What Are the Differences Between Various Fourier Transformations?

In summary, the conversation discusses the differences between Fourier transforms with variable f, e^{jw}, and Fourier series. It also touches on the conditions for when convolution in the time domain becomes multiplication in the frequency domain for these transforms. The conversation also mentions the use of z-transforms in signal processing and the relationship between z-transforms and Fourier transforms. Finally, the question is raised about why Fourier transforms of some discrete signals yield a continuous function while in discrete Fourier transform it yields a discrete function.
  • #1
EngWiPy
1,368
61
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
 
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  • #2
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
pmsrw3 said:
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
S_David said:
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
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
S_David said:
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
pmsrw3 said:
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
S_David said:
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
pmsrw3 said:
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
S_David said:
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
S_David said:
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).
 

Related to What Are the Differences Between Various Fourier Transformations?

1. What is a Fourier Transformation?

A Fourier Transformation is a mathematical technique used to decompose a complex signal into its individual frequency components. It converts a signal from its original time or spatial domain to a representation in the frequency domain.

2. How does a Fourier Transformation work?

A Fourier Transformation works by breaking down a signal into its individual frequency components through a series of mathematical operations. It involves converting a signal from its original domain to a representation in the frequency domain, where the signal can be expressed as a sum of sine and cosine waves with different amplitudes and frequencies.

3. What is the purpose of using a Fourier Transformation?

The purpose of using a Fourier Transformation is to analyze signals and understand their frequency components. It is commonly used in fields such as signal processing, image processing, and data analysis to extract relevant information from a complex signal. It is also used for solving differential equations and filtering out noise from a signal.

4. What are some applications of Fourier Transformations?

Fourier Transformations have a wide range of applications in various fields such as engineering, physics, and mathematics. Some common applications include image and audio compression, spectral analysis, pattern recognition, and signal filtering.

5. What are the differences between Fourier Transformations and Fourier Series?

The main difference between Fourier Transformations and Fourier Series is that a Fourier Transformation is used for continuous signals, while a Fourier Series is used for periodic signals. A Fourier Series decomposes a signal into a series of sine and cosine functions with different amplitudes and frequencies, while a Fourier Transformation converts a signal from its original domain to a representation in the frequency domain.

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