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Fourier Series/transform and eigenvalues

  1. May 28, 2013 #1
    Hello Physics Forums community,

    I'm afraid I really need a hand in understanding Why are the Fourier Series for continuous and periodic signals using diferent notation of the Fourier Series for discrete and periodic Signals.

    I have been following the book " Signals and Systems " by Alan V. Oppenheim, but I find it hard to understand when it comes to:

    why in LTI systems :

    in countinuous time: e^(st)----->H(s)e^(st)
    and in discrete time: z^n ------> H(z)z^n

    and why do we try to make it look like complex exponentials by making:

    in countinuous time: s=jw
    and in discrete time: z=e^(jw)

    Does this has anything to do with eigenvalues or eigenfuctions?

    and so far I put this in my head like:

    Continuous time periodic signals ---> Fourier Series to represent it in a sum of exponentials, wich can be then evaluated in frequency ending up maintaining the coeficients and substituting (Fourier transforming) the complex exponentials sum into delta functions.

    Continuous time aperiodic Signals ----> the Fourier Series ends up being the Fourier transform but extending the period to infinity.

    Discrete time periodic and aperiodic signals ---> trying to establish relation to countinuous signals but my brain its pointing to NULL due to the notation... :)

    So this leaves me wondering... is the Fourier Series "the Fourier Transform of Periodic Signals" Or is the Fourier Series just a way to represent x(t) has a sum of exponentials so it can be easily translated into the sum of dirac delta functions in Frequency?

    Because the Fourier Series is just a tool to represent a function as a sum of exponentials. And the Fourier Transform is a tool to transform the functions in time to frequency. I know its in front of my eyes... but I'm not seeing it. And I want to put it in my head the right way.

    I would be really greatfull,

    Thanks in advance
     
  2. jcsd
  3. May 29, 2013 #2

    Simon Bridge

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    The notation is historical - but it is useful to distinguish discrete from continuous forms.
    Try applying the discrete transform to a continuous signal for example.

    We could make the discrete signal into a continuous one by using dirac delta functions.

    The Fourier transform uses complex exponentials by definition - the s-form is just a way to make them look like they are not complex exponentials (more like Laplace transforms).

    It looks like you are trying to fit Fourier analysis into some previous idea about signals instead of just letting it be itself. It is just changing the domain of your calculations. Some things make more sense in the changed domain than others.
     
  4. May 29, 2013 #3

    pwsnafu

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    Completely correct. Well done. This is studied in the theory of Pontryagin duality. It says if you take something with compact domain*, the transform with be discrete.

    *Remember, a periodic function ##f : \mathbb{R}\to\mathbb{R}## is the same as ##f : \mathbb{T} \to \mathbb{R}## where T is the unit circle.
     
  5. May 29, 2013 #4
    Thanks Alot for the responses!!
    But... One thing I realize is that when you use the Fourier series on a periodic signal in continuous time, you end up having the coeficients and the exponentials and you can perfectly work with just using the coefficients due to the superposition property of the LTI systems but then, why would you still apply a continuous time Fourier transform to it, and get those delta functions?

    Thanks again! :)
     
  6. May 29, 2013 #5
    @pwsnafu Oh boy, I really have to learn alot of math to understand that Pontryagin duality ^^

    I just had my first contact with abelian groups and subgroups and isomorphism, endomorphism and homomorphism (alien life to me)
    Can you give me a hand with a quick understanding what that morphism is?
    Sorry to bother you and take your time,

    Thanks again.
     
    Last edited: May 29, 2013
  7. May 29, 2013 #6

    pwsnafu

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    At it's core, abstract algebra boils down to this: a mathematician looks at a concrete mathematical object (the real numbers, matrices, continuous functions etc) and says "Hmm, that's a useful thing. What other structures have these properties?" and then studies them together. A mathematical structure is just a set with a bunch of operations (usually binary operations).

    Now we can study functions between these structures to see their relationships. But any old function won't work. They don't give us useful information. This is where "structure preservation" comes in. If your domain is a group you want the "group-iness" preserved. That's a group homomorphism. A linear transform preserves vector space-ness. And so on. A "morphism" is a just a term used for "structure preserving function".
     
  8. May 29, 2013 #7
    Wow, sounds really perceptible, so, (obviously, I guess) Fourier transform is a "morphism"( and there is structure preservation otherwise you wouldn't be able to trace back where you came from? and be able not to lose information?), once you can just transform it back to the same place again. But if so, there is pattern and there is a way that takes me back to where I came from. So if I transform a signal x(t) to a Fourier series representation of the signal and although that is enought to give me the frequency information why is there the 't' in the exponentials? if that is already working has the transform, and the transform gets rid of the 't' domain and gets me in the frequency domain, why do we still have to apply the transform again to really get rid of the 't' ?

