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[[Jodi Hodge| Back to my home page]] |
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[[Jodi Hodge| Back to my home page]] |
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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> |
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Find the Fourier transform of <math> cos(2\pi f_0t)g(t)= \!</math> |
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\mathcal{F}[cos(2\pi f_0t)g(t)]= |
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Using Euler's cosine identity |
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Using Euler's cosine identity |
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<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}+\frac{1}{2}e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math> |
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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> |
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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> |
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<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}+\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math> |
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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> |
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<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 ft} dt + \int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math> |
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<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> |
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<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-f_0)t}g(t)dt \ + \ \frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi (f+f_0)t}g(t)dt \!</math> |
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<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> |
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Identifying that the above equation contains Fourier Transforms the solution is |
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Identifying that the above equation contains Fourier Transforms the solution is |
Revision as of 10:07, 19 December 2009
Back to my home page
Find the Fourier transform of 
\mathcal{F}[cos(2\pi f_0t)g(t)]=
Using Euler's cosine identity
![{\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]=\int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2141863fea59d36f0fb6be6452b53a62cc58accd)
![{\displaystyle =\int _{-\infty }^{\infty }[{\frac {1}{2}}e^{j2\pi f_{0}t}+{\frac {1}{2}}e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0064fdf1ad3b4066363ede60e3407a87b286e017)
Identifying that the above equation contains Fourier Transforms the solution is
![{\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}G(f-f_{0})+{\frac {1}{2}}[G(f+f_{0})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3860455af7a9378c44f2e45201bbca36c3ffb589)