My goal is to find the Fourier transform of the following functions.

$$y_1(t) = (t-2)e^{-t} u(t-2)$$ $$y_2(t) = y_1(t)\times \cos(2\pi mt)$$ $$y_3(t) =y_2(t)\times A\cos (2\pi mt +\phi))$$

$m$, $A$ and $\phi$ are just unknowns.

I have found:

$\mathcal{F}\{y_1(t)\} = \frac{e^{-2(1+j2\pi f)} }{(1+j2\pi f)^2}$
$\mathcal{F}\{\cos(2\pi mt)\} = \frac{1}{2}[\delta(f-m)+\delta(f+m))]$

I am aware of the modulation property of the Fourier transform, so I believe I need to use the Convolution Theorem. However, I am unsure how to sub these transforms into the convolution formula and solve further.

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Let $\mathcal{F}\{y_1(t)\}=Y_1(f)$ Then using the convolution theorem, $$\mathcal{F}\{y_2(t)\}=Y_2(f)=\frac{1}{2}\left(Y_1(f-m)+Y_1(f+m)\right)$$ and assuming$$\mathcal{F}\{\cos(2\pi mt+\phi)\}=\frac{1}{2}\left(e^{j\phi}\delta(f-m)+e^{-j\phi}\delta(f+m)\right)$$ we have $$\mathcal{F}\{y_3(t)\}=Y_3(f)=\frac{A}{2}\left(e^{j\phi}Y_2(f-m)+e^{-j\phi}Y_2(f+m)\right)$$

This is generally due to the shifting property of delta function: $\delta(f\pm f_0)*F(f)=F(f\pm f_0)$

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  • $\begingroup$ Thanks a lot for this! Wouldn't the transform of y3 be Y2(f-m)... not Y1(f-m).. Or am I mistaken? $\endgroup$ – tinefihu Sep 21 '16 at 15:25

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