### Inverse Fourier Transform help

Posted:

**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

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