2
$\begingroup$

I understand the Fourier transform pretty well, and that I'm finding the frequency domain of a signal in the time domain.

I also understand that the Fourier transform is a special case of the Laplace transform where $\sigma=0$ for $s=\sigma+j\omega$

If $\omega$ represents the frequency of the sinusoids that make up a time signal, then what does $\sigma$ represent?

Furthermore, why does that relationship change in the Z transform? The DTFT is a special case of the Z transform not when the real portion is zero, but when $r=1$ in $z=re^{j\omega}$, or when $|z|=1$. So in this case, what does the radius represent?

And what does one tell us over the other? If the Fourier transform contains all the information we need about a signal, why bother with Laplace or Z? If Z contains more information, why do we bother with Fourier?

$\endgroup$
1
  • $\begingroup$ The Laplace transform is the Fourier transform on the imaginary axis. The Z transform is engineers' jargon for the ordinary generating function. $\endgroup$ Commented Apr 26, 2017 at 19:13

1 Answer 1

2
$\begingroup$

A signal $x(t)$ has a Fourier transform only if it is absolutely integrable, that is if $$\int_{-\infty}^{\infty} |x(t)| \mathrm{d}t < \infty.$$ Let's look at the definition of the Laplace transform now. The Laplace transfrom $H(s)$ of $x(t)$ is given by $$H(s) = \int_{-\infty}^{\infty} x(t)e^{-st}\mathrm{d}t.$$ Because $s = \sigma + j\omega$, this can be written as $$H(s) = \int_{-\infty}^{\infty} x(t)e^{-\sigma t}e^{-j\omega t}\mathrm{d}t.$$ What does it tell us? This indicates that the Laplace transform of $x(t)$ is nothing else that the Fourier transform of $x(t)e^{-\sigma t}$. For this Fourier transform, the necessary convergence condition becomes $$\int_{-\infty}^{\infty} |x(t)e^{-\sigma t}|\mathrm{d}t.$$ Even if $x(t)$ is not absolutely integrable, there exists a range of values of $\sigma$ for which $x(t)e^{-\sigma t}$ is absolutely integrable, this range is called the region of convergence (ROC). Hence, the Laplace transform exists for signals that do not have a Fourier transform. This should answer your question about the meaning of $\sigma$ and why do we have to bother about Laplace.

An identical reasoning for the z-transform will lead you to the same conclusion.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .