ASN4 -Fourier Transform property: Difference between revisions

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<math>\mathcal{F}[cos(2\pi f_0t)g(t)]= \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt  \!</math>
Find the Fourier transform of <math> cos(2\pi f_0t)g(t) \!</math>


Using Euler's cosine identity
<math>\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math>


<math>\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math>
<math> \mathcal{F}[cos(2\pi f_0t)g(t)] \!</math>


<math> \mathcal{F}[cos(2\pi f_0t)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 \!</math>
Applying the forward Fourier transform


Identifying that the above equation contains Fourier Transforms the solution is
<math> =\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt \!</math>
 
Applying Euler's cosine identity
 
<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>
 
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>
 
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)