## Figure of merit of FM

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$S_{Nd}(f)=0$    elsewhere.

the power spectral density functions are drawn in the following figure

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

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

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

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

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

Calculation of $(SNR)_{output}$:-

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

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

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

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

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

From Equations(I) and (II)

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

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

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

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

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

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

from Equations (III) and (IV)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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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## Energy Spectal Density of rectangular pulse and solved problems

As we all know the Rectangular pulse is defined as $x(t)&space;=&space;rect(\frac{t}{T})$, exists for a duration of T sec symmetrical with respect to y-axis as shown

Fourier Transform is $X(f)$

$X(f)&space;=&space;\int_{\frac{-T}{2}}^{\frac{T}{2}}x(t)e^{-j2\pi&space;ft}dt$

$X(f)&space;=&space;\int_{\frac{-T}{2}}^{\frac{T}{2}}&space;1.e^{-j2\pi&space;ft}dt$

$X(f)&space;=&space;\left&space;[&space;\frac{e^{-j2\pi&space;ft}}{j2\pi&space;f}&space;\right&space;]&space;^{\frac{T}{2}}_{\frac{-T}{2}}$

$X(f)&space;=&space;\frac{1}{\pi&space;f}\left&space;[&space;\frac{e^{j\pi&space;ft}-e^{-j\pi&space;ft}}{2j}&space;\right&space;]$

$X(f)&space;=&space;\frac{T&space;\sin&space;\pi&space;fT}{\pi&space;fT}$

$X(f)&space;=&space;X(f)&space;=&space;sinc&space;(\pi&space;fT)$

$\therefore$ The Energy Spectral Density of the given signal $x(t)$ will be $S(f)&space;=&space;\left&space;|&space;X(f)&space;\right&space;|^{2}$

$s(f)=&space;T^{2}sinc^{2}(\pi&space;fT)$.

Pb2. Let $x(t)&space;=&space;sinc(T\pi&space;t)$ and  $y(t)&space;=&space;\frac{dx(t)}{dt}$ find the Energy Spectral Density $S(f)$ of $y(t)$.

Ans:-  The Energy Spectral Density of $y(t)$is $S(f)$

i.e,  $S(f)&space;=&space;\left&space;|&space;Y(f)&space;\right&space;|^{2}$

$y(t)&space;\leftrightarrow&space;\frac{dx(t)}{dt}$, then Fourier Tansform of  $y(t)$is

$Y(f)&space;\leftrightarrow&space;j2\pi&space;f&space;X(f)$

as $x(t)&space;=&space;sinc(\pi&space;Tt)\leftrightarrow&space;X(f)=&space;\frac{1}{T}rect(\frac{f}{T})$

then $Y(f)&space;\leftrightarrow&space;j2\pi&space;f&space;X(f)$

$\left&space;|&space;Y(f)&space;\right&space;|^{2}&space;\leftrightarrow&space;\left&space;|&space;j2\pi&space;f&space;\frac{1}{T}rect(\frac{f}{T})&space;\right&space;|&space;^{2}$

$\left&space;|&space;Y(f)&space;\right&space;|^{2}\leftrightarrow&space;\frac{4\pi&space;^{2}f^{2}}{T^{2}}rect^{2}(\frac{f}{T})$

$\therefore&space;S(f)&space;=&space;\frac{4\pi&space;^{2}f^{2}}{T^{2}}rect^{2}(\frac{f}{T})$

Pb3. The Energy contained with in the band $[0,&space;f]$, $f>0$ $E(f)&space;=&space;\frac{1}{\sigma&space;^{3}}(&space;{2-(2+2\sigma&space;f+\sigma&space;^{2}f^{2})e^{-\sigma&space;f}})$ then find the ESD $S(f)$, for any $f>0$ Energy of a signal can be defined $\int_{-\infty&space;}^{\infty&space;}\left&space;|&space;S(f)&space;\right&space;|&space;df$

Ans:- Given $E(f)&space;=&space;\int_{0}^{f}\left&space;|&space;S(f)&space;\right&space;|&space;df$

$S(f&space;)&space;=&space;\frac{dE(f)}{df}$

$E(f)&space;=&space;\frac{1}{\sigma&space;^{3}}(&space;{2-(2+2\sigma&space;f+\sigma&space;^{2}f^{2})e^{-\sigma&space;f}})$

$\frac{dE(f)}{df}&space;=&space;0-&space;\frac{1}{\sigma&space;^{3}}[2(-\sigma&space;)e^{-\sigma&space;f}+2\sigma&space;f(-\sigma&space;)e^{-\sigma&space;f}+2\sigma&space;e^{-\sigma&space;f}+\sigma&space;^{2}f^{2}(-\sigma&space;)e^{-\sigma&space;f}+\sigma&space;^{2}(2f&space;)e^{-\sigma&space;f}]$

$\frac{dE(f)}{df}&space;=&space;-&space;\frac{1}{\sigma&space;^{3}}[-2\sigma&space;e^{-\sigma&space;f}-2\sigma^{2}&space;fe^{-\sigma&space;f}+2\sigma&space;e^{-\sigma&space;f}+\sigma&space;^{3}f^{2}e^{-\sigma&space;f}+2\sigma&space;^{2}fe^{-\sigma&space;f}]$

$\frac{dE(f)}{df}&space;=&space;-&space;\frac{1}{\sigma&space;^{3}}\sigma&space;^{3}f^{2}e^{-\sigma&space;f}$

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

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

Ans:-  Auto Correlation Function $\leftrightarrow$Energy spectral Density

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

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

By using the above rule

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

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

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

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

Now the Energy Spectral Density $S(f)&space;=&space;\left&space;|&space;X(f)&space;\right&space;|^{2}$

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

by finding the inverse Fourier Transform of $S(f)$

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

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

Auto Correlation Function = $\frac{1}{2\sqrt{\pi&space;}}e^{\frac{-t^{2}}{4}}$

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