I've been solving my homework and I got to the last question and I can't figure out whether it's a simple question or that I completely don't understand what I'm being asked: "Consider the equation x=sin(2*pi*t) + sin(2*pit+pi/4). Find the values of a and p such that x = asin(2*pi*t+p). (hint, check the power and phase of the DFT)" isn't a=1 and p=0? please explain and help me understand. Thank you!

  • $\begingroup$ you should find a $a, p$ such that $\sin(2\pi t)+\sin(2\pi t+\pi/4)=a\sin(2\pi t+p)$. You could either Fourier-Transform both sides and perform coefficient comparison, or you apply trigonometric theorems (adding two sines). $\endgroup$ – Maximilian Matthé Jun 16 '18 at 18:17

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