Requesting for some insights about Fourier transform basic

Jan 2017
209
3
Toronto
I am studying the Fourier transform and trying to understand the following equation

w is frequency..
t is time..
A(w) is a a signal amplified function
delay(w) is a time delay function

Trig Identity: sin( a + b ) = sin(a)cos(b) + sin(b)cos(a)
Therefore:
A(w)sin((wt + delay(w))) = A(w)cos(delay(w))sin(wt) + A(w)sin(delay(w))cos(wt) - eq.1

cos(a) = sin(a + pi/2) The cosine is just a time-advanced sine,
it follows that the response to the input cos(wt) is just:

A(w)cos(delay(w))cos(wt) - A(w)sin(delay(w))sin(wt) - eq.2

I don't understand how eq.1 get converted into eq.2... Any insights would be much appreciated.

Thanks.
 
Last edited:

SDK

Sep 2016
743
497
USA
Just apply your identity twice. $\sin(a + \pi) = -\sin(a)$.
 
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romsek

Math Team
Sep 2015
2,885
1,609
USA
you talk about the response to the input $\cos(\omega t)$

the response of what? I don't see any system mentioned in this post.

eq 2. can be backed out to produce $A(\omega) \cos(\omega t + delay(\omega))$

but that is not the same as what you start with in eq. 1

if you expand out $A(\omega)\sin(\omega t + delay(\omega) + \pi/2)$ you should eventually end up with eq. 2
 
Jan 2017
209
3
Toronto
May I ask how you come up an extra term of \(\displaystyle \pi/2\)??

you talk about the response to the input $\cos(\omega t)$

the response of what? I don't see any system mentioned in this post.

eq 2. can be backed out to produce $A(\omega) \cos(\omega t + delay(\omega))$

but that is not the same as what you start with in eq. 1

if you expand out $A(\omega)\sin(\omega t + delay(\omega) + \pi/2)$ you should eventually end up with eq. 2
 

romsek

Math Team
Sep 2015
2,885
1,609
USA
You state that $\cos(\omega t ) = \sin(\omega t + \pi/2)$ which is correct.

let $x = \omega t + delay(\omega)$

$\cos(x) = \sin(x + \pi/2)$

$\cos(\omega t + delay(\omega)) = \sin(\omega t + delay(\omega) + \pi/2)$
 
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