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Given the system/filter $H(\omega)=\frac{1}{5-j\omega}$, find $h(t)$, it's magnitude response and phase and identify what type of filter it is.

Now clearly given it's form, $h(t)=e^{5t}u(-t)$, but I'm confused on finding it's magnitude and phase as every example uses the form $\frac{1}{a+j\omega}$.

Another confusing function in regards to its magnitude, phase and type is $H(\omega)=\pi\delta(\omega)+\frac{1}{j\omega}$, again this is simply $h(t)=u(t)$ but has infinite magnitude at $\omega=0$ with $\omega$ approaching positive and negative infinity the manitude approaches 0 while it's phase is $\frac{\pi}{2}$, correct? Would this be a high-pass filter?

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The response is non-causal so $h(t)$ looks correct as done.

For the frequency response, assuming a non-causal input, don’t use the step function u(t) as the input but use an impulse (as we seek the FT or the impulse response not the step response, and the FT of the impulse response is the freq response you seek ). Since the FT of an impulse is 1, thus reduces to simply finding the magnitude and phase of $H(\omega)$. When $\omega$ is 5 rad/sec the magnitude will be at -3 dB and the phase will be at -45 degrees consistent with the expected 3 dB cutoff for this non-causal first order low pass filter.

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  • $\begingroup$ This is in reference to the FT of the unit function or the first portion? I now understand the first portion is the same as 1/(5+jw), just flipped in the time domain, correct? $\endgroup$ Commented Apr 13, 2022 at 21:00
  • $\begingroup$ Right it is non-causal and I agree with your h(t). The magnitude and phase response is just the magnitude and phase of $H(\omega)$ as you sweep $\omega$, assuming a time reversed input. $\endgroup$ Commented Apr 13, 2022 at 21:12
  • $\begingroup$ Alright, thanks $\endgroup$ Commented Apr 13, 2022 at 21:15
  • $\begingroup$ Thanks so much. With the second portion with the FT of the unit step function, is it possible to classify it as a filter since when w=0, it has infinite magnitude but trends toward 0 as w approaches positive or negative infinity. I was under the assumption that a filter needed some actual value to filter in a high or low pass filter. $\endgroup$ Commented Apr 13, 2022 at 21:18
  • $\begingroup$ It’s non causal so will not converge with any causal input. Consider the set of non-causal sinusoids that start at t=0 and then continue as time goes backwards but are 0 for all positive time. (Or change the sign in the denominator to get a causal 5 rad/sec low pass filter- either way it is a low pass in the frequency domain and we must define our sense of time under which this will converge. $\endgroup$ Commented Apr 13, 2022 at 21:22

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