ASN4 -Fourier Transform property: Difference between revisions

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[[Jodi Hodge| Back to my home page]]
[[Jodi Hodge| Back to my home page]]


'''Find '''<math>\mathcal{F}[cos(2\pi f_0t)g(t)]\!</math>
<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>


Recall <math> w_0 = 2\pi f_0\!</math>, so
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>


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>


Now <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>
<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>


So <math>\mathcal{F}[cos(2\pi f_0t)g(t)] = \frac{1}{2}[G(f-f_0)+ G(f+f_0)]\!</math>
Identifying that the above equation contains Fourier Transforms the solution 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>

Revision as of 23:54, 18 December 2009

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[cos(2πf0t)g(t)]=cos(2πf0t)g(t)ej2πftdt

Using Euler's cosine identity cos(2πf0t)g(t)ej2πftdt=12[ej2πf0t+ej2πf0t]g(t)ej2πftdt

cos(2πf0t)g(t)ej2πftdt=12[ej2πf0t+ej2πf0t]g(t)ej2πftdt

[cos(2πf0t)g(t)]=12ej2π(ff0)tg(t)dt+12ej2π(f+f0)tg(t)dt

Identifying that the above equation contains Fourier Transforms the solution is

[cos(2πf0t)g(t)]=12G(ff0)+12[G(f+f0)