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Anyone know about Fourier Transforms?

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RaynorWolfcastle

Diamond Member
Feb 8, 2001
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Yeah it was something like that, it's been a couple of years since I've taken signals and systems but that looks about right. You've got to use the fact that a translation in frequency a is a multiplication in time by exp(at). You can also use the fact that integrating in time is the same as multiplying by 1/jw in frequency.

So you start out with an f(t) = 1 => F(w) = delta(t)
then integrate twice to get g(t) = t*u(t) => G(w)=1/(jw^2)
and then take the frequency domain translation by 5 and you get h(t) = exp(-5t)*t*u(t) => H(w) = 1/(jw+5)^2

Done like dinner.
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
8,968
16
81
Originally posted by: PurdueRy
oh trust me, I know I need to understand it.
While I have someone here with experience, got any clues about:
((jw+2)(jw+5))/(jw-1)

I'd use a partial fractions expansion
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: RaynorWolfcastle
Originally posted by: PurdueRy
oh trust me, I know I need to understand it.
While I have someone here with experience, got any clues about:
((jw+2)(jw+5))/(jw-1)

I'd use a partial fractions expansion

Except you can't because the order of the numerator is greater than the denominator.
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
8,968
16
81
Originally posted by: hypn0tik
Originally posted by: RaynorWolfcastle
Originally posted by: PurdueRy
oh trust me, I know I need to understand it.
While I have someone here with experience, got any clues about:
((jw+2)(jw+5))/(jw-1)

I'd use a partial fractions expansion

Except you can't because the order of the numerator is greater than the denominator.

exclude the w^2 term and try it ;)


edit: actually you don't even need that, expand it and solve each of the three separately, shouldn't be too complicated
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: RaynorWolfcastle
Originally posted by: hypn0tik
Originally posted by: RaynorWolfcastle
Originally posted by: PurdueRy
oh trust me, I know I need to understand it.
While I have someone here with experience, got any clues about:
((jw+2)(jw+5))/(jw-1)

I'd use a partial fractions expansion

Except you can't because the order of the numerator is greater than the denominator.

exclude the w^2 term and try it ;)


edit: actually you don't even need that, expand it and solve each of the three separately, shouldn't be too complicated

Well, you get:

[(jw)^2 + 7jw + 10] / (jw-1)

10/(jw-1) can be done easily. It's the other 2 that are troublesome.
 

slpaulson

Diamond Member
Jun 5, 2000
4,414
14
81
Originally posted by: PurdueRy
Originally posted by: hypn0tik
Make sure you understand how they got that result. Or I guess you could just memorize it, lol.

Good luck.

Anything else, just ask.


oh trust me, I know I need to understand it.

While I have someone here with experience, got any clues about:

((jw+2)(jw+5))/(jw-1)

I'd try expanding the numerator...
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
8,968
16
81
ok fine so you rearrange and add terms so that you can refactor the top, that makes sense

just to be clear on how they got it

(jw+5)(jw+2) = -w^2 + 7jw + 10 = -w^2 -2jw +1 + 9jw + 9 = (jw - 1)^2 + 9(jw - 1) + 18

divide the whole thing by (jw - 1) and you get

jw - 1 + 9 + 18/(jw-1) = jw + 8 + 18/(jw-1)

F(w) = d/dt(delta(t) + 8*delta(t) + 18exp(-t)*u(t)

 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: RaynorWolfcastle
ok fine so you rearrange and add terms so that you can refactor the top, that makes sense

just to be clear on how they got it

(jw+5)(jw+2) = -w^2 + 7jw + 10 = -w^2 -2jw +1 + 9jw + 9 = (jw - 1)^2 + 9(jw - 1) + 18

divide the whole thing by (jw - 1) and you get

jw - 1 + 9 + 18/(jw-1) = jw + 8 + 18/(jw-1)

F(w) = d/dt(delta(t) + 8*delta(t) + 18exp(-t)*u(t)

Oh ok. I understand how they broke it up.

The only painful transform was that of jw. I had to look it up in a table of Laplace transforms to see that it was d/dt (delta(t)). There's no way I would have come up with that on my own.
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
8,968
16
81
Originally posted by: hypn0tik
Oh ok. I understand how they broke it up.

The only painful transform was that of jw. I had to look it up in a table of Laplace transforms to see that it was d/dt (delta(t)). There's no way I would have come up with that on my own.
:confused: multiplying by jw in the frequency domain is always the same as differentiating in time.
 

PurdueRy

Lifer
Nov 12, 2004
13,837
4
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Thanks all yet again. Excellent explanation Raynor, I am still going into office hours to get cleared up on more of this though. Hey hypn0tik, make your 2000th post!
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: RaynorWolfcastle
Originally posted by: hypn0tik
Oh ok. I understand how they broke it up.

The only painful transform was that of jw. I had to look it up in a table of Laplace transforms to see that it was d/dt (delta(t)). There's no way I would have come up with that on my own.
:confused: multiplying by jw in the frequency domain is always the same as differentiating in time.

OMFG, you're right. I sir, am an idiot. I had it right in front of my too...

Bah.

Edit: Hehehe, Purdue, here it is. I get to bash myself, w00t!
 

PurdueRy

Lifer
Nov 12, 2004
13,837
4
0
Originally posted by: hypn0tik
Originally posted by: RaynorWolfcastle
Originally posted by: hypn0tik
Oh ok. I understand how they broke it up.

The only painful transform was that of jw. I had to look it up in a table of Laplace transforms to see that it was d/dt (delta(t)). There's no way I would have come up with that on my own.
:confused: multiplying by jw in the frequency domain is always the same as differentiating in time.

OMFG, you're right. I sir, am an idiot. I had it right in front of my too...

Bah.


I shoulda realized that too...too much thinking bah!
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
8,968
16
81
alright, I'm glad I could help. It reminded me of when I took Signals 1. That class was brutal, the prof failed like half the class, but you can bet your ass the rest of us could do Fourier transforms like it's going out of style.
 

Mday

Lifer
Oct 14, 1999
18,647
1
81
ah memories.. such an EASY A that signals class was... =D
i think i even used the same text lol