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JohnSimpson
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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?
Eidos said:The other thing which the Fourier transform does is to ignore any transients and give only steady state behaviour of your system.
rbj said: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.
rbj said: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 said: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.
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 said: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.
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 said: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.
Eidos said: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 said:[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.
Eidos said: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 said:[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.
Eidos said: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
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?
rbj said:[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.
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?
Fourier and Laplace transforms are mathematical operations that transform a function in the time or spatial domain into a function in the frequency or complex domain. They are used to analyze signals or systems in various fields of science and engineering, such as physics, mathematics, and electrical engineering.
The main difference between Fourier and Laplace transforms is the type of function they operate on. Fourier transforms are used for periodic and non-periodic signals in the time domain, while Laplace transforms are used for non-periodic signals or functions with exponential growth in the time domain.
Fourier and Laplace transforms have numerous applications in various fields, including signal processing, control systems, heat transfer, image processing, and quantum mechanics. They are also used in solving differential equations and analyzing the behavior of linear systems.
Laplace transforms are an extension of Fourier transforms, as they can be used to solve more complex problems that cannot be solved using Fourier transforms alone. In fact, the Fourier transform can be obtained from the Laplace transform by setting the complex variable to zero.
The Fourier transform is calculated by taking the integral of the function with respect to frequency, while the Laplace transform is calculated by taking the integral of the function multiplied by a decaying exponential. These integrals can be solved analytically or numerically using various techniques, such as the Fast Fourier Transform (FFT) algorithm.