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## Main Question or Discussion Point

How does it come about that the laplace transform requires that you specify initial conditions whereas the fourier transform does not?

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How does it come about that the laplace transform requires that you specify initial conditions whereas the fourier transform does not?

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HallsofIvy

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If the laplace transform of y(x) is Y(s), what is the Laplace transform of y'(x), in terms of Y(s)?

- #3

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[tex] X(s) = \int_{0}^{+\infty} x(t) e^{-st} dt [/tex]

vs. the

[tex] X(s) = \int_{-\infty}^{+\infty} x(t) e^{-st} dt [/tex]

which is

[tex] X(s) = \int_{-\infty}^{0} x(t) e^{-st} dt \ + \ \int_{0}^{+\infty} x(t) e^{-st} dt [/tex]

it turns out that the "left side" of the double-sided L.T. contains the same information as the initial conditions at

i don't think i've ever seen the F.T. expressed in a single-sided expression. the F.T. can be considered a less general expression or a "special case" of the double-sided L.T.

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Okay, I think that most answers my question. A couple of related follow ups.

1.Take the harmonic oscillator as an example.

[tex]

\frac{d^2y}{dt^2} + \frac{k}{m} y = 0

[/tex]

I know how to solve this given initial conditions [tex] y(0) [/tex] and [tex] y'(0) [/tex], yielding

[tex]

Y(s) = \frac{sy(0) + y'(0)}{s^2 + \frac{k}{m}}

[/tex]

When I try to do a similar solution using fourier transforms, i get

[tex]

(i\omega)^2 Y(\omega) + \frac{k}{m} Y(\omega) = 0

[/tex]

which I can't seem to make sense of. What am i doing wrong?

2.

I can see how the fourier transform is a special case of the laplace transform, where [tex] s = a + bi [/tex]. What do we really lose when we set a = 0 for the fourier transform? What generality is lost, I can't really see it.

Hope what I wrote is correct, did it quickly, thanks!

1.Take the harmonic oscillator as an example.

[tex]

\frac{d^2y}{dt^2} + \frac{k}{m} y = 0

[/tex]

I know how to solve this given initial conditions [tex] y(0) [/tex] and [tex] y'(0) [/tex], yielding

[tex]

Y(s) = \frac{sy(0) + y'(0)}{s^2 + \frac{k}{m}}

[/tex]

When I try to do a similar solution using fourier transforms, i get

[tex]

(i\omega)^2 Y(\omega) + \frac{k}{m} Y(\omega) = 0

[/tex]

which I can't seem to make sense of. What am i doing wrong?

2.

I can see how the fourier transform is a special case of the laplace transform, where [tex] s = a + bi [/tex]. What do we really lose when we set a = 0 for the fourier transform? What generality is lost, I can't really see it.

Hope what I wrote is correct, did it quickly, thanks!

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now, the problem

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Very insightful, thanks!

- #7

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Thats what the information that [tex]\sigma[/tex] in [tex]s=\sigma + j\omega[/tex] contains, the transient behaviour of your system.

Incidentally, this is also the reason why your poles need to be in the left half of the s plane.

Consider only the real part of a pole (the imaginary part basically adds an oscillation)

[tex]\mathcal{L}^{-1}(\frac{1}{s+a})=e^{-at}[/tex] which dies down as

[tex]t\rightarrow\infty[/tex] whereas

[tex]\mathcal{L}^{-1}(\frac{1}{s-a})=e^{at}[/tex] does not.

In other words, the Laplace transform can tell you if your linear, time invariant system is stable or not whereas the Fourier transform can only tell you what steady state behaviour of your system is like.

- #8

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i'm sorry, but i have to completely disagree with this notion. the Fourier Transform (sometimes called the "Fourier Integral") can be and essentiallyThe other thing which the Fourier transform does is to ignore any transients and give only steady state behaviour of your system.

because the inverse F.T. is an integral, not a discrete summation, the adjacent frequency components are infinitesimally close to each other. that means that there is no necessary periodicity (a steady-stateness) in any of the signals that are transformed by the F.T.

because of boundary theorems (was it called "Green's Theorem"?), if you have an analytic function in

- #9

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Would you mind clarifying what you mean here? The Fourier Transform tells us the spectral content of our signal i.e. what frequencies are present. In what way is frequency not periodic? Case in point: we have a rect function in time (which is not periodic). This gives us a sinc function in frequency, it has infinite frequency content. Meaning if you added an infinite number of cosine waves with differential frequencies apart from one another at the amplitude and phase given by the F.T then you would have a rect function in time exactly.because the inverse F.T. is an integral, not a discrete summation, the adjacent frequency components are infinitesimally close to each other. that means that there is no necessary periodicity (a steady-stateness) in any of the signals that are transformed by the F.T.

Green's Theorem is about relating the curl of a vector field in the plane to a closed path integral around the domains boundry. I'd be interested to find the name of the theorem you're referring to here please.because of boundary theorems (was it called "Green's Theorem"?), if you have an analytic function ins(the Laplace space) in all but the left half-plane, knowledge of whatH(s)does on the [itex]i \omega[/itex] axis (which is all the inverse F.T. sees) is sufficient to tell you whatH(s)does for all others. just because the [itex]i \omega[/itex] axis is associated with steady-state sinusoids, does not mean that it contains only information of the steady state.

