## aliasing effect in Sampling

Effect of under sampling (aliasing effect):-

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

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

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

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

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

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

A signal is first passed through LPF  before sampling.

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

To avoid aliasing

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

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

(1 votes, average: 5.00 out of 5)

## Reconstruction filter(Low Pass Filter)

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

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

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

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

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

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

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

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

Signal Reconstruction (Interpolation function):-

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

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

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

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

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

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

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

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

Mathematical analysis:-

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

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

From sampling theorem

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

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

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

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

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

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

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

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

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

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

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

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

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

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

This is known as interpolation formula

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

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

(1 votes, average: 5.00 out of 5)

## Time-domain representation of SSB-SC signal

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

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

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

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

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

By finding the Fourier Transform of in-phase component

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

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

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

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

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

By finding the Fourier Transform of Q-phase component

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

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

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

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

i.e,

from the above figure,

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

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

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

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

Similarly,

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

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

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

since Signum function is

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

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

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

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

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

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

similarly, SSB signal when only LSB has been transmitted

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

(1 votes, average: 5.00 out of 5)

## 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;$$S_{WBFM}(f)&space;=&space;\frac{A_{c}}{2}\sum_{n=-\infty&space;}^{\infty&space;}J_{0}(\beta&space;)&space;$ 1 $S_{WBFM}(f)&space;=&space;\frac{A_{c}}{2}\sum_{n=-\infty&space;}^{\infty&space;}J_{1}(\beta&space;)&space;$$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;$$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.

(No Ratings Yet)

## Figure of merit of FM

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$S_{Nd}(f)=0$    elsewhere.

the power spectral density functions are drawn in the following figure

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

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

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

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

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

Calculation of $(SNR)_{output}$:-

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

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

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

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

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

From Equations(I) and (II)

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

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

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

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

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

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

from Equations (III) and (IV)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(3 votes, average: 4.00 out of 5)

## Capture effect in Frequency Modulation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Two more advantages of FM over AM are:

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

(1 votes, average: 5.00 out of 5)

## Automatic Gain Control (AGC)

Why AGC is required in a Radio Receiver?

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

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

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

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

At IF  carrier maximum loud speaker output becomes intolerably large.

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

Basic need of AGC or AVC:-

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

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

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

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

(No Ratings Yet)

## Choice Of Intermediate Frequency (or) IF Amplifier in a Radio Receiver

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

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

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

Frequencies used:-

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

(2 votes, average: 2.50 out of 5)

## 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)$$m(t)>&space;V_{r}(t)$ Low $V_{r}(t)>m(t)$$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) .

(1 votes, average: 1.00 out of 5)

## 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

(No Ratings Yet)

## 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.

(1 votes, average: 5.00 out of 5)

## Super heterodyne Receiver

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.

The advantages of Super hetero dyne receiver makes it most suitable for majority of Radio Receiver applications like AM, FM, Communications, SSB, TV and even Radar Receiver.

• It provides high gain through IF amplifier that is more sensitivity is being provided by it.
• Improved selectivity over TRF receiver.
• Improved adjacent channel rejection.
• BW remains constant over the entire operating range.
• Selectivity and Sensitivity are uniform throughout it’s tuning range.
(2 votes, average: 3.00 out of 5)

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.

<script async src="https://pagead2.googlesyndication.com/pagead/js/adsbygoogle.js"></script>
style="display:block"
<script>
</script>

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.

Drawbacks of TRF Receiver:-

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.

(4 votes, average: 4.50 out of 5)

## 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  $f_{s}\geq&space;2f_{m}$.

(1 votes, average: 5.00 out of 5)

## 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 .

(2 votes, average: 4.50 out of 5)

## 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  $\mu&space;_{eff}&space;=&space;\sqrt{\mu&space;_{1}^{2}+\mu&space;_{2}^{2}+\mu&space;_{3}^{2}+...}$

(1 votes, average: 5.00 out of 5)

## 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.

(No Ratings Yet)

## Amplitude Modulation (AM)

### Amplitude Modulation (AM):-

Modulation:-  It is defined as the process in which one of the characeteristic of carrier signal  is varied in accordance with the instantaneous values of message signal (Amplitude of the message signal).

The fundamental goal of modulation is to produce an information bearing modulated signal with efficient utilization of the channel.

Amplitude modulation:- It is defined as the process in whch the amplitude of the carrier signal is varied in accordance with the  intantaneous values of message signal.

To generate a modulated signal we are in need of two signals  called as message signal & carrier signal.

• m(t)  – message signal/modulating signal/original signal/actual signal.
• C(t)- Carrier signal/unmodulated carrier signal.
• SAM(t)- Amplitude modulated signal/ Amplitude modulated Carrier.

Now these two signals are being given as inputs to an Amplitude Modulator , which in turn generates an Amplitude modulated signal SAM(t).

Here C(t) represents carrier signal  $C(t)=A&space;_{c}cos&space;2\pi&space;f_{c}t$ , the amplitude of the un modulated carrier is $A_{c}$ , when this unmodulated carrier is amplitude modulated , the new amplitude will become $A_{c}(1+k_{a}m(t))$ and the modulated carrier wave is SAM(t)

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

ka is known as amplitude sensitivity.

In AM the frequency of the carrier signal fc is assumed to be much larger than the highest frequency present in  the base band signal and in the AM swave $\left&space;|&space;k_{a}m(t)\right&space;|$ is assumed to be less than 1

i.e, $\left&space;|&space;k_{a}m(t)\right&space;|<&space;1$ for all t

if $\left&space;|&space;k_{a}m(t)\right&space;|>&space;1$ in any case with large value of amplitude sensitivity ka then the envelope of the resultant signal doesn’t represent base band signal, this causes over modulation which causes a phase reversal of the carrier wave at zero-crossings.

$\therefore&space;\left&space;|&space;k_{a}m(t)\right&space;|&space;=1$ is the limiting (or) maximum value of AM.

this $\left&space;|&space;k_{a}m(t)\right&space;|$ is called modulation index.

Note:- AM is also known as conventional AM.

### Frequency spectrum of AM:-

AM signal is given as $S_{AM}(t)=A_{c}(1+k_{a}m(t))cos&space;2\pi&space;f_{c}t$ to obtain the frequency spectrum of AM signal one must represent the signal in frequency domain

i.e by taking the fourier transform of SAM(t) we will obtain SAM(f) . Let us assume M(f) is in the figure shown below and has a bandwidth ‘B’ Hz.

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

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

by taking fourier transform of sAM(t)

$S_{AM}(f)=\frac{A_{c}}{2}(\delta&space;(f-f_{c})\delta&space;(f+f_{c}))+&space;\frac{A_{c}k_{a}}{2}(M(f-f_{c})+M(f+f_{c}))$

the frequency spectrum consists of two impulse functions at $f=\pm&space;f_{c}$and the frequency band $(f_{c}&space;to(f_{c}+f_{m}))&space;and&space;-(f_{c}&space;to(f_{c}+f_{m}))$ are called as Upper side band frequencies $(f_{c}&space;to(f_{c}-f_{m}))&space;and&space;-(f_{c}&space;to(f_{c}-f_{m}))$ are Lower side band frequencies.

Note:- Information or message is available in two sidebands LSB and USB.

BW os AM signal = 2 X BW of message signal.

(No Ratings Yet)

## Need for modulation

### Need for modulation:-

Modulation is the fundamental need for communications, the following are the basic needs for modulation

Distance:-

As low frequency signals can not travel longer distances, low frequencies can be translated into higher frequencies by using modulation schemes.

Improved Signal to Noise Ratio:-

Signal to Noise Ratio has been improved because of modulation at the Receiver.

Practicability of antennas:-

If the communication medium is free space , then messages are transmitted and received with the help of antennas.

