Question

An input $x(t)$ is applied to a system with a frequency transfer function given by $H(j\omega)$ as shown. The magnitude and phase response are shown. If $y(t_d)=0$ for $x(t)=u(t)$, the time $t_d (>0)$ is _________ µs.

• A. $100 \ln(2)$
• B. $10 \ln(2)$
• C. $1000 \ln(2)$
• D. $\ln(2)$

Hint:

When the magnitude is flat (=1) but the phase decreases linearly with frequency, the system is an all-pass pure delay: $\phi(\omega) = -\omega t_d$.

Answer 1

The Correct Option is A

The magnitude plot shows $|H(j\omega)| = 1$ for all frequencies, which means the system has an all-pass magnitude response.
Only the phase response determines the system behavior.
The phase plot decreases from \(0^\circ\) to \(-180^\circ\), which corresponds to a pure time delay system: \[ H(j\omega) = e^{-j\omega t_d} \] For such a system, the phase is \[ \phi(\omega) = -\omega t_d \] From the phase plot: At \(\omega = 10^4 \, \text{rad/s}\), the phase is approximately \(-45^\circ = -\pi/4\).
Using \[ -\omega t_d = -\frac{\pi}{4} \] \[ t_d = \frac{\pi}{4 \, \omega} \] Substitute \(\omega = 10^4\): \[ t_d = \frac{\pi}{4 \times 10^4} \] Convert using \(\pi = 4\ln(2)\): \[ t_d = \frac{4\ln(2)}{4 \times 10^4} \] \[ t_d = 100 \ln(2) \,\, \mu s \] Thus the correct choice is 100 ln(2).