Question

What is the inverse fourier transform of $X\left(\omega \right)=4\sum _{n=-\infty }^{\infty }\delta (\omega -\frac{\pi }{2}n)?$

• A. $x\left(t\right)=\frac{8}{\pi }\sum _{n=-\infty }^{\infty }\delta (t-2n)$
• B. $x\left(t\right)=\frac{8}{\pi }\sum _{n=-\infty }^{\infty }\delta (t-4n)$
• C. $x\left(t\right)=\frac{2}{\pi }\sum _{n=-\infty }^{\infty }\delta (t-2n)$
• D. $x\left(t\right)=\frac{2}{\pi }\sum _{n=-\infty }^{\infty }\delta (t-4n)$

Answer 1

The correct option is B $x\left(t\right)=\frac{8}{\pi }\sum _{n=-\infty }^{\infty }\delta (t-4n)$

Fourier transform of periodic impulse train is also a periodic impulse train

$x\left(t\right)=\sum _{n=-\infty }^{\infty }\delta (t-n{T}_s)$

$X\left(\omega \right)=\frac{2\pi }{T_s}\sum _{n=-\infty }^{\infty }\delta (\omega -n{\omega }_s)$

Where, ${\omega }_s=\frac{2\pi }{T_s}$

Let us assume, $T_s=4$

${\omega }_s=\frac{\pi }{2}$

Then, $X\left(\omega \right)=\frac{\pi }{2}\sum _{n=-\infty }^{\infty }\delta (\omega -\frac{n\pi }{2})$

So, $\frac{2}{\pi }\sum _{n=-\infty }^{\infty }\delta (t-4n)\stackrel{F.T.}{⟷}\sum _{n=-\infty }^{\infty }\delta (\omega -\frac{n\pi }{2})$

$\frac{2}{\pi }\sum _{n=-\infty }^{\infty }\delta (t-4n)\stackrel{F.T.}{⟷}\sum _{n=-\infty }^{\infty }\delta (\omega -\frac{n\pi }{2})$

$\frac{8}{\pi }\sum _{n=-\infty }^{\infty }\delta (t-4n)\stackrel{F.T.}{⟷}4\sum _{n=-\infty }^{\infty }\delta (\omega -\frac{n\pi }{2})$