An ideal low pass filter has frequency response given by
$H(j\omega) = \begin{cases} 1, & |\omega| \le 200\pi \\ 0, & \text{otherwise} \end{cases}$
Let $h(t)$ be its time domain representation. Then $h(0) = $ ___________ (round off to the nearest integer)
Ideal Low Pass Filter Impulse Response Calculation
The problem asks for the time-domain impulse response $h(t)$ evaluated at $t=0$ for an ideal low-pass filter whose frequency response $H(j\omega)$ is given.
The frequency response is defined as:
$ H(j\omega) = \begin{cases} 1, & |\omega| \le 200\pi \\ 0, & \text{otherwise} \end{cases} $This represents an ideal low-pass filter with a cutoff angular frequency $\omega_c = 200\pi$ rad/s.
Calculating Time Domain Representation h(t)
The time-domain impulse response $h(t)$ is the inverse Fourier Transform (IFT) of the frequency response $H(j\omega)$. The formula for the IFT is:
$ h(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} H(j\omega) e^{j\omega t} d\omega $Substituting the given $H(j\omega)$, the integral becomes:
$ h(t) = \frac{1}{2\pi} \int_{-200\pi}^{200\pi} (1) e^{j\omega t} d\omega $Evaluating h(0)
To find $h(0)$, we substitute $t=0$ into the integral:
$ h(0) = \frac{1}{2\pi} \int_{-200\pi}^{200\pi} e^{j\omega (0)} d\omega $ $ h(0) = \frac{1}{2\pi} \int_{-200\pi}^{200\pi} 1 d\omega $Now, we evaluate the definite integral:
$ h(0) = \frac{1}{2\pi} [\omega]_{-200\pi}^{200\pi} $ $ h(0) = \frac{1}{2\pi} (200\pi - (-200\pi)) $ $ h(0) = \frac{1}{2\pi} (400\pi) $ $ h(0) = 200 $Final Result and Rounding
The calculated value for $h(0)$ is exactly 200.
The question asks to round the result to the nearest integer. Since 200 is already an integer, rounding it gives 200.
Therefore, $h(0) = 200$.
Important Questions from Introduction To Signals and Systems
All energy signals will have an average Power of
Arrange the following system types in ascending order of their intersection with 0 dB axis.
A. Type N system
B. Type 3 system
C. Type 2 system
D. Type 4 system
Choose the correct answer from the options given below
Given below are two statements
Statement I: The scaling theorem relates scale changes in frequency domain to the consequent changes in scale in the time domain.
Statement II: The scaling theorem is not applicable to an impulse signal.
In light of the above statements, choose the correct answer from the options given below
The period of the signal X(t) = cos (2 πt) cos (4 πt) – sin (2πt) sin (4πt) is
The time period of a 50 kHz sine wave is
