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