation in signal processing, a Continuous Time (CT) signal can be converted into a Discrete Time (DT) signal using Sampling process. Sampling is required since the advancement in both signals and systems which are digitized i.e, Digital systems operates only on digital signals only.

Sampling Theorem:-

A CT signal is first converted into DT signal by Sampling process. The sufficient number of samples must be taken so that the original signal is represented in it’s samples completely, and also the signal is represented from it’s samples, these two conditions representation and reconstruction depends on the sampling process ‘fs‘ Hz.

Sampling theorem can be given into two parts

i. A band limited signal of finite energy, which has no frequency component higher than ‘fm‘ Hz, is completely described by it’s sample values at uniform intervals less than (or) equal to 1/2fm seconds apart.

i.e, $T_{s}\leq&space;\frac{1}{2f_{m}}$  Seconds.

ii. A Band limited signal of finite energy, which has no frequency component higher than fm Hz may be completely recovered from the knowledge of it’s samples if samples are taken at the rate of 2fm samples/second.

i.e, $f_{s}\geq&space;2f_{m}$ Hz.

Statement:- A Continuous Time signal can be completely represented in it’s samples and recovered from it’s samples if the sampling frequency $f_{s}\geq&space;2f_{m}Hz.$

where $f_{s}$ is the sampling frequency.

$f_{m}$ is the highest frequency present in the original signal / Band width of the signal.

proof of Sampling theorem:-

Let us consider a CT signal x(t), which is a band limited to $f_{m}$ Hz as shown

To prove Sampling theorem, it should be shown a signal whose spectrum is band limited to fm Hz can be reconstructed exactly without any error from it’s samples taken uniformly at a rate of $f_{s}>&space;2f_{m}$ Hz.

The circuit shows the sampler

Now sampling of x(t) at a rate of fs may be achieved by multiplying x(t) with a train of impulses  $\delta&space;T_{s}(t)$ with a period ‘Ts‘ seconds.

The sampling signal is an ideal (or) instantaneous signal. This is also known as ideal (or) instantaneous sampling.

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

As $\delta&space;T_{s}(t)$ is a periodic impulse train it can be expressed in it’s Fourier Series expansion as follows

Exponential Fourier Series is

$\delta&space;T_{s}(t)&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}F_{n}e^{jnw_{s}t}$

$F_{n}=&space;\frac{1}{T_{s}}\int_{\frac{-T_{s}]}{2}}^{\frac{T_{s}}{2}}\delta&space;T_{s}(t)e^{-jn\omega&space;_{s}t}dt$

$F_{n}=\frac{1}{T_{s}}$

$F_{n}=f_{s}$

∴ Exponential Fourier Series is $\delta&space;T_{s}(t)=\sum_{n=-\infty&space;}^{\infty&space;}f_{s}e^{jn\omega&space;_{s}t}$

now the sampled signal $g(t)&space;=&space;x(t).\delta&space;T_{s}(t)$

$g(t)=x(t)\sum_{n=&space;-\infty&space;}^{\infty&space;}f_{s}e^{jn\omega&space;_{s}t}$

$g(t)=\sum_{n=&space;-\infty&space;}^{\infty&space;}f_{s}x(t)e^{jn\omega&space;_{s}t}$

By finding Fourier Transform of g(t) is G(f)

$G(f)=\sum_{n=&space;-\infty&space;}^{\infty&space;}f_{s}X(f-nf_{s})$

Now the frequency spectrum of the sampled signal G(f) is of the form

From G(f) spectrum the original spectrum of X(f) has been shifted to different center frequencies

i.e, when n=0  center frequency is 0.

n=1  center frequency is fs

n=-1 center frequency is -fs etc

Some important conclusions from frequency spectrum of sampled signal:-

1. The spectrum of sampled signal G(f)/G(w) will repeat periodically if $f_{s}>&space;2f_{m}$ without any overlapping.
2. G(f) is extending up to infinity and the Band width is infinity as well, out of G(f) , X(f) need to be recovered , which is band limited to fm Hz.
3. X(f) is centered at f=0 and has fm as the highest frequency, X(f) may be recovered by passing it through a Loe Pass filter with cutoff frequency approximately equals to fm  Hz.
4. to reconstruct x(t) from g(t) the condition that must be satisfied is  $f_{s}\geq&space;2f_{m}$.