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an individual sinusoidal component is periodic. the infinite sum of sinusoids, with infinitesimally close frequencies, as indicated in the Fourier integral is not necessarily periodic.Would you mind clarifying what you mean here? The Fourier Transform tells us the spectral content of our signal i.e. what frequencies are present. In what way is frequency not periodic?

sure, and the rect() function is not periodic nor represents a steady-state solution to anything. instead of a rect, how aboutCase in point: we have a rect function in time (which is not periodic). This gives us a sinc function in frequency, it has infinite frequency content. Meaning if you added an infinite number of cosine waves with differential frequencies apart from one another at the amplitude and phase given by the F.T then you would have a rect function in time exactly.

[tex] y(t) = e^{-\alpha t} u(t) [/tex]

you can use the Fourier Transform and solve for that transient response out of a simple 1-pole LTI system driven by a dirac impulse:

[tex] x(t) = \delta(t) [/tex]

you're using the F.T., not the L.T. and the problem is not about steady state. it's about transient response. and the F.T. got you to the same result that the L.T. would if you set them both up correspondingly. they are both legitimate methods to solve the same problem that can be expressed in the proper manner for each.

and it was mis-cited. too much cannabis for this 52-year-old brain. i mean Cauchy's Integral Formula. If the values of anGreen's Theorem is about relating the curl of a vector field in the plane to a closed path integral around the domains boundry. I'd be interested to find the name of the theorem you're referring to here please.

[tex] H(s_0) = \frac{1}{2 \pi i} \int_C \frac{H(s)}{s-s_0} ds [/tex]

where

if the poles are all in the left half plane and there is a finite number of them, by just knowing what [itex]H(i\omega)[/itex] for real [itex]\omega[/itex] is enough to tell you what [itex]H(s)[/itex] is for all other

- #11

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Ahh I see what you were getting at now, thanksyou're using the F.T., not the L.T. and the problem is not about steady state. it's about transient response. and the F.T. got you to the same result that the L.T. would if you set them both up correspondingly. they are both legitimate methods to solve the same problem that can be expressed in the proper manner for each.

Provided that your contour includes a pole, or else the contour integral is zero (By Cauchy-Goursat Theorem). I still don't see how you know the value of the function at every interiour point. Sorry to be a pain but can you please elaborate?i mean Cauchy's Integral Formula. If the values of ananalyticcomplex function are known on the boundary of a closed curve, the values of the function is known at every interior point.

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no, no, no! you absolutely doProvided that your contour includes a pole, or else the contour integral is zero (By Cauchy-Goursat Theorem).

it's explicit in that Cauchy integral formula. it explicitly says whatI still don't see how you know the value of the function at every interiour point.

[tex] H(s_0) = \frac{1}{2 \pi i} \int_C \frac{H(s)}{s-s_0} ds [/tex]

where

it's not a pain, it's actually pretty straight forward. i can tell that you've already been exposed to this (probably far more recently than me, so please check whatever i say out), but it's really this simple.Sorry to be a pain but can you please elaborate?

of course, if

[tex] \int_C H(s) ds = 0 [/tex]

now considering a single simple pole (we can put it at 0 really without loss of generality, but hey, let's put it at

[tex] \int_C \frac{1}{s-s_0} ds = 2 \pi i [/tex]

as long as

now, because of linearity and scaling, we know also that

[tex] \int_C \frac{A}{s-s_0} ds = 2A \pi i [/tex]

where

now consider, for decently well-behaved

[tex] \int_C \frac{H(s)}{s-s_0} ds = 2 \pi i \ \times \ \mathrm{what??}[/tex]

the "what" has to be whatever

- #13

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The only reason you can tighten up the contour is because of the cross-cut you use from the original contour and knowing that the function is analytic in the contour everywhere except at the pole/s. Its quite a cool trick, but we don't need to spell it out here.[tex] \int_C \frac{1}{s-s_0} ds = 2 \pi i [/tex]

as long asCis a closed path arounds_{0}. so we can tighten up that closed path so it is a vanishing little circle going arounds_{0}. keep that in mind.

The point is, if we have a contour which encloses a pole on its interiour then the contour integral is [tex]2\pi i[/tex] times the sum of residues inside the contour. That says nothing of the values that the function takes on inside the contour.

- #14

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fine by me. you asked me to elaborate.The only reason you can tighten up the contour is because of the cross-cut you use from the original contour and knowing that the function is analytic in the contour everywhere except at the pole/s. Its quite a cool trick, but we don't need to spell it out here.

Eidos, please heed theThe point is, if we have a contour which encloses a pole on its interiour then the contour integral is [tex]2\pi i[/tex] times the sum of residues inside the contour. That says nothing of the values that the function takes on inside the contour.

