Fourier Transform vs z Transform

by fayaazhussain
Tags: fourier transform, signal processing, z transform
 P: 12 Both fourier transform and z transform can convert discrete time domain to frequency spectrum domain. Then why do we use fourier transform rather than z transform? What is the reason behind it? Both give us the frequency spectrum we want.
 P: 951 Fourier transform is concentrated and was originally made for continuous functions. Z-transform works better that Fourier transform in discrete systems. In fact, what is Fourier transform for continuous systems, that is z-transform for discrete systems.
 P: 12 Then what about Discrete Fourier Transform? How is it different from z transform?
P: 951

Fourier Transform vs z Transform

 Quote by fayaazhussain Then what about Discrete Fourier Transform? How is it different from z transform?
How will you do continuous Fourier transform on a computer? You need to discrete it. That is why DFT came into place.

In DFT you get from discrete input discrete spectre.

In z transfrom you take discrete signal(input) and translate it into complex continuous function. At least thats what I know for now. I am too currently studying z transform.

And z transform is powerful for manipulating discrete signals. If you are studying z transform, this should be clear.
 P: 9 First of all, seeing we're obviously discussing discrete signals let's make a few things clear. The Fourier Transform of a 1D signal can be defined over $\mathbb{R}$, unlike the Discrete Fourier Transform which results in a discrete function. On the other hand, the Z-Transform is a function defined on the complex plane. Your question is actually very pertinent. The Fourier-Transform of a discrete signal, if it exists, is its own Z-Transform evaluated at $z=\mathbb{e}^{j w}$. On the other hand, the DFT of a signal of length N is simply the sampling of its Z-Transform in the same unit circle as the Fourier Transform. This however, doesn't make the DTFT our the DFT useless. The information content of the frequency exists in the Fourer Transform and not in all of the Z-Transform, so if you want to study the frequency response of a signal, the DTFT is all you need. However, a problem arises: how can you store a DTFT of a signal on a computer if it has infinite entries? This is where the discrete fourier transform is useful, but the signal has to be of limited length. This is not the only reason to use the DFT, however. For these reasons, the Z-Transform is actually rarely used in favour of the other two. However, it still has some uses due to it's relationship with Laurent Series. Using the residue theorem, you can compute certain instances of a signal based on its Z-Transform, but this is generally not very useful.
 P: 2,237 I might suggest taking this question to comp.dsp. DTFT, DFT are all either specific or restricted versions of the Fourier transform regarding sampled signals or sampled signals with finite length, and the Z-transform, Laplace Transform are generalizations of the the DTFT or FT respectively. they're all the same thing, with different degrees of focus.
 HW Helper P: 6,164 That is what I agree with. Mathematically they are all equivalent. The DFT, DTFT, continuous FT, and Fourier series are all variations of the theorem that says that functions can be written as the sum of a set of sines and cosines. The sines and cosines form a so called "basis" of the vector space of functions. The DTFT is equivalent to the z-transform. The z-transform could be (but isn't as far as I know) mathematically extended to match the other flavors of Fourier. As I see it, it's matter of convention (and what is most intuitive) in which area they are applied.
 P: 5,462 The purpose of all transforms is to reduce the mathematical description to a simpler algebraic expression. The z transform is no exception. Use of the Laplace transform in the s plane representation for frequency analysis of sampled (digital or discrete) data is made difficult by the the need for infinite polynomials with infinite numbers of poles/zeros. The z plane and transform reduces this to a finite (and therfore manageable) number.
HW Helper
P: 6,164
 Quote by Studiot The purpose of all transforms is to reduce the mathematical description to a simpler algebraic expression.
Exactly!

 Quote by Studiot Use of the Laplace transform in the s plane representation for frequency analysis of sampled (digital or discrete) data is made difficult by the the need for infinite polynomials with infinite numbers of poles/zeros.
The Laplace transform is equivalent to the (continuous) Fourier transform.
(It is identical to the one-sided Fourier transform with just a different choice of frequency variable.)

Both are badly suited for discrete signals, because, as you say, they yield expressions that are hard to manage then.
P: 2,237
 Quote by I like Serena The Laplace transform is equivalent to the (continuous) Fourier transform. (It is identical to the one-sided Fourier transform with just a different choice of frequency variable.)
there is a double-sided Laplace transform that is identical with the continuous Fourier transform with the substitution of $s = j \omega = j 2 \pi f \$. the only difference with the single-sided LT is that the integral from $-\infty$ to 0 in the double-sided becomes the initial conditions in the single-sided.

 Both are badly suited for discrete signals, because, as you say, they yield expressions that are hard to manage then.
not too bad. so it's $z=e^{sT}$ instead of s. it's just that discrete-time systems are built out of adders, scalers, and delay elements. so it's adding signals (their transforms add), scaling signals by a constant (same for the transforms of the signals) and delays (that's scaling by $z^{-1}=e^{-sT}$). so the transfer functions of such a system is a rational function of z.

for continuous-time systems, they're built outa adders (transforms also add), scalers (no big deal), and integrators (scale by $s^{-1}$). then the transfer function of these systems are rational functions of s.

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