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Say I learned the theoretical result of continuous time Fourier transform. And I want to extends some results(say "convolution rule") to

Lapace transform, Z transform, DTFT, DFT, Fourier sequence etc.

Then really, I just need to remember how one works and derive the rest. How to do this as a general principle? Is there any textbook that teaches this way? Thanks.

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    $\begingroup$ really, the bilateral Laplace Transform is the most general and the Fourier Transform is a special case of that where $s=j\omega$. but you have the right idea. Z Transform is merely the Laplace Transform of the sampled (with a Dirac comb) continuous-time function. The DTFT can be viewed two equivalent ways: either the Z Transform evaluated at $z=e^{j \omega}$ or the Fourier Transform of the sampled continuous-time function. The DFT is either the DTFT evaluated at $z=e^{j 2 \pi k/N}$ or the FT of the sampled and periodically extended input. $\endgroup$ – robert bristow-johnson Mar 20 '15 at 20:08
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    $\begingroup$ @robertbristow-johnson: The idea that the bilateral Laplace transform is the most general, and that the Fourier transform is just a special case is quite misleading. E.g., given a sinc function in the time domain, what does the bilateral LT tell us about it? Nothing, because it doesn't even exist. The FT, on the other hand, is a nice ideal lowpass response, exactly what it should be. $\endgroup$ – Matt L. Mar 21 '15 at 11:32
  • $\begingroup$ The Fourier transform is a very general idea in mathematics, you can recreate the $L^2$ theory with very minimal assumptions on the underlying space. amazon.com/Fourier-Analysis-Groups-Interscience-Mathematics/dp/… $\endgroup$ – dls Mar 22 '15 at 3:59
  • $\begingroup$ well, @MattL., "misleading" is in the eye of the beholder. you can't plug $\operatorname{sinc}(t)$ directly into the F.T. and get the $\operatorname{rect}(f)$ coming out. you have to be a little circuitous about it (use F.T. duality). but if the function is equivalently defined for both F.T. and (bilateral) L.T., if you can transform it with L.T., you can always transform it to F.T. by substituting $s \leftarrow j \omega = j 2 \pi f$. but not always the other way around. at least, not directly. e.g.: the unit step function. $\endgroup$ – robert bristow-johnson Mar 22 '15 at 21:20
  • $\begingroup$ @robertbristow-johnson: I agree that you need to know some tricks to make the FT work for idealized filter responses and some other useful functions, but that's no real problem of the FT. It is a fact that is often overlooked that the FT can do things that you can't do with the LT (such as impulse responses of idealized filters or periodic functions). As for your last statement, using $s=j\omega$ to go from bilateral LT to FT only works if the imaginary axis is inside the ROC, otherwise it doesn't. $\endgroup$ – Matt L. Mar 22 '15 at 21:40
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Yes you can, somehow. But that won't be the easiest road.

When teaching basic DSP, I often see students afraid to "learn all the formulae" and properties pertaining to the Fourier transform, Fourier series, discrete-time Fourier transform, $z$-transform and the Discrete Fourier transform with the fast Fourier transformation or FFT, up to time-frequency transformations.

They share a lot of common traits: energy conservation, sorts of orthogonality, scalar product formulae, convolution turning into multiplication, Parseval-Plancherel formulae, etc. but also important (and often subtle) differences. . One tool to unite such concepts is sometimes called the Pontryagin duality. For a basic overview, the wiki page is ok:

Pontryagin duality places in a unified context a number of observations about functions on the real line or on finite abelian groups

For instance, you can view the four primal and dual domains for Fourier "transformations" as in the table:

\begin{array}{lcc} \textrm{Transform} & \textrm{Original domain}& \textrm{Transform domain}\\ \textrm{Fourier transform} & \mathbb {R} &\mathbb {R} \\ \textrm{Fourier series}& \mathbb {T} & \mathbb {Z} \\ \textrm{Discrete-time Fourier transform (DTFT)} & \mathbb {Z} & \mathbb {T} \\ \textrm{Discrete Fourier transform (DFT)}& {\displaystyle \mathbb {Z} /(n)} & {\displaystyle \mathbb {Z} /(n)} \end{array}

Its provides several ways to derive properties from one domain to another. But this tool dives deep into complicated concepts such as category theory, see for instance Notes 2: The Fourier transform (Terry Tao) or Category theory applied to Pontryagin duality.

Knowing properties about one transformation, you can derive properties of others, but for that you will have to master some advanced mathematics. There is no free lunch, unfortunately.

You would better learn how to remember a few formulae for a handful of transformations

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If you really want to do that, study the Laplace transform because:

  1. Z-Transform can be viewed as a mapping with $z = e^{s/f_s}$, where $f_s$ is the sample frequency.
  2. Fourier transform (and sequence) is the a special case when $s = j\omega$ (although you can't calculate the transform by just substituting UNLESS the ROC of the Laplace Transform encompasses the imaginary axis.)
  3. DTFT has an analogous relation to the Z-Transform that the Fourier Transform has to the Laplace Transform (evaluated on $ z = e^{j\omega}$)
  4. DFT is a "frequency domain sampled version" of the DTFT

But I don't recommend it.

Also, if you really want to understand everything in a general framework, study Linear Algebra. Since all of this is nothing more than projections on to vector spaces (or function spaces, if you will).

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Probably it would be more suited as a comment than a full answer but I don't have enough points yet.

I personally found very instructive the book from E. O. Brigham, The Fast Fourier Transform.

Doesn't cover Laplace and Z, but IMHO it explains really well how the DTFT and DFT stem from continuous Fourier Transform when you introduce sampling, just using continuous math, convolutions and distribution theory.

It also does this using a pictorial, easy to understand and remember approach.

Definitely not enough to cover everything, but I believe it's a really good starting point.

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    $\begingroup$ Love anonymous downvoters who don't have 30 seconds to spend to explain their reasons, especially when the other end is a low reputation newcomer. $\endgroup$ – filippo Jan 25 '16 at 12:36

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