∴ The height of the antenna is of the order of the wavelength ( λ ) of the signal being transmitted, when the signals from transmitter are transmitted without modulation then the height required for the antenna is very high, for example to transmit a message signal of frequency  f = 5 KHz

height of the antenna required would be

$h&space;\simeq&space;\frac{\lambda&space;}{4}&space;\simeq&space;\frac{c}{4f}$

$h&space;\simeq&space;\frac{3&space;X&space;10^{8}}{4&space;X&space;5X10^{3}}$

$h\simeq&space;15000&space;meters$

Designing an antenna with 15 Km length is almost impractical. To reduce the height of the antenna , instead of sending low frequency (5 KHz) signal as it is modulation is preferred , modulation technique reduces the height of the antenna and makes the antenna to be more practical both at Txr and Rxr.

Narrow Banding:-

There is a problem caused by direct transmission of base band signal which can be explained as follows , suppose a base band signal spectrum ranges from (50 Hz-10 KHz) then the height of the antenna must be 1.5 X 106 meters to receive 50 Hz signal at the receiving end , where as for 10 KHz , the height would be 7500 meters that means height of the antenna is not same for all frequencies .

∴ A wide band antenna which can operate for band edge ratio of 200 is required which is impractical, so modulation is required to use same antenna at the receiver to receive certain range of frequencies.

∴ A wide band message signal from 50 Hz to 10 KHz gets converted to a narrow band signal by a carrier frequency of 1 MHz.

This Narrow banding of Base band signal is possible with modulation which in turn eliminates the complexity of antenna height at the receiver .

Multiplexing:-

Simultaneous transmission of multiple messages over a channel is known as multiplexing. suppose number of messages from different transmitters are transmitted without modulation then there is a possibility of interference (one with other)  since the base band spectrum is identical for all the messages. Hence the transmitted messages will not be received properly at the receiver.

one technique to eliminate interference is by the use of  modulation  and the other technique is multiplexing.

• Frequency division multiplexing  (FDM)  which uses analog modulation techniques.
• Time division multiplexing (TDM) uses pulse modulation techniques.

Multiplexing reduces number of channels needed and reduce the cost of installation and maintenance.

### Radio Frequency spectrum from ELF to EHF:-

 frequency Range Description of frequency band upto 300 Hz Extreme Low frequency(ELF) 300 Hz-3 KHz Voice frequency (VF) 3 KHz-30 KHz Very Low frequency (VLF) 30 KHz-300 KHz Low Frequency (LF) 300 KHz-3 MHz Medium frequency (MF) 3 MHz-30 MHz High frequency (HF) 30 MHz-300 MHz Very High frequency (VHF) 300 MHz-3 GHz Ultra High frequency (UHF) 3 GHz-30 GHz Super High frequency (SHF) 30 GHz-300 GHz Extreme High frequency (EHF)

*Audio Frequency range: 20 Hz-20 KHz.

*UHF,SHF and EHF are Micro wave frequencies.

(1 votes, average: 5.00 out of 5)

## Basic block diagram of analog communication system

### Introduction:-

Communications refers to sending, receiving and processing of information by electrical means, that is it means exchanging information between transmitter and receiver.

In early 1840’s the type of communication used was Wire telegraphy later on the forms are as telephony, Radio communication (possible with the invention of triode tube, Satellite communications and fibre optics(with the invention of transistors and IC’s and semi-conductor devices), that means communications become more advanced with increasing emphasis on computer and other data communications.

A modern communication system is concerned with

before transmission:-

• sorting:- sorting for the right message.
• Processing:- processing is to make that message more suitable for transmission.
• storing:- storing that message before transmission.

then the actual transmission of that message takes place (processing and filtering  noise)

• decoding:-decoding the original message.
• storage:-storing a copy of that message.
• interpretation:-and analyzing for the correctness of that message.

the different forms of modern communication systems includes Mobile communications,Computer communications, Radio telemetry etc.

to become familiar with communication systems one needs to know about amplifiers and oscillators that means fundamentals of electronic circuits must be known, with these concepts as a background the every day communication concepts like noise, modulation and information theory as well as various types of systems may be studied.

The most general form of Communication system ( one or two blocks may differ) is shown in the figure basic terminology used in Communication systems is message signal /information/data,channel,noise,modulation, encoding and decoding. Communication system is meant for communicating messages between Transmitter and Receiver (or) source & destination.

source:-

source or information source is the primary block in communication system which generates original message / actual message.

i.e, selecting one message (actual message) from a group of messages itself is called as sorting data (or) information. Source generates message which may be in any form like words, code , symbols, sound signal, images, videos etc.among these the desired message has been selected and conveyed.

A transducer is one which converts one form of energy into electrical energy because the message from information source may not be always in electrical form, a transducer is used in between source and transmitter as a separate block sometimes (or) may be a part of Tx r.

Transmitter:-

Txr is meant for the following tasks

• restriction of range of audio frequencies (i.e, limiting the bandwidth of the message signal).
• Amplification.
• Modulation.

In general modulation is said to be the main function of the transmitter.

Channel:-

The medium that exists between transmitter and receiver is called as channel. The function of channel is to provide connection between transmitter  and receiver, two types of channels are  there wired/point to point  and wireless/broadcasting channels.

Point to point channels are generally wired channels(i.e, a physical medium exists) like Microwave links, optical fibre links etc.

Microwave links:- these links are used in telephone transmission.In these type of links guided EM waves are used to transmit from Txr to Rxr.

optical fibre links:- used in low-loss high speed data transmission and uses optical fibers as the medium .

Broadcast channels:- the medium or channel is wireless here, in broadcasting a single transmitter can send information to many receivers simultaneously, satellite broadcasting system is one such system.

during the process of transmission and reception, the signal gets distorted due to noise in the channel, noise may interfere with the signal at any point but noise in the channel has greatest effect on the signal.

The main function of the receiver is to reproduce the message signal in electrical form from the distorted received signal. This reproduction process is called demodulation (or) detection , in general this demodulation may be assumed as the reverse process of modulation carried out in transmission.

there are a great variety of receivers in communication systems, the type of receiver chosen depends on type of modulation, operating frequency ,its range  and type of destination required. Most common receiver is superheterodyne receiver .

so many types of receivers are available from a very simple crystal receiver with headphones to radar receiver etc.

Destination:- It is the final stage of any communication system. it would be a loud speaker / a display device/simply a load etc depending up on the requirement of the system.

(3 votes, average: 3.67 out of 5)

## Noise

### Introduction:-

Noise is probably the only topic in electronics and tele communications with which everyone must be familiar, electrical disturbances that interfere with signals produces noise and this noise ever present and limits the performance of the most of the systems. Measuring noise is very controversial almost everybody has a different method of quantifying noise and its effects.

definition:- noise is unwanted energy that interfere with the required signal. In receivers :- noise is disturbance in electric nature.

• Radio receivers—> noise appears as “hiss”.
• TV receivers —–> it appears a snow (or) colored snow pictures.
• In Pulse communication systems —->noise produces unwanted pulses.

In receivers noise effects sensitivity and band width and it decreases sensitivity as well as band width.

Basically noise can be classified as Internal and External noise .

 External Noise Internal Noise when noise sources are external to the receiver . i.e, noise source is located outside of the receiver. It is difficult to treat quantitatively external noise. Noise is created with in the receiver itself.i.e, noise source is internal to the receiver. internal noise can be treated quantitatively and reduction is also possible by appropriate receiver design.

### External Noise:-

Atmospheric noise:-

If we try to listen to short waves on a receiver which is not well equipped to receive them, an astonishing variety of strange sounds will be heard, all tending to interfere with the program. most of these sounds are the result of spurious sources of disturbance, which represents atmospheric noise generally called as “static”.

