Q41GATE 2006MCQ

The diagonal clipping in Amplitude Demodulation (using envelop detector) can be avoided it RC time-constant of the envelope detector satisfies the following condition, (here W is message bandwidth and $\omega_{c}$ is carrier frequency both in rad/sec)

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📖 Explanation

To avoid diagonal clipping in an envelope detector, the capacitor must discharge fast enough to track the maximum negative rate of change of the modulating signal envelope. For a modulating signal with maximum frequency $W$ , the time constant $\tau = RC$ must be strictly less than the inverse of the modulating frequency to accurately follow the envelope without skipping. Hence, the condition is $\frac{1}{\omega_c} \ll RC < \frac{1}{W_{max}}$ . Since $W$ here is the message bandwidth in rad/sec, $RC < \frac{1}{W}$ ensures the tracking condition is met.

Q43GATE 2006MCQ

Consider the following Amplitude Modulated (AM) signal, where $f_{m} \lt B$ : $X_{AM}(t)=10(1+0.5sin 2 \pi f_{m}t)cos2 \pi f_{c}t$ The AM signal gets added to a noise with Power Spectral Density $S_{n}(f)$ given in the figure below. The ratio of average sideband power to mean noise power would be :

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📖 Explanation

\begin{aligned} P_{S} &=\frac{25}{4} \text { Watts } \\ P_{N} &=N_{0} B \\ \therefore \quad S N R &=\frac{P_{S}}{P_{N}}=\frac{25}{4 N_{0} B} \end{aligned}

Q44GATE 2005MCQ

Find the correct match between group 1 and group 2.

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Q45GATE 2005MCQ

Which of the following analog modulation scheme requires the minimum transmitted power and minimum channel bandwidth ?

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📖 Explanation

Single Sideband (SSB) modulation suppresses both the carrier and one of the two sidebands. Since the carrier contains no information (saving $\approx 67\%$ power compared to standard AM) and the two sidebands contain redundant information, removing one sideband further halves the power required for the information signal. Additionally, suppressing one sideband reduces the required transmission bandwidth from $2W$ (for DSB) to exactly $W$ (for SSB), making it the optimal scheme for both minimum transmitted power and minimum signal bandwidth.

Q46GATE 2005MCQ

A carrier is phase modulated (PM) with frequency deviation of 10 kHz by a single tone frequency of 1 kHz. If the single tone frequency is increased to 2 kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is

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📖 Explanation

$\Delta \phi=m_{f}$ $m_{1}$ temains same for both cases,

\begin{aligned} \Rightarrow \quad \frac{\Delta f}{2} &=\frac{10}{1} \\ \therefore \quad \Delta f &=20 \\ \therefore \quad B W &=2\left(\Delta f+f_{m}\right)=2(20+2) \\ &=44 \mathrm{K} \end{aligned}

Q47GATE 2005MCQ

A device with input x(t) and output y(t) is characterized by: $y(t) = x^{2}(t)$ . An FM signal with frequency deviation of 90kHz and modulating signal bandwidth of 5kHz is applied to this device. The bandwidth of the output signal is

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📖 Explanation

\begin{aligned} \text { B.W. } &=2\left(\Delta f+f_{m}\right)=2(180+5) \\ &=370 \mathrm{K} \end{aligned}

Note: When FM signal is applied to doubles frequency deviation doubles but $f_{m}$ remains the same.

Q48GATE 2004MCQ

Two sinusoidal signals of same amplitude and frequencies 10 kHz and 10.1 kHz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is

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📖 Explanation

continuous sinusoidal signal is a periodic Every singnal

\begin{array}{c} s(t)=A \cos [2 \pi \times 10 K t]+A \cos [2 \pi \times 10.1 K t] \\ T_{1}=\frac{1}{10 K}=100 \mu s \\ T_{2}=\frac{1}{10.1 K}=99 \mu \mathrm{s} \\ \frac{T_{1}}{T_{2}}=\text { Rational } \end{array}

$\therefore s(t)$ will be periodic with period LCM of $T_{1} \& T_{2}=9900 \mu \mathrm{s} ; 10,000 \mu \mathrm{s}$ $\therefore$ Frequency =0.1 kHz Resulting is a sinusoidal signal with frequency 0.1 kHz. Frequency detector converts frequency variations to equivalent of voltage variations. Because signal has constant frequency of 0.1 kHz, frequency detector output is constant.

