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GATE 2023 EC – Question 47

Networks, Signals and Systems · Discrete-time Signals · 2 marks · Multiple choice

Consider a discrete-time periodic signal with period $N=5$. Let the discrete-time Fourier series (DTFS) representation be $x[n]=\sum_{k=0}^{4}a_ke^{j\frac{2\pi kn}{5}}$, where $a_0=1,\ a_1=3j,\ a_2=2j,\ a_3=-2j$ and $a_4=-3j$. The value of the sum $\sum_{n=0}^{4}x[n]\sin\frac{4\pi n}5$ is

  1. −10
  2. 10
  3. −2
  4. 2

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Correct answer: (A) −10

Explanation

$\sin\frac{4\pi n}{5}=\dfrac{e^{j2\pi\cdot2n/5}-e^{-j2\pi\cdot2n/5}}{2j}$ and $\sum_{n=0}^4x[n]e^{-j2\pi kn/5}=5a_k$. So the sum equals $\dfrac{5}{2j}\left(a_{-2}-a_{2}\right)=\dfrac5{2j}(a_3-a_2)=\dfrac5{2j}(-2j-2j)=-10$.