The GATE Grind

GATE 2022 EE – Question 49

Signals and Systems · Fourier series representation of continuous and discrete time periodic signals, Sampling theorem · 2 marks · Multiple choice

The discrete time Fourier series representation of a signal $x[n]$ with period $N$ is written as $x[n]=\sum_{k=0}^{N-1}a_ke^{j(2kn\pi/N)}$. A discrete time periodic signal with period $N=3$, has the non-zero Fourier series coefficients: $a_{-3}=2$ and $a_4=1$. The signal is

  1. $2+2e^{-j\left(\frac{2\pi}{6}n\right)}\cos\left(\frac{2\pi}{6}n\right)$
  2. $1+2e^{j\left(\frac{2\pi}{6}n\right)}\cos\left(\frac{2\pi}{6}n\right)$
  3. $1+2e^{j\left(\frac{2\pi}{3}n\right)}\cos\left(\frac{2\pi}{6}n\right)$
  4. $2+2e^{j\left(\frac{2\pi}{6}n\right)}\cos\left(\frac{2\pi}{6}n\right)$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (B) $1+2e^{j\left(\frac{2\pi}{6}n\right)}\cos\left(\frac{2\pi}{6}n\right)$

Explanation

The coefficients repeat with period 3, so $a_{-3}=a_0=2$ and $a_4=a_1=1$. Hence $x[n]=2+e^{j\frac{2\pi}{3}n}$. Using $2e^{j\theta}\cos\theta=e^{j2\theta}+1$ with $\theta=\frac{2\pi}{6}n$, option B equals $1+\left(e^{j\frac{2\pi}{3}n}+1\right)=2+e^{j\frac{2\pi}{3}n}$.