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August 18th, 2018, 03:18 PM   #1
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Requesting for some insights about Fourier transform basic

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.

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Last edited by zollen; August 18th, 2018 at 03:21 PM.
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August 18th, 2018, 07:14 PM   #2
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Just apply your identity twice. $\sin(a + \pi) = -\sin(a)$.
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August 18th, 2018, 07:16 PM   #3
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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
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August 19th, 2018, 06:28 AM   #4
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May I ask how you come up an extra term of $\displaystyle \pi/2$??

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Originally Posted by romsek View Post
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
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August 19th, 2018, 09:22 AM   #5
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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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