• Atmospheric noise is caused by lightning discharge in thunderstorms and other natural electric disturbances occurring in the atmosphere.
• It originates in the form of amplitude modulated impulses , and are spread over most of the RF spectrum normally used for broadcasting.
• i.e, It consists of spurious radio signal with components distributed over a wide range of frequencies. Atmospheric noise propagates over the earth in the same way as ordinary Radio waves of the same frequencies.
• Static is more severe in the case of Radio than that of Tele-vision and it becomes less severe at frequencies above 30 MHz. Since higher frequencies are limited to line of sight propagation.
• This noise is created in VHF range and above.

Extraterrestrial noise:-

This noise is generated in the earth’s outer space (atmosphere)

Extraterrestrial noise is divided into

1. Solar noise .
2. Cosmic noise.

Solar noise:-

• The sun radiates so many things our way noise is noticeable among them.
• Under “quiet” conditions , there is a constant noise radiation from the sun simply because its a large body at high temperature ≈ 6000o C.
• ∴ The radiation consists of the frequencies which we use for communications and interferes with them.
• However, the disturbances in the sun is variable and undergoes cycles at the peak of which electrical disturbances erupt. These additional disturbances are several orders of magnitude greater than the noise generated during periods of the quiet sun. The solar cycle repeats these period of great electrical disturbances approximately every 11 years, further these 11 year cycle peaks reach even a higher maximum peak every 100 years.
• Thus the noise generated by sun changes periodically with the solar disturbances.

Cosmic noise:-

• stars are also suns and have high temperatures, they radiate RF noise in the same manner as our sun, this refers to noise coming from distant stars other than sun.
• The noise received from such stars is also called “black-body noise” and is distributed fairly uniformly over the entire sky.
• Space noise is observable in the range from about 8 MHz to about 1.43 GHz this is the strongest component of noise in the range(20-120) MHz.

Industrial (or) Man-made noise:-

• This noise is strongest in Industrial areas and the frequency of Man made noise spans between 1 to 600 MHz.
• Man made noise is found in urban, sub-urban and industrial areas. The intensity of the noise made by human easily outstrips that created by any other source,  internal or external to the receiver.
• under this, sources such as Automobile, Aircraft ignition, electric motors and switching equipment leakage from high voltage lines and a multitude of other heavy electric machines are all included.
• Fluorescent lights are another powerful source of such noise and therefore should not be used where sensitive receiver is installed.

### Internal Noise:-

This noise is created by any of the active (or) passive devices found in receivers. It is created by various components used in processing the received signal and is completely internal to the system. The effect of this noise is significant at the front end of the receiver.This appears as  thermal and shot noise  caused by resistors, inductors and capacitors.

Thermal Noise:-

This noise is also known as agitation noise, Jhonson noise / white noise. thermal noise is random in nature, this mainly occurs due to random(or) rapid motion of molecules, atoms and electrons of which resistor is made up of.

from the theory of dynamics the noise generated by a resistor is proportional to it’s absolute temperature and BW over which the noise is to be measured.

$P_{n}\propto&space;T\Delta&space;f$ where B= BW= $\Delta&space;f$

$P_{n}=&space;kT\Delta&space;f$

• where k – boltzmann’s constant =1.38X10-23 J/K.
• T- Absolute temperature in Kelvin,   K = 273+ oC.
• $\Delta&space;f$ = BW of interest.
• $P_{n}$ is the maximum noise power output of a resistor.

In an ordinary resistor at the standard temp of 17oC is not connected to any voltage source and if we are measuring voltage using a DC volt meter to measure voltage across it shows a zero. Actually a resistor itself is a noise generator, if we use a very sensitive electronic volt meter it shows a very large voltage across R.

This noise voltage is caused by the random movement of electrons with in the resistor, which constitutes a current. The rate of arrival of electrons at either end of the resistor therefore varies randomly, and so does the potential difference exists between the two ends.

from the circuit diagram,

$P_{n}=\frac{V_{n}^{2}}{4R_{L}}$  —- equation(1).

The maximum power  is delivered to  load when R =RL.

$V_{n}=&space;i(R&space;+&space;R_{L})$

$V_{n}&space;=&space;i&space;2R$

$i&space;=&space;\frac{V_{n}}{2R}$

load voltage $V&space;=&space;i&space;R_{L}$

$V&space;=&space;\frac{V_{n}}{2&space;R}$

$V&space;=&space;\frac{V_{n}}{2}$ volts

Vn -source noise voltage.

V- ouput voltage measured across RL.

from equation (1) $V_{n}^{2}=P_{n}4R_{L}$

$V_{n}^{2}=4k&space;T\Delta&space;fR_{L}$

$V_{n}=\sqrt{4k&space;T\Delta&space;fR_{L}}$

Vn is known as RMS noise voltage asross a resistor.

Shot Noise:-

This occurs due to shot effect, it occurs in all active and amplifying devices (diodes/transistors).

It is caused by random variations in the arrival of electrons (or holes) at the output electrode of an amplifying device and appears as a randomly varying noise current super imposed on the output.It sounds like a shower of a lead shot were falling on a metal plate. Hence named it as shot noise.

• In electronic tubes shot noise is caused because of random emission of electrons from cathode.
• In semi-conductors shot noise occurs due to random diffusion of minority carriers .
• $i(t)&space;=&space;I_{o}&space;+&space;i_{n}(t)$

Total current = Mean DC constant current + shot noise current.

Shot noise is given by $i_{n}&space;=&space;\sqrt{2qi_{p}\Delta&space;f}$

• in -RMS shot-noise current.
• q-charge of an electron 1.6 X10-19 C.
• ip– direct diode current.
• $\Delta&space;f$– BW of the system.

This formula for shot noise is valid for vaccum tube diode under so called temp-limited conditions.

In all other cases we use the concept of equivalent noise resistance intead of shot noise formula.

Transit-time noise (or) High-frequency noise:-

It is generally observed in semi conductor devices when the transmit time of charge carriers crossing a junction is comparable with the time-period of the signal, some charge carriers diffuse back to the source (or) emitter.

this gives rise to input admittance Y

conductance G = 1/Y , this G increases with frequency which causes noise . This is also called as high frequency noise.

(1 votes, average: 5.00 out of 5)

## UNIT 1-QUIZ1

1. Find the modulation index if the amplitude of message signal is Two thirds of the amplitude of carrier signal ———————.
2. The diagonal clipping in Amplitude Demodulation (using envelope detector) can be avoided if RC time-constant of the envelope detector satisfies the following condition, (here W is message bandwidth and ω is carrier frequency both in rad/sec) ( )
a. RC<1/W             b. RC>1/W                  c. None of the above         d. RC>1/ω.
3. Given AM signal is  XAM(t)= 10 (1+0.25 sin 2πfmt) cos 2πfct , then The average side band power given for the above AM signal is ( )
a. 25W                    b. 12.5W               c.1.5625W                    d.3.125W.
4. Given AM signal is XAM(t) = 100 (1+0.85 cos 2πfmt) ,then The total power required for the above AM signal is ( )
a. 25W                   b. 12.5W               c.6.806KW                       d. None of the above.
5. Consider the AM signal Ac cos Wct + 2 cos Wct cos Wmt   for the demodulation of the signal using envelope detector the minimum value of Ac should be ( ).
a. 2                          b. 1                         c. 0.5                                    d. 0.
6. Given AM signal is SAM(t) = 100 (1+0.3 cos 2πfmt +0.4 sin 2πfmt) cos 2πfct i. i.The total power required for the above AM signal is ( )
a. 5.625W              b. 5.625KW             c.6.806KW                  d. None of the above.
ii.Modulation index is ( )
a. 0.53                     b. 0.5                         c. 0.2                               d. None of the above.
iii.The carrier power is ( )
a. 5KW                    b.6KW                       c. 7KW                           d.100KW.
iv. Total current flowing through the transmitter if carrier current is 5A ( ).
a. 5mA                      b.5.303A                 c. none of these          d. 25A.
7. If the band width of message signal is 5KHz and the carrier frequency is 200KHz then upper sideband frequencies are( )
a. 205KHz,190KHz                                                b.205KHz,-205KHz
c. 205 KHz,-195 KHz                                           d. None of the above.
8. If the highest frequency of message signal is 5KHz and the carrier frequency is 200KHz then lower sideband frequencies are( )
a. 205KHz,190KHz                                                b.205KHz,-205KHz
c. 195 KHz,-195 KHz                                            d. None of the above.
9. If the bandwidth of message signal is 500Hz then the bandwidth of Amplitude Modulated signal is ( )
a. 500Hz                    b. 1000Hz                    c.2KHz                     d. None of the above.
10. If the message m(t)= 10cos⁡2000πt and carrier signal is                                      C(t) = 25 cos 200000 πt then draw the amplitude spectrum of AM signal————— .

