I've been able to wrap my head around the continuous and discrete Fourier transform just fine. I understand that the Fourier transform brings you from the time domain into frequency domain, and that the Fourier transform is just the Laplace transform but where $\sigma$, the real valued portion of $s = \sigma + j\omega$, is set to $0$. So if the imaginary portion, $\omega$, is the frequency, what does the real $\sigma$ represent?
Furthermore, why is it not like this between the DTFT and the Z transform? The DTFT is a specialized case not where $\sigma=0$, but where $r$ in $z=re^{j\omega}$ is set to $0$, i.e when $|z|=1$. Do the real and imaginary parts of the signal change what they represent in continuous and discrete signals?
Best Answer
I've also been trying to find an intuitive meaning for the sigma parameter, and I believe I found one. What follows is what I've gathered after reading The Scientist and Engineer's Guide to Digital Signal Processing, by Steven W. Smith, Chapter 32, from page 9, paragraph 4 onwards (although I recommend reading the whole section)
The Fourier transform can also be called the frequency response. What we mean by this is: if we feed a sinusoidal signal into the system (i.e. a frequency) what will be the relation between the amplitude of the input and that of the output?
When we switch to Laplace transforms we are getting this information for another kind of input. The Laplace transform of a system informs you of the gain when you introduce a sinusoidal signal of frequency omega multiplied by an decaying exponential of parameter sigma.
Alternatively you can view both the Fourier and Laplace transforms as the coefficients of a decomposition of the original function into a base of sinusoidals or exponentially weighted sinusoidals, respectively. I think the other view, being more related to physics, is more relevant to your question, though.