Fourier Transform question (kind of Urgent)

In summary, the conversation discusses the system response to the input signal x(t) = 2 + 2*cos(50*t + pi/2), where the system's transfer function is H(w) = 10/(jw + 10). The conversation explores different methods of finding the system's output, including using the Fourier transform and the phasor method. It is determined that the output can be calculated by finding the system's response to the input of '2' and '2*cos(50*t + pi/2)', and then adding them together. The final expression for the output is y(t) = 20/(10j.w + 10) at w = 0.
  • #36
toneboy1 said:
Just experimenting, I noticed if w = 3, width = 2 then
(2/w)*sin(2*w/2) is ALMOST the same as (2.pi/w)*sin(2*w/2.pi),
so you seem right about it being a notational thing, but what I can't understand is that the one with the pi in it seems more accurate because in one instance the number was finite and the other kept going. Can you see why this would be?

Not sure what you're getting at here.

In the below, where on Earth did you get that equation? You're apparently trying to do an inverse transform on sin(wt)?? On a function containing t ?

H'mm, I'm still not clear, I'll elaborate on where I got stuck:

x(t) = (1/2pi)* ∫ (exp(jwt)-exp(-jwt) /2j)*exp(jwt) dw -∞ to ∞

Cheers for any input!

I think I need to go to bed! :uhh:
Cheers!
 
Physics news on Phys.org
  • #37
rude man said:
Not sure what you're getting at here.

In the below, where on Earth did you get that equation? You're apparently trying to do an inverse transform on sin(wt)?? On a function containing t ?



I think I need to go to bed! :uhh:
Cheers!

[I was saying
(2/w)*sin(2*w/2) ≈ (2.pi/w)*sin(2*w/2.pi)
for reasons I cannot reckon]

Woops, good call, it should have been 't'less, I.e:
x(t) = (1/2pi)* ∫ (exp(jw)-exp(-jw) /2j)*exp(jwt) dw -∞ to ∞


Heh, heh, fairenough, Night.
 
  • #38
toneboy1 said:
[I was saying
(2/w)*sin(2*w/2) ≈ (2.pi/w)*sin(2*w/2.pi)
for reasons I cannot reckon]

Woops, good call, it should have been 't'less, I.e:
x(t) = (1/2pi)* ∫ (exp(jw)-exp(-jw) /2j)*exp(jwt) dw -∞ to ∞


Heh, heh, fairenough, Night.

Why are you trying to determine F-1{(sin(w)} anyway? Its inverse is not at all a common time function ...

If I had to do it I would study the way sin(wt) was transformed, then try to apply that technique to the inversion integral.
 

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
744
  • Calculus and Beyond Homework Help
Replies
5
Views
324
  • Calculus and Beyond Homework Help
Replies
1
Views
776
Back
Top