GATE 2018 EC Q.28: solution and answer

GATE 2018 · ECQ.28 · 2 marks · MCQ

Consider a white Gaussian noise process $N (t)$ with two-sided power spectral density $S_{N} (f) = 0.5$ W/Hz as input to a filter with impulse response $0.5 e^{- t^{2} /2}$ (where $t$ is in seconds) resulting in output $Y (t)$ . The power in $Y (t)$ in watts is

(A)
0.11
(B)
0.22
(C)
0.33
(D)
0.44

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Answer and solution

ANSWER

The output power is $π /8 \approx 0.22 W$ , so the answer is (B).

Official answer: BSource: Official answer key, GATE 2018 EC (IIT Guwahati)

The noise is white with two-sided PSD $S_{N} (f) = \frac{N _{0}}{2} = 0.5 W/Hz$ . For an LTI filter, $S_{Y} (f) = S_{N} (f) ∣ H (f) ∣^{2}$ , so the output power is $P_{Y} = \int_{- \infty}^{\infty} S_{Y} (f) df$ .
The Gaussian transform pair, with $f$ in Hz, is $e^{- t^{2} /2} \leftrightarrow 2 π e^{- 2 π^{2} f^{2}}$ .
Hence $H (f) = 0.5 2 π e^{- 2 π^{2} f^{2}}$ and $∣ H (f) ∣^{2} = 0.25 \cdot 2 π e^{- 4 π^{2} f^{2}} = 0.5 π e^{- 4 π^{2} f^{2}}$ .
The Gaussian integral gives $\int_{- \infty}^{\infty} e^{- 4 π^{2} f^{2}} df = \frac{π}{4 π ^{2}} = \frac{1}{2 π}$ .
$P_{Y} = 0.5 \times 0.5 π \times \frac{1}{2 π} = \frac{π}{8} \approx 0.2216 W$
Cross-check in the time domain, using Parseval's theorem: $P_{Y} = \frac{N _{0}}{2} \int_{- \infty}^{\infty} h^{2} (t) d t = 0.5 \times 0.25 \int_{- \infty}^{\infty} e^{- t^{2}} d t = 0.5 \times 0.25 π = \frac{π}{8} \approx 0.2216 W$ . The two methods agree.
The nearest option is $0.22 W$ , which is option (B).

KEY IDEA

For white noise through an LTI filter, the output power is $\frac{N _{0}}{2} \int h^{2} (t) d t$ (Parseval), where $N_{0} /2$ is the two-sided PSD. There is no need to find $H (f)$ .

WATCH OUT FOR

Using the one-sided PSD $N_{0} = 0.5$ instead of the two-sided value $N_{0} /2 = 0.5$ doubles or halves the result. Another slip is forgetting the factor $0.5$ in $h (t)$ , which must be squared to give $0.25$ . A third is using the wrong Gaussian integral, since $\int e^{- t^{2}} d t = π$ , not $2 π$ .

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GATE 2018 EC Q.28: solution and answer

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