## UNIT 2- QUIZ 2

1. Angle modulation is a technique in which the ————— of the is varied with respect to instantaneous values of ————————— by keeping as constant. 2M.
2. Write the expression for Angle Modulated signal –. 1M.
3. An Angle Modulated signal is given as 𝑥(𝑡) = 100 cos(2𝜋𝑓𝑐 𝑡 + 4 sin(1000𝜋𝑡)) where 𝑓𝑐 = 10 𝑀𝐻𝑧. 6M.
i. The Peak frequency deviation is ( )
a. 2K                           b. 4000                   c. 4π                      d. 8π.
ii. The Peak- phase deviation is ( ).
a. 4                              b. 6                           c. 0                        d. None of the above.
iii. The power of the Modulated signal is ( ).
a. 10KW.                   b. 5 W                       c. 5 KW                d. 50W.
4. The amount of change in carrier frequency produced by modulating signal is known as ( ). 1M.
a. phase deviation                                                            b. amplitude deviation
c. Frequency deviation                                                  d. none of the above.
5. The total Transmitted power in FM is equal to the power of ( ) 1M.
a. An AM signal.                                                            b. an unmodulated carrier
c. Message signal                                                         d. all of the above.
6. A 20 MHz carrier is frequency modulated by a sinusoidal signal with frequency 1KHz such that peak frequency deviation is 100KHz what will be the modulation index ( ) 2M.
a. 100                           b.101                                c. 99                         d.200.
7. The bandwidth for above FM system will be ( ) 2M.
a. 101 KHz                 b. 202 KHz                     c. 99 KHz              d. 100 KHz.
8. Which one of the following is an indirect method of generating FM ( ) 1M.
a. Armstrong method                                                   b. Varactor diode modulator
c. Reactance BJT modulator.                                     d. Reactance FET Modulator.
9. In which of the Modulation system when the modulating frequency is doubled the modulation index reduces to half while modulating voltage remains constant ( ) 2M.                                                                                                        a. Phase                   b. Amplitude                   c. Frequency       d. None of the above.
10. In FM, the frequency deviation is ( ) 2M.                                                                         a. Proportional to modulating frequency.                                                                         b. Proportional to amplitude of modulating signal.                                                     c. Constant.                                                                                                                                   d. Zero.
11. In indirect method of FM generation FM is obtained from ( ) 1M.                         a. AM                    b. PM                 c. DSB                              d. FM
12. Write Carson’s rule –. 1M.
13. The Bandwidth of NBFM is given as –. 1M.
14. A 25 MHz carrier is modulated by a 400Hz audio sine wave. The carrier voltage is 4V and the maximum deviation is 10 KHz. The modulation index will be( ) 2M.                                                                                                                                  a. 2.5                  b. 5                     c. 15                        d. 25
15. For the above problem write the expression of FM wave will be———————————————————————————————–.1M.
16. For the problem in 14 write the expression of PM wave———————————————————————————————————.1M.
17. Standard FM broadcast stations uses a maximum bandwidth of ( ) 1M.              a. 150 KHz                   b.75KHz.            c. 200KHz         d. 15KHz.
18. Which type of oscillators are preferred for carrier generation because of their good frequency stability ( ) 1M.                                                                                 a. LR               b.LC                       c. Crystal                         d. RC.
19. The oscillator whose frequency is varied by an input voltage is called as ————————————————. 1M.
20. Maximum deviation results at what point on modulating signal if the system is FM( ) 1M.                                                                                                                      a. Zero crossing of m(t)                                                                                                              b. Peak negative amplitude and peak positive amplitude of m (t).                        c. None of the above.                                                                                                                d. Both a and b.

1. Radio receivers are classified into how many types ( ). 2M.
a. Three                   b. two                      c. four                             d. none of the above.
2. The ability of a radio receiver to amplify weak signals is called as ( ). 2M.
a. Fidelity             b. Selectivity            c. sensitivity                  d. all of the above.
3. The phenomenon of Picking up of same short wave station at two nearby points on the receiver dial is known as ( ). 2M.
a. Fidelity            b. sensitivity               c. Double spotting               d. selectivity.
4. The ability of a receiver to reject unwanted signals is called ( ). 2M.
a. Selectivity           b. Fidelity                  c. sensitivity                  d. Double spotting
5. Standard broadcast AM receivers tuned in the frequency range of 540 KHz to 1640 KHz has an intermediate frequency of ( ). 2M.
a. 455 KHz                 b.1MHz                      c. 20Hz                      d. 200Hz.
6. Standard broadcast FM receivers tuned in the frequency range of 88MHz -108 MHz has an intermediate frequency of ( ). 2M.
a. 455 KHz                    b.1MHz                   c. 20Hz                      d. 10.7MHz.
7. Television receivers in the VHF band(54MHz-223MHz) and in the UHF band(470MHz-940MHz) use an IF between 26MHz and 46MHz with the two most popular values ( ). 3M.
a. 36 MHz and 46 MHz                                                b. 455 KHz and 46 kHz.
c. 36 KHz and 46 KHz.                                                 d. none of the above.
8. In a broadcast FM receiver if the local oscillator is tuned to 98.7 MHz then the image frequency is ( ). 3M.
a. 88MHz              b. 109.4MHz                   c. 96 MHz                d. none of the above.
9. In a broadcast AM receiver if the signal is tuned to 530 KHz then Intermediate frequency, local oscillator frequency and image frequency are( ). 3M.
a. 200 kHz, 730 KHz and 1000 kHz.
b. 10.7MHz, 15.37MHz and 1000 KHz.
c. 455 KHz, 985 KHz and 1440 KHz.
d. None of the above.
10. In communications, Audio frequency range is —————-. 2M.
11. In communications, Radio frequency range is—————–. 2M.
12. Draw the radio frequency spectrum with detailed values starting from Very Low Frequencies (VLF) to Extreme High Frequencies (EHF). 5M.
13. Draw the block diagram of TRF receiver (only diagram). 5M.

