Q21GATE 2015NAT

The input X to the Binary Symmetric Channel (BSC) shown in the figure is ' 1' with probability 0.8.The cross-over probability is 1/7. If the received bit Y = 0, the conditional probability that '1' was transmitted is _______.

🚀 Solve in Practice Mode

📖 Explanation

\begin{array}{l} P(X=1 / Y=0) \\ =\frac{P(Y=0 / X=1) P(X=1)}{P(Y=0 / X=1) P(X=1)+P(Y=0 / X=0) P(X=0)} \\ =\frac{\left(\frac{1}{7}\right)(0.8)}{\left(\frac{1}{7}\right)(0.8)+\left(\frac{6}{7}\right)(0.2)}=0.4 \end{array}

Q23GATE 2015MCQ

Consider a binary, digital communication system which uses pulses g(t) and -g(t) for transmitting bits over an AWGN channel. If the receiver uses a matched filter, which one of the following pulses will give the minimum probability of bit error?

🚀 Solve in Practice Mode

📖 Explanation

since $\quad \mathrm{SNR}=\frac{E}{2 \eta}$ Thus for having minimum error we have to select signal with maximum energy. Energy of signal 1=1 $\therefore$ Energy of signal $2=\frac{1}{3}$ Energy of signal $3=\frac{1}{3}$ Energy of signal $4=\frac{1}{3}$

Q24GATE 2015MCQ

A source emits bit 0 with probability 1/3 and bit 1 with probability 2/3. The emitted bits are communicated to the receiver. The receiver decides for either 0 or 1 based on the received value R . It is given that the conditional density functions of R are as

f_{(R|0)}(r)=\left\{\begin{matrix} \frac{1}{4}, & -3\leq x\leq 1,\\ 0, & otherwise, \end{matrix}\right.

and

f_{(R|1)}(r)=\left\{\begin{matrix} \frac{1}{6}, &-1\leq x\leq 5, \\ 0,& otherwise. \end{matrix}\right.

The minimum decision error probability is

🚀 Solve in Practice Mode

📖 Explanation

According to MAP criteria, for optimum threshold, $f_{R}\left(r \mid s_{0}\right) P\left(s_{0}\right) \sum_{k_{1}}^{r_{0}} f_{B}\left(r \mid s_{1}\right) P\left(s_{1}\right)$ Given that, $P\left(s_{0}\right)=\frac{1}{3} \text { and } P\left(s_{1}\right)=\frac{2}{3}$ From the above figure, it is clear that,

\begin{array}{l} f_{B}\left(r \mid s_{0}\right) P\left(s_{0}\right)>f_{R}\left(r \mid s_{1}\right) P\left(s_{1}\right) \Rightarrow \text { for } r<-1 \\ f_{A}\left(r \mid s_{1}\right) P\left(s_{1}\right)>f_{R}\left(r \mid s_{0}\right) P\left(s_{0}\right) \Rightarrow \text { for } r>-1 \end{array}

So, optimum threshold will be, $r_{\mathrm{th}}=-1$

\begin{aligned} \text{ For }r_{\mathrm{th}}=-1& \\ P\left(r_{0} \mid s_{1}\right) P\left(s_{1}\right)&=0\\ P\left(r_{1} \mid s_{0}\right) P\left(s_{0}\right)&=\frac{1}{12}[1-(-1)]=\frac{1}{6} \\ P_{e(\mathrm{min})}&=P\left(r_{0} \mid s_{1}\right) P\left(s_{1}\right)+P\left(q \mid s_{0}\right) P\left(s_{0}\right)=\frac{1}{6} \end{aligned}

Q25GATE 2015MCQ

A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size $\Delta$ of the delta modulator are 20,000 samples per second and 0.1 V, respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is

🚀 Solve in Practice Mode

📖 Explanation

Slope signal = Slope of delta modulator

\begin{aligned} A_{m}\left(2 \pi f_{m}\right) &=\Delta f_{s} \\ A_{m}\left(2 \pi 2 \times 10^{3}\right) &=20,000 \times 0.1 \\ A_{m} &=\frac{1}{2 \pi} \end{aligned}

Q26GATE 2015NAT

A sinusoidal signal of amplitude A is quantized by a uniform quantizer. Assume that the signal utilizes all the representation levels of the quantizer. If the signal to quantization noise ratio is 31.8 dB, the number of levels in the quantizer is _______.

