ASN4 -Fourier Transform property

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Find the Fourier transform of $cos(2\pi f_{0}t)g(t)\!$

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

Applying the forward Fourier transform

$=\int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt\!$

Applying Euler's cosine identity

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

Distribting to both terms in side the brackets

$=\int _{-\infty }^{\infty }{\frac {1}{2}}e^{j2\pi f_{0}t}e^{-j2\pi ft}dt+\int _{-\infty }^{\infty }{\frac {1}{2}}e^{-j2\pi f_{0}t}g(t)e^{-j2\pi ft}dt\!$

Combining exponential terms

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

Note that there are forward Fourier Transform expressions in the above equation. With substitution the result is

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