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$$x(t)= A+ A_1\sin (2 \pi f t + \theta ) + A_2\cos (2 \pi f_1 t + \theta )$$ I want to find the Fourier transform $|\mathcal{F}[x(t)]|^2$ . Is this possible by hand? I can find the Fourier transform but then raise to power seems difficult.Is there any shortcut for this ?

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Have you tried the various formulae for the various products $\cos x \sin y,$ etc? –  copper.hat Oct 3 '12 at 0:47
    
I'm confused. Are you trying to calculate $\mathcal{F}\{x(t)\}$ and then square it? –  Pragabhava Oct 3 '12 at 1:13
    
@Pragabhava yes but the real part :) –  Parhs Oct 3 '12 at 10:15
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1 Answer 1

Try using convolution rule (specific for this case): $\mathcal{F}[x(t)]\cdot\mathcal{F}[x(t)]=\frac{1}{2\pi}\mathcal{F}[x(t)\ast x(t)]$.

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