Inverse Fourier Transform help

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sfabriz
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Inverse Fourier Transform help

Post by sfabriz » Tue May 12, 2009 3:22 pm

Hi guys, I need some help with a inverse fourier transform:

Given s(t) a signal and S(f) its fourier transform, we define:
S(f) = Integrate[-inf, +inf, s(t) * exp(-j2pi * f * t) dt] = fourier transform of s(t)
and
s(t) = Integrate[-inf, +inf, S(f) * exp(j2pi * f * t) df] = inverse fourier transform of S(f)

Consider a low pass filter, as the one that is depicted in fig. 1 at this page: http://en.wikipedia.org/wiki/Low-pass_filter.

To solve the filter we put into a system these 2 equations (Z are impedances; Vi, Vo are voltages and I is the current):
Vi - (Zr + Zc)*I = 0
Vi - Zr*I = Vc = Vo

Zr = R, Zc = 1/(sC) with s = j2pi * f

They give this frequency response: G(f) = Vo/Vi = 1 / (1 + sRC) = 1 / (1 + j2pi * f * RC)
Now, if we call a = 1/(RC) we get this equation to describe the lpf:
G(f) = Vo/Vi = a/(a + j2pi * f)

Now, to get the inpulsive response of this filter, I have to do the inverse fourier transform of the G(f) function.
My book gives me this result: g(t) = a * 1(t) * exp(-at), with a=1/(RC) and 1(t) being the heaviside step function, which is 1 for t>0 and 0 for t<0.

So, I really can't get to that result on my own (maybe I'm dumber in may...), but I can prove it is correct going from g(t) to G(f), in fact:

G(f) = Integrate[-inf, +inf, g(t)] = a * Integrate[0, +inf, exp[-(j2pi * f + a)t] dt] =
=[-a/(j2pi * f + a)]* exp[-(j2pi * f + a)t]{t,0,+inf} and this gives exactly:
a/(j2pi * f + a) and when we substitute a with 1/(RC) we get back to: G(f) = 1/(1 + j2pi * f * RC)

So, since the fourier transformation is unique, there must be a method to get from G(f) to g(t) making an inverse tranformation.

Any of you happens to know it?
Thanks in advance for any help,

cheers,
sfabriz
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daniel.is.fischer
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Re: Inverse Fourier Transform help

Post by daniel.is.fischer » Tue May 12, 2009 4:05 pm

1. Look it up (Bron&scaron;tein/Semendyayev, Grad&scaron;tein/Ryzhik or similar).
2. For functions like this, the Residue Theorem comes in handy.
R and C are real, aren't they?, so a = RC is real, too (anyway, a must have nonzero real part, otherwise the function wouldn't be square integrable), and [frac]a,a+2[pi]*i*f[/frac] (sorry, I can't write the imaginary unit as j) has a simple pole at [frac]ia,2[pi][/frac] with residue [frac]-ia,2[pi][/frac].
Let's call the integrand of the inverse Fourier integral K, so K(t,f) = [frac]a,a+2[pi]*i*f[/frac]*exp(2[pi]i*t*f)
Now consider the two cases t < 0 and t > 0 separately.
a) t < 0:
We have |exp(2[pi]i*t*f)| = exp(-2[pi]*t*Im(f)), which decreases nicely for Im(f) &rarr; -&infin;. So integrating along the boundary of a large enough rectangle [-R, R, R-iS, -R-iS], by the Residue Theorem, this integral is -2[pi]*i*[sum](residues of K(t,_) in the lower half plane). The integrals of K(t,_) over the vertical edges and the edge below the real axis vanish for R, S &rarr; &infin;, leaving
&int;-&infin;&infin;K(t,f) df = -2[pi]i*(sum of residues).
Assuming a > 0, the only pole is in the upper half plane, sum of residues is 0.

b) t > 0:
Now we consider K(t,_) in the upper half plane, take the rectangle [-R, R, R+iS, -R+iS]. Now the boundary is positively oriented, so the integral is 2[pi]*i*(sum of enclosed residues). One pole is enclosed, the residue is [frac]-ia,2[pi][/frac]*exp(2[pi]i*t*[frac]ia,2[pi][/frac]) = [frac]-ia,2[pi][/frac]*exp(-t*a), giving the overall result
&int;-&infin;&infin;K(t,f) df = a*exp(-t*a) for t > 0.

Patching both cases together, s(t) = a*H(t)*exp(-ta).

3. Isn't this kind of problem more of a case for the Laplace transform (where the method of 2. always works)?
Il faut respecter la montagne -- c'est pourquoi les gypa&egrave;tes sont l&agrave;.

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sfabriz
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Re: Inverse Fourier Transform help

Post by sfabriz » Wed May 13, 2009 9:42 am

Ah, very nice.
Yes, this kind of problem usually gets done with the Laplace transform when you get the transfer function of a system, but in this case it is done with the fourier transform because it is a study of a simple low-pass filter and we're not considering the transfer function but the frequency response of the system, of which we calc other details after the impulse response. I'm a little bit rusty with these things and couldn't recall how to get that.

Thank you very much Daniel,
sfabriz
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sfabriz
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Re: Inverse Fourier Transform help

Post by sfabriz » Wed May 13, 2009 9:53 am

Oh, by the way, I don't like the j notation too for the imaginary unit, but here we've got a convention: if you're doing math you use i, if you're doing other things, you use j. That's because "i" could also mean a generic current that you use when you analyze a circuit and it is easy to get confused when you do complex analysis along with current equations.
So, usually in uni books you find j on electronic books and i in pure math books.
This fourier problem I asked for, is inside a complex analysis of a circuit that is used as a blackbox filter on a wider signal transmission system chain, and therefore all equations use j instead of i, and I didn't even think about that when I wrote the post. :wink:

Thanks again,
sfabriz
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daniel.is.fischer
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Re: Inverse Fourier Transform help

Post by daniel.is.fischer » Wed May 13, 2009 12:46 pm

sfabriz wrote:Oh, by the way, I don't like the j notation too for the imaginary unit, but here we've got a convention: if you're doing math you use i, if you're doing other things, you use j. That's because "i" could also mean a generic current that you use when you analyze a circuit and it is easy to get confused when you do complex analysis along with current equations.
I know. And hereabouts it's an old custom that mathematicians annoy electric engineers by using i and electric engineers annoy mathematicians by using j, each side pretending that they're utterly shocked by the other's convention 8-)
Il faut respecter la montagne -- c'est pourquoi les gypa&egrave;tes sont l&agrave;.

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sfabriz
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Re: Inverse Fourier Transform help

Post by sfabriz » Wed May 13, 2009 1:03 pm

Ahahahah, yeah I guess it's truly like that!

I feel like a hybrid though, since being an i.t. engineer I'm midway between electric engineers and mathematicians, with the result that I neither know math as mathematicians nor electronic as electric engineers!
:lol:
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