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📖 Explanation

SNR of uniform quantized sinusoidal signal is given by

\begin{aligned} \mathrm{SNR} &=1.8+6 n \\ 31.8 &=1.8+6 n \\ n &=5 \end{aligned}

Thus total number of levels $=2^{5}=32$

Q27GATE 2015MCQ

The modulation scheme commonly used for transmission from GSM mobile terminals is

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Q28GATE 2014NAT

Let $Q(\sqrt{\gamma })$ be the BER of a BPSK system over an AWGN channel with twosided noise power spectral density $N_{0}/2$ . The parameter $\gamma$ is a function of bit energy and noise power spectral density. A system with two independent and identical AWGN channels with noise power spectral density $N_{0}/2$ is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels. If the BER of this system is $Q(b\sqrt{\gamma })$ , then the value of $b$ is

🚀 Solve in Practice Mode

📖 Explanation

since the same signal is passed through two parallel channels and added at the receiver input, the effective distance between the adjacent message points will be $d_{\min }=4 \sqrt{E_{b}}$ and the effective noise PSD(one-sided) will be $N_{\alpha \text { ett }}=2 N_{o}$

\begin{aligned} \text{So},\quad B E R &=Q(\sqrt{\frac{d_{\min }^{2}}{2 N_{o(\mathrm{etl})}}})=Q(\sqrt{\frac{16 E_{D}}{4 N_{o}}}) \\ &=Q(\sqrt{4 \frac{E_{b}}{N_{0}}})=Q(\sqrt{2} \sqrt{\frac{2 E_{b}}{N_{0}}}) \end{aligned}

For single channel case. $\mathrm{BER}=\alpha(\sqrt{\gamma})=Q(\sqrt{\frac{2 E_{b}}{N_{o}}})$ So, for the given problem,

\begin{aligned} B E R=Q(\sqrt{2} \sqrt{\frac{2 E_{b}}{N_{0}}})&=Q(\sqrt{2} \sqrt{\gamma})=Q(b \sqrt{\gamma}) \\ \text{So, }\quad b=\sqrt{2}&=1.414 \end{aligned}

Q29GATE 2014NAT

In a PCM system, the signal $m(t)=\{sin(100\pi t)+cos(100\pi t)\}$ V is sampled at the Nyquist rate. The samples are processed by a uniform quantizer with step size 0.75 V. The minimum data rate of the PCM system in bits per second is ____.

🚀 Solve in Practice Mode

📖 Explanation

$m(t)=\sin (100 \pi t)+\cos (100 \pi t)$ Maximum peak of m(t) is $\sqrt{2}$

\begin{aligned} \Delta &=\frac{V_{\max }-V_{\min }}{2^{n}} \\ 0.75 &=\frac{2 \sqrt{2}}{2^{n}} \\ 2^{n} &=\frac{2 \sqrt{2}}{0.75} \\ 2^{n} &=3.8 \rightarrow n=2 \\ R_{b} &=n f_{s}=2 \times 100=200 \mathrm{bits} / \mathrm{sec} \end{aligned}

Q30GATE 2014NAT

An M-level PSK modulation scheme is used to transmit independent binary digits over a band-pass channel with bandwidth 100 kHZ. The bit rate is 200 kbps and the system characteristic is a raised-cosine spectrum with 100% excess bandwidth. The minimum value of M is ________.

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📖 Explanation

\begin{aligned} B &=\frac{R_{b}}{\log _{2} M}(1+\alpha) \Rightarrow 100=\frac{200 \times 2}{\log _{2} M} \\ \log _{2} M &=4 \\ \text { So, } M &=16 \end{aligned}

Q31GATE 2014NAT

An analog voltage in the range 0 to 8 V is divided in 16 equal intervals for conversion to 4-bit digital output. The maximum quantization error (in V) is_____.

