Complex frequency of a pole
If we have the transfer function of a LC high pass filter:
$H(s) = \frac{sL}{sL+\frac{1}{sC}}$
If we want to find the pole of that filter in the end we get:
$s = \frac{j}{\sqrt{LC}}$
and for a sinuisodal input signal $s = j\omega$ the pole exists at the resonant frequency $\omega_{r}$
However if we dont apply a sinuisodal signal at the input s may become a complex number->the frequency of the pole may be complex.But what physical meaning does it have?
3 answers
Your question doesn't make a lot of sense.
for a sinusoidal input signal s = jω the pole exists at the resonant frequency ωr.
Actually the pole exists regardless of what the input signal is. The transfer function is a description of what happens with any* input signal.
However if we don't apply a sinusoidal signal at the input s may become a complex number->the frequency of the pole may be complex.
This is the same misconception. The pole is part of the system. It is always there. It has nothing to do with what input might be thrown at that system. The pole doesn't change as a function of the input. The only thing that changes as a function of the input signal is the output signal.
* Any real world signal. Technically, any analytic signal, which means that it is infinitely differentiable. That in turn means the signal or any of its derivatives can't jump instantly. Real world signal always have some finite bandwidth, so can't jump instantly. Such real world signals can then be decomposed into a set of sines. Since we're talking about linear systems, contributions from multiple superimposed input signals can be analyzed separately and their results added to determine the output signal. Therefore, an input signal can be decomposed into a sum of sines, the result of each sine computed separately, then those results added to find the overall output signal. (Credit to user LvW for pointing out "any" should be qualified).
This is mostly another view on @Olin's answer.
For different input signals the frequency of the pole will have different values ,the pole as a pole exists for some specific values of L,C s but the frequency of the pole changes.
The poles of an LTI system do not change. Since you're giving an example of a highpass, look at it from this perspective: all the values are given by the consituent inductors and capacitors. If those do not change over time then the system has fixed values. If not, the system is no longer LTI. Also, the pole in your example is at $\pm j/\sqrt{LC}$, since it's a quadratic.
0 comment threads
if we don't apply a sinusoidal signal at the input s may become a complex number
$$$$
All real-world signals exist only on the $j\omega$ axis.
Consider that a complicated signal can be broken down into a fundamental sinewave and its harmonics. The harmonics are still sinusoidal and, for a linear filter circuit, they can be treated as individual sinewaves.
0 comment threads
0 comment threads