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Taking the inverse Fourier Transform and changing the order of intgration |
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Taking the inverse Fourier Transform and changing the order of intgration |
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<math> \int_{- \infty}^{t} s(\lambda ) \,d\lambda = \int_{- \infty}^{t} e^{j2 \pi f t} \,dt \int_{- \infty}^{\infty} S(f)\,df = \frac{ e^{j2 \pi f t}} {j2 \pi f }\int_{- \infty}^{\infty} S(f) \,df =\! </math> |
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<math> \int_{- \infty}^{t} s(\lambda ) \,d\lambda = \int_{- \infty}^{t} e^{j2 \pi f t} \,dt \int_{- \infty}^{\infty} S(f)\,df = \frac{ e^{j2 \pi f t}} {j2 \pi f }\int_{- \infty}^{\infty} S(f) \,df \! </math> |
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Then |
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Then |
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b) If <math>S(f_0)\neq 0</math> |
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b) If <math>S(f_0)\neq 0</math> |
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Then |
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Then <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}^{-1}\left[ S (f)- S(f_0) \right] \,d\lambda = \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t} [S (f)- S(f_0)] \,d\lambda\,d\lambda \! </math>
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<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}^{-1}\left[ S (f)- S(f_0) \right] \,d\lambda = \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t} [S (f)- S(f_0)] \,d\lambda\,d\lambda \! </math> |
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<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{\infty} e^{j2 \pi f t}S (f)\,d\lambda\,d\lambda - \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t} S(f_0) \,d\lambda\,d\lambda \! </math> |
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<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t}S (f)\,d\lambda\,d\lambda - \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t} S(f_0) \,d\lambda\,d\lambda \! </math> |
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<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{\infty}\frac{ e^{j2 \pi f t}} {j2 \pi f } S (f)\,d\lambda - - \int_{- \infty}^{\infty}\frac{ e^{j2 \pi f t}} {j2 \pi f } S (f_0)\,d\lambda \! </math> |
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<math> dt=d\lambda</math> and taking the Fourier transform of the equation |
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<math> dt=d\lambda</math> and taking the Fourier transform of the equation |
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answer is <math>\mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = S(f) - S(f_0) \! </math> |
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answer is <math>\mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{ e^{j2 \pi f t}} {j2 \pi f }S(f) - \frac{ e^{j2 \pi f t}} {j2 \pi f }S(f_0) \! </math> |
Revision as of 22:29, 18 December 2009
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Problem Statement
6(a) Show ![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}{\mbox{ if }}S(f_{0})=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b0fd565cfe32ef4c48175f2aa079fab985576cc)
.
6(b) If 
can you find ![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0af3b7173a594a013131920839a34842568e41f3)
in terms of 
?
Answer
a)Remember that dummy variable 
was used in substitution such that 
Then ![{\displaystyle s(\lambda )=s(t-t_{0})={\mathcal {F}}\left[S(f)-S(f_{0})\right]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a698d9b80c3ed2d1235e2c81c6f71635c9e722d)
and ![{\displaystyle \int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}\left[S(f)-S(f_{0})\right]\,d\lambda \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/426689bcd3afea3043dbafdd065ae7f95a1a90cd)
The problem statement says to make 
that makes the above equation simplify to
![{\displaystyle \int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}\left[S(f)\right]\,dt\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d31315811e9778e11a05fa2750ba76d55b513dd0)
Taking the inverse Fourier Transform and changing the order of intgration
Then
![{\displaystyle \int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{\infty }^{\infty }S(f){\frac {e^{j2\pi ft}}{j2\pi f}}\,df={\mathcal {F}}^{-1}\left[{\frac {S(f)}{j2\pi f}}\right]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a55ae349fb07fecbca1c1baf4c66ed8408702d6)
Therefore it is demonstrated that ![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27447ac8285d60d653060b8d93c9c08046a3069b)
b) If
Then
![{\displaystyle \int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}^{-1}\left[S(f)-S(f_{0})\right]\,d\lambda =\int _{-\infty }^{t}\int _{-\infty }^{\infty }e^{j2\pi ft}[S(f)-S(f_{0})]\,d\lambda \,d\lambda \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d848131ae6c557ee33f85e2d6f7643bbb7f1e320)
and taking the Fourier transform of the equation
answer is ![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {e^{j2\pi ft}}{j2\pi f}}S(f)-{\frac {e^{j2\pi ft}}{j2\pi f}}S(f_{0})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93d1706787189713de1901a98b9b0ecad59cc163)