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Indirect method of generation of FM signal

Indirect method of generation of FM signal is also known as Armstrong method .Here a crystal oscillator generates carrier signal , which provides very high stability compared to Direct method. this method  generates a  WBFM signal, i.e a phase modulator  generates a NBFM signal in the first step , then in the second step NBFM will be converted to WBFM signal using a frequency multiplier.

In NBFM modulation index is small and the distortion is very low in NBFM ,here we prefer phase modulator to generate NBFM as it’s generation is easy, the frequency multiplier multiplies incoming frequency along with frequency deviation $\Delta&space;f$ . Hence NBFM will be converted into WBFM with large frequency deviation as well.

Frequency multiplier:-

The frequency multiplier consists of a non-linear device followed by a Band Pass Filter, the non-linear device is a memory less device.

If the input to a non-linear device is an FM wave with frequency $f_{c}$ and  deviation $\Delta&space;f$ then output consists of DC component and ‘n’ frequency modulated waves with carrier frequencies $f_{c},2f_{c},3f_{c},......nf_{c}$ and frequency deviations $\Delta&space;f,2\Delta&space;f,3\Delta&space;f,4\Delta&space;f.....n\Delta&space;f$ . The BPF designing is in such a way that it passes the FM wave centered at the frequency $nf_{c}$with frequency deviation $n\Delta&space;f$ and to suppress all other FM components. Thus a frequency multiplier generates a WBFM wave from a NBFM wave.

Generation of WBFM by Armstrong’s method:-

This Armstrong’s method is indirect method used to generate WBFM signal.It is used to generate FM signal having both the desired frequency deviation and carrier frequency.

The block diagram consists of two stage multiplier and an intermediate stage of frequency translator .

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Adaptive Delta Modulation, a modification of Linear Delta Modulation (LDM) is a scheme that circumvents the deficiency of DM. In ADM step size Δ of the Quantizer  is not a constant but varies with time , we shall express Δ as $\Delta&space;(n)=&space;2&space;\delta&space;(n)$ .

where $\delta&space;(n)$ increases during a steep segment of input and decreases for a slowly varying segment of input.

The adaptive step size control which forms the basis of an ADM scheme can be classified in various ways such as

• Discrete or Continuous.
• instantaneous (or) syllabic (fairly gradual change).
• forward (or) backward.

we shall describe an adaptation scheme that is backward, instantaneous and discrete in practical implementation , the step size $\delta&space;(n)$ is constrained in between some pre-determined minimum and maximum values.

$\delta&space;_{min}&space;\leq&space;\delta&space;(n)\leq&space;\delta&space;_{max}$

The upper limit $\delta&space;_{max}$ controls the amount of Slope over load distortion and the lower limit $\delta&space;_{min}$ controls the granular noise (or) Idle noise.

The adaptive rule for  $\delta&space;(n)$can be expressed in the general form $\delta&space;(n)&space;=&space;g(n)&space;\delta&space;(n-1)$

where the time varying gain g(n) depends on the present binary output b(n) and M previous values b(n-1),b(n-2) ……….b(n-M). The algorithm is usually initiated with $\delta&space;_{min}$.

when M=1, b(n) and b(n-1) are compared to detect probable slope over load {b(n) = b(n-1)} (or) probable granularity  {b(n) ≠ b(n-1)} then g(n) is

• $g(n)&space;=&space;P$ if     $b(n)&space;=b(n-1)$.
• $g(n)=\frac{1}{P}$ if   $b(n)&space;\neq&space;b(n-1)$.

when $b(n)&space;=b(n-1)$ Slope overload distortion is detected and when $b(n)&space;\neq&space;b(n-1)$ Idle noise is detected.

where $P\geq&space;1$, note that $P=1$ represents LDM. $P_{optimum}=1.5$ minimizes the Quantization noise for speech signal, where $1<&space;P<&space;2$ is for broad class of signals.

