Cheers, I have a response function of: $$H(\omega) = \begin{cases} \frac{1}{2}j\omega, |\omega| < \omega_c \\ 0, \text{elsewhere}\end{cases},$$ and I am asked to find draw its magnitude and phase. For the phase function, my professor rewrote this as: $H(\omega)= \frac{1}{2}\omega e^{j\frac{\pi}{2}} \Pi(\frac{\omega }{2 \omega_c})$. From that he concluded that it should be drawn as $-\frac{\pi}{2}, -\omega_c<\omega < 0$ and $\frac{\pi}{2}, 0 < \omega < \omega_c$, but I don't see why that's the case. I see that the phase is steady and not dependent on $\omega$, so why would it get affected by $\omega$ and turn negative, before 0? Thanks for any help.
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asked Feb 10, 2022 at 22:19
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1 Answer
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Consider a frequency $\omega_0$ such that $-\omega_c < \omega_0 < 0$. We can write $\omega_0 = |\omega_0|e^{j\pi}$; in other words, the phase of a negative number is $\pi$.
Then, we can write $$H(\omega_0) = 0.5|\omega_0|e^{j\pi}e^{j\pi/2} = 0.5|\omega_0|e^{j3\pi/2} = 0.5|\omega_0|e^{-j\pi/2}.$$ So, the phase of $H(\omega_0)$ is $-\pi/2$.
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$\begingroup$ Thanks for your answer. Would there be a way for me to find its phase with some calculations starting from what I am starting? Because you used an assumption, but in a test for example, I wouldn't know that before hand. Thanks again =) $\endgroup$ Commented Feb 10, 2022 at 22:54
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1$\begingroup$ The key strategy is to write $H(\omega)$ as $|H(\omega)| e^{\phi(\omega)}$, where $|H(\omega)|$ is always positive. In your case, the sign depends on $\omega$, so when $\omega<0$, you must replace it with $|\omega|e^{j\pi}$. $\endgroup$– MBazCommented Feb 10, 2022 at 23:02