Sum of Sines of Arbitrary AmplitudesDate: 01/19/2012 at 19:15:17 From: Bob Subject: Sum of sines with unequal amplitudes Given two sine functions: A*sin(x) B*sin(y) What is their sum? A*sin(x) + B*sin(y) The identity for sin(x) + sin(y) is readily available, but I can't find an expression for the case where the amplitudes are not equal. I'm most interested in the case where the frequency is the same for both of the sines but their amplitudes and phases are arbitrary. Thank you for your help. Bob Date: 01/21/2012 at 16:40:46 From: Doctor Rick Subject: Re: Sum of sines with unequal amplitudes Hi, Bob. So you've got something like this: f(t) = A*sin(wt) + B*sin(wt + phi) For simplicity, I chose the initial time so that the phase of the first sinusoid is 0. Now, I suppose you'd like to represent this as a single sinusoid, finding its amplitude and phase. Here is a way to do it. First, I'll expand the second sinusoid: [1] sin(wt + phi) = sin(wt)*cos(phi) + cos(wt)*sin(phi) Then we have [2] f(t) = (A + B*cos(phi))*sin(wt) + B*sin(phi)*cos(wt) We can put this into the form of a single sinusoid if we can find C and alpha such that [3] A + B*cos(phi) = C*cos(alpha) [4] B*sin(phi) = C*sin(alpha) Squaring equations [3] and [4] and adding, we get (A + B*cos(phi))^2 + (B*sin(phi))^2 = C^2 A^2 + 2AB*cos(phi) + B^2 = C^2 [5] C = sqrt(A^2 + 2AB*cos(phi) + B^2) Dividing [4] by [3], we get B*sin(phi)/(A + B*cos(phi)) = tan(alpha) [6] alpha = tan^-1(B*sin(phi)/(A + B*cos(phi))) With the values for C and alpha from [5] and [6], we have f(t) = C*sin(wt)*cos(alpha) + C*cos(wt)*sin(alpha) [7] f(t) = C*sin(wt + alpha) Is that what you're after? There's a little detail I glossed over: each of[5] and [6] is really only one of two solutions to the preceding equation. This can be handled by use of the atan2() function in software. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 01/21/2012 at 21:54:08 From: Bob Subject: Thank you (Sum of sines with unequal amplitudes) Thank you very much for your help. I really appreciate this. -- Bob |
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