GATE 2020 EC Q.24: solution and answer

GATE 2020 · ECQ.24 · 1 mark · NAT

The random variable $Y = \int_{- \infty}^{\infty} W (t) ϕ (t) d t$ , where $ϕ (t) = 1$ for $5 \leq t \leq 7$ and $0$ otherwise, and $W (t)$ is a real white Gaussian noise process with two-sided power spectral density $S_{W} (f) = 3 W/Hz$ for all $f$ . The variance of $Y$ is ______.

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Answer and solution

ANSWER

The variance of $Y$ is $6$ .

Official answer: 6Source: Official answer key, GATE 2020 EC (IIT Delhi)

The two-sided power spectral density is $S_{W} (f) = 3 W/Hz$ for all $f$ , so $N_{0} /2 = 3$ . The autocorrelation is the inverse Fourier transform of this flat spectrum: $R_{W} (τ) = 3 δ (τ)$ .
$W (t)$ is zero-mean white Gaussian noise, so $E [Y] = \int ϕ (t) E [W (t)] d t = 0$ . Therefore $Var (Y) = E [Y^{2}]$ .
Write $Y^{2}$ as a double integral and take the expectation: $E [Y^{2}] = \int \int ϕ (t) ϕ (s) R_{W} (t - s) d t d s = 3 \int \int ϕ (t) ϕ (s) δ (t - s) d t d s$
The delta function sifts out $s = t$ in the inner integral, which gives $E [Y^{2}] = 3 \int_{- \infty}^{\infty} ϕ^{2} (t) d t$
$ϕ (t)$ equals 1 on $5 \leq t \leq 7$ and 0 elsewhere, so $ϕ^{2} (t) = ϕ (t)$ . Then $\int ϕ^{2} d t = 7 - 5 = 2$ , the energy of the pulse.
$Var (Y) = 3 \times 2 = 6$

KEY IDEA

Integrating white noise against a deterministic window $ϕ (t)$ gives a zero-mean random variable. Its variance is the two-sided PSD level times the energy of the window: $Var (Y) = S_{W} \int ϕ^{2} (t) d t$ .

WATCH OUT FOR

Using $N_{0}$ in place of $N_{0} /2$ . The 3 W/Hz given here is already the two-sided level, so using $2 \times 3 = 6$ as the delta weight would give 12. Another slip is to take the window length as 7 instead of $7 - 5 = 2$ .

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GATE 2020 EC Q.24: solution and answer

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