Fourier Transformations

  • Thread starter EngWiPy
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  • #1
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Hi,

I am confused about a lot of Fourier transformations:

  1. A Fourier transform with variable [tex]f[/tex]
  2. A Fourier transform with variable [tex]e^{jw}[/tex]
  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.

Thanks in advance
 

Answers and Replies

  • #2
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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
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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 .
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:

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

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

[tex]Y[n]=H[n]X[n][/tex]

Otherwise at least one of the sequences must be periodic.

Regards
 
  • #4
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The second one, that does not seem right to you, is encountered especially in signal processing.
"A Fourier transform with variable [itex]e^{jw}[/itex]" is encountered frequently in signal processing? Really? Give me an example.

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

About the convolution, what I meant is, I heard once, for example, in discrete convolution such as:

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

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

[tex]Y[n]=H[n]X[n][/tex]

Otherwise at least one of the sequences must be periodic.
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
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For example it is said that the frequency response of a discrete time system is:

[tex]H(e^{jw})=\sum_{k=-\infty}^{\infty}h[k]e^{-jwk}[/tex]
 
  • #6
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For example it is said that the frequency response of a discrete time system is:

[tex]H(e^{jw})=\sum_{k=-\infty}^{\infty}h[k]e^{-jwk}[/tex]
OK. But this is not the Fourier transform of [itex]e^{jw}[/itex]. It's the Fourier transform (or a series, to be precise, but ignore that) of h[k].
 
  • #7
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OK. But this is not the Fourier transform of [itex]e^{jw}[/itex]. It's the Fourier transform (or a series, to be precise, but ignore that) of h[k].
What I meant is that the Fourier transform is a function of [tex]e^{jw}[/tex], why?
 
  • #8
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What I meant is that the Fourier transform is a function of [tex]e^{jw}[/tex], why?
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 [itex]\log\omega[/itex] 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 = [itex]e^{j\omega}[/itex], 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 [itex]H(e^{j\omega})[/itex] instead of the more usual [itex]H(\omega)[/itex]. 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
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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 [itex]\log\omega[/itex] 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 = [itex]e^{j\omega}[/itex], 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 [itex]H(e^{j\omega})[/itex] instead of the more usual [itex]H(\omega)[/itex]. 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.
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
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why does the Fourier transform for some discrete signals yield a continuous function, while in DFT it yields a sequence?
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
pwsnafu
Science Advisor
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why does the Fourier transform for some discrete signals yield a continuous function, while in DFT it yields a sequence?
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).
 

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