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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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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)] \!</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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Applying the forward Fourier transform |
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<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> |
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<math> =\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt \!</math> |
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Identifying that the above equation contains Fourier Transforms the solution is |
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Applying Euler's cosine identity |
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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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Distribting to both terms in side the brackets |
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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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Combining exponential terms |
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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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Note that there are forward Fourier Transform expressions in the above equation. With substitution the result is |
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<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> |
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<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 10:37, 19 December 2009
Back to my home page
Find the Fourier transform of 
![{\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)g(t)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e2df42cc264afce1eae7d5bdcdcdda53516f72c)
Applying the forward Fourier transform
Applying Euler's cosine identity
![{\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)
Distribting to both terms in side the brackets
Combining exponential terms
Note that there are forward Fourier Transform expressions in the above equation. With substitution the result 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)