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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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'''Find '''<math>\mathcal{F}[cos(2\pi f_0t)g(t)]\!</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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Using Euler's cosine identity |
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Recall <math> w_0 = 2\pi f_0\!</math>, so |
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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 f_0t}]g(t)e^{-j2\pi ft}dt\!</math> |
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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>
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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 f_0t}]g(t)e^{-j2\pi ft}dt\!</math> |
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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>
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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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Identifying that the above equation contains Fourier Transforms the solution is |
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So <math>\mathcal{F}[cos(2\pi f_0t)g(t)] = \frac{1}{2} [G(f-f_0)+ 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> |
Revision as of 22:54, 18 December 2009
Back to my home page
![{\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)
Using Euler's cosine identity
![{\displaystyle \int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }{\frac {1}{2}}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa67da17dc1d9f04fa8b72c6d0cf3a712f40c70f)
![{\displaystyle {\mathcal {F}}[cos(2\pi f_{0}t)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\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/553ec68dad91d376d485594e651e85d3840e58d9)
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)