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# GATE-2012 ECE Q2 (communication)

by on December 15, 2012

Question 52 on communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Solution

For a wide sense stationary function, the auto-correlation with delay $\tau$ is defined as,

$R$$\tau$$=E$X(t+\tau)X^*(t)$$

From the Weiner-Kinchin theorem, the auto-correlation function $R$$\tau$$$ and power spectral density $S(f)$ are Fourier Transform pairs, i.e.

$R$$\tau$$=\int_{-\infty}^{\infty}S$$f$$e^{j2\pi f\tau}df$

Expressing in terms of $\omega$, where $\omega=2\pi f$

$R$$\tau$$=\frac{1}{2\pi}\int_{-\infty}^{\infty}S$$w$$e^{jw\tau}dw$.

When the delay $\tau=0$, the above equations simplifies to

$\begin{array}R$$0$$&=&E$X(t)X^*(t)$&=&E$|X(t)|^2$&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}S$$w$$dw\end{array}$.

Applying this to the problem at hand,

$\begin{array}{lll}E$|X(t)|^2$&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}S$$w$$dw\\&=&\frac{2}{2\pi}\int_{0}^{\infty}S$$w$$dw\\&=&\frac{2}{2\pi}$$\frac{1}{2}6*2*10^3 + 400$$\\&=&\Large{\frac{6400}{\pi}}\end{array}$.

Further, since the power spectral density $S(f)$ does not have any dc component, the mean of the signal $\|E$X(t)$\|=0$

Based on the above, the right choice is (B) $\Large{6400/\pi, \mbox{ } 0}$

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

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