GATE 2026 EC Q.37: solution and answer

GATE 2026 · ECQ.37 · 2 marks · MCQ

The continuous time signal $x (t)$ is real, periodic with period $T$ and satisfies the Dirichlet conditions.

The Fourier series representation of $x (t) = \sum_{- \infty}^{\infty} a_{n} e^{j (\frac{2 π n t}{T})}$ and $x (t)$ satisfies the following: $x (t - \frac{T}{2}) = - x (t)$ .

For any integer $m$ , which of the following options is correct?

(A)
$a_{2 m} = 0$
(B)
$a_{2 m} = 1$
(C)
$a_{2 m} = a_{2 m + 1}$
(D)
$a_{2 m} = - 1$

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Answer and solution

ANSWER

(A) $a_{2 m} = 0$ . Half-wave symmetry, $x (t - \frac{T}{2}) = - x (t)$ , makes every even-indexed Fourier coefficient zero, including the DC term.

Official answer: ASource: Official answer key, GATE 2026 EC (IIT Guwahati)

Time-shift property of the Fourier series. Let $ω_{0} = \frac{2 π}{T}$ . If $x (t) \leftrightarrow a_{n}$ , then $x (t - t_{0}) \leftrightarrow a_{n} e^{- j n ω_{0} t_{0}}$ . This follows directly from the series: $x (t - t_{0}) = n = - \infty \sum \infty a_{n} e^{j n ω_{0} (t - t_{0})} = n = - \infty \sum \infty (a_{n} e^{- j n ω_{0} t_{0}}) e^{j n ω_{0} t}$
Put $t_{0} = \frac{T}{2}$ : $e^{- j n ω_{0} T /2} = e^{- j nπ} = (- 1)^{n}$ so $x (t - \frac{T}{2}) \leftrightarrow (- 1)^{n} a_{n}$ .
Match coefficients. The Fourier coefficients of a signal are unique. The given condition $x (t - \frac{T}{2}) = - x (t)$ therefore means that, for every $n$ , $(- 1)^{n} a_{n} = - a_{n} \Rightarrow [1 + (- 1)^{n}] a_{n} = 0$
Odd $n$ : $1 + (- 1)^{n} = 0$ , so the equation holds whatever $a_{n}$ is. Odd harmonics can be present.
Even $n = 2 m$ : $1 + (- 1)^{2 m} = 2 \neq = 0$ , so $a_{2 m} = 0$ for every integer $m$ . This includes $m = 0$ , so $a_{0} = 0$ and the signal has zero average.
Check the options. (B) and (D) contradict $a_{2 m} = 0$ . If (C) held for every $m$ , it would force $a_{2 m + 1} = a_{2 m} = 0$ as well. Every coefficient would then be zero and $x (t) \equiv 0$ . A nonzero signal can satisfy the condition, for example $x (t) = cos (\frac{2 π t}{T})$ : here $a_{1} = \frac{1}{2} \neq = 0 = a_{0}$ , so (C) fails. Answer: (A), $a_{2 m} = 0$ .

KEY IDEA

Shifting by half a period multiplies the $n$ -th coefficient by $e^{- j nπ} = (- 1)^{n}$ . With half-wave symmetry, $x (t - \frac{T}{2}) = - x (t)$ , you need $(- 1)^{n} a_{n} = - a_{n}$ , so only odd harmonics survive. Every even-indexed coefficient is zero, and that includes $a_{0}$ .

WATCH OUT FOR

Students often mix up half-wave symmetry, $x (t - \frac{T}{2}) = - x (t)$ , with odd symmetry, $x (- t) = - x (t)$ . Odd symmetry only makes the coefficients purely imaginary. It does not remove the even harmonics. Another slip is to use a shift of a full period $T$ instead of $T /2$ , which gives $e^{- j 2 π n} = 1$ and seems to tell you nothing. Some students also forget that $m = 0$ is included, so the DC value $a_{0}$ is zero as well.

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GATE 2026 EC Q.37: solution and answer

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