🚀 Solve in Practice Mode

📖 Explanation

The maximum quantization error is

\begin{aligned} Q_{e} &=\frac{\Delta}{2} \\ \text{where,}\quad \Delta &=\frac{\text { Dynamic range }}{2^{n} \text { or } L}\\ \Delta &=\frac{8}{2^{4}}=\frac{8}{16}=0.5 \\ \therefore \quad Q_{e} &=\frac{0.5}{2}=0.25 \end{aligned}

Q32GATE 2014MCQ

Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms $s_{1}(t)= \alpha cos 2 \pi f_{1}t$ and [/latex] S_{2}(t)=\alpha cos 2\pi f_{2}t[/latex] , where $\alpha$ = 4 mV. Assume an AWGN channel with two sided noise power spectral density. $\frac{N_{0}}{2}=0.5 \times 10^{-12}$ W/Hz. Using an optimal receiver and the relation $Q(v)=\frac{1}{\sqrt{2\pi }}\int_{v}^{\infty }e^{-u^{2}/2}du$ , the bit error probability for a data rate of 500 kbps is

🚀 Solve in Practice Mode

📖 Explanation

\begin{array}{c} \text { For FSK } \rightarrow \text { Bit error Probability }=Q(\sqrt{\frac{A_{c}^{2} T_{b}}{2 N_{o}}}) \\ A_{c}=4 \mathrm{mV}, T_{b}=\frac{1}{R_{b}}=\frac{1}{500 \times 10^{3}} \\ \frac{N_{o}}{2}=0.5 \times 10^{-12} \mathrm{W} / \mathrm{Hz}\\ N_{o}=10^{-12} \mathrm{W} / \mathrm{Hz} \\ \text{Bit error Proability} =Q(4) \end{array}\\

Q33GATE 2013MCQ

The bit rate of a digital communication system is R kbits/s. The modulation used is 32-QAM. The minimum bandwidth required for ISI free transmission is

🚀 Solve in Practice Mode

📖 Explanation

Nyquist pulses should be used for baseband pulse shaping to avoid ISI, with minimum bandwidth requirement. In this case the bandwidth of the base band signal will be,

\begin{aligned} (B W)_{\text {baseband }}&=\frac{R}{2} \mathrm{kHz} \\ (\mathrm{BW})_{\text {MOAM }} &=\frac{2(\mathrm{BW})_{\text {baseband }}}{\log _{2} M}\\ \text{For }32-\mathrm{QAM} \\ (\mathrm{BW})_{32 \mathrm{QAM}}&=\frac{2(\mathrm{BW})_{\text {baseband }}}{\log _{2}(32)}=\frac{R}{5} \mathrm{kHz} \end{aligned}

Q34GATE 2013MCQ

Let $g(t)=e^{-\pi t^{2}}$ , and $h(t)$ is a filter matched to $g(t)$ . If $g(t)$ is applied as input to $h(t)$ , then the Fourier transform of the output is

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📖 Explanation

\begin{array}{l} g(t)=e^{-\pi t^{2}} \\ G(f)=e^{-\pi f^{2}} \\ H(f)=e^{-\pi f^{2}} \\ Y(f)=G(f) H(f) \\ Y(t)=e^{-\pi f^{2}} \cdot e^{-\pi f^{2}}=e^{-2 \pi f^{2}} \end{array}

Q35GATE 2013MCQ

Bits 1 and 0 are transmitted with equal probability. At the receiver, the pdf of the respective received signals for both bits are as shown below. The optimum threshold to achieve minimum bit error rate (BER) is

🚀 Solve in Practice Mode

📖 Explanation

Optimum threshold is given by the point of intersection of two pdf curves. $f(z / 0)=1-|z| \quad ; \quad|z| \leq 1$ $f(z / 1)=z / 4 \quad ; \quad 0 \lt z \lt 2$ The point of intersection which decides optimum threshold

\begin{aligned} 1-z&=\frac{1}{4} \\ 1&=z+\frac{z}{4}\\ \Rightarrow \quad z&=\frac{4}{5} \end{aligned}

Q36GATE 2012MCQ

A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X=0) = 9/10, then the probability of error for an optimum receiver will be

🚀 Solve in Practice Mode

📖 Explanation

\begin{array}{l} P\left(y_{0} \mid x_{0}\right) P\left(x_{0}\right)=\frac{7}{8} \times \frac{9}{10}=\frac{63}{80} \\ P\left(y_{0} \mid x_{1}\right) P\left(x_{1}\right)=\frac{1}{8} \times \frac{1}{10}=\frac{1}{80} \\ P\left(y_{1} \mid x_{0}\right) P\left(x_{0}\right)=\frac{1}{8} \times \frac{9}{10}=\frac{9}{80} \\ P\left(y_{1} \mid x_{1}\right) P\left(x_{1}\right)=\frac{7}{8} \times \frac{1}{10}=\frac{7}{80} \end{array}