[tex] \int_C \frac{H(s)}{s-s_0} ds = \int_C \frac{H(s_0)}{s-s_0} ds = H(s_0) \int_C \frac{1}{s-s_0} ds = H(s_0) \ \times \ 2 \pi i [/tex]

for

well, if we return to the main topic of the thread, there is a little more to the story than that and that is that, given some decent conditions for

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- #15

quadraphonics

In my mind, the main thing that LT gets you is a framework for handling unstable systems. It also leads to the pole/zero description of LTI systems, which is extremely useful for systems analysis and design.

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Please, not that, we only transform the signals, not the systems!

H(iw) is only meaningful if we mean the impulse response of a system at rest, from a given input channel to the output channel. Then we can multiply with the arbitrary input signal u(iw)...

If we want to see the effects of the initial conditions also, from linearity, you have to apply an input function u(t) and an impulse, say imp(t) together. Then together, you should consider FT of the overall response.

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Thanks for being patient with me[tex] \int_C \frac{H(s)}{s-s_0} ds = H(s_0) \ \times \ 2 \pi i [/tex]

forCbeing a simple closed curve going counter-clockwize arounds_{0}andH(s)being analytic (no poles) for allsinside ofC. that's the story, Eidos. nothing more than that.

I do see your and other peoples point that you can reformulate the F.T to solve for transients, sorry about my blunder. I thought that the days of 'lies to children' were over in coming to university, I was obviously wrong :tongue:

My issue is a bit off topic, but I suppose here is as good a place as any to bring it up.

I've found a better way of asking my question:

If the function [tex]H(i \omega)[/tex] has a finite number of poles, how do I distribute them in the plane knowing only their position on the imaginary line? You say this is possible from the Cauchy-Integral Theorem. How so?

To get the Fourier Transform from the Laplace Transform we let [tex]s \rightarrow i\omega[/tex]. In doing so we have lost information of the real part of our poles. So if I give you a Fourier Transform [tex]H(i \omega)[/tex], without you knowing the Laplace transform which I used to get it, how do you get [tex]H(s)[/tex] from [tex]H(i \omega)[/tex]?

Lets say we have a pole, [tex]s-(\alpha+i \beta)[/tex]. [tex]s \rightarrow i\omega[/tex] means we get [tex]i \omega-(\alpha+i \beta)=i (\omega-\beta)-\alpha[/tex]. Its easy to see how to go back from here, is this what you meant?

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well it's better than high school, where they try to brainwash you with the State Religion. i dunno what your prof said to you, but it's possible that you (or your prof) took a truth about the F.T. and extended it to where it wasn't true.I do see your and other peoples point that you can reformulate the F.T to solve for transients, sorry about my blunder. I thought that the days of 'lies to children' were over in coming to university, I was obviously wrong :tongue:

evaluating the F.T. (of an LTI impulse response

well, it's not really what i said. i said that knowledge of [itex]H(i \omega)[/itex] for all real [itex]\omega[/itex] is enough to tell you what [itex]H(s)[/itex] for allIf the function [tex]H(i \omega)[/tex] has a finite number of poles, how do I distribute them in the plane knowing only their position on the imaginary line? You say this is possible from the Cauchy-Integral Theorem. How so?

[tex] H(s_0) = \frac{1}{2 \pi i} \int_C \frac{H(s)}{s-s_0} ds [/tex]

whereCis the closed path arounds_{0}.

if the poles are all in the left half plane and there is a finite number of them, by just knowing what [itex]H(i\omega)[/itex] for real [itex]\omega[/itex] is enough to tell you what [itex]H(s)[/itex] is for all others,except at the poles. so even though the Fourier Transform only has the information of [itex]H(s)[/itex] for [itex]s=i\omega[/itex] (the imaginary axis), it need not know the information for [itex]H(s)[/itex] for others. it can figger it out. it knows enough.

now here is another way to look at it, if you know the entire F.T. [itex]H(i \omega)[/itex], then you know the impulse response

how to get the locations of the poles? well, in DSP we have techniques for

i am not saying that you had a parametric representation ofTo get the Fourier Transform from the Laplace Transform we let [tex]s \rightarrow i\omega[/tex]. In doing so we have lost information of the real part of our poles.

[tex] H(s) = \frac{A}{s - p} [/tex]

and say you have a pole at

[tex] H(i \omega) = \frac{A}{i \omega - p} [/tex]

but the actual numbers that go into the Cauchy contour integral do not have

well, you can, if you want, do "analytical extension" the inverse of [itex]s \rightarrow i \omega [/itex]. so in your F.T., wherever you see [itex]\omega[/itex], substitute [itex]s/i[/itex] in for it. then you gotSo if I give you a Fourier Transform [tex]H(i \omega)[/tex], without you knowing the Laplace transform which I used to get it, how do you get [tex]H(s)[/tex] from [tex]H(i \omega)[/tex]?

for that it's easy, but if your F.T. is of a sinc() or rect(), it's not so easy.Lets say we have a pole, [tex]s-(\alpha+i \beta)[/tex]. [tex]s \rightarrow i\omega[/tex] means we get [tex]i \omega-(\alpha+i \beta)=i (\omega-\beta)-\alpha[/tex]. Its easy to see how to go back from here, is this what you meant?

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