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 August 18th, 2018, 03:18 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 193 Thanks: 2 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. Thanks. Last edited by zollen; August 18th, 2018 at 03:21 PM.
 August 18th, 2018, 07:14 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 471 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics Just apply your identity twice. $\sin(a + \pi) = -\sin(a)$. Thanks from zollen
 August 18th, 2018, 07:16 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,121 Thanks: 1101 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
August 19th, 2018, 06:28 AM   #4
Senior Member

Joined: Jan 2017
From: Toronto

Posts: 193
Thanks: 2

May I ask how you come up an extra term of $\displaystyle \pi/2$??

Quote:
 Originally Posted by romsek 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

 August 19th, 2018, 09:22 AM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,121 Thanks: 1101 You state that $\cos(\omega t ) = \sin(\omega t + \pi/2)$ which is correct. let $x = \omega t + delay(\omega)$ $\cos(x) = \sin(x + \pi/2)$ $\cos(\omega t + delay(\omega)) = \sin(\omega t + delay(\omega) + \pi/2)$ Thanks from zollen

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