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Why in these type of 3D plots does the gain value (z axis) come back down from infinity for sigma values that are greater than the sigma value of a pole?

Low pass filter response

I thought that these plots are the value of the a systems impulse response multiplied by a probing waveform (exponentially decaying or growing sine wave for all values of sigma and jw)

A pole is when the probing waveform is one that grows at exactly the same rate as the the impulse response is decaying. And the multiple of the 2 waveforms produces a waveform that is has an infinite area.

Pole pole

If the sigma value is then increased the area of the resulting waveform will be greater than infinity (undefined, in a region of non convergence). So why does the value go back down to finite values in this graph?

Sigma greater than pole undefined

John Spence
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  • having exactly the same question, still waiting for a good answer. This question is asking: in the region of divergence in the laplace transform (a sigma on the left of the pole), fourier transform at that sigma is no longer defined, how can we still have a shape in the 3D plot? – eliu Apr 03 '23 at 13:27

2 Answers2

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A "pole" is defined mathematically by an equation whose denominator becomes zero at a particular or several values of s; where s = \$\sigma +j\omega\$. A particular example is a 2nd order function like this: -

enter image description here

This happens to be a low pass filter and, if you set the denominator to zero and solved it you would find one or two values for s that tell you where the response is infinite. At all other values of s the response is non infinite and clearly forms the peaking shape in your question.

The shape has nothing to do with the driving waveform (probing waveform?)

Here's a picture that might help: -

enter image description here

It assumes a 2nd order low pass filter and has solutions for s shown in the bottom right picture.

Andy aka
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  • Is the 3D plot not the solution to the Laplace transform of the impulse response of a low pass filter? I thought that there was a region of non-convergence where the solution becomes greater than infinity (solution does not exist). Mentioned here if you scroll down a bit http://www.ece.utah.edu/~ece3500/notes/class15.html – John Spence Feb 06 '16 at 12:16
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not really an answer, but I am tormented by the same question you had. My own personal guess of what happened:

first, I share your exact understanding of what lapace transform is trying to do, the sigma probing and taking fourier transform for every sigma. In the region of divergence there shouldn't be any fourier transform, I thought the graph should be cut off as well.

let's use the simple example of 1 / (s + 1) I believe what happened is that, the plot you see is not ploting the Laplace transform, but it is ploting the F(S) itself, which is simply 1 / (x + jy + 1).

in our shared understanding, when sigma is -2, fourier transform shouldn't exist. but 1 / (-2 + jy + 1) exists quite fine, and we can plot it quite fine too.

eliu
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