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Fourier/Laplace transforms

by JohnSimpson
Tags: fourier or laplace, transforms
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JohnSimpson
#1
May26-08, 01:53 PM
P: 91
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
#2
May27-08, 05:10 PM
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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)?
rbj
#3
May27-08, 08:05 PM
P: 2,251
i think Halls, that the direct answer to John's question has to do with the single-sided Laplace Transform:

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

vs. the double-sided Laplace Transform:

[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 t=0 of the singled-sided.

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.

JohnSimpson
#4
May27-08, 10:11 PM
P: 91
Fourier/Laplace transforms

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!
rbj
#5
May28-08, 01:15 PM
P: 2,251
okay, John, what you're doing is spelling out why we have and use both the Fourier Transform and Laplace Transform, but in different applications. you are seeing which one is more suitable for the problem you have which is a Linear, Time-Invariant system (we EEs call that an "LTI") that has no input but does have two internal states that are not both zero at t=0. (if both states were zero, then both of your initial conditions would be zero and your output y(t) would always be zero.) in addition, your problem doesn't say diddley about what the output or states were for any t<0. it tells you what the states are at t=0, what the relationship of the states are with each other and the output for t>0, and from that asks you what, specifically, is the output for t>0. the single-sided L.T. is better for that kind of problem.

now, the problem can be re-formed or restated in such a way that the double-sided Fourier Transform can be used to solve it. restating would go something like this: "We have an undampled 2nd-order harmonic oscillator that is completely at rest for all time before t=0 and at t=0 some driving function, x(t), is such that displaces y(0) from its rest position and also imposes some initial velocity, y'(0) to it at t=0. For all later times, t>0, this driving function is zero." then the right-hand side of your F.T. equation would not be zero.
JohnSimpson
#6
May28-08, 02:07 PM
P: 91
Very insightful, thanks!
Eidos
#7
May29-08, 12:58 AM
P: 108
The other thing which the Fourier transform does is to ignore any transients and give only steady state behaviour of your system.

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.
rbj
#8
May29-08, 11:31 AM
P: 2,251
Quote Quote by Eidos View Post
The other thing which the Fourier transform does is to ignore any transients and give only steady state behaviour of your system.
i'm sorry, but i have to completely disagree with this notion. the Fourier Transform (sometimes called the "Fourier Integral") can be and essentially is used to find a total solution to the system output. both transient and steady-state. in fact, unlike solving ordinary diff. eqs. classically, where the "homogeneous solution" and "particular solution" are separately solved and added together to be the final solution, neither the F.T. nor L.T. differentiate between the two. it just solves for the solution which may or may not contain either the homogeneous (a.k.a. transient or "natural" response) and particular (a.k.a. steady-state or "driven" response) solutions.

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 s (the Laplace space) in all but the left half-plane, knowledge of what H(s) does on the [itex]i \omega[/itex] axis (which is all the inverse F.T. sees) is sufficient to tell you what H(s) does for all other s. 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.
Eidos
#9
May29-08, 12:33 PM
P: 108
Quote Quote by rbj View Post
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.
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.

Quote Quote by rbj View Post
because of boundary theorems (was it called "Green's Theorem"?), if you have an analytic function in s (the Laplace space) in all but the left half-plane, knowledge of what H(s) does on the [itex]i \omega[/itex] axis (which is all the inverse F.T. sees) is sufficient to tell you what H(s) does for all other s. 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.
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.
rbj
#10
May29-08, 07:13 PM
P: 2,251
Quote Quote by Eidos View Post
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?
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.

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.
sure, and the rect() function is not periodic nor represents a steady-state solution to anything. instead of a rect, how about

[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.

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.
and it was mis-cited. too much cannabis for this 52-year-old brain. i mean Cauchy's Integral Formula. If the values of an analytic complex function are known on the boundary of a closed curve, the values of the function is known at every interior point.

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

where C is the closed path around s0.

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 s, 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 other s. it can figger it out. it knows enough.
Eidos
#11
May30-08, 06:45 AM
P: 108
Quote Quote by rbj View Post
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.
Ahh I see what you were getting at now, thanks

Quote Quote by rbj View Post
i mean Cauchy's Integral Formula. If the values of an analytic complex function are known on the boundary of a closed curve, the values of the function is known at every interior point.
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?
rbj
#12
May30-08, 12:38 PM
P: 2,251
Quote Quote by Eidos View Post
Provided that your contour includes a pole, or else the contour integral is zero (By Cauchy-Goursat Theorem).
no, no, no! you absolutely do not want a pole of H(s) in there (otherwise you'll have to subtract the residue you get from it). there is already the pole that we toss in with the 1/(s-s0) in the integral and that is the only pole we want in there.

I still don't see how you know the value of the function at every interiour point.
it's explicit in that Cauchy integral formula. it explicitly says what H(s0) in terms of the contour integration around it. it's a constructive proof. explicitly:

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

where C is any closed path going counter-clockwize around s0 and where H(s) is entirely analytic inside.