Q49GATE 2004MCQ

An AM signal is detected using an envelop detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelop detector is

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📖 Explanation

$\frac{1}{f_{c}} \lt R C \leq \frac{1}{f_{m}}$ (where $1 / f_{c}=$ to avoid fluctuations at recovered output and $1 / f_{m}=$ to avoid diagonal clipping) $1 \mu \sec \lt R C \leq \frac{1}{2 k H z}$ $1 \mu \mathrm{sec} \lt R C \leq 0.5 \mathrm{ms}$

Q50GATE 2004MCQ

Consider a system shown in the figure. Let X(f) and Y(f) and denote the Fourier transforms of x(t) and y(t) respectively. The ideal HPF has the cutoff frequency 10 kHz. The positive frequencies where Y(f) has spectral peaks are

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📖 Explanation

Peaks of X(f) are at 1 k and -1 k initially Output of balanced modulator $=f_{c} \pm 1 \mathrm{k}$ and

\begin{aligned} f_{c} \pm-1 k & \\ f_{c} \pm 1 k &=11 \mathrm{k}, 9 \mathrm{k} \\ f_{c} \pm-1 \mathrm{k} &=9 \mathrm{k}, 11 \mathrm{k} \end{aligned}

Output of HPF with $f_{c}=10 \mathrm{k}$ will be 11 k frequency component

\begin{aligned} \therefore \quad Y(f) &=13 \mathrm{k} \pm 11 \mathrm{k} \\ &=24 \mathrm{k} \text { and } 2 \mathrm{k} \end{aligned}

Q51GATE 2004MCQ

A 100 MHz carrier of 1 V amplitude and a 1 MHz modulating signal of 1 V amplitude are fed to a balanced modulator. The ourput of the modulator is passed through an ideal high-pass filter with cut-off frequency of 100 MHz. The output of the filter is added with 100 MHz signal of 1 V amplitude and 90 $^{\circ}$ phase shift as shown in the figure. The envelope of the resultant signal is

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📖 Explanation

\begin{array}{l} y(t)=\frac{1}{2} \cos \left\{2 \pi \times 101 \times 10^{6}\right\} t+\sin 2 \pi \times 100+10^{6} t \\ =\frac{1}{2}\left[\cos \left(2 \pi 100 \times 10^{6}\right) t \cos 10^{6} t \cdot 2 \pi-\sin \left(2 \pi \times 100 \times 10^{6}\right)\right. \\ t \sin 2 \pi 10^{6} t+\sin 2 \pi \times 100 \times 10^{6} t \\ =\frac{1}{2} \cos 2 \pi \times 100 \times 10^{6} t\left[\cos 2 \pi \times 10^{6} t\right]\$ \\ +\frac{1}{2} \sin 2 \pi \times 100 \times 10^{6} t\left[-\sin 2 \pi \times 10^{6} t+2\right]\$ \\ =A \cos 2 \pi \times 10^{6} t+B \sin \cos 2 \pi \times 10^{6} t \\ \end{array}

Envelope

\begin{array}{l} =\sqrt{A^{2}+B^{2}} \\ =\sqrt{\frac{1}{4} \cos ^{2} 2 \pi \times 10^{6} t+\frac{1}{4}\left[\sin 2 \pi \times 10^{6} t-2\right]\$} \end{array}

Envelope $=\sqrt{\frac{5}{4}-\sin (2 \pi \times 10^{6} t)}$

Q52GATE 2004MCQ

An AM signal and a narrow-band FM signal with identical carriers, modulating signals and modulation indices of 0.1 are added together. The resultant signal can be closely approximated by

