ASN4 -Fourier Transform property

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

${ℱ}[cos(2\pi f_{0}t)g(t)]$

Applying the forward Fourier transform

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

Applying Euler's cosine identity

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

Distribting to both terms in side the brackets

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

Combining exponential terms

$$\begin{gathered} =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{2}{e}^{-j2\pi (f-f_0)t}g(t)dt \\ +\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{2}{e}^{-j2\pi (f+f_0)t}g(t)dt \end{gathered}$$

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

${ℱ}[cos(2\pi f_{0}t)g(t)]=\frac{1}{2}G(f-f_0)+\frac{1}{2}[G(f+f_0)$