$P\left(y_{0} \mid x_{0}\right) P\left(x_{0}\right)>P\left(y_{0} \mid x_{1}\right) P\left(x_{1}\right);$ So, when a symbol is received as $y_{0}$ , the decision can be made in favour of $x_{0}$ in an optimum way. $P\left(y, \mid x_{0}\right) P\left(x_{0}\right)>P\left(y_{1} \mid x_{1}\right) P\left(x_{1}\right) ;$ So, when a symbol is received as $y_{1}$ , the decision can be made in favour of $x_{0}$ in an optimum way. So, for the given BSC, with optimum receiver, both the received symbols will be decoded as $x_{0}$ Hence, the probability of error is equal to the probability of transmitting $x_{1}$ So, $P_{e}=P\left(x_{1}\right)=1 / 10$

Q37GATE 2012MCQ

In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \mathrm{BW} &=\frac{R_{b}}{2}(1+\alpha) \\ 3500 &=\frac{R_{b}}{2}(1+0.75) \\ R_{b} &=4000 \mathrm{bits} / \mathrm{sec} \end{aligned}

In base band transmission $\rightarrow$ Symbol rate = Bit rate =4000 symbols/sec

Q38GATE 2012MCQ

A BPSK scheme operating over an AWGN channel with noise power spectral density of $N_{0}/2$ , uses equiprobable signals $s_{1}(t)=\sqrt{\frac{2E}{T}}sin(\omega _{c}t)$ and $s_{2}(t)=-\sqrt{\frac{2E}{T}}sin(\omega _{c}t)$ over the symbol interval (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45 $^{\circ}$ with respect to the received signal, the probability of error in the resulting system is

🚀 Solve in Practice Mode

📖 Explanation

For a BPSK system in the presence of AWGN, the standard probability of error with coherent detection is given by $P_e = Q\left(\sqrt{\frac{2E_b}{N_0}}\right)$ . However, if there is a phase error $\phi$ between the received carrier and the local oscillator, the effective received signal amplitude is scaled by $\cos(\phi)$ .\nTherefore, the effective energy becomes $E_{eff} = E \cos^2(\phi)$ .\nGiven $\phi = 45^{\circ}$ , we have $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$ , so $\cos^2(45^{\circ}) = \frac{1}{2}$ .\nSubstituting this into the error probability formula:\n $P_e = Q\left(\sqrt{\frac{2E \cos^2(45^{\circ})}{N_0}}\right) = Q\left(\sqrt{\frac{2E (1/2)}{N_0}}\right) = Q\left(\sqrt{\frac{E}{N_0}}\right)$ This aligns perfectly with Option B.

Q39GATE 2011MCQ

A four-phase and an eight-phase signal constellation are shown in the figure below. Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is

🚀 Solve in Practice Mode

📖 Explanation

$P_{e} \propto \sqrt{E_{s}}$ for both cases $SO, \frac{\sqrt{E_{s}}}{\sqrt{E_{s_{2}}}}=\frac{r_{1}}{r_{2}}=\frac{0.707 d}{1.307 d}$ $\Rightarrow \quad \frac{E_{s_{2}}}{E_{s_{1}}}=\left(\frac{1.307}{0.707}\right)^{2}=3.42$ To achieve same error, $2^{\text {nd }}$ must have 3.42 time than $1^{\text {st. }}$ The value in $\mathrm{dB}=10 \log (3.42)=5.33 \mathrm{dB}$

Q40GATE 2011MCQ

An analog signal is band-limited to 4 kHz, sampled at the Nyquist rate and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is ________ bits / second.

🚀 Solve in Practice Mode

📖 Explanation

Quantized levels are equiprobable; hence

\begin{aligned} H&=\log _{2} 4=2 \text { bits/sample }\\ r&=2 \text{samples/sec} \\ \end{aligned}

Hence information rate $R=r \cdot H=2 \text{ samples/sec}$ $\times 2\text{ bits/sample}$ $\Rightarrow R=4 \mathrm{bits} / \mathrm{sec}$