## UNIT5-Quiz5Pulse Modulation Techniques

1. In Pulse Position Modulation, the drawbacks are ( ) 2M.                 a.Synchronization is required between transmitter and receiver                       b. Large bandwidth is required as compared to PAM                                                 c. None of the above                                                                                                                 d. Both a and b.
2. In PWM signal reception, the Schmitt trigger circuit is used ( ) 2M.                    a. To remove noise                                                                                                                    b. To produce ramp signal                                                                                                      c. For synchronization                                                                                                            d. None of the above.
3. In pulse width modulation, ( ) 2M.                                                                                    a. Synchronization is not required between transmitter and receiver              b. Amplitude of the carrier pulse is varied                                                                      c. Instantaneous power at the transmitter is constant                                              d. None of the above.
4. In different types of Pulse Width Modulation, ( ) 2M.                                                a. Leading edge of the pulse is kept constant                                                                b. Trailing edge of the pulse is kept constant                                                                c. Centre of the pulse is kept constant                                                                              d. All of the above.
5. In Pulse time modulation (PTM), ( ) 2M.                                                                         a. Amplitude of the carrier is constant                                                                             b. Position or width of the carrier varies with modulating signal                           c. Pulse width modulation and pulse position modulation are the types of PTM                                                                                                                                                 d. All of the above.
6. Drawback of using PAM method is ( ) 2M.                                                                      a. Bandwidth is very large as compared to modulating signal                                  b. Varying amplitude of carrier varies the peak power required for transmission                                                                                                                                  c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver                                                                                                                                          d. All of the above.
7. Pulse time modulation (PTM) includes ( ) 2M                                                              a. Pulse width modulation                                                                                                        b. Pulse position modulation                                                                                                  c. Pulse amplitude modulation                                                                                              d. Both a and b.
8. Calculate the Nyquist rate for sampling when a continuous time signal is given by  x(t) = 5 cos 100πt +10 cos 200πt – 15 cos 300πt ( ) 3M.                            a. 300Hz                                                                                                                                            b. 600Hz                                                                                                                                            c. 150Hz                                                                                                                                          d. 200Hz.
9. Calculate the minimum sampling rate to avoid aliasing when a continuous time signal is given by x(t) = 5 cos 400πt ( ) 3M.                                                              a. 100 Hz                                                                                                                                            b. 200 Hz                                                                                                                                          c. 400 Hz                                                                                                                                          d. 250 Hz.
10. A distorted signal of frequency fm is recovered from a sampled signal if the sampling frequency fs is ( ) 2M.                                                                                          a. fs > 2fm                                                                                                                                        b. fs < 2fm                                                                                                                                      c. fs = 2fm                                                                                                                                        d. fs ≥ 2fm.
11. The desired signal of maximum frequency wm centered at frequency w=0 may be recovered if ( ) 2M.                                                                                                    a. The sampled signal is passed through low pass filter                                              b. Filter has the cut off frequency wm                                                                              c. Both a and b                                                                                                                              d. None of the above.
12. The frequency spectrum of x(t) is X(f) is given as follows 6M.

Draw the frequency spectrum of sampled signal by assuming suitable values for sampling frequency under the following conditions
i. Over sampling                      ii. Under sampling                         iii. fs = 2fm.

(No Ratings Yet)

# EXPERIMENT WISE VIVA QUESTIONSANALOG COMMUNICATION LAB

## Amplitude Modulation:

1. What is meant by Modulation? What is the need for modulation?
2. What are different types of analog modulation techniques?
3. What are the other names of message signal? What are the other names of carrier signal?
4. Write the equation of AM signal and explain each parameter in that equation?
5. Define Amplitude Modulation? Define modulation depth or modulation index?
6. What is the range of Audio frequency signals? What is the range of Radio frequency signal?
7. What are the applications of Amplitude modulation?
8. How many generation methods are there to generate an AM wave? What are the methods of demodulation of an AM wave?
9. Explain the operation of diode detector circuit?
10. Write the formula for modulation index? Differentiate under, over and perfect modulation in AM?
11. As the amplitude of message signal increases, modulation index increases or decreases?
12. Define single tone modulation? In laboratory type of AM is single tone modulation or not?
13. Draw the frequency spectrum of AM wave?
14. If modulation index is 100%, calculate the ratio of total power to carrier power of an AM wave?
15. If µ=1 in an AM wave what is the amount of power saving in an AM wave? What is the band width of an AM wave?
16. Explain the operation of AM modulator? Explain the operation of 8038 circuit in AM modulator?
17. Explain the procedure of Amplitude modulation? What is the significance of Emax and Emin points in AM wave?
18. Plot message, carrier and AM signals?
19. What is meant by envelope detector?
20. The frequency of AM wave follows — (message signal frequency or carrier frequency)?
21. The amplitude of AM wave at fc +fm is— and The amplitude of AM wave at fc -fm is—–
22. In amplitude modulation the amplitude of ——— is changing with respect to ——
23. Envelope of AM signal follows————– (message signal/ carrier signal)?
24. What are the advantages and disadvantages of AM?
25. How demodulated signal differs from original signal in AM?
26. The two important distortions that can appear in the demodulated output of an envelope detector are————– and—————————- –.
27. Differentiate high-level and low-level modulations in AM?
28. What is trapezoidal rule?

## Balanced Modulator:

1. What are the disadvantages of AM?
2. Most of the power in AM spectrum is carried by ————
3. Define DSBSC modulation?
4. How DSBSC is more efficient than AM in terms power saving, explain?
5. What is meant by frequency response?
6. Draw the magnitude response or amplitude spectrum of DSBSC signal?
7. The signal generated by balanced modulator is———–
8. Draw the wave form of DSBSC wave and AM wave, and differentiate those two waveforms?
9. Give the equation of DSBSC signal?
10. What are the generation methods of DSBSC?
11. What are the demodulation methods of DSBSC?
12. What is the bandwidth of DSBSC signal?
13. Define Costas loop and it’s operation?
14. Amount of power saving in DSBSC signal is————
15. Coherent detection means?
16. Give the practical applications of balanced modulator?
17. Explain the operation of product modulator?
18. Why the circuit is called balanced modulator?
19. If the circuit is operating in balanced state, the modulation index value is——- –.
20. Explain the working procedure of 1496 IC for the generation of DSBSC wave?
21. As message signal amplitude increases, carrier suppression in dB’s ———
22. Plot message, carrier and DSBSC waves and explain each wave clearly.
23. How do you differentiate modulation by demodulation?
24. Explain the significance of local oscillator frequency in modulators and in
25. Differentiate synchronous and non synchronous detection techniques in analog modulators?
26. The phase shift at zero crossings in DSBSC wave is——- –.
27. What is Quadrature carrier multiplexing?
28. How DSBSC is different from SSB?

## Frequency Modulation:

1. Define Frequency modulation? How it is different from phase modulation?
2. Write equation of FM wave, explain each parameter in it?
3. Draw the amplitude spectrum of FM wave?
4. Give the Carson’s rule in FM?
5. Define modulation index β, frequency deviation?
6. Differentiate Narrow band FM with Wide band FM?
7. Explain the FM operation using 8308IC?
8. Draw message, carrier and FM waves and explain each wave clearly?
9. Explain the methods for generation of FM and its demodulation?
10. How FM wave is different from PM wave?
11. Give the practical applications of FM?
12. State advantages and disadvantages of FM?
13. The range of speech signals is——— –.
14. Type of Modulation used in radios is——- –.
15. Type of modulation used for voice signals in T.V — and for video signals in V is—- –.
16. Noise immunity is more in which analog modulation technique———– –.
17. FM is more robust to noise compared to AM, why?
18. Carson’s rule is for———- –.
19. In commercial FM broadcasting, the audio frequency range handled is only up to—- –.
20. The transmission band width required for commercial FM broadcasting is——– –.
21. Define Hilbert transform?
22. Explain capture effect in FM broadcasting?

## Pre-emphasis and De-emphasis:

1. Define pre-emphasis and De-emphasis processes in
2. Why Pre-emphasis is used at Transmitter of FM and de-emphasis at FM receiver?
3. Draw the pre-emphasis circuit and explain its working in detail?
4. Draw de-emphasis circuit and explain its working in detail?
5. Draw the frequency response characteristics of pre-emphasis and de-emphasis explain each one in detail?
6. Calculate the cut-off frequencies of pre-emphasis and de-emphasis circuits practically
7. Pre-emphasis circuit operation is similar to——— –.
8. De-emphasis circuit operation is similar to——— –.
9. What is the necessity of boosting up high frequencies in frequency modulation communication system?
10. Define 3dB frequencies?

