Q41GATE 2010MCQ

The Nyquist sampling rate for the sign $s(t)=\frac{sin(500\pi t)}{\pi t} \times \frac{sin(700)\pi t}{\pi t}$ is given by

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📖 Explanation

\begin{aligned} s(t) &=\frac{1}{2}\left[\frac{2 \sin (500 \pi t) \sin (700 \pi t)}{\pi^{2} t^{2}}\right]\$ \\ s(t) &=\frac{1}{2 \pi^{2} t^{2}}[\cos (700 \pi t-500 \pi t)\\ &-\cos (700 \pi t+500 \pi t)] \\ s(t) &=\frac{1}{2 \pi^{2} t^{2}}[\cos (200 \pi t)-\cos (1200 \pi t)] \end{aligned}

Maximum frequency component, $f_{m}=\frac{1200 \pi}{2 \pi}=600 \mathrm{Hz}$ Nyquist sampling rate, $f_{s \min }=2 f_{m}=1200 \mathrm{Hz}$

Q43GATE 2009MCQ

The amplitude of a random signal is uniformly distributed between -5 V and 5 V. If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative values are uniformly quantized with a step size of 0.1 V, the resulting signal to quantization noise ration is approximately

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📖 Explanation

\begin{aligned} \mathbf{\Delta}_{1}&=0.05 \mathrm{V} \text { (for positive values) }\\ \Delta_{2}&=0.1 \mathrm{V} \text{ (for negative values)} \end{aligned}

The amplitude of the input signal is symmetrically distributed about zero. Hence, $P\left(\Delta_{1}\right)=P\left(\Delta_{2}\right)=0.50$ So, the quantization noise power will be

\begin{aligned} N_{0} &=\frac{\Delta_{1}^{2}}{12} P\left(\Delta_{1}\right)+\frac{\Delta_{2}^{2}}{12} P\left(\Delta_{2}\right) \\ &=\frac{1}{24}\left(\Delta_{1}^{2}+\Delta_{2}^{2}\right)=\frac{125}{24 \times 10000}=\frac{1}{1920} \end{aligned}

Signal power,

\begin{aligned} S &=\frac{25}{3} \\ S Q N R &=\frac{S}{N_{0}}=\frac{1920 \times 25}{3}=16000 \end{aligned}

In decilogs, $[\mathrm{SQNR}]=10 \log _{10}(\mathrm{SONR})=42 \mathrm{dB}$

Q44GATE 2008MCQ

Four messages band limited to W, W, 2W and 3W respectively are to be multiplexed using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is

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📖 Explanation

\begin{array}{l} f_{s_{1}}=2 \times W=2 W \\ f_{s_{2}}=2 \times W=2 W \\ f_{s_{3}}=2 \times 2 W=4 \mathrm{W} \\ f_{s_{4}}=2 \times 3 \mathrm{W}=6 \mathrm{W} \\ f_{\mathrm{s}}=f_{\mathrm{s}_{1}}+f_{\mathrm{s}_{2}}+f_{s_{3}}+f_{s_{4}} \\ t_{\mathrm{s}}=14 \mathrm{W} \end{array}

For minimum bandwidth n=1

\begin{aligned} R_{D}&=n f_{s}\\ r_{b}&=1 \times 14 \mathrm{W}=14 \mathrm{W} \\ \text { (B.W.) }_{\min } &=\frac{R_{b}}{2} \\ \text { (B.W.) }_{\min } &=\frac{14 \mathrm{W}}{2}=7 \mathrm{W} \end{aligned}

Q45GATE 2008MCQ

A speech signal, band limited to 4kHz and peak voltage varying between +5V and -5V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. The number of quantization levels required to reduce the quantization noise by a factor of 4 would be

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📖 Explanation

Quantization noise, $N_{q}=\frac{S^{2}}{12}$ where S= Step size of quantization level

\begin{aligned} N_{q}^{\prime}&=\frac{N_{q}}{4} \\ \Rightarrow \quad \frac{S^{2}}{12}&=\frac{S^{2}}{12 \times 4} \\ \Rightarrow \quad s&=\frac{S}{2} \\ \Rightarrow \quad S&=\frac{V_{P p}}{2^{n}} \\ \text{where,} V_{p}&= \text{ peak-to-peak value}\\ n&= \text{no. of bits/sample} \\ \Rightarrow \quad \frac{v_{\mathrm{rP}}}{2^{n}}&=\frac{V_{p 0}}{2.2^{n}}=\frac{V_{p p}}{2.2^{8}} \\ \Rightarrow \quad 2^{\pi}&=2^{9}=512 \end{aligned}

Therefore, number of quantization levels required to reduce the quantization noise by a factor 4 would be 512 .

