Electrical Engineering Asked by Phae Leung on November 4, 2020
I hope I can help solve a little bit of confusion here.
$$H(s=-omega_1) = frac{1}{-omega_1 + omega_1} rightarrow infty$$
$$H(s) = frac{1}{s^2 + omega^2_1}$$
You just have to input s to be $$s = pm jomega_1$$ and that would yield:
$$H(s=pm jomega_1) = frac{1}{(pm jomega_1)^2 + omega^2_1} = frac{1}{-omega^2_1 + omega^2_1} rightarrow infty$$
So as you see in the second case you have two complex poles that are on the imaginary axis of the s plane and the system is border line stable. But it is not a real freaquency. Great job though I think you understand it well but maybe this will solve some confussion.
Answered by Nikola Vulinovic on November 4, 2020
You are mixing Laplace-domain transfer functions with Fourier-domain transfer functions.
Either you start with the Laplace version: $$H(s) = frac{omega_1}{s + omega_1}$$ and the system response to a sinusoid $x(t) = cos omega t$ is found by evaluating $H(s)$ at $s = jomega$.
Or you use the somewhat more obscure Fourier version: $$H(omega) = frac{omega_1}{jomega + omega_1}$$ and the system response to a sinusoid $x(t) = cos omega t$ is found by evaluating $H(omega)$ at, well, $omega$.
There's some mathematical 'i's and 't's that get dotted and crossed differently with Laplace vs. Fourier, so there's a few cases where one is valid and the other isn't. Mostly, if you have to ask, they're equivalent ways of saying the same thing.
Answered by TimWescott on November 4, 2020
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