# Sampling and reconstruction:

1. Define sampling theorem? What is the need for sampling?
2. What are the necessary and sufficient condition for sampling and reconstruction of a signal?
3. Define Nyquist rate and Nyquist interval in sampling theorem?
4. If message frequency is 2 KHz and sampling frequency is 2 KHz,4 KHz, 8 KHz and 16 KHz in each case the number of samples are—————————– –.
5. What are different types of sampling techniques?
6. What was the effect on sampled signal if fs < 2 fm ?
7. Draw the amplitude spectrum of sampled signal if fs < 2 fm, fs =2 fm, fs > 2 fm.
8. What is aliasing effect in sampling? How to avoid it?
9. Why do we use pre-filtering in sampling?
10. What do you mean by reconstruction of sampling theorem?
11. What are the types of filters used in reconstruction?
12. Define sample and hold process?
13. Differentiate second order, fourth order and sixth order low pass filters in reconstruction process.
14. Explain the sampling and reconstruction process in detail by using the trainer
15. Define band pass sampling?
16. How sampling is different from PAM?
17. Define a continuous time signal or an analog signal. Give some examples of analog signals.
18. Define a discrete time signal. Give some examples of discrete
19. What is the difference between discrete and a digital signal?
20. Define a digital signal? Give some
21. What is the need for converting a continuous signal into a discrete
22. Explain about zero-order hold circuit.
23. How to convert an analog signal into digital signals?

Digital signal processors operates———— as inputs.As the number of samples increases, the reconstruction of original signal becomes?

# Pulse Amplitude Modulation:

1. What is the basic principle of PAM?
2. Name some Pulse Modulation techniques?
3. Define PAM?
4. How PAM is different from AM?
5. Can we produce a PAM signal using a sampling circuit?
6. Differentiate PAM output with sampling output?
7. Does PAM come under Analog modulation technique or Digital Modulation technique?
8. What is the Bandwidth of PAM?
9. Compare BW of PAM and AM?
10. Draw waveforms of PAM. explain each one briefly.
11. What are the advantages of PAM over AM?
12. What are the advantages and Disadvantages of PAM?
13. Explain the working procedure PAM kit?
14. Can we use PAM technique in TDM?
15. Differentiate uni-polar and bi-polar PAM.
16. What do you mean by zero order holding? And draw the circuit diagram of zero-order hold circuits?
17. What are the drawbacks of PAM?
18. Explain the working of PAM demodulation circuit?
19. Define Flat-Top sampling?
20. Draw the circuit diagram of Flat-Top Sampled circuit?
21. What was the roll off characteristics of sinc pulse?

## Pulse Width Modulation (PWM):

1. Define PWM?
2. Differentiate PWM, PAM and PPM?
3. Name the applications of Mono-stable multivibrator?
4. What is a Multivibrator?
5. Differentiate Monostable, Bi stable and Astable Multivibrators?
6. How a Monostable Multivibrator produces a PWM signal?
7. What are the other names of PWM?
8. Define Pulse Duration Modulation?
9. What is Pulse Time Modulation?
10. Draw PAM and PWM signals and each one in detail.
11. Draw PWM signal with respect to message signal?
12. In PWM —————- of Pulse carrier signal is changing with respect to message signal.
13. Explain the operation of PWM circuit.
14. 555 timer in Monostable mode produces————.
15. 555 timer in Astable mode produces——————.
16. What are the advantages of PWM over PAM?
17. What is the difference between PWM and FM?
18. Which type of noise is affecting the amplitude of PWM signal?
19. Which system is more immune to noise (PWM or PAM)?
20. What are the disadvantages of PWM?
21. What are the applications of PWM?
22. Band Width of PWM is—————-.
23. Band width of PAM is—————-.

## Pulse Position Modulation (PPM):

1. Define PPM?
2. The information is conveyed by ————- of Pulses in PPM.
3. In PWM information is conveyed by————— of pulses.
4. In PAM information is conveyed by————— of pulses.
5. What are the advantages and disadvantages of PPM?
6. Compare PPM with Phase Modulation.
7. PAM is similar to————————–.
8. PWM is similar to————————–.
9. PPM is similar to————————–.
10. Differentiate Analog Modulation Techniques with Pulse Modulation Techniques.
11. What are the applications of PPM?
12. Draw PPM signal with respect to message signal.
13. Draw PPM signal with respect to PWM signal.
14. Explain the operation of PPM Modulator?

## Phase Locked Loop (PLL):

1. What are the applications of PLL?
2. Why this circuit is called Phase Locked Loop?
3. What are the three components of PLL circuitry?
4. Explain the operation of PLL by using a Block Diagram?
5. Define free-running frequency?
6. Define Lock range and Capture range of a PLL?
7. What is meant by Frequency synthesizer?
8. Why PLL is used in FM Receivers/
9. How PLL is used in FSK demodulation circuits?
10. What do you mean by Lock state in a PLL?
11. What is meant by Pull in time in PLL?
12. Phase Detector or Phase Comparator is used for ——————.
13. Why VCO is used in feedback loop of PLL?
14. What are the input and Output signals of a VCO in a PLL?
15. What are the advantages and Disadvantages of PLL?
16. Why Lock range is greater than Capture range in a PLL?

## Time Division Multiplexing (TDM):

1. Define the concept of Time Division Multiplexing?
2. Differentiate Multiplexing and Sampling?
3. What are the different types of Multiplexing Techniques?
4. Does TDM come under analog Multiplexing or Digital Multiplexing?
5. Define a frame in a TDM?
6. Why synchronization is required in TDM?
7. Why multiplexing is required?
8. What do you mean by inter-leaving gaps in TDM frame?
9. If two signals of frequencies 2KHz and 4 KHz are multiplexed in time –domain then draw TDM signal
1. Without inter-leaving gap.
2. with inter-leaving gap of 10 ms.
10. What are the advantages and Disadvantages of TDM?
11. Differentiate TDM with FDM?
12. What are the advantages of TDM over FDM?
13. Explain the operation of TDM using trainer kit used.
14. Synchronous TDM means?

1. What is a filter?
2. Differentiate Active and Passive Filters?
3. Why LPF is used in Demodulation Circuits?
4. Why pre-filtering is required in sampling?
5. Anti-aliasing is achieved by using———– in Sampling Circuits?
6. Define Single-tone modulation in AM and FM?
7. Why FM receivers are more immune to AM receivers?
8. BW of Narrow band FM is ——————.
9. BW of AM is ———————————-.
10. How to calculate Image frequency of a Radio Receiver?
11. Define Power Spectral Density.
12. Define AWGN noise.
13. Define SNR.
14. Where do we use Hilbert Transform?
15. Over modulation in AM means….
16. What is µ value when AM wave is similar to DSBSC wave?

(No Ratings Yet)

## Analog Communications Important Questions unit 2 (FAQ)

UNIT 2:

1. What is meant by Carson’s rule? Problem on calculation of Bandwidth of FM signal using Carson’s rule.(imp)
2. Distinguish between NBFM and WBFM.(imp)
3. Problems on instantaneous phase and frequency.
4. What is the basic principle involved in Phase discriminator?
5. Define capture range and lock range of a PLL.
6. Problem on Armstrong method of generation of a WBFM signal?(imp)
7. Capture effect and threshold effects in FM (very very imp).
8. Problems on carrier swing, frequency deviation, BW, modulation index etc.
9. Phasor diagrams of NBFM and AM.
10. What is the need for limiter in FM receivers? (Very very imp).
11. Find the maximum Frequency deviation of Frequency Modulated signal given by  S(t) = 10 cos(1000000 *pi*t + 5 sin(2000*pi*t)).
12. A carrier signal 10 cos(8x 106πt) is modulated by a modulating signal               5 cos(30x 103πt)  then i. Find the Band width of Frequency Modulated signal assuming  kf  =15KHz.   ii.     Calculate Highest Frequency and lowest Frequency of FM signal.  iii. Find  Modulation index of FM signal.
13. Compare Frequency Modulation with Phase Modulation.

