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
B_{T}+W | |
\frac{3}{2} B_{T} | |
2 B_{T}+W | |
\frac{5}{2} B_{T} |
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}
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).
2 | |
8 | |
10 | |
12 |
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}
\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}
.
850 | |
1825 | |
1400 | |
1250 |
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}
.
1011313.7 | |
12543.3 | |
124534.2 | |
1025361.2 |
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}
\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 _____
2.54 | |
5.45 | |
7.82 | |
3.14 |
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.