 My Math Forum Requesting for some insights about Fourier transform basic

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 August 18th, 2018, 04:18 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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 04:21 PM. August 18th, 2018, 08:14 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 680 Thanks: 454 Math Focus: Dynamical systems, analytic function theory, numerics Just apply your identity twice. $\sin(a + \pi) = -\sin(a)$. Thanks from zollen August 18th, 2018, 08:16 PM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,630 Thanks: 1469 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, 07:28 AM   #4
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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, 10:22 AM #5 Senior Member   Joined: Sep 2015 From: USA Posts: 2,630 Thanks: 1469 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 Tags basic, fourier, insights, requesting, transform Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post aows61 Complex Analysis 10 August 1st, 2017 12:45 AM szz Applied Math 0 December 16th, 2015 02:03 PM asa2012 Calculus 0 October 19th, 2012 01:47 AM beckie Real Analysis 3 June 20th, 2010 01:58 PM Lononeer Real Analysis 1 September 3rd, 2009 02:47 PM

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