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Maxwell’s First Equation in Electrostatics

From the Divergence theorem, we have

$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{D}=&space;div&space;\overrightarrow{D}=(&space;\frac{\partial&space;D&space;_{x}}{\partial&space;x}+&space;\frac{\partial&space;D&space;_{y}}{\partial&space;y}+&space;\frac{\partial&space;D&space;_{z}}{\partial&space;z})$

$\lim_{dv->0}\frac{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}{dv}=\lim_{dv->0}(&space;\frac{\partial&space;D&space;_{x}}{\partial&space;x}+&space;\frac{\partial&space;D&space;_{y}}{\partial&space;y}+&space;\frac{\partial&space;D&space;_{z}}{\partial&space;z})$

${\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}=\(&space;\frac{\partial&space;D&space;_{x}}{\partial&space;x}+&space;\frac{\partial&space;D&space;_{y}}{\partial&space;y}+&space;\frac{\partial&space;D&space;_{z}}{\partial&space;z})dv$

from Gauss’s law $\oint_{s}\overrightarrow{D}.\overrightarrow{ds}&space;=Q_{enclosed}$

${\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}=\(&space;\frac{\partial&space;D&space;_{x}}{\partial&space;x}+&space;\frac{\partial&space;D&space;_{y}}{\partial&space;y}+&space;\frac{\partial&space;D&space;_{z}}{\partial&space;z})dv&space;=&space;Q_{enclosed}$

dividing it by $dv(or)&space;\Delta&space;v$ differential volume on both sides

$\frac{{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}}{dv}&space;=&space;\frac{Q_{enclosed}}{dv}$

by applying limit on both  sides

$\lim_{dv->0}\frac{{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}}{dv}&space;=&space;\lim_{dv->0}&space;\frac{Q_{enclosed}}{dv}$

$div\overrightarrow{D}&space;=&space;\rho&space;_{v}$

$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{D}&space;=&space;\rho&space;_{v}$

This equation is known as Maxwell’s first equation and is also known as point form of Gauss’s law /Differential form of Gauss’s law.

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Tutorial problems Analog communications

Pb1. A broadcast radio transmitter radiates 5KW power when the modulation percentage is 60% , how much is the carier power?

pb2. A 400 Watts carrier is modulated to a depth of 75%, calculate the total power in the modulated wave by assuming the modulating wave as sinusoidal signal. .

Pb3. The antenna current of an AM transmitter is 8 A when only carrier is being transmitted , but is increases to 8.96 A , when the carrier is modulated by a single-tone sinusoid, find the percentage of modulation? find the antenna current when the depth of modulation changes to 0.8.  .

Pb4. A 300 Watts carrier is simultaneously modulated by two audio waves with modulation percentages of 50 and 60 respectively. what will be the total side band power radiated? .

pb5. Find the power of the signal $V(t)=&space;cos\omega&space;_{c}t&space;+&space;cos\omega&space;_{c}tcos\omega&space;_{m}t$  .

pb6. Find the power of the signal $V(t)=&space;cos\omega&space;_{l}t&space;+&space;cos\omega&space;_{c}tcos\omega&space;_{m}t$  .

Effective Modulation index of a Multi-tone AM signal

In a single-tone AM, message signal is a single-tone $i.e,&space;m(t)&space;=&space;A_{m}cos&space;2\pi&space;f_{m}t$ being modulated by a carrier signal and generates a single-tone modulated signal, where as in Multi-tone environment  message signal is a composite signal formed by number of frequencies f1,f2,f3 …..fn … being modulated by a carrier signal to generate an Amplitude  Modulated signal.

i.e, Multi-tone message signal is

$\therefore&space;m(t)&space;=&space;A_{1}cos&space;2\pi&space;f_{1}t&space;+A_{2}cos&space;2\pi&space;f_{2}t+A_{3}cos&space;2\pi&space;f_{3}t+....+A_{n}cos&space;2\pi&space;f_{n}t+....$

Now from the equation of General AM signal $S_{AM}(t)=A_{c}(1+k_{a}m(t))cos&space;2\pi&space;f_{c}t$

the Multi-tone modulated signal can be obtained as

$S_{AM}(t)=A_{c}(1+k_{a}(A_{1}cos&space;2\pi&space;f_{1}t&space;+A_{2}cos&space;2\pi&space;f_{2}t+A_{3}cos&space;2\pi&space;f_{3}t+....+A_{n}cos&space;2\pi&space;f_{n}t+....))cos&space;2\pi&space;f_{c}t$

$S_{AM}(t)=A_{c}(1+k_{a}A_{1}cos&space;2\pi&space;f_{1}t&space;+k_{a}A_{2}cos&space;2\pi&space;f_{2}t+k_{a}A_{3}cos&space;2\pi&space;f_{3}t+....+k_{a}A_{n}cos&space;2\pi&space;f_{n}t+....)cos&space;2\pi&space;f_{c}t$

