GATE 2026 EC Q.58: solution and answer

Official answer: A, C, DSource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Frequencies. $x_1$ has frequency $n$ (period $1/n$). $x_2$ has frequency $2n$ (period $1/(2n)$). For real signals, two signals are orthogonal over an interval when their inner product $\int x_1(t)x_2(t)dt$ over that interval is zero.

2. Product-to-sum. Using $2\cos A\sin B=\sin (B+A)+\sin (B-A)$ with $A=2\pi nt$ and $B=4\pi nt$: $$x_1{x}_2=2\cos (2\pi nt)\sin (4\pi nt)=\sin (6\pi nt)+\sin (2\pi nt).$$

3. (A) Interval $[0,1/n)$. Over this interval $\sin (2\pi nt)$ completes exactly 1 cycle and $\sin (6\pi nt)$ completes exactly 3 cycles, so each integrates to zero: $${\int }_0^{1/n}{x}_1{x}_{2}dt=0.$$ The signals are orthogonal over this interval, so (A) is valid.

4. (B) Orthonormal also needs unit energy. $${\int }_0^{1/n}{\cos }^2(2\pi nt)dt=\frac{1}{2n},{\int }_0^{1/n}4{\sin }^2(4\pi nt)dt=4\cdot \frac{1}{2n}=\frac{2}{n}.$$ Both would have to equal 1, which needs $n=\frac{1}{2}$ for $x_1$ and $n=2$ for $x_2$ at the same time. No positive integer $n$ satisfies both (the first condition alone already fails for every positive integer), so (B) is not valid.

5. (C) Harmonic relation. The frequency of $x_2$ is $2n=2\times$ (the frequency of $x_1$), an integer multiple. So $x_2$ is the second harmonic of $x_1$; its amplitude (2) and its phase (sine rather than cosine) do not change that. (C) is valid.

6. (D) Interval $[0,1/(2n))$. This is only half a period of $x_1$: $\sin (2\pi nt)$ completes half a cycle and $\sin (6\pi nt)$ completes one and a half cycles, so neither term integrates to zero: $${\int }_0^{1/(2n)}\sin (2\pi nt)dt=\frac{1-\cos \pi }{2\pi n}=\frac{1}{\pi n},{\int }_0^{1/(2n)}\sin (6\pi nt)dt=\frac{1-\cos 3\pi }{6\pi n}=\frac{1}{3\pi n}.$$

7. Adding the two terms gives $${\int }_0^{1/(2n)}{x}_1{x}_{2}dt=\frac{1}{\pi n}+\frac{1}{3\pi n}=\frac{4}{3\pi n}\ne 0.$$ The signals are not orthogonal over this interval, so (D) is valid.

8. Answer: (A), (C) and (D) are valid.