HopkinsMedStudent
Thinker
- Joined
- Jul 23, 2003
- Messages
- 210
Prove that inverse fourier of (rect(w/z) x e^(-w/b))= z sinc (zt + iz/b)
Assume that w in the frequency domain translates to t in the time domain.
The way I am approaching this problem is by using the following property:
F(f1 x f2) = F(f1) * F(f2), convolution property
I know what the Fourier of rect(w/z) is, thats z sinc (zt)
However, I'm having trouble with the Fourier of the exponential.
Looking ahead, I'm assuming that I'm supposed to show that Fourier of the exponential is something like delta(zt + iz/b) where delta is the unit impulse function.
Then once you get that, you have z sinc (zt) * delta(zt + iz/b) which would give you the final result (convolution of any function with an impulse is simply the shifted original function by the argument of the impulse).
But the Fourier of an exponential is unbounded unless you apply a unit step function right? So for example, F(e^-at) is undefined but F(e^-at) u(t) is defined because it has the step function attached.
Assume that w in the frequency domain translates to t in the time domain.
The way I am approaching this problem is by using the following property:
F(f1 x f2) = F(f1) * F(f2), convolution property
I know what the Fourier of rect(w/z) is, thats z sinc (zt)
However, I'm having trouble with the Fourier of the exponential.
Looking ahead, I'm assuming that I'm supposed to show that Fourier of the exponential is something like delta(zt + iz/b) where delta is the unit impulse function.
Then once you get that, you have z sinc (zt) * delta(zt + iz/b) which would give you the final result (convolution of any function with an impulse is simply the shifted original function by the argument of the impulse).
But the Fourier of an exponential is unbounded unless you apply a unit step function right? So for example, F(e^-at) is undefined but F(e^-at) u(t) is defined because it has the step function attached.