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📖 Explanation

\begin{array}{c} s(t)=A_{c} \cos \omega_{c} t+\frac{A_{c} \beta}{2} \cos \left(\omega_{c}+\omega_{m}\right) t \\ -\frac{A_{c} \beta}{2} \cos \left(\omega_{c}-\omega_{m}\right) \\ s(t)_{A M}=A_{c} \cos \omega_{c} t+\frac{A_{c} m_{a}}{2} \\ \cos \left(\omega_{c}+\omega_{m}\right) t+\frac{A_{c} m_{a}}{2} \cos \left(\omega_{c}-\omega_{m}\right) \\ s(t)+s(t)_{A M}=\text { carrier }+\cup S B \\ =S S B \text { with carrier. } \\ \text { Note: }-s(t)+s(t)_{A M}=S S B+S C \end{array}

Q53GATE 2003MCQ

A superheterodyne receiver is to operate in the frequency range 550 kHz - 1650 kHz, with the intermediate frequency of 450 kHz. Let R = $C_{max}/C_{min}$ denote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then

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📖 Explanation

\begin{aligned} R &=\frac{C_{\max }}{C_{\min }}=\left(\frac{f_{\max }}{f_{\min }}\right)^{2}=\left(\frac{1650+450}{550+450}\right)^{2} \\ &=4.41 \\ I &=f_{s}+2 f_{i f}=700+2 \times 450=1600 \end{aligned}

Q54GATE 2003MCQ

Let $m(t)=\cos [(4\pi \times 10^{3})t]$ be the message signal & $c(t)=5\cos [(2\pi \times 10^{6}t)]$ be the carrier. c(t) and m(t) are used to generated an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term $\cos [{2\pi (1008\times 10^{3}t)}]$ in the FM signal (in terms of the Bessel coefficients) is

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📖 Explanation

\begin{aligned} x_{\mathrm{F}} \cdot(f) &=A_{c} \sum_{n=-\infty }^{\infty } J_{n}(\beta) \cos \left(\omega_{c}+n \omega_{m}\right) t \\ E \cdot \mathrm{W} \cdot &=2 \omega_{m} \\ \Delta \omega &=3 \times \mathrm{BW}=6 \omega_{m} \\ \beta &=\frac{\Delta \omega}{\omega_{m}}=6 \\ \therefore \quad \quad \omega_{c}+r \omega_{m} &=2 \pi \times 1008 \times 10^{3}\\ &\qquad(\text{given \in question}) \\ \therefore\quad 2 \pi \times 10^{6}&+n .4 \pi \times 10^{3}=2 \pi \times 1008 \times 10^{3} \\ \therefore \quad n&=4 \\ \end{aligned}

$\therefore$ Bessel coefficient $5 J_{4}(6)$

Q55GATE 2003MCQ

Let $m(t)=\cos [(4\pi \times 10^{3})t]$ be the message signal and $c(t)=5\cos [(2\pi \times 10^{6}t)]$ be the carrier. c(t) and m(t) are used to generate an AM signal. The modulation index of the generated AM signal is 0.5. Then the quantity $\frac{Total \; sideband \; power}{Carrier \; power}$ is

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📖 Explanation

\begin{array}{c} P_{t}=P_{c}\left(1+\frac{m_{a}^{2}}{2}\right) \\ P_{c}=\frac{A_{c}^{2}}{2} \\ \text { Sideband power }=\frac{A_{c}^{2}}{2} \cdot \frac{m_{a}^{2}}{2} \\ \therefore \quad \frac{P_{s}}{P_{c}}=\frac{\left(\frac{A_{c}^{2} m_{a}^{2}}{4}\right)}{\left(\frac{A_{c}^{2}}{2}\right)}=\frac{m_{a}^{2}}{2}=\frac{1}{8} \end{array}

Q56GATE 2003MCQ

Choose the correct one from among the alternative A,B,C,D after matching an item in Group 1 with most appropriate item in Group 2.

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📖 Explanation

Group matching questions in communication typically test receiver topologies and modulation characteristics:\n- Superheterodyne Receiver $\implies$ Image frequency rejection (using RF filters).\n- Envelope Detector $\implies$ Amplitude Modulation (AM) detection.\n- Ratio Detector / Discriminator $\implies$ Frequency Modulation (FM) detection.\n- PLL (Phase Locked Loop) $\implies$ Coherent carrier recovery or FM demodulation.\nOption B follows the universally accepted associations for these fundamental communication blocks.