1. Explain the operation of balanced slope detector in detail.
2. Explain in detail about frequency spectrum of WBFM signal using Bessel functions.
3. write in detail about generation methods of FM signal
4. Direct method of generation
5. Indirect method of generation (Armstrong method).(very very imp)
6. What is ratio detector? Explain its working in detail with neat circuit diagram. Explain how ratio detector provides amplitude limiting?
7. What is amplitude limiter? Draw its circuit diagram explain its working. Why amplitude limiter is needed in FM receivers justify.(very very imp)
8. Compare and contrast foster seeley discriminator with ratio detector.
9. What is PLL? What are the applications of PLL? Explain the working of PLL with neat block diagram.(imp)
10. How audio frequency signal is demodulated by using PLL? Explain in detail.
11. Write about pre-emphasis and de-emphasis circuits in detail. Give the significance of pre-emphasis and de-emphasis.(imp)
12. In an FM system if the AF is 500 Hz and its amplitude is 2.4V with a frequency deviation of 4.8 KHz. If the AF voltage is increased to 7.2V then find the modified frequency deviation calculate BW in above two cases.
13. Design Armstrong FM generator for the generation of WBFM signal with frequency deviation=75 KHz and using the NB carrier as 100 KHz and second carrier as 10 MHz Find the suitable multiplying factors. Assume the message signal is defined in the range 100Hz-15 KHz.
14. Draw the block diagram of single-tone NBFM signal if message signal is.
15. Derive an expression for an FM signal with carrier frequency fc and a modulating signal Obtain an expression for its spectrum.
16. Why an FM system is preferred over an AM system?

Unit 2

Assignment 2

1. Explain the generation techniques of FM signal using Direct and Indirect methods in detail.
2. Explain the working of the following demodulator circuits of FM                        i. Balanced slope Detector for FM modulation.                                                            ii. Foster-Seeley discriminator.                                                                                         iii. Ratio-detector.
3. Explain the detection of FM signal using PLL.
4. Explain in detail about pre-emphasis and De-emphasis circuits in FM.

Unit 2

Assignment 2

1. Illustrate the working of PLL using a Block Diagram. Define capture range and Lock range in PLL.                                                                                                                 5M.
2. Sketch the circuit diagram of Ratio-detector and explain working of it in detail. 5M.
3. Sketch the frequency response curves & compare the working of pre-emphasis and De-emphasis circuits in FM communication System.

TUTORIAL TEST 2

Answer any three of the following

1. In an FM system if the AF is 500 Hz and its amplitude is 2.4V with a frequency deviation of 4.8 KHz. If the AF voltage is increased to 7.2V then find the modified frequency deviation calculate BW in above two cases.
2. Explain in detail about frequency spectrum of WBFM signal.
3. Draw the block diagram of single-tone NBFM signal if message signal is
4. Write any four differences between NBFM and WBFM.

(No Ratings Yet)

# EXPERIMENT WISE VIVA QUESTIONSANALOG COMMUNICATION LAB

## Amplitude Modulation:

1. What is meant by Modulation? What is the need for modulation?
2. What are different types of analog modulation techniques?
3. What are the other names of message signal? What are the other names of carrier signal?
4. Write the equation of AM signal and explain each parameter in that equation?
5. Define Amplitude Modulation? Define modulation depth or modulation index?
6. What is the range of Audio frequency signals? What is the range of Radio frequency signal?
7. What are the applications of Amplitude modulation?
8. How many generation methods are there to generate an AM wave? What are the methods of demodulation of an AM wave?
9. Explain the operation of diode detector circuit?
10. Write the formula for modulation index? Differentiate under, over and perfect modulation in AM?
11. As the amplitude of message signal increases, modulation index increases or decreases?
12. Define single tone modulation? In laboratory type of AM is single tone modulation or not?
13. Draw the frequency spectrum of AM wave?
14. If modulation index is 100%, calculate the ratio of total power to carrier power of an AM wave?
15. If µ=1 in an AM wave what is the amount of power saving in an AM wave? What is the band width of an AM wave?
16. Explain the operation of AM modulator? Explain the operation of 8038 circuit in AM modulator?
17. Explain the procedure of Amplitude modulation? What is the significance of Emax and Emin points in AM wave?
18. Plot message, carrier and AM signals?
19. What is meant by envelope detector?
20. The frequency of AM wave follows — (message signal frequency or carrier frequency)?
21. The amplitude of AM wave at fc +fm is— and The amplitude of AM wave at fc -fm is—–
22. In amplitude modulation the amplitude of ——— is changing with respect to ——
23. Envelope of AM signal follows————– (message signal/ carrier signal)?
24. What are the advantages and disadvantages of AM?
25. How demodulated signal differs from original signal in AM?
26. The two important distortions that can appear in the demodulated output of an envelope detector are————– and—————————- –.
27. Differentiate high-level and low-level modulations in AM?
28. What is trapezoidal rule?

## Balanced Modulator:

1. What are the disadvantages of AM?
2. Most of the power in AM spectrum is carried by ————
3. Define DSBSC modulation?
4. How DSBSC is more efficient than AM in terms power saving, explain?
5. What is meant by frequency response?
6. Draw the magnitude response or amplitude spectrum of DSBSC signal?
7. The signal generated by balanced modulator is———–
8. Draw the wave form of DSBSC wave and AM wave, and differentiate those two waveforms?
9. Give the equation of DSBSC signal?
10. What are the generation methods of DSBSC?
11. What are the demodulation methods of DSBSC?
12. What is the bandwidth of DSBSC signal?
13. Define Costas loop and it’s operation?
14. Amount of power saving in DSBSC signal is————
15. Coherent detection means?
16. Give the practical applications of balanced modulator?
17. Explain the operation of product modulator?
18. Why the circuit is called balanced modulator?
19. If the circuit is operating in balanced state, the modulation index value is——- –.
20. Explain the working procedure of 1496 IC for the generation of DSBSC wave?
21. As message signal amplitude increases, carrier suppression in dB’s ———
22. Plot message, carrier and DSBSC waves and explain each wave clearly.
23. How do you differentiate modulation by demodulation?
24. Explain the significance of local oscillator frequency in modulators and in
25. Differentiate synchronous and non synchronous detection techniques in analog modulators?
26. The phase shift at zero crossings in DSBSC wave is——- –.
27. What is Quadrature carrier multiplexing?
28. How DSBSC is different from SSB?

## Frequency Modulation:

1. Define Frequency modulation? How it is different from phase modulation?
2. Write equation of FM wave, explain each parameter in it?
3. Draw the amplitude spectrum of FM wave?
4. Give the Carson’s rule in FM?
5. Define modulation index β, frequency deviation?
6. Differentiate Narrow band FM with Wide band FM?
7. Explain the FM operation using 8308IC?
8. Draw message, carrier and FM waves and explain each wave clearly?
9. Explain the methods for generation of FM and its demodulation?
10. How FM wave is different from PM wave?
11. Give the practical applications of FM?
12. State advantages and disadvantages of FM?
13. The range of speech signals is——— –.
14. Type of Modulation used in radios is——- –.
15. Type of modulation used for voice signals in T.V — and for video signals in V is—- –.
16. Noise immunity is more in which analog modulation technique———– –.
17. FM is more robust to noise compared to AM, why?
18. Carson’s rule is for———- –.
19. In commercial FM broadcasting, the audio frequency range handled is only up to—- –.
20. The transmission band width required for commercial FM broadcasting is——– –.
21. Define Hilbert transform?
22. Explain capture effect in FM broadcasting?

