Question

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

• A. $-10$
• B. $10$
• C. $-2$
• D. $2$

Hint:

When summing $x[n]$ times sines/cosines over one period, convert to exponentials and use $\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi mn}{N}}=N a_m$ from the DTFS pair.

Answer 1

The Correct Option is A

Step 1: Use DTFS orthogonality.
Given $x[n]=\sum_{k=0}^{4} a_k e^{j\frac{2\pi kn}{5}}$, we have \[ \sum_{n=0}^{4} x[n]\,e^{-j\frac{2\pi mn}{5}}=5\,a_m,\qquad m=0,1,2,3,4. \] Step 2: Express the sine term in exponentials.
\[ \sin\!\left(\frac{4\pi n}{5}\right)=\frac{e^{j\frac{4\pi n}{5}}-e^{-j\frac{4\pi n}{5}}}{2j}. \] Hence \[ S=\sum_{n=0}^{4}x[n]\sin\!\left(\frac{4\pi n}{5}\right) =\frac{1}{2j}\!\left(\sum_{n=0}^{4}x[n]e^{j\frac{4\pi n}{5}}-\sum_{n=0}^{4}x[n]e^{-j\frac{4\pi n}{5}}\right). \] Step 3: Evaluate the two sums via Step 1.
Note $e^{j\frac{4\pi n}{5}}=e^{-j\frac{2\pi(-2)n}{5}}$ corresponds to $m\equiv-2\pmod{5}=3$, while $e^{-j\frac{4\pi n}{5}}$ corresponds to $m=2$. Therefore \[ S=\frac{1}{2j}\big(5a_3-5a_2\big)=\frac{5}{2j}(a_3-a_2). \] Step 4: Substitute coefficients.
$a_3=-2j$, $a_2=2j \Rightarrow a_3-a_2=-4j$. Thus \[ S=\frac{5}{2j}.(-4j)=-10. \] \[ \boxed{-10} \]