GATE 2026 EE Q.36: solution and answer

GATE 2026 · EEQ.36 · 2 marks · MCQ

A time-limited waveform $g (x)$ is specified as follows:

g (x) = ⎩ ⎨ ⎧ - k, + k, 0, - π < x \leq 0 0 < x \leq π otherwise

A new waveform $f (x)$ is constructed from $g (x)$ as follows:

f (x) = m = - \infty \sum \infty g (x + 2 π m), for all x \in R

The sum of the coefficients of the third harmonics of the sine and cosine terms in the trigonometric Fourier series expansion of $f (x)$ is $\frac{2}{3 π}$ .

What is the value of $k$ ?

(A)
$1$
(B)
$\frac{1}{2}$
(C)
$\frac{1}{3}$
(D)
$\frac{1}{4}$

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Answer and solution

ANSWER

The periodic extension $f (x)$ is an odd square wave of amplitude $k$ and period $2 π$ . Its third harmonic has $a_{3} = 0$ and $b_{3} = \frac{4 k}{3 π}$ . Setting $a_{3} + b_{3} = \frac{2}{3 π}$ gives $k = \frac{1}{2}$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2026 EE (IIT Guwahati)

Periodic extension: the shifted copies $g (x + 2 π m)$ occupy the intervals $(- π + 2 π m, π + 2 π m]$ . These do not overlap and together cover the whole real line, so at every $x$ exactly one term of the sum is non-zero. So $f (x)$ is a square wave with period $T_{0} = 2 π$ ( $ω_{0} = 1$ ). It equals $+ k$ on $(0, π]$ and $- k$ on $(- π, 0]$ , and repeats every $2 π$ .
Symmetry: $f (- x) = - f (x)$ for every $x$ except the isolated jump points $x = nπ$ , and single points do not change a Fourier integral. So $f$ is odd over a period. The dc term and every cosine coefficient vanish: $a_{0} = 0$ and $a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) cos (n x) d x = 0$ . In particular $a_{3} = 0$ .
Sine coefficients: $f (x) sin (n x)$ is even, so the full-period integral is twice the half-range integral: $b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) sin (n x) d x = \frac{2}{π} \int_{0}^{π} k sin (n x) d x = \frac{2 k}{nπ} (1 - cos nπ) = ⎩ ⎨ ⎧ \frac{4 k}{nπ}, 0, n odd n even$
Third harmonic ( $n = 3$ , which is odd): $b_{3} = \frac{4 k}{3 π}$ and $a_{3} = 0$ .
Apply the given condition: $a_{3} + b_{3} = 0 + \frac{4 k}{3 π} = \frac{2}{3 π} \Rightarrow k = \frac{1}{2}$
Option (B) is correct.

KEY IDEA

Summing $g (x + 2 π m)$ over all $m$ is simply the $2 π$ -periodic extension of $g$ , which here is an odd square wave. Odd symmetry makes every cosine coefficient zero. The sine series of this square wave is standard: $b_{n} = \frac{4 k}{nπ}$ for odd $n$ and $0$ for even $n$ . The \"sum of the third-harmonic coefficients\" is therefore just $b_{3}$ .

WATCH OUT FOR

Using the peak-to-peak value $2 k$ as the amplitude, which gives $b_{3} = \frac{8 k}{3 π}$ and $k = \frac{1}{4}$ (option D). Another slip is dropping the factor 2 from the half-range integral, which gives $b_{3} = \frac{2 k}{3 π}$ and $k = 1$ (option A). A third is assuming $a_{3} \neq = 0$ without checking the symmetry. The waveform swings between $- k$ and $+ k$ , so its amplitude is $k$ , and it is odd, so all $a_{n} = 0$ .

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GATE 2026 EE Q.36: solution and answer

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