## Pre-emphasis and De-emphasis:

1. Define pre-emphasis and De-emphasis processes in
2. Why Pre-emphasis is used at Transmitter of FM and de-emphasis at FM receiver?
3. Draw the pre-emphasis circuit and explain its working in detail?
4. Draw de-emphasis circuit and explain its working in detail?
5. Draw the frequency response characteristics of pre-emphasis and de-emphasis explain each one in detail?
6. Calculate the cut-off frequencies of pre-emphasis and de-emphasis circuits practically
7. Pre-emphasis circuit operation is similar to——— –.
8. De-emphasis circuit operation is similar to——— –.
9. What is the necessity of boosting up high frequencies in frequency modulation communication system?
10. Define 3dB frequencies?

# Sampling and reconstruction:

1. Define sampling theorem? What is the need for sampling?
2. What are the necessary and sufficient condition for sampling and reconstruction of a signal?
3. Define Nyquist rate and Nyquist interval in sampling theorem?
4. If message frequency is 2 KHz and sampling frequency is 2 KHz,4 KHz, 8 KHz and 16 KHz in each case the number of samples are—————————– –.
5. What are different types of sampling techniques?
6. What was the effect on sampled signal if fs < 2 fm ?
7. Draw the amplitude spectrum of sampled signal if fs < 2 fm, fs =2 fm, fs > 2 fm.
8. What is aliasing effect in sampling? How to avoid it?
9. Why do we use pre-filtering in sampling?
10. What do you mean by reconstruction of sampling theorem?
11. What are the types of filters used in reconstruction?
12. Define sample and hold process?
13. Differentiate second order, fourth order and sixth order low pass filters in reconstruction process.
14. Explain the sampling and reconstruction process in detail by using the trainer
15. Define band pass sampling?
16. How sampling is different from PAM?
17. Define a continuous time signal or an analog signal. Give some examples of analog signals.
18. Define a discrete time signal. Give some examples of discrete
19. What is the difference between discrete and a digital signal?
20. Define a digital signal? Give some
21. What is the need for converting a continuous signal into a discrete
22. Explain about zero-order hold circuit.
23. How to convert an analog signal into digital signals?

Digital signal processors operates———— as inputs.As the number of samples increases, the reconstruction of original signal becomes?

# Pulse Amplitude Modulation:

1. What is the basic principle of PAM?
2. Name some Pulse Modulation techniques?
3. Define PAM?
4. How PAM is different from AM?
5. Can we produce a PAM signal using a sampling circuit?
6. Differentiate PAM output with sampling output?
7. Does PAM come under Analog modulation technique or Digital Modulation technique?
8. What is the Bandwidth of PAM?
9. Compare BW of PAM and AM?
10. Draw waveforms of PAM. explain each one briefly.
11. What are the advantages of PAM over AM?
12. What are the advantages and Disadvantages of PAM?
13. Explain the working procedure PAM kit?
14. Can we use PAM technique in TDM?
15. Differentiate uni-polar and bi-polar PAM.
16. What do you mean by zero order holding? And draw the circuit diagram of zero-order hold circuits?
17. What are the drawbacks of PAM?
18. Explain the working of PAM demodulation circuit?
19. Define Flat-Top sampling?
20. Draw the circuit diagram of Flat-Top Sampled circuit?
21. What was the roll off characteristics of sinc pulse?

## Pulse Width Modulation (PWM):

1. Define PWM?
2. Differentiate PWM, PAM and PPM?
3. Name the applications of Mono-stable multivibrator?
4. What is a Multivibrator?
5. Differentiate Monostable, Bi stable and Astable Multivibrators?
6. How a Monostable Multivibrator produces a PWM signal?
7. What are the other names of PWM?
8. Define Pulse Duration Modulation?
9. What is Pulse Time Modulation?
10. Draw PAM and PWM signals and each one in detail.
11. Draw PWM signal with respect to message signal?
12. In PWM —————- of Pulse carrier signal is changing with respect to message signal.
13. Explain the operation of PWM circuit.
14. 555 timer in Monostable mode produces————.
15. 555 timer in Astable mode produces——————.
16. What are the advantages of PWM over PAM?
17. What is the difference between PWM and FM?
18. Which type of noise is affecting the amplitude of PWM signal?
19. Which system is more immune to noise (PWM or PAM)?
20. What are the disadvantages of PWM?
21. What are the applications of PWM?
22. Band Width of PWM is—————-.
23. Band width of PAM is—————-.

## Pulse Position Modulation (PPM):

1. Define PPM?
2. The information is conveyed by ————- of Pulses in PPM.
3. In PWM information is conveyed by————— of pulses.
4. In PAM information is conveyed by————— of pulses.
5. What are the advantages and disadvantages of PPM?
6. Compare PPM with Phase Modulation.
7. PAM is similar to————————–.
8. PWM is similar to————————–.
9. PPM is similar to————————–.
10. Differentiate Analog Modulation Techniques with Pulse Modulation Techniques.
11. What are the applications of PPM?
12. Draw PPM signal with respect to message signal.
13. Draw PPM signal with respect to PWM signal.
14. Explain the operation of PPM Modulator?

## Phase Locked Loop (PLL):

1. What are the applications of PLL?
2. Why this circuit is called Phase Locked Loop?
3. What are the three components of PLL circuitry?
4. Explain the operation of PLL by using a Block Diagram?
5. Define free-running frequency?
6. Define Lock range and Capture range of a PLL?
7. What is meant by Frequency synthesizer?
8. Why PLL is used in FM Receivers/
9. How PLL is used in FSK demodulation circuits?
10. What do you mean by Lock state in a PLL?
11. What is meant by Pull in time in PLL?
12. Phase Detector or Phase Comparator is used for ——————.
13. Why VCO is used in feedback loop of PLL?
14. What are the input and Output signals of a VCO in a PLL?
15. What are the advantages and Disadvantages of PLL?
16. Why Lock range is greater than Capture range in a PLL?

## Time Division Multiplexing (TDM):

1. Define the concept of Time Division Multiplexing?
2. Differentiate Multiplexing and Sampling?
3. What are the different types of Multiplexing Techniques?
4. Does TDM come under analog Multiplexing or Digital Multiplexing?
5. Define a frame in a TDM?
6. Why synchronization is required in TDM?
7. Why multiplexing is required?
8. What do you mean by inter-leaving gaps in TDM frame?
9. If two signals of frequencies 2KHz and 4 KHz are multiplexed in time –domain then draw TDM signal
1. Without inter-leaving gap.
2. with inter-leaving gap of 10 ms.
10. What are the advantages and Disadvantages of TDM?
11. Differentiate TDM with FDM?
12. What are the advantages of TDM over FDM?
13. Explain the operation of TDM using trainer kit used.
14. Synchronous TDM means?

1. What is a filter?
2. Differentiate Active and Passive Filters?
3. Why LPF is used in Demodulation Circuits?
4. Why pre-filtering is required in sampling?
5. Anti-aliasing is achieved by using———– in Sampling Circuits?
6. Define Single-tone modulation in AM and FM?
7. Why FM receivers are more immune to AM receivers?
8. BW of Narrow band FM is ——————.
9. BW of AM is ———————————-.
10. How to calculate Image frequency of a Radio Receiver?
11. Define Power Spectral Density.
12. Define AWGN noise.
13. Define SNR.
14. Where do we use Hilbert Transform?
15. Over modulation in AM means….
16. What is µ value when AM wave is similar to DSBSC wave?

(No Ratings Yet)