Fourier Transform Properties: Difference between revisions

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So <math>\mathcal{F}[cos(w_0t)g(t)] = \frac{1}{2}[G(f-f_0)+ G(f+f_0)]\!</math>
So <math>\mathcal{F}[cos(w_0t)g(t)] = \frac{1}{2}[G(f-f_0)+ G(f+f_0)]\!</math>


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[[Nick Christman|<b><u>Nick Christman</u></b>]]<br><br>
[[Nick Christman|<b><u>Nick Christman</u></b>]]<br><br>
'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''

To begin, we know that<br/>

<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>

<br/>

But recall that, <math>e^{j2 \pi f(t_{0}-t} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math><br>
Also recall <math> cos(\theta) = \frac{1}{2}(e^{j\theta} + e^{-j\theta})\!</math>,so <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><br>
Now <math>\int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt = \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 = \frac{1}{2}G(f-f_0)+ \frac{1}{2}G(f+f_0)\!</math><br>
So <math>\mathcal{F}[cos(w_0t)g(t)] = \frac{1}{2}[G(f-f_0)+ G(f+f_0)]\!</math>


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Revision as of 14:05, 16 October 2009

Max Woesner

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Nick Christman

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