    Thanks,
    you are helping me alot!
     
  9. May 29, 2013 #8

    pwsnafu

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    There are three "levels" here.

    First is the level of elements: in our case numbers 2, 4, πi etc. These are related to each other by addition, subtraction, etc. This is what the "locally compact abelian group" in the article refers to. Let's pick one of these sets and call it G.

    The second level are functions, specifically functions from the domain (above) to the complex numbers (or reals, but reals are a special case of C anyway). Usually we want the functions to be square integrable (it helps) so ##f \in L^2(G)##.

    The third level is the Fourier transform itself. It takes a function (f from above) and returns a new function, F. However, the domain of F might not be G. For example, if f was periodic, we get a Fourier series hence the domain is now the integers. This new domain is called the "dual of G", and on the Wikipedia page it's written as G^. The Fourier transform is now the morphism ##\mathcal{F} : L^2(G) \to L^2(\hat G)## and preserves the vector space (remember square integrable functions are a vector space).

    Oh, in the Wikipedia article, when they talk about characters and all that? They are talking about the morphism of ##G \to \hat G##. Notice how it sit underneath the Fourier transform. The whole point of Pontryagin theory is to write the Fourier transform in terms of this second morphism. That way, predicting G^ becomes possible. Also the properties of the Fourier transform are related to the properties this other morphism. Eg the reason why Fourier is invertible is because (G^)^ = G.

    Could you write that out in symbols. Where specifically is t?
     
    Last edited: May 29, 2013
  10. Jun 2, 2013 #9

    jasonRF

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    Mr. Albot,

    I just wanted to address this question. It is a standard convention in signals and systems texts, including (at least the first edition of) Oppenheim's "Signals and Systems," for Fourier transforms to be a function of a real argument [itex]\omega[/itex], and Laplace transforms to be a function of a complex argument [itex]s[/itex]. Likewise, for discrete time [itex]\Omega[/itex] is real and [itex]z[/itex] is complex.

    Consider the continuous case. It turns out that you cannot always put [itex]s=j\omega[/itex] as you wrote. I will use the notation
    [tex]
    \begin{eqnarray}
    \hat{F}(\omega) & = & \int_{-\infty}^{\infty} dt\, e^{-j \omega t} \, f(t) \\
    F(s) & = & \int_{-\infty}^{\infty} dt\, e^{- s t} \, f(t)
    \end{eqnarray}
    [/tex]
    to represent Fourier and Laplace transforms, respectively. Recall that the Laplace transform only converges for a region of convergence (ROC) that in general looks like [itex] \sigma_a < Re(s) < \sigma_b [/itex] (that is, a vertical strip in the complex plane). There are three cases:

    Case 1: [itex] \sigma_a < 0 < \sigma_b [/itex], so that the ROC of the Laplace transform includes the imaginary axis. In this case, it is indeed true that [itex]\hat{F}(\omega) = F(j\omega) [/itex] as you wrote. Example: [itex]f(t) = e^{-|t|} \cos(t) [/itex].

    Case 2: Imaginary axis is outside of ROC ([itex] 0 < \sigma_a [/itex] or [itex] \sigma_b < 0[/itex]). In this cases there is no Fourier transform since we are using the convention that [itex]\omega[/itex] is real. Example: [itex] f(t) = e^t [/itex]. This case is one reason why Laplace transforms are used for analyzing systems that may be unstable.

    case 3: [itex]\sigma_a = 0[/itex] or [itex]\sigma_b = 0[/itex]. In this case, [itex]\hat{F}(\omega) \neq F(j\omega) [/itex], even though the Fourier transform exists. HOwever, the Fourier transform seems to include generalized functions for this case. Example: [itex]f(t) = u(t)[/itex] where [itex]u(t)[/itex] is the Heaviside step function. Here, [itex]F(s) = 1/s[/itex] with ROC [itex]Re(s)>0[/itex]. However, [itex]\hat{F}(\omega) = \frac{1}{j\omega} + \pi \delta (\omega)[/itex].

    For the discrete case a similar result holds, where now the Z transform and discrete-time Fourier transform "coincide" if the ROC of the z transform includes the unit circle.

    Fourier is usually used for signals, and Laplace/Z for systems analysis. The main reason is the Laplace/Z will exist even for unstable things, even when Fourier doesn't.

    Again, this is somewhat due to the "signals and systems" convention. Especially if you look at literature/books outside of electrical engineering, you may find folks using Fourier transforms with a complex argument.

    Hope that helped a little!

    jason
     
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