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This is an example from my text book of a continuous signal: $$x_{in}(t)=\sin \left( 2\pi \cdot 1000 \cdot t\right) + 0.5\sin \left( 2\pi \cdot 2000 \cdot t + \dfrac{3\pi}{4} \right) $$ So to perform a fourier transform on this signal, how to do that, isn't it a bit funny, since it has two sine components. Shouldn't complex numbers have a sine term and a cosine term? And it's got a scalar term applied to only one component, don't those usually apply across both terms of a complex component? And it's phase shifted, what to do about that?

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  • $\begingroup$ actually that equation is inconsequential isn't it? since we don't perform an FT on that, we'll just have a bunch of numbers, supposedly generated from those, and it's on those # that we perform the foueier transform operation isn't it? $\endgroup$ – user8769 May 29 '14 at 2:48
  • $\begingroup$ Have you tried writing out the Fourier transform and evaluating the integral? $\endgroup$ – geometrikal May 29 '14 at 5:31
  • $\begingroup$ Another hint: $\cos(w t + \phi) = \cos(\phi) \cos(wt)-\sin(\phi) \sin(wt)$ $\endgroup$ – geometrikal May 29 '14 at 5:34
  • $\begingroup$ yeah I just need to actually plug in for N and calculate everything out isn't it? $\endgroup$ – user8769 May 29 '14 at 8:16
  • $\begingroup$ yea sometimes working through the maths can give you a real feel for what is going on $\endgroup$ – geometrikal May 29 '14 at 9:12
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Fourier Transform is a linear one, so you can make use of superposition principle:

$$ \mathscr{F} [\alpha x(t) + \beta y(t)] = \alpha \mathscr{F}[x(t)] + \beta \mathscr{F}[y(t)] $$

So for the first component $$x(t) = \sin \left( 2\pi \cdot 1000 \cdot t\right)$$

by definition:

$$\mathscr{F}\left[\sin(2\pi f_0 t + \phi) \right] = \dfrac{i}{2} \left[ e^{-i \phi}\delta(f+f_0) - e^{i \phi}\delta(f-f_0) \right] $$

you get:

$$ \mathscr{F}[x(t)]=\dfrac{i}{2} \left[ \delta(f+1000) - \delta(f-1000) \right] $$

Second component is a sinusoid with shifted phase, so the complex exponent represents that:

$$y(t) = \dfrac{1}{2} \sin \left( 2\pi \cdot 2000 \cdot t + \dfrac{3\pi}{4} \right)$$

has following Fourier Transform:

$$\mathscr{F}[y(t)] = \dfrac{1}{2}\dfrac{i}{2} \left[ e^{\dfrac{-3\pi i}{4}}\delta(f+2000) - e^{\dfrac{3\pi i}{4}}\delta(f-2000) \right] $$

By summing both results you get the Fourier Transform of your signal.

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