$S_{AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+k_{a}A_{1}cos&space;2\pi&space;f_{1}t&space;cos&space;2\pi&space;f_{c}t&space;+k_{a}A_{2}cos&space;2\pi&space;f_{2}t&space;cos&space;2\pi&space;f_{c}t+.......$

$S_{AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+A_{c}\mu&space;_{1}cos&space;2\pi&space;f_{1}t&space;cos&space;2\pi&space;f_{c}t&space;+A_{c}\mu&space;_{2}cos&space;2\pi&space;f_{2}t&space;cos&space;2\pi&space;f_{c}t+.......$

$S_{AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+\frac{A_{c}\mu&space;_{1}}{2}cos&space;2\pi&space;(f_{c}+f_{1})t+&space;\frac{A_{c}\mu&space;_{1}}{2}cos&space;2\pi&space;(f_{c}-f_{1})t&space;+&space;\frac{A_{c}\mu&space;_{2}}{2}cos&space;2\pi&space;(f_{c}+f_{2})t&space;+&space;\frac{A_{c}\mu&space;_{2}}{2}cos&space;2\pi&space;(f_{c}-f_{2})t&space;+&space;..........$

from the above signal the total power can be obtained as

$P_{Total}=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}\mu_{1}&space;^{2}}{8}+\frac{A_{c}^{2}\mu_{1}&space;^{2}}{8}+\frac{A_{c}^{2}\mu_{2}&space;^{2}}{8}+\frac{A_{c}^{2}\mu_{2}&space;^{2}}{8}+......$

$P_{Total}=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}\mu_{1}&space;^{2}}{4}+\frac{A_{c}^{2}\mu_{2}&space;^{2}}{4}+......$

$P_{Total}=\frac{A_{c}^{2}}{2}(1+\frac{\mu_{1}&space;^{2}}{2}+\frac{\mu_{2}&space;^{2}}{2}+......)$

This expression can further represented in terms of effective modulation index $\mu&space;_{eff}$  as   $P_{Total}=\frac{A_{c}^{2}}{2}(1+\frac{\mu_{eff}&space;^{2}}{2})$ where  $\mu&space;_{eff}&space;=&space;\sqrt{\mu&space;_{1}^{2}+\mu&space;_{2}^{2}+\mu&space;_{3}^{2}+...}$

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Single-tone AM

single tone AM:-

The expression for conventional AM is $S_{AM}(t)=A_{c}(1+k_{a}m(t))cos&space;2\pi&space;f_{c}t$

now if the message signal is a single-tone    $i.e,&space;m(t)&space;=&space;A_{m}cos&space;2\pi&space;f_{m}t$

$S_{Single-tone&space;AM}(t)=A_{c}(1+k_{a}A_{m}cos2\pi&space;f_{m}t)cos&space;2\pi&space;f_{c}t$

where $\mu&space;=k_{a}A_{m}$ is called as modulation index

$S_{Single-tone&space;AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+\mu&space;A_{c}cos2\pi&space;f_{m}tcos&space;2\pi&space;f_{c}t$

this equation can be further simplified as follows     $S_{Single-tone&space;AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+\frac{\mu&space;A_{c}}{2}cos2\pi&space;(f_{c}+f_{m})t+\frac{\mu&space;A_{c}}{2}cos2\pi&space;(f_{c}-f_{m})t$

that is by taking the fourier transform

$\dpi{150}&space;S_{Single-tone&space;AM}(f)=\frac{A_{c}}{2}\left&space;\{&space;\delta&space;(f-f_{c})+\delta&space;(f+f_{c})&space;\right&space;\}+\frac{\mu&space;A_{c}}{4}\left&space;\{&space;\delta&space;(f-(f_{c}+f_{m}))+\delta&space;(f+(f_{c}+f_{m}))&space;\right&space;\}+\frac{\mu&space;A_{c}}{4}\left&space;\{&space;\delta&space;(f-(f_{c}-f_{m}))+\delta&space;(f+(f_{c}-f_{m}))&space;\right&space;\}$

from the above expression the amplitude spectrum can be drawn as follows

from the spectrum single tone AM consists of 6 impulse functions located at frequencies $\pm&space;f_{c}$ , $\pm&space;(f_{c}&space;+&space;f_{m})$ and $\pm&space;(f_{c}&space;-&space;f_{m})$ respectively.

