# Fourier Transformations

1. Jul 9, 2011

### EngWiPy

Hi,

I am confused about a lot of Fourier transformations:

1. A Fourier transform with variable $$f$$
2. A Fourier transform with variable $$e^{jw}$$
3. Fourier series

What is the difference between these different Fouriers?

Another thing, when does the convolution in the time domain become a multiplication in the frequency domain in the above Fouriers? I read that there are some conditions for that.

2. Jul 10, 2011

### pmsrw3

1. Commonly one takes a Fourier transform of a function of space or time: f(x) or f(t). You get out of that a function F(omega) of frequency. (In the case of a function of space, it's a spatial frequency.)

2. This doesn't look right to me.

3. You use a Fourier transform (an integral) when the function you're transforming covers all space or all time (-inf, +inf). When you have a restricted domain, e.g. [0, 1], you can use a Fourier series.

For your final question, see http://mathworld.wolfram.com/ConvolutionTheorem.html .

3. Jul 10, 2011

### EngWiPy

The second one, that does not seem right to you, is encountered especially in signal processing. About the convolution, what I meant is, I heard once, for example, in discrete convolution such as:

$$y[n]=\sum_{k=-\infty}^{\infty}h[k]x[n-k]$$

the bounds must be infinity such that it equals to the product of their transforms, i.e.:

$$Y[n]=H[n]X[n]$$

Otherwise at least one of the sequences must be periodic.

Regards

4. Jul 10, 2011

### pmsrw3

"A Fourier transform with variable $e^{jw}$" is encountered frequently in signal processing? Really? Give me an example.

Now, if you had said "A Fourier transform with kernel $e^{j \omega t}$", I would have some idea what you were talking about.

You can get around these limitations by doing cyclic convolutions and/or by padding the finite sequences in various clever ways. What I mean by a cyclic convolution is that, if you have a sequence h[k] that goes from k=0 to N-1, say, you pretend it is an infinite sequence whose value at a general K is h[K mod N]. That is, you just pretend it repeats forever to +-infinity. Combining this trick with appending a string of zeroes to one or both ends (padding), then throwing out the appropriate part of the result, you can do most of the finite convolutions you would want to.

5. Jul 10, 2011

### EngWiPy

For example it is said that the frequency response of a discrete time system is:

$$H(e^{jw})=\sum_{k=-\infty}^{\infty}h[k]e^{-jwk}$$

6. Jul 10, 2011

### pmsrw3

OK. But this is not the Fourier transform of $e^{jw}$. It's the Fourier transform (or a series, to be precise, but ignore that) of h[k].

7. Jul 10, 2011

### EngWiPy

What I meant is that the Fourier transform is a function of $$e^{jw}$$, why?

8. Jul 10, 2011

### pmsrw3

Well, why not? Any function of frequency can also be expressed as a function of something that is related to frequency. You could, for instance, equate a function of $\log\omega$ to the FT. As I understand from the Wikipedia article, LTI engineers find it convenient to use something called a Z-transform, related to Laplace and Fourier transforms, probably because in a discrete time system it just becomes a power series in z. The relationship is z = $e^{j\omega}$, so since they're in the habit of writing their Z-transform as H(z), they find it convenient, when they're thinking in terms of frequency response, to express the FT as $H(e^{j\omega})$ instead of the more usual $H(\omega)$. Maybe there's a fundamental reason for doing it that way, or maybe they're just used to it. I would bet on the latter.

9. Jul 10, 2011

### EngWiPy

You are right, in digital signal processing z-transform is used frequently, which is basically a generalization of the Fourier transform of a discrete signal. In particular, z-transform reduces to Fourier transform on the unit circle on the complex plan, i.e.: |z|=1, where z=|z|exp(jA).

why does the Fourier transform for some discrete signals yield a continuous function, while in DFT it yields a sequence?

10. Jul 10, 2011

### pmsrw3

The FT on a finite domain, or of a periodic function on an infinite domain, yields a series. The FT of an aperiodic function on a infinite domain yields a continuous function.

11. Jul 10, 2011

### pwsnafu

The "why" question can only answered in a course on locally compact abelian groups (which includes the real numbers, the integers, the unit circle and roots of unity as special cases).

The Pontryagin duality theorem states: the dual of a compact abelian group is a discrete abelian group. And vice-versa. This means that if the domain of f is a compact abelian group (ie f is periodic), then the Fourier transform of f is defined on a discrete group (ie series).

Now with the DFT, the domain of f is both compact and discrete, so applying the theorem twice we get the transform is also compact and discrete (ie finite series).