Q46GATE 2008MCQ

Consider a Binary Symmetric Channel (BSC) with probability of error being p. To transmit a bit, say 1, we transmit a sequence of three 1s. The receiver will interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is

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📖 Explanation

There will be error if all the three received bits are 0 or two of the three received bits are 0. Therefore, probability of error in output

\begin{array}{l} ={ }^{3} C_{3} \cdot p \cdot p \cdot p+{ }^{3} C_{2} \cdot p \cdot p \cdot(1-p) \\ =p^{3}+3 p^{2}(1-p) \end{array}

Q47GATE 2007MCQ

Two 4-ray signal constellations are shown. It is given that $\phi_{1}$ and $\phi_{2}$ constitute an orthonormal basis for the two constellations. Assume that the four symbols in both the constellations are equiprobable. Let $\frac{N_{0}}{2}$ denote the power spectral density of white Gaussian noise. If these constellations are used for digital communications over an AWGN channel, then which of the following statements is true ?

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Q48GATE 2007MCQ

An input to a 6-level quantizer has the probability density function f(x) as shown in the figure. Decision boundaries of the quantizer are chosen so as t maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are '-1', '0' and '1' . Assuming that the reconstruction levels of the quantizer are the mid-points of the decision boundaries, the ratio of signal power to quantization noise power is

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📖 Explanation

Signal power,

\begin{aligned} S &=E\left[X^{2}\right]\$=\int_{-\infty }^{\infty } x^{2} f_{X}(x) d x \\ &=2\left[\int_{0}^{1} a x^{2} d x+\int_{1}^{5} b x^{2} d x\right]\$=2\left[a\left(\frac{x^{3}}{3}\right)^{1}+b\left(\frac{x^{3}}{3}\right)^{5}\right]_{1} \\ &=2\left[\frac{a}{3}+\frac{124 b}{3}\right]=2\left[\frac{1}{6 \times 3}+\frac{124}{12 \times 3}\right]=7 \end{aligned}

Quantization noise power,

\begin{aligned} N_{Q}&=\sum_{i=1}^{2} \frac{\Delta_{i}^{2}}{12} P\left(\Delta_{i}\right) \\ \Delta_{1}&=1 \vee ; \\ for 1 \lt x \lt 1 \\ \Delta_{2}&=2 V ; \\ for 1 \lt |x| \lt 5 \\ P\left(\Delta_{1}\right)&=P\{-1 \lt x \lt 1\}=2 a=\frac{1}{3} \\ P\left(\Delta_{2}\right)&=1-P\left(\Delta_{1}\right)=\frac{2}{3} \\ So, N_{Q}&=\frac{1}{3}\left(\frac{1^{2}}{12}\right)+\frac{2}{3}\left(\frac{2^{2}}{12}\right)=\frac{1}{4} \\ \text{Hence, SQNR}&=\frac{S}{N_{Q}}=28 \end{aligned}

Q49GATE 2007MCQ

In a Direct Sequence CDMA system the chip rate is $1.2288 \times 10^{6}$ chips per second. If the processing gain is desired to be AT LEAST 100, the data rate

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📖 Explanation

Processing gain $=\frac{R_{c}}{R_{b}}$

\begin{aligned} \frac{R_{c}}{R_{b}} &\geq 100 \\ R_{b} &\leq \frac{R_{c}}{100} \\ R_{b} &\leq 12.288 \times 10^{3} \text{ bits per sec.} \end{aligned}

Q50GATE 2007MCQ

Three functions f1(t), f2(t) and f3(t), which are zero outside the interval [0, T], are shown in the figure. Which of the following statements is correct?

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📖 Explanation

Two functions $f(x) \& g(x)$ are said to be orthogonal

\begin{aligned} \text { if } \int_{-\infty }^{\infty } f(x) g(x) d x=0 & \\ \int_{-\infty }^{\infty } f_{2}(t) f_{3}(t) d t &=\int_{0}^{T} f_{2}(t) f_{3}(t) d t \\ &=\frac{T}{3}-T+\frac{2 T}{3} \\ &=0 \end{aligned}

Q51GATE 2007MCQ

During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of AT MOST one bit in error in a block of n bits is given by

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📖 Explanation

Prob of no error in n bits $=(1-p)^{n}$ Prob of one error in n bits $=n \cdot p(1-p)^{n-1}$ $\therefore$ Total probability $=n p(1-p)^{n-1}+(1-p)^{n}$

Q52GATE 2007MCQ

In a GSM system, 8 channels can co-exist in 200 kHz bandwidth using TDMA. A GSM based cellular operator is allocated 5 MHz bandwidth. Assuming a frequency reuse factor of $\frac{1}{5}$ , i.e. a five-cell repeat pattern, the maximum number of simultaneous channels that can exist in one cell is

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📖 Explanation

Allocated Bandwidth $=5 \mathrm{MHz}$ Frequency reuse factor $=\frac{1}{5}$ Bandwidth allocated for 1 cell

\begin{array}{rl} & =5 \times \frac{1}{5}=1 \mathrm{MHz} \\ \text { To allocate } 8 \mathrm{channels} & 200 \mathrm{kHz} \text { bandwidth required. } \end{array}

Number of channels in $1 \mathrm{MHz}=\frac{1 \mathrm{MHz}}{200 \mathrm{kHz}} \times 8=40$

Q53GATE 2007MCQ

In delta modulation, the slope overload distortion can be reduced by

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📖 Explanation

Condition to avoid slope overload in DM $\frac{\Delta}{T_{s}} \geq \frac{d}{d t} m(t)$ So, by increasing the step size, slope overload distortion can be avoided.

