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> |
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'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>''' |
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To begin, we know that<br/> |
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<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> |
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<br/> |
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But recall that, <math>e^{j2 \pi f(t_{0}-t} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math><br> |
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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> |
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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> |
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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> |
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Revision as of 14:05, 16 October 2009
Max Woesner
Find
Recall , so
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Nick Christman
Find
To begin, we know that
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