ASN4 -Fourier Transform property

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Find ${\mathcal {F}}[cos(2\pi f_{0}t)g(t)]\!$

Recall $w_{0}=2\pi f_{0}\!$, so

Using Euler's cosine identity $\int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }{\frac {1}{2}}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt\!$

Now ${\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f-f_{0})t}g(t)dt\ +\ {\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f+f_{0})t}g(t)dt\!$

So ${\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}[G(f-f_{0})+G(f+f_{0})]\!$