Sorry to be a pain but can you please elaborate?
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.

of course, if H(s) is entirely analytic inside C, then we know that

[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 s0), we know that where it is analytic everywhere else but the pole, then

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

as long as C is a closed path around s0. so we can tighten up that closed path so it is a vanishing little circle going around s0. keep that in mind.

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

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

where A is some number that i pull out of my butt. can be any number. now if A wasn't constant but somehow varies a little with s, it's only what value that A takes over s0 that matters since that contour integral is true even if the contour path is a teeny-weeny tiny circle of vanishing radius of [itex]\epsilon>0[/itex] around s0. we might hope that A is pretty much the same value for all points on C, if it's so small. (not a very rigorous argument.)

now consider, for decently well-behaved H(s):

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

the "what" has to be whatever H(s) is at (and around) point s0.
Eidos
#13
May30-08, 01:07 PM
P: 108
Quote Quote by rbj View Post
[tex] \int_C \frac{1}{s-s_0} ds = 2 \pi i [/tex]

as long as C is a closed path around s0. so we can tighten up that closed path so it is a vanishing little circle going around s0. keep that in mind.
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.

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.
rbj
#14
May30-08, 01:30 PM
P: 2,251
Quote Quote by Eidos View Post
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.
fine by me. you asked me to elaborate.

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.
Eidos, please heed the given conditions: H(s) itself is analytic inside of C. no poles of H(s) inside of C. not so for for H(s)/(s-s0). that has one simple pole at s0 and you can think of H(s) as approaching the constant H(s0) for that tightening little circle around s0. so

[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 C being a simple closed curve going counter-clockwize around s0 and H(s) being analytic (no poles) for all s inside of C. that's the story, Eidos. nothing more than that.

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 H(s) (like the poles are all in the left half-plane and it's analytic for the right half-plane and imaginary axis, and maybe there has to be at least one more pole than zeros) knowledge of H(s) on the imaginary axis (which is all the Fourier Transform sees) is enough to tell you what H(s) is doing for all other s, except for when s is right on top of one of them poles. you can prove that from this Cauchy integral thingie above with a path that goes down the imaginary axis and makes an ever-widening semi-circle in the right half-plane to swing back. the contribution to the line-integral that the semi-circle makes can be shown to go to zero as the semi-circle gets infinitely large in radius. so the integral along the [itex]i \omega[/itex] axis can tell you what H(s) is for all the other s except for at the poles.
quadraphonics
#15
May30-08, 05:51 PM
P: 270
My impression is that the old "LT for transient response/FT for steady-state" canard had to do with the inconvenience of dealing with initial conditions with the FT approach. It can be done, but you have to jump through a lot of hoops to modify the input function so that it just happens to produce the initial conditions you want. So, rather than explaining these finer points, profs in lower-division circuits classes just tell students not to use FT for transient response.

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.
trambolin
#16
May31-08, 11:17 AM
P: 341
Fourier Transform can distribute the initial condition/transient effects onto the steady state/zero initial condition responses!

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.
Eidos
#17
Jun1-08, 02:14 AM
P: 108
Quote Quote by rbj View Post
[tex] \int_C \frac{H(s)}{s-s_0} ds = H(s_0) \ \times \ 2 \pi i [/tex]

for C being a simple closed curve going counter-clockwize around s0 and H(s) being analytic (no poles) for all s inside of C. that's the story, Eidos. nothing more than that.
Thanks for being patient with me
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

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?
rbj
#18
Jun1-08, 11:20 AM
P: 2,251
Quote Quote by Eidos View Post
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
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.

evaluating the F.T. (of an LTI impulse response h(t)) at a single frequency (or a finite number of frequencies), [itex]H(i \omega)[/itex] does only tells you about the steady-state performance of that LTI system. it's when you have knowledge of [itex]H(i \omega)[/itex] for an infinite number of frequencies all spaced infinitesimally close to each other, that's when you know about the entire response, transient and steady-state, of the LTI system.

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?
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 all s, except at the poles.

Quote Quote by rbj View Post

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

where C is the closed path around s0.

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 s, 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 other s. 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 h(t), and from that, you can calculate the Laplace Transform at every s, except at the poles (which are infinite).

how to get the locations of the poles? well, in DSP we have techniques for estimating parameters of H(s) or h(t) (like auto-regressive analysis), but i only said you could, with the Cauchy integral formula, compute the value of H(s0) for some s0 that is not sitting directly atop a pole location.


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.
i am not saying that you had a parametric representation of H(s). sure, by examination, you can look at the expression

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

and say you have a pole at p. same with the Fourier representation

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

but the actual numbers that go into the Cauchy contour integral do not have p in them.

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]?
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 got H(s), but i can't ascribe easy meaning to it if it happens to be a sinc() function or rect() function. it's with these kinda functions where the F.T. and the double-sided L.T. are hard to interchange.

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?
for that it's easy, but if your F.T. is of a sinc() or rect(), it's not so easy.


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