GATE 2024 EC Q.20: solution and answer

GATE 2024 · ECQ.20 · 1 mark · MCQ

A white Gaussian noise $w (t)$ with zero mean and power spectral density $\frac{N _{0}}{2}$ , when applied to a first-order RC low pass filter produces an output $n (t)$ . At a particular time $t = t_{k}$ , the variance of the random variable $n (t_{k})$ is ________.

(A)
$\frac{N _{0}}{4 R C}$
(B)
$\frac{N _{0}}{2 R C}$
(C)
$\frac{N _{0}}{R C}$
(D)
$\frac{2 N _{0}}{R C}$

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Answer and solution

ANSWER

The variance of $n (t_{k})$ is $\frac{N _{0}}{4 R C}$ , option (A).

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

The input has zero mean and the filter is linear and time-invariant, so the output $n (t)$ also has zero mean. Hence $Var [n (t_{k})] = E [n^{2} (t_{k})] = P_{n}$ , the output noise power.
The two-sided input PSD is $S_{w} (f) = \frac{N _{0}}{2}$ . The first-order RC low pass filter has $H (f) = \frac{1}{1 + j 2 π f R C}$ .
The output PSD is $S_{n} (f) = S_{w} (f) ∣ H (f) ∣^{2} = \frac{N _{0}}{2} \cdot \frac{1}{1 + ( 2 π f R C ) ^{2}}$ .
The output power is $P_{n} = \int_{- \infty}^{\infty} S_{n} (f) df = \frac{N _{0}}{2} \int_{- \infty}^{\infty} \frac{df}{1 + ( 2 π f R C ) ^{2}} .$
Substitute $x = 2 π f R C$ , so $df = \frac{d x}{2 π R C}$ . Since $\int_{- \infty}^{\infty} \frac{d x}{1 + x ^{2}} = π$ , we get $\int_{- \infty}^{\infty} ∣ H (f) ∣^{2} df = \frac{π}{2 π R C} = \frac{1}{2 R C}$ .
Time-domain check: $h (t) = \frac{1}{R C} e^{- t / R C} u (t)$ , so $\int_{0}^{\infty} h^{2} (t) d t = \frac{1}{( R C ) ^{2}} \cdot \frac{R C}{2} = \frac{1}{2 R C}$ . This agrees with the frequency-domain value (Parseval).
Therefore $Var [n (t_{k})] = \frac{N _{0}}{2} \cdot \frac{1}{2 R C} = \frac{N _{0}}{4 R C}$ , which is option (A).

KEY IDEA

For zero-mean noise through an LTI filter, variance equals output power. That power is the input PSD (two-sided $N_{0} /2$ ) times $\int ∣ H (f) ∣^{2} df$ , which is $\frac{1}{2 R C}$ for an RC low pass filter. This integral is the noise-equivalent bandwidth of the RC filter, $\frac{1}{4 R C}$ Hz one-sided.

WATCH OUT FOR

Taking the 3 dB bandwidth $\frac{1}{2 π R C}$ in Hz as the noise bandwidth, or using the one-sided PSD $N_{0}$ over the wrong band. The noise-equivalent bandwidth is $\frac{π}{2}$ times the 3 dB bandwidth, and the integral must run over both positive and negative frequencies when the PSD is two-sided.

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GATE 2024 EC Q.20: solution and answer

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