GATE 2025 EC Q.55: solution and answer

GATE 2025 · ECQ.55 · 2 marks · MSQ

Let $f (t)$ be a periodic signal with fundamental period $T_{0} > 0$ . Consider the signal $y (t) = f (α t)$ , where $α > 1$ .

The Fourier series expansions of $f (t)$ and $y (t)$ are given by

f (t) = k = - \infty \sum \infty c_{k} e^{j \frac{2 π}{T _{0}} k t} and y (t) = k = - \infty \sum \infty d_{k} e^{j \frac{2 π}{T _{0}} α k t} .

Which of the following statements is/are TRUE?

(A)
$c_{k} = d_{k}$ for all $k$
(B)
$y (t)$ is periodic with a fundamental period $α T_{0}$
(C)
$c_{k} = d_{k} / α$ for all $k$
(D)
$y (t)$ is periodic with a fundamental period $T_{0} / α$

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Answer and solution

ANSWER

(A) and (D): $c_{k} = d_{k}$ for all $k$ , and $y (t)$ has fundamental period $T_{0} / α$ .

Official answer: A, DSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Fundamental period of $y (t)$ (time scaling). $y (t + T) = f (α t + α T)$ , so $y (t + T) = y (t)$ for every $t$ exactly when $α T$ is a period of $f$ . The smallest positive period of $f$ is $T_{0}$ , so the smallest positive $T$ is given by $α T = T_{0}$ , that is $T = T_{0} / α$ . Compressing a signal in time ( $α > 1$ ) shortens its period; it does not lengthen it. So (D) is TRUE and (B) is FALSE.
Substitute $α t$ into the series for $f$ . Replace $t$ by $α t$ in $f (t) = \sum_{k} c_{k} e^{j \frac{2 π}{T _{0}} k t}$ : $y (t) = f (α t) = k = - \infty \sum \infty c_{k} e^{j \frac{2 π}{T _{0}} k (α t)} = k = - \infty \sum \infty c_{k} e^{j \frac{2 π}{T _{0}} α k t} .$
Compare with the given expansion of $y (t)$ . The problem writes $y (t) = \sum_{k} d_{k} e^{j \frac{2 π}{T _{0}} α k t}$ . This uses the same basis functions $e^{j k ω_{y} t}$ , with $ω_{y} = \frac{2 π α}{T _{0}} = \frac{2 π}{T _{0} / α}$ , which is the fundamental frequency of $y$ from Step 1. These basis functions are orthogonal over one period $T_{0} / α$ , so the Fourier series of $y$ is unique. Therefore $d_{k} = c_{k}$ for every $k$ , and (A) is TRUE.
Check with the analysis formula. $d_{k} = \frac{α}{T _{0}} \int_{0}^{T_{0} / α} f (α t) e^{- j \frac{2 π}{T _{0}} α k t} d t$ . Put $τ = α t$ , so $d t = d τ / α$ and the limits become $0$ to $T_{0}$ : $d_{k} = \frac{1}{T _{0}} \int_{0}^{T_{0}} f (τ) e^{- j \frac{2 π}{T _{0}} k τ} d τ = c_{k} .$ The factor $1/ α$ from $d t$ cancels the $α$ in $1/ (T_{0} / α)$ , so no $α$ appears in the coefficients.
Why (C) is false. If (C) held as well, then $c_{k} = d_{k} / α = c_{k} / α$ , so $c_{k} (1 - 1/ α) = 0$ . Since $α > 1$ , every $c_{k}$ would be $0$ and $f (t) \equiv 0$ . The zero signal has no fundamental period, so it cannot be the given $f (t)$ . (C) is FALSE.
Result. Time scaling by $α$ leaves the Fourier series coefficients unchanged, multiplies the fundamental frequency by $α$ and divides the fundamental period by $α$ . The correct options are (A) and (D).

KEY IDEA

Time scaling $y (t) = f (α t)$ compresses the period to $T_{0} / α$ and raises the fundamental frequency to $α ω_{0}$ , but the Fourier series coefficients stay the same. The same harmonic weights now sit at harmonics of a higher fundamental frequency. In the analysis integral, the $1/ α$ from the change of variable cancels the $α$ in $1/ (T_{0} / α)$ .

WATCH OUT FOR

Carrying over the Fourier transform scaling rule $f (α t) \leftrightarrow \frac{1}{∣ α ∣} F (ω / α)$ and picking (C). The $1/ α$ appears in the transform because the transform integrates over all time. The series averages over one period, and the shorter period cancels it. A second slip is thinking compression lengthens the period and choosing (B).

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GATE 2025 EC Q.55: solution and answer

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