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Fourier/Laplace transforms 
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#1
May2608, 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?



#2
May2708, 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)?



#3
May2708, 08:05 PM

P: 2,251

i think Halls, that the direct answer to John's question has to do with the singlesided Laplace Transform:
[tex] X(s) = \int_{0}^{+\infty} x(t) e^{st} dt [/tex] vs. the doublesided 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 doublesided L.T. contains the same information as the initial conditions at t=0 of the singledsided. i don't think i've ever seen the F.T. expressed in a singlesided expression. the F.T. can be considered a less general expression or a "special case" of the doublesided L.T. 


#4
May2708, 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! 


#5
May2808, 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, TimeInvariant 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 singlesided L.T. is better for that kind of problem.
now, the problem can be reformed or restated in such a way that the doublesided Fourier Transform can be used to solve it. restating would go something like this: "We have an undampled 2^{nd}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 righthand side of your F.T. equation would not be zero. 


#6
May2808, 02:07 PM

P: 91

Very insightful, thanks!



#7
May2908, 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}{sa})=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
May2908, 11:31 AM

P: 2,251

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 steadystateness) 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 halfplane, 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 steadystate sinusoids, does not mean that it contains only information of the steady state. 


#9
May2908, 12:33 PM

P: 108




#10
May2908, 07:13 PM

P: 2,251

[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 1pole 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. [tex] H(s_0) = \frac{1}{2 \pi i} \int_C \frac{H(s)}{ss_0} ds [/tex] where C is the closed path around s_{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 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. 


#11
May3008, 06:45 AM

P: 108




#12
May3008, 12:38 PM

P: 2,251

[tex] H(s_0) = \frac{1}{2 \pi i} \int_C \frac{H(s)}{ss_0} ds [/tex] where C is any closed path going counterclockwize around s_{0} and where H(s) is entirely analytic inside. 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 s_{0}), we know that where it is analytic everywhere else but the pole, then [tex] \int_C \frac{1}{ss_0} ds = 2 \pi i [/tex] as long as C is a closed path around s_{0}. so we can tighten up that closed path so it is a vanishing little circle going around s_{0}. keep that in mind. now, because of linearity and scaling, we know also that [tex] \int_C \frac{A}{ss_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 s_{0} that matters since that contour integral is true even if the contour path is a teenyweeny tiny circle of vanishing radius of [itex]\epsilon>0[/itex] around s_{0}. 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 wellbehaved H(s): [tex] \int_C \frac{H(s)}{ss_0} ds = 2 \pi i \ \times \ \mathrm{what??}[/tex] the "what" has to be whatever H(s) is at (and around) point s_{0}. 


#13
May3008, 01:07 PM

P: 108

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
May3008, 01:30 PM

P: 2,251

[tex] \int_C \frac{H(s)}{ss_0} ds = \int_C \frac{H(s_0)}{ss_0} ds = H(s_0) \int_C \frac{1}{ss_0} ds = H(s_0) \ \times \ 2 \pi i [/tex] for C being a simple closed curve going counterclockwize around s_{0} 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 halfplane and it's analytic for the right halfplane 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 everwidening semicircle in the right halfplane to swing back. the contribution to the lineintegral that the semicircle makes can be shown to go to zero as the semicircle 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. 


#15
May3008, 05:51 PM

P: 270

My impression is that the old "LT for transient response/FT for steadystate" 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 lowerdivision 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. 


#16
May3108, 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. 


#17
Jun108, 02:14 AM

P: 108

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 CauchyIntegral 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? 


#18
Jun108, 11:20 AM

P: 2,251

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 steadystate 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 steadystate, of the LTI system. 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 autoregressive analysis), but i only said you could, with the Cauchy integral formula, compute the value of H(s_{0}) for some s_{0} that is not sitting directly atop a pole location. [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. 


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