Q54GATE 2007MCQ

The raised cosine pulse p(t) is used for zero ISI in digital communications. The expression for p(t) with unity roll-off factor is given by $p(t)=\frac{sin4 \pi Wt}{4\pi Wt(1-16 W^{2}t^{2})}$ . The value of p(t) at $t=\frac{1}{4W}$ is

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📖 Explanation

\begin{aligned} \rho(t) &=\frac{\sin 4 \pi \omega t}{4 \pi M\left(1-16 W^{2} t^{2}\right)} \\ \rho\left(\frac{1}{4 W}\right) &=\frac{\sin 4 \pi W \times \frac{1}{4 W}}{4 \pi W \times \frac{1}{4 W}\left(1-16 W^{2} \frac{1}{16 W^{2}}\right)} \end{aligned}

As it comes in the form of 0/0 so applying LH rule

\begin{aligned} P\left(\frac{1}{4 W}\right)&=\frac{\frac{d}{d t} \sin 4 \pi w t}{\frac{d}{d t} 4 \pi W t\left(1-16 W^{2} t^{2}\right)} \\ &=\frac{\cos 4 \pi w t}{1-48 W^{2} t^{2}} \\ \text { Putting } \quad t &=\frac{1}{4 \mathrm{W}} \\ P(t) &=\frac{\cos \pi}{1-3}=0.5 \end{aligned}

Q55GATE 2006MCQ

The minimum sampling frequency (in samples/sec) required to reconstruct the following signal from its samples without distortion. $x(t)=5(\frac{sin 2\pi 1000t}{\pi t})^{3}+7(\frac{sin2\pi 1000t}{\pi t})^{2}$ would be

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📖 Explanation

$f_{s}=\left(2 f_{m}\right) \times 3=(2 \times 1000) \times 3=6 \times 10^{3} \mathrm{Hz}$

Q56GATE 2006MCQ

The minimum step-size required for a Delta-Modulator operating at 32k samples/sec to track the signal (here u(t) is the unit-step function) x(t) = 125(u(t) - u(t - 1)) + (250-125t)(u(t - 1) - u(t - 2)) so that slope-overload is avoided, would be

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📖 Explanation

To avoid slope overload

\begin{aligned} \frac{\Delta}{T_{s}} &\geq m^{\prime}(t) \\ \Delta .32 \times 1024 &\geq 125 \\ \Rightarrow \quad \Delta .2^{15}&=125 \\ \Delta &\geq \frac{2^{7}}{2^{15}} \\ \Rightarrow \quad \Delta &\geq 2^{-8} \end{aligned}

Q57GATE 2006MCQ

Let g(t)=p(t)*p(t) where * denotes convolution and p(t)=u(t)-u(t-1) with u(t) being the unit step function. An Amplitude Modulated signal is given as $X_{AM} (t) = 100(p(t) + 0.5 g(t)) cos\omega _{c}t$ in the interval 0 $\leq$ t $\leq$ 1. One set of possible values of modulating signal and modulation index would be

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📖 Explanation

$X_{A M}(t)=100[\rho(t)+0.5 g(t)] \cos \omega_{c} t$ $s(t)=100(1+m(t)) \cos \left(u_{c} t, p(t)=1\right.$ $m(t)=0.5 g(t)$ , comparing AM equation with (i) $=0.5 t$ $\Rightarrow$ Modulating signal = t(from equation of line) Modulation index =0.5

Q58GATE 2006MCQ

In the following figure the minimum value of the constant "C", which is to be added to $y_{1}$ (t) such that $y_{1}$ (t) and $y_{2}$ (t) are different , is

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📖 Explanation

For $Y_{1}(t)$ and $Y_{2}(t)$ to be different minimum step size of $\Delta/2 A,$ is needed else they will be same.

Q59GATE 2005MCQ

A signal as shown in the figure is applied to a matched filter. Which of the following does represent the output of this matched filter ?

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📖 Explanation

The impulse response of a matched filter is exactly a time-reversed and delayed version of the input signal: $h(t) = s(T - t)$ .\nThe output of the matched filter is the convolution of the input signal $s(t)$ with $h(t)$ , which corresponds to the auto-correlation function of the input signal.\nThe auto-correlation function of any finite-energy real signal is always locally symmetric (even) around its peak at the optimal sampling instant $t = T$ . Therefore, the corresponding symmetric auto-correlation waveform (Option C) represents the correct matched filter output.

Q60GATE 2004MCQ

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor

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📖 Explanation

For every 1 bit $\mathrm{SNR}_{9}$ improves by a factor of 4. Thus for 2 bits increase in message the $\mathrm{SNR}_{9}$ increases by a factor of 16.