0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The Laplace transform is the Fourier transform on the imaginary axis. The Z transform is engineers' jargon for the ordinary generating function. $\endgroup$ – Lord Shark the Unknown Apr 26 '17 at 19:13
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.