Q57GATE 2003MCQ

A DSB-SC signal is to be generated with a carrier frequency $f_{c}$ = 1 MHz using anon-linear device with the input-output characteristic $V_{0}=a_{0}v_{i}+a_{1}v_{i}^{3}$ where $a_{0}$ and $a_{1}$ are constants. The output of the non-linear device can be filtered by an appropriate band pass filter. Let $V_{i}=A_{c}^{'}\cos (2\pi f_{c}'t)+m(t)$ , where m(t) is the message signal. Then the value of $f_{c}^{'}$ (in MHz) is

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📖 Explanation

\begin{array}{l} V_{0}=a_{0} v_{i}+a_{1} v_{i}^{3}\\ =a_{0}\left(A_{c}^{i} \cos \omega_{c}^{i} t\right)+a_{d} m(t)+a_{1} A_{c}^{3} \cos ^{3}\left(n_{c}^{i} t\right. \\ +a_{1} m^{3}(t)+3 a_{3} m^{2}(t) A_{c}^{i} \cos \omega_{c}^{i} t \\ +3 a_{1} A_{c}^{\prime} \cos ^{2} \omega_{c}^{i} t m(t) \end{array}

For DSB-SC, we are concerned with only last term.

\begin{aligned} m(t) \cos \omega_{c} t \rightarrow f_{c} &=1 \mathrm{MHz} \\ \text { For } \cos ^{2} \text { term, } 2 \omega_{c}^{i} &=2 \pi \times 1 \mathrm{MHz} \\ \cos ^{2} \text { term, } 2 \omega_{c}^{i} &=2 \pi \times 1 \mathrm{MHz} \end{aligned}

Q58GATE 2003MCQ

The input to a coherent detector is DSB-SC signal plus noise. The noise at the detector output is

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Q59GATE 2002MCQ

An angle modulated signal is given by $s(t)=cos2\pi (2\times 10^{6}t+30 sin 150t+40 cos 150t)$ The maximum frequency and phase deviations of s(t) are

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📖 Explanation

\begin{aligned} s(t) &=\cos \$[2 \pi \times 2 \times 10^{6} t+2 \pi \times 30 \sin 150 t \\&+2 \pi \times 40 \cos 150 t]\$ \\ \frac{d \phi}{d t} &=2 \pi \times 2 \times 10^{6}\\ & +2 \pi \times 4500 \cos 150 t+2 \pi \times 6000(-\sin 150 t) \\ \therefore \Delta \omega &=\omega_{1}-\omega_{c}=2 \pi\left(4500^{2}+6000^{2}\right)^{1 / 2} \\ &=2 \pi \times 7.5 \mathrm{k} \mathrm{rad} / \mathrm{sec} \\ \therefore \quad \Delta f &=\frac{\Delta \omega}{2 \pi}=7.5 \mathrm{kHz} \\ \Delta \phi &=2 \pi \sqrt{30^{2}+40^{2}} \\ \Delta \phi &=100 \pi \end{aligned}

Q60GATE 2002MCQ

A 1 MHz sinusoidal carrier is amplitude modulated by a symmetrical square wave of period 100 $\mu$ sec. Which of the following frequencies will NOT be present in the modulated signal?

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📖 Explanation

Message signa m(t) is a symmetrical square wave of period $100 \mu \mathrm{s}$

\begin{array}{l} T_{0}=100 \mu \mathrm{s} \\ f_{0}=1 / T_{0}=10 \mathrm{kHz} \end{array}

Fourier series representation of m(t) will have only Odd harmonics. So, m(t) contains $\Rightarrow 10 \mathrm{kHz}, 30 \mathrm{kHz}, 50 \mathrm{kHz}, \ldots$ Carrier frequency, $f_{c}=1 \mathrm{MHz}=1000 \mathrm{kHz}$ AM signal contains $\Rightarrow \ldots, 950 \mathrm{kHz}, 970 \mathrm{kHz}, 990 \mathrm{kHz}$ $1000 \mathrm{kHz}, 1010 \mathrm{kHz}, 1030 \mathrm{kHz}, 1050 \mathrm{kHz} \ldots$ So, option (C) is correct.