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

## Inverse Fourier Transform help

- daniel.is.fischer
**Posts:**2400**Joined:**Sun Sep 02, 2007 10:15 pm**Location:**Bremen, Germany

### Re: Inverse Fourier Transform help

1. Look it up (Bronštein/Semendyayev, Gradštein/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) → -∞. 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 → ∞, leaving

∫

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

∫

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.

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) → -∞. 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 → ∞, leaving

∫

_{-∞}^{∞}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

∫

_{-∞}^{∞}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 gypaètes sont là.

### Re: Inverse Fourier Transform help

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

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

### Re: Inverse Fourier Transform help

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.

Thanks again,

sfabriz

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.

Thanks again,

sfabriz

- daniel.is.fischer
**Posts:**2400**Joined:**Sun Sep 02, 2007 10:15 pm**Location:**Bremen, Germany

### Re: Inverse Fourier Transform help

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 conventionsfabriz 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.

Il faut respecter la montagne -- c'est pourquoi les gypaètes sont là.

### Re: Inverse Fourier Transform help

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!

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!