Q61GATE 2006MCQ

The following question refer to wide sense stationary stochastic processes The parameters of the system obtained in first order highpass R-c filter would be

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📖 Explanation

$H(j \omega)=\frac{R}{R+j \omega L}$ Comparing with (i) and (ii) $R=4, L=1$

Q63GATE 2005MCQ

An output of a communication channel is a random variable v with the probability density function as shown in the figure. The mean square value of \v is

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📖 Explanation

\begin{aligned} E\left[X^{2}\right]\$&=\int_{0}^{4} x^{2} f_{X}(x) d x \\ &\qquad\left[\begin{array}{l} f_{X}(x)=y=m x+C=\frac{x}{8} \\ \frac{K \times 4}{2}=\text { Area }=1 \\ K=\frac{1}{2} \end{array}\right] \\ E\left[X^{2}\right]\$&=\int_{0}^{4} x^{2}\left(\frac{x}{8}\right) d x=\left.\frac{x^{4}}{4 \times 8}\right|_{0} ^{4} \\ E\left[X^{2}\right]\$&=8 \end{aligned}

Q64GATE 2005MCQ

Noise with uniform power spectral density of $N_{0}$ W/Hz is passed though a filter $H(\omega )=2 exp(-j\omega t_{d})$ followed by an ideal pass filter of bandwidth B Hz. The output noise power in Watts is

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📖 Explanation

\begin{array}{l} P S D_{a t}=|H(\omega)|^{2} P S D_{i n}=4 N_{0} \\ P_{N}=B \cdot W \cdot x S N_{\text {out }}=4 N_{0} B \end{array}

Q65GATE 2004MCQ

The distribution function $F_{X}(X)$ of a random variable X is shown in the figure. The probability that X=1 is

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📖 Explanation

\begin{array}{l} P(x=1)=F_{x}\left(x=1^{+}\right)-F_{x}(x=1) \\ =0.55-0.25=0.30 \end{array}

Q66GATE 2004MCQ

A random variable X with uniform density in the interval 0 to 1 is quantized as follows : If 0 $\leq$ X $\leq$ 0.3, $x_{q}$ = 0 If 0.3 $\lt$ X $\leq$ 1, $x_{q}$ = 0.7 where $x_{q}$ is the quantized value of X. The root-mean square value of the quantization noise is

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📖 Explanation

$Q_{e}=\mathrm{s.v.}-\mathrm{q} \cdot \mathrm{v}=x-x_{q}$ Mean square value of quantization noise

\begin{array}{l} =E\left[\left(x-x_{q}\right)^{2}\right]\$ \\ =\int_{0}^{1}\left(x-x_{q}\right)^{2} f_{X}(x) d x \\ =\int_{0}^{1}\left(x-x_{q}\right)^{2} \cdot 1 d x \\ \int_{0}^{0.3}(x-0)^{2} d x+\int_{0.3}^{1}(x-0.7)^{2} d x \\ =0.039 \end{array}

Root mean square value of the quantization arise $=\sqrt{0.039}=0.198$

Q67GATE 2004MCQ

A 1 mW video signal having a bandwidth of 100 MHz is transmitted to a receiver through cable that has 40 dB loss. If the effective one-side noise spectral density at the receiver is $10^{-20}$ Watt/Hz, then the signal-to-noise ratio at the receiver is

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📖 Explanation

\begin{aligned} \mathrm{SNR} &=\frac{P_{s}}{N_{0} B} \\ &=\frac{10^{-3}}{10^{-20} \times 100 \times 10^{6}}=10^{9} \\ (\mathrm{SNR})_{\mathrm{dB}} &=10 \log 10^{9}=90 \mathrm{dB} \\ \text { Cable loss } &=40 \mathrm{dB}\\ \therefore \quad \text{ SNR at }R_{x}&=(90-40) d B=50 \mathrm{dB} \end{aligned}\\

Q68GATE 2003MCQ

X(t) is a random process with a constant mean value of 2 and the auto correlation function $R_{x}(\tau )=4(e^{-0.2\left | \tau \right |}+1)$ . Let Y and Z be the random variable obtained by sampling X(t) at t=2 and t=4 respectively. Let W=Y-Z. The variance of W is

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📖 Explanation

\begin{aligned} W &=Y-Z \\ \sigma_{w}^{2} &=E\left[\gamma^{2}\right]\$+E\left[Z^{2}\right]\$-2 E[Y \cdot Z] \\ \text{Y is sampled }&\text{at t=2 and Z at t=4}\\ \therefore \quad \sigma_{W}^{2}&=\sigma_{Y}^{2}+\sigma_{Z}^{2}-2 R_{x x}(2) \\ &=8+8-2\left[4 e^{-0.212 t}+1\right]\$ \\ \Rightarrow \quad \sigma_{W}^{2} &=2.64 \end{aligned}

Q69GATE 2003MCQ

Let X and Y be two statistically independent random variables uniformly distributed in the ranges (-1,1) and (-2,1) respectively. Let Z=X+Y. Then the probability that (Z $\leq$ -1) is

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📖 Explanation

$\Rightarrow \quad K^{11}=\frac{1}{6}$ since it lies in middle of $\frac{1}{3}$

\begin{aligned} \therefore \quad \text{Prob}[z \leq-2] \\ & =\text { Area of graph }[z \leq-2] \\ &=\frac{1}{2} \times \frac{1}{6} \times 1=\frac{1}{12} \end{aligned}

Q70GATE 2003MCQ

The noise at the input to an ideal frequency detector is white. The detector is operating above threshold. The power spectral density of the noise at the output is

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📖 Explanation

Frequency detector can be modelled as a differentiator. $S_{N}(f)=\frac{N_{0}}{2} 4 \pi^{2} f^{2} \Rightarrow \text { Parabolic }$

Q71GATE 2002MCQ

If the variance $\sigma _{x}^{2}$ of d(n)=x(n)-x(n-1) is one-tenth the variance $\sigma _{x}^{2}$ of a stationary zero mean discrete time signal x(n), then the normalized autocorrelation function $\frac{R_{xx}(k)}{\sigma _{x}^{2}}$ at k=1 is

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📖 Explanation

\begin{array}{l} E\left[d^{2}(n)\right]\$=E\left[\{x(n)-x(n-1)\}^{2}\right]\$ \\ \sigma_{d}^{2}=E[x(n)]^{2}+E[x(n-1)]^{2}-2 E[x(n) x(n-1)] \\ \Rightarrow \frac{\sigma_{X}^{2}}{10}=\sigma_{X}^{2}+\sigma_{X}^{2}-2 R_{X X}(1) \rightarrow \end{array}

Auto correlation at, k=1

\begin{aligned} 2 R_{x x}(1)&=\frac{19}{10} \sigma_{x}^{2} \\ \therefore \quad \frac{R_{x x}(1)}{\sigma_{x}^{2}}&=\frac{19}{20}=0.95 \end{aligned}

Q72GATE 2001MCQ

The PDF of a Gaussian random variable X is given by $P_{X}(X)=\frac{1}{3\sqrt{2\pi}}e^{\frac{-(x-4)^{2}}{18}}$ . The probability of the event {X=4} is

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Q73GATE 2001MCQ

The PSD and the power of a signal g(t) are, respectively, $S_g(\omega ) \; and \; P_g$ . The PSD and the power of the signal g(t) are, respectively

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📖 Explanation

$S_X(\omega )\rightarrow H(\omega )\rightarrow S_Y(\omega )$ $S_Y(\omega )= |H(\omega )|^2 S_X(\omega )$