# frequency response fourier

Discussion in 'Homework Help' started by darkknight, Nov 19, 2013.

1. ### darkknight Thread Starter Active Member

Oct 7, 2009
41
0
i have this problem but i am slight confused on how to solve hopefully someone can help me

I have

h(t) = e^(-3t)
x(t) = 1+cos (4pi/3)t

find y(t) frequency response

so i took the integral of

e^(-3t)*e^(-jwt) dt from 0 to oo and i got an answer of 1/(3+jw) however i am stuck on what to do now, if someone could explain i would appreciate it

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,507
512
Ok. Let me see if I remember these stuff.

In time domain you would have to convolve h(t) with x(t) to get y(t).

But in frequency domain it is simply multiplication. Y(w)=H(w)X(w).
So. Find X(w).
Also. I think you got H(w) wrong. On page 3 of this: http://uspas.fnal.gov/materials/11ODU/FourierTransformPairs.pdf Look for the definition of e^(-at) in the table. It looks like H(w) should have been 2(-3)/(9+w^2)

3. ### darkknight Thread Starter Active Member

Oct 7, 2009
41
0
thanks for pointing that out for me, i have to see what i did wrong however thats where i am stuck at finding X(w), i see that through my notes that my teacher gets some w values but i am not sure how he got them, was hoping someone could help me learning this

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,507
512
The simplest frequency response can be demonstrated using the extreme cases. Case 1, very very low frequency. Case 2, very very very high frequency.

You don't have the X(w). So we can not do Y(w).

So let us look at just H(w) by itself.
If I am right, then H(w)=2(-3)/(9+w^2)=-6/(9+w^2)

Case 1. Low Frequencies. w->0
H(0)=-6/(9+0)=-2/3
So at very low frequencies your H(w) has a value.

Case 2. High Frequencies. w->infinity
H(infinity)=-6/(9+infinity^2) Obviously infinity^2 is a huge number, so the 9 is not really significant, so we can discard it. -6 divided by a huge number is pretty much a zero. So.
H(infinity)=0

This is for H(w). But this is example of how to do it. It is not actually the solution to your problem. To find out real frequency response you need to do this procedure on Y(w). To have Y(w) you need H(w) and X(w). Y(w)=H(w)X(w)