ASN4 -Fourier Transform property: Difference between revisions

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<math> = \int_{-\infty}^{\infty} [\frac{1}{2}e^{j2\pi f_0t}+\frac{1}{2}e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math>
<math> = \int_{-\infty}^{\infty} [\frac{1}{2}e^{j2\pi f_0t}+\frac{1}{2}e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math>
Distribting to both terms in side the brackets


<math> = \int_{-\infty}^{\infty} \frac{1}{2}e^{j2\pi f_0t}e^{-j2\pi ft} dt + \int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math>
<math> = \int_{-\infty}^{\infty} \frac{1}{2}e^{j2\pi f_0t}e^{-j2\pi ft} dt + \int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math>
Combining exponential terms


<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>

Latest revision as of 11:37, 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

Distribting to both terms in side the brackets

=12ej2πf0tej2πftdt+12ej2πf0tg(t)ej2πftdt

Combining exponential terms

=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)