GATE EC 2026 Set 1 — Question 37

Given the Fourier series representation of $x(t)$ as $x(t) = \sum_{n=-\infty}^{\infty} a_n e^{j\frac{2\pi n t}{T}}$.
We are also given the condition $x \left(t - \frac{T}{2}\right) = -x(t)$.Let's substitute $t - \frac{T}{2}$ into the Fourier series expression:
$x \left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^{\infty} a_n e^{j\frac{2\pi n (t - T/2)}{T}}$
$x \left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^{\infty} a_n e^{j\frac{2\pi n t}{T}} e^{j\frac{2\pi n (-T/2)}{T}}$
$x \left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^{\infty} a_n e^{j\frac{2\pi n t}{T}} e^{-j\pi n}$We know that $e^{-j\pi n} = (e^{-j\pi})^n = (-1)^n$.
So, $x \left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^{\infty} a_n (-1)^n e^{j\frac{2\pi n t}{T}}$.From the given condition, $x \left(t - \frac{T}{2}\right) = -x(t)$.
Substituting the Fourier series for $x(t)$:
$-x(t) = - \sum_{n=-\infty}^{\infty} a_n e^{j\frac{2\pi n t}{T}} = \sum_{n=-\infty}^{\infty} (-a_n) e^{j\frac{2\pi n t}{T}}$.Equating the two expressions for $x \left(t - \frac{T}{2}\right)$:
$\sum_{n=-\infty}^{\infty} a_n (-1)^n e^{j\frac{2\pi n t}{T}} = \sum_{n=-\infty}^{\infty} (-a_n) e^{j\frac{2\pi n t}{T}}$.By the uniqueness of Fourier series coefficients, the coefficients of corresponding exponential terms must be equal:
$a_n (-1)^n = -a_n$
$a_n (-1)^n + a_n = 0$
$a_n ((-1)^n + 1) = 0$.Now, let's analyze this condition for even and odd values of $n$:Case 1: $n$ is an even integer.
Let $n = 2m$ for some integer $m$. Then $(-1)^n = (-1)^{2m} = 1$.
Substituting this into the condition:
$a_{2m} (1 + 1) = 0$
$2a_{2m} = 0$
$a_{2m} = 0$.Case 2: $n$ is an odd integer.
Let $n = 2m+1$ for some integer $m$. Then $(-1)^n = (-1)^{2m+1} = -1$.
Substituting this into the condition:
$a_{2m+1} (-1 + 1) = 0$
$a_{2m+1} (0) = 0$.
This equation is satisfied for any value of $a_{2m+1}$, meaning it does not impose a constraint on odd coefficients.The question asks for the correct option for $a_{2m}$ for any integer $m$. Since $2m$ always represents an even integer, from Case 1, we conclude that $a_{2m} = 0$.The final answer is $\boxed{\text{A}}$