ASN4 -Fourier Transform property: Difference between revisions

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<math>  =\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 \!</math>
<math>  =\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 \!</math>


Note the Inverse Fourier Transform expressions in the above equation. With substitution the result is
Note that there are forward Fourier Transform expressions in the above equation. With substitution the result is
   
   
<math>\mathcal{F}[cos(2\pi f_0t)g(t)] = \frac{1}{2}G(f-f_0)+ \frac{1}{2}[G(f+f_0)\!</math>
<math>\mathcal{F}[cos(2\pi f_0t)g(t)] = \frac{1}{2}G(f-f_0)+ \frac{1}{2}[G(f+f_0)\!</math>

Revision as of 11:28, 19 December 2009

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Find the Fourier transform of cos(2πf0t)g(t)


[cos(2πf0t)g(t)]

Applying the forward Fourier transform

=cos(2πf0t)g(t)ej2πftdt

Applying Euler's cosine identity

=[12ej2πf0t+12ej2πf0t]g(t)ej2πftdt

=12ej2πf0tej2πftdt+12ej2πf0tg(t)ej2πftdt

=12ej2π(ff0)tg(t)dt+12ej2π(f+f0)tg(t)dt

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

[cos(2πf0t)g(t)]=12G(ff0)+12[G(f+f0)