GATE 2020 EE Q.6: solution and answer

GATE 2020 · EEQ.6 · 1 mark · MCQ

$x_{R}$ and $x_{A}$ are, respectively, the rms and average values of $x (t) = x (t - T)$ , and similarly, $y_{R}$ and $y_{A}$ are, respectively, the rms and average values of $y (t) = k x (t)$ . $k, T$ are independent of $t$ . Which of the following is true?

(A)
$y_{A} = k x_{A}; y_{R} = k x_{R}$
(B)
$y_{A} = k x_{A}; y_{R} \neq = k x_{R}$
(C)
$y_{A} \neq = k x_{A}; y_{R} = k x_{R}$
(D)
$y_{A} \neq = k x_{A}; y_{R} \neq = k x_{R}$

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Answer and solution

ANSWER

(A): $y_{A} = k x_{A}$ and $y_{R} = k x_{R}$ . Scaling a periodic signal by a constant gain $k \geq 0$ scales both its average and its rms value by $k$ . (Strictly, the rms scales by $∣ k ∣$ .)

Official answer: ASource: Official answer key, GATE 2020 EE (IIT Delhi)

Periodicity. $x (t) = x (t - T)$ says $x (t)$ is periodic with period $T$ . Its average and rms values are therefore taken over one period. $y (t) = k x (t)$ , with $k$ a constant, has the same period $T$ .
Average value (linearity of the integral). $y_{A} = \frac{1}{T} \int_{0}^{T} k x (t) d t = k \cdot \frac{1}{T} \int_{0}^{T} x (t) d t = k x_{A}$ This holds for any constant $k$ , positive, negative or zero. That rules out (C) and (D).
RMS value (definition). $y_{R} = \frac{1}{T} \int_{0}^{T} k^{2} x^{2} (t) d t = k^{2} \frac{1}{T} \int_{0}^{T} x^{2} (t) d t = ∣ k ∣ x_{R}$ Squaring inside the rms does not stop it scaling with $k$ . The constant $k^{2}$ comes out of the integral, and its square root comes out of the radical.
Pick the option. $k$ is a constant scale factor (a gain). For $k \geq 0$ , $∣ k ∣ = k$ , so $y_{R} = k x_{R}$ . Both relations in (A) hold. Option (B) needs $y_{R} \neq = k x_{R}$ , which is false for every $k \geq 0$ .
A caveat on the sign. Strictly, $y_{R} = ∣ k ∣ x_{R}$ , because an rms value can never be negative. If $k$ were negative (and $x_{R} \neq = 0$ ), then $y_{R} = - k x_{R} \neq = k x_{R}$ , and the statement in (B) would be the true one. The question treats $k$ as a plain non-negative scaling constant, and the official key (A) is the answer on that reading.

KEY IDEA

The average is a linear operation, so it scales by exactly $k$ . The rms is the square root of a mean square: $k^{2}$ comes out of the integral and $k^{2} = ∣ k ∣$ comes out of the radical, so the rms scales by $∣ k ∣$ . That equals $k$ for a non-negative gain. The rms is not additive, $rms (x_{1} + x_{2}) \neq = rms (x_{1}) + rms (x_{2})$ in general, but it does scale with a constant multiplier.

WATCH OUT FOR

Reasoning that the rms is 'nonlinear because of the squaring' and so cannot scale by $k$ , which leads to (B). The rms fails additivity, not scaling: multiplying the signal by $k$ multiplies the rms by $∣ k ∣$ . The opposite slip is forgetting that the rms is never negative, so $y_{R} = ∣ k ∣ x_{R}$ and not $k x_{R}$ when $k < 0$ . The question takes $k$ as a non-negative gain.

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GATE 2020 EE Q.6: solution and answer

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