# Analog Communication Systems

 Question 1
Let a frequency modulated (FM) signal $x(t)=A \cos \left(\omega_{c} t+k_{f} \int_{-\infty}^{t} m(\lambda) d \lambda\right)$, where $m(t)$ is a message signal of bandwidth $W$. It is passed through a non-linear system with output $y(t)=2 x(t)+5(x(t))^{2}$. Let $B_{T}$ denote the FM bandwidth. The minimum value of $\omega_{c}$ required to recover $x(t)$ from $y(t)$ is
 A $B_{T}+W$ B $\frac{3}{2} B_{T}$ C $2 B_{T}+W$ D $\frac{5}{2} B_{T}$
GATE EC 2023   Communication Systems
Question 1 Explanation:
$x(t)=A \cos \left[\omega_{c} t+K_{f} \int_{-\infty}^{t} m(\lambda) d \lambda\right]$
B.W. $[x(t)] \rightarrow B T=2[\Delta f+\omega]$

$\left.x^{2}(t) \rightarrow \begin{array}{c} \Delta f^{\prime}=2 \Delta f \\ \omega_{c}^{\prime}=2 \omega_{c} \end{array}\right\}$

$\mathrm{BW}\left[x^{2}(t)\right]=2\left[\Delta f^{\prime}+\omega\right]$ $=2[2 \Delta f+\omega]=B T+2 \Delta f$
$y(t)=2x(t)+5x^2(t)$

To recover $x(t) \rightarrow \quad 2 \omega_{C}-\frac{B_{T}}{2}-\Delta f \gt \omega_{C}+\frac{B_{T}}{2}$
$\omega_{c} \gt \Delta f+B T$
$\omega_{c} \gt \Delta f+2 \Delta f+2 \omega$
$\omega_{c} \gt 3 \Delta f+2 \omega$
$\omega_{c} \gt \frac{3}{2}\{2[\Delta f+\omega]\}-\omega$
$\omega_{c} \gt \frac{3}{2} B_{T}-\omega$
Compared to FM BW, message BW is very small. So, that it can be ignored.
\begin{aligned} \omega_{C} & \gt \frac{3}{2} B_{T} \\ \left[\omega_{C}\right]_{min } & =\frac{3}{2} B_{T} \end{aligned}
 Question 2
The signal-to-noise ratio (SNR) of an ADC with a full-scale sinusoidal input is given to be $61.96 \mathrm{~dB}$. The resolution of the ADC is _____ bits. (rounded off to the nearest integer).
 A 2 B 8 C 10 D 12
GATE EC 2023   Communication Systems
Question 2 Explanation:
We know that for sinusoidal input, the signal to noise ratio (SNR) is given as,
\begin{aligned} \mathrm{SNR} & =1.76+6.02 \mathrm{n} \mathrm{dB} \\ 61.96 \mathrm{~dB} & =1.76+6.02 \mathrm{n} \mathrm{dB} \\ 6.02 \mathrm{n} & =61.96-1.76 \\ n & =10 \text { bits } \end{aligned}

 Question 3
Consider a superheterodyne receiver tuned to $600 \text{ kHz}$. If the local oscillator feeds a $1000 \text{ kHz}$ signal to the mixer, the image frequency (in integer) is ____________________ $\text{kHz}$ .
 A 850 B 1825 C 1400 D 1250
GATE EC 2021   Communication Systems
Question 3 Explanation:
\begin{aligned} f_{s} &=600 \mathrm{kHz} \\ f_{l} &=1000 \mathrm{kHz} \\ f_{s i} &=? \\ \text { Default } f_{l}>f_{s} & \\ \mathrm{IF} &=f_{l}-f_{s}=400 \mathrm{kHz} \\ f_{\mathrm{si}} &=f_{s}+2 \mathrm{IF}=600 \mathrm{~K}+800 \mathrm{~K}=1400 \mathrm{kHz} \end{aligned}
 Question 4
A sinusoidal message signal having root mean square value of $4\:V$ and frequency of $\text{1 kHz}$ is fed to a phase modulator with phase deviation constant 2 rad/ volt. If the carrier signal is $c\left ( t \right )=2\cos\left ( 2\pi 10^{6} t\right )$, the maximum instantaneous frequency of the phase modulated signal (rounded off to one decimal place) is ______ $\text{Hz}$ .
 A 1.01131e+06 B 12543.3 C 124534 D 1.02536e+06
GATE EC 2021   Communication Systems
Question 4 Explanation:
Given sinusoidal message signal $\rightarrow V_{\mathrm{rms}}=4 \mathrm{~V}$
\begin{aligned} f_{m} &=1 \mathrm{kHz} \\ V_{\text {peak }} &=\sqrt{2} \times V_{\mathrm{rms}}=\sqrt{2} \times 4 \mathrm{~V} \\ m(t) &=4 \sqrt{2} \sin 2 \pi \times 10^{3} t\\ K_{p}&=2 \mathrm{rad} / \mathrm{volt}\\ c(t) &=2 \cos 2 \pi \times 10^{6} \mathrm{t} \\ \left(f_{i}\right)_{\max } &=f_{c}+\frac{K_{p}}{2 \pi}\left[\frac{d}{d t} m(t)\right]_{\max } \\ \frac{d}{d t} m(t) &=4 \sqrt{2} \times\left(2 \pi \times 10^{3}\right) \cos 2 \pi \times 10^{3} t \\ \left[\frac{d}{d t} m(t)\right]_{\max } &=4 \sqrt{2} \times 2 \pi \times 10^{3} \\ \left(f_{i}\right)_{\max } &=10^{6}+\frac{2}{2 \pi}\left(4 \sqrt{2} \times 2 \pi \times 10^{3}\right)=1011313.7 \mathrm{~Hz} \end{aligned}
 Question 5
A $4\:kHz$ sinusoidal message signal having amplitude $4\:V$ is fed to a delta modulator (DM) operating at a sampling rate of $32\:\text{kHz}$ . The minimum step size required to avoid slope overload noise in the DM (rounded off to two decimal places) is _____
 A 2.54 B 5.45 C 7.82 D 3.14
GATE EC 2021   Communication Systems
Question 5 Explanation:
\begin{aligned} f_{m}&=4 \mathrm{kHz} \\ A_{m}&=4 \mathrm{~V} \\ f_{s}&=32 \mathrm{kHz}\\ \text { To avoid } \text{SOE} \rightarrow &\frac{\Delta}{T_{s}} \geq 2 \pi f_{m} A_{m} \\ \Delta \times 32 k &\geq 2 \pi \times 4 k \times 4 \mathrm{~V} \\ \Delta &\geq \pi\\ (\Delta)_{\min }&=\pi volts\\ (\Delta)_{\min }&=3.14 volts \end{aligned}

There are 5 questions to complete.