Question

Fourier transform of a periodic signal is given as,

$X\left(j\omega \right)=j\delta (\omega -\frac{\pi }{3})+2\delta (\omega -\frac{\pi }{7})$

The Fourier series coefficients of $C_7=j$ ___.

• A. 0.16

Answer 1

The correct option is A 0.16
Given Fourier transform of a periodic signal as,

$X\left(j\omega \right)=j\delta (\omega -\frac{\pi }{3})+2\delta (\omega -\frac{\pi }{7})$

$\delta \left(t\right)\stackrel{F.T}{\rightleftharpoons }1$

$1\stackrel{}{\rightleftharpoons }2\pi \delta \left(\omega \right)$

According to frequency shifting property,

$e^{j{\omega }_{0}t}x\left(t\right)\rightleftharpoons X(\omega -{\omega }_0)$

$e^{j\left(\frac{x}{3}\right)t}\rightleftharpoons 2\pi \delta (\omega -\frac{\pi }{3})$

$j{e}^{j\left(\frac{\pi }{3}\right)t}\rightleftharpoons j2\pi \delta (\omega -\frac{\pi }{3})$

Similarly, $2e^{j\left(\frac{\pi }{7}\right)t}\rightleftharpoons 4\pi \delta (\omega -\frac{\pi }{7})$

$x\left(t\right)\rightleftharpoons X\left(j\omega \right)$

$x\left(t\right)=\frac{j}{2\pi }{e}^{f\left(\frac{\pi }{3}\right)t}+\frac{1}{\pi }{e}^{f\left(\frac{\pi }{7}\right)t}$

Fundamental angular frequency,

${\omega }_0=GCD\{\frac{\pi }{3},\frac{\pi }{7}\}=\frac{\pi }{21}rad/s$

x(t) can be represented as,

$x\left(t\right)={\sum }_{k=-\infty }^{\infty }{C}_n{e}^{j\pi {\omega }_{0}t}$

By comparing (i) and (ii)

$x\left(t\right)=\frac{j}{2\pi }{e}^{j7{\omega }_{0}t}+\frac{1}{\pi }{e}^{js{\omega }_{0}t}$

So, Fourier series coefficient are,

$C_3=\frac{1}{\pi },C_7=\frac{j}{2\pi }=j0.16$