Question

Let x(n) be a periodic signal with period N = 8 and Fourier series coefficients $X_k=-X_k-4$. An another signal $y\left(n\right)=\left(\frac{1+(-1{)}^n}{2}\right)x(n-1)$ with period N = 8 is generated. Let Fourier series coefficient of y(n) be $Y_k$. The relation between $Y_k$ and $X_k$ is given $Y_k=F\left(k\right)X_k$

• A. $-e^{-\frac{\pi }{2}k}$
• B. $e^{-\frac{\pi }{4}k}$
• C. 0
• D. $\frac{1}{2}{e}^{-\frac{\pi }{2}k}$

Answer 1

The correct option is B $e^{-\frac{\pi }{4}k}$

$x\left(n\right)⟷X_k$

From frequency shifting property

$(-1{)}^{n}x(n)=e^{j\pi n}x(n)$

Here, ${\omega }_0=\frac{2\pi }{N}$

$e^{j\pi n}=e^{(j\times \frac{2\pi }{N}\times \frac{Nn}{2})}=e^{j\omega 0\frac{N}{2}n}$

So, $e^{j{\omega }_{0}Mn}x\left(n\right)⟷X_{k-M}$

$e^{j{\omega }_0\frac{N}{2}n}x\left(n\right)⟷X_{k-\frac{N}{2}}$

N = 8.

$(-1{)}^{n}x(n)⟷X_{k-4}$

Now as it is given,

$x_K=-X_{k-4}$

$(-1{)}^{n}x(n)⟷-X_k$

$(-1{)}^{n}x(n)⟷-x(n)$

$(-1{)}^{n}x(n)⟷x(n)$

$(-1{)}^{n+1}x(n)⟷x(n)$

$n\to n-1$

$(-1{)}^{n}x(n-1)⟷x(n-1)$

Now, $y\left(n\right)=\left[\frac{1+(-1{)}^n}{2}\right]x(n-1)$

$y\left(n\right)=\frac{x(n-1)}{2}+\frac{x(n-1)}{2}=x(n-1)$

$y\left(n\right)⟷Y_k$

$x(n-1)⟷X_k{e}^{-j{\omega }_{0}k}$

$Y_k=e^{\frac{2\pi }{S}k}{X}_k$

$F\left(k\right)=e^{-\frac{\pi }{4}k}$