Question

An ideal low pass filter has frequency response given by
\[ H(j\omega) = \begin{cases} 1, & |\omega| \leq 200\pi \\ 0, & \text{otherwise} \end{cases} \] Let \( h(t) \) be its time domain representation. Then \( h(0) = \underline{\hspace{2cm}} \) (round off to the nearest integer). 

Hint:

The impulse response of an ideal low-pass filter with cutoff frequency \( \omega_c \) is a sinc function. Its value at \( t = 0 \) is equal to \( \frac{\omega_c}{\pi} \), since \( h(0) = \frac{1}{2\pi} \cdot 2\omega_c = \frac{\omega_c}{\pi} \).

Answer 1

Correct Answer: 200

We are given the frequency response \( H(j\omega) \) of an ideal low pass filter. This is a rectangular function in frequency domain, so its time domain response is a sinc function. \[ h(t) = \frac{1}{2\pi} \int_{-200\pi}^{200\pi} e^{j\omega t} d\omega \] Step 1: Evaluate the inverse Fourier transform at \( t = 0 \): \[ h(0) = \frac{1}{2\pi} \int_{-200\pi}^{200\pi} 1 \cdot d\omega = \frac{1}{2\pi} \cdot (200\pi - (-200\pi)) = \frac{1}{2\pi} \cdot 400\pi = \boxed{200} \] Hence, \[ \boxed{h(0) = 200} \]