Power content in AM/ Conventional AM:-

$S_{AM}(t)=A_{c}(1+k_{a}m(t))cos&space;2\pi&space;f_{c}t$ represents the AM signal , here m(t) is  some arbitrary signal , then the power of this signal can be calculated from its Mean Square value $\overline{}{m^{2}(t)}$

i.e, message signal power = $\overline{}{m^{2}(t)}$ Watts.

Carrier signal is  $C(t)=A_{c}cos&space;2\pi&space;f_{c}t$ and it’s power is $\frac{A_{c}^{2}}{2}$ Watts.

Now the total power available in the signal $S_{AM}(t)=A_{c}(1+k_{a}m(t))cos&space;2\pi&space;f_{c}t$   will be  $P_{TOTAL}$ .

$S_{AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t&space;+A_{c}k_{a}cos&space;2\pi&space;f_{c}t&space;.&space;m(t)$

$P_{TOTAL}&space;=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}k_{a}^{2}}{2}&space;X&space;message&space;signal&space;power$

$P_{TOTAL}&space;=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}k_{a}^{2}}{2}&space;X\overline{m(t)^{2}}$ Watts.

Total Side Band power can be calculated from the term   $A_{c}k_{a}cos&space;2\pi&space;f_{c}t&space;.&space;m(t)$ can be denoted as $P_{SB}$ that would be $\frac{A_{c}^{2}k_{a}^{2}}{2}&space;X\overline{m(t)^{2}}$ Watts.

from these power calculations transmission efficiency of AM can be obtained as $\eta&space;=&space;\frac{P_{SB}}{P_{Total}}&space;X100$ %

$\eta&space;=&space;\frac{\frac{A_{c}^{2}k_{a}^{2}}{2}&space;.\overline{m(t)^{2}}}{\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}k_{a}^{2}}{2}&space;.\overline{m(t)^{2}}}$ X 100%.

$\eta&space;=&space;\frac{k_{a}^{2}.\overline{m(t)^{2}}}{1+k_{a}^{2}.\overline{m(t)^{2}}}$ X 100%.

Power content in Single-tone AM:-

In single tone AM message signal is $i.e,&space;m(t)&space;=&space;A_{m}cos&space;2\pi&space;f_{m}t$, then power of the message signal is $\frac{A_{m}^{2}}{2}$ watts

carrier signal is $C(t)&space;=&space;A_{c}cos&space;2\pi&space;f_{c}t$ implies the carrier power is $\frac{A_{c}^{2}}{2}$ watts.

$S_{Single-tone&space;AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+\frac{\mu&space;A_{c}}{2}cos2\pi&space;(f_{c}+f_{m})t+\frac{\mu&space;A_{c}}{2}cos2\pi&space;(f_{c}-f_{m})t$

then the  total power of the  single-tone AM signal is from the  above equation given as

$P_{Total}=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}\mu&space;^{2}}{8}+\frac{A_{c}^{2}\mu&space;^{2}}{8}$

PTotal = Pc +PUSB+PLSB

$P_{Total}=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}\mu&space;^{2}}{4}$

$P_{Total}=\frac{A_{c}^{2}}{2}(1+\frac{\mu&space;^{2}}{2})$ Watts.

USB and LSB has same power $P_{USB}=P_{LSB}=\frac{A_{c}^{2}\mu&space;^{2}}{8}$ watts.

Now total side band power is $P_{SB}=P_{USB}+P_{LSB}=\frac{A_{c}^{2}\mu&space;^{2}}{4}$

from these power calculations transmission efficiency of AM can be obtained as $\eta&space;=&space;\frac{P_{SB}}{P_{Total}}&space;X100$ %

$\eta&space;=&space;\frac{\frac{A_{c}^{2}\mu&space;^{2}}{4}}{\frac{A_{c}^{2}}{2}(1+\frac{\mu&space;^{2}}{2})}&space;X&space;100$%

$\eta&space;=&space;\frac{\mu&space;^{2}}{(\mu&space;^{2}+2)}&space;X100$%.

Note:- Effeciency (or) Transmission efficiency of AM is only 33.3% only i.e, $\eta$ value  calculated when $\mu$ =1.

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