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'''Problem Statement''' |
'''Problem Statement''' |
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6(a) Show <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S( |
6(a) Show <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S(f_0) = 0 </math>. |
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6(b) If <math> S( |
6(b) If <math> S(f_0) \neq 0 </math> can you find <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] </math> in terms of <math> \displaystyle S(0) </math>? |
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'''Answer''' |
'''Answer''' |
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a)Remember that dummy variable <math> \lambda \!</math> was used in substitution such that <math> \lambda= t-t_0 \! </math> |
a)Remember that dummy variable <math> \lambda \!</math> was used in substitution such that <math> \lambda= t-t_0 \! </math> |
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Then <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!</math> |
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In the problemstatement it says to make <math>S(f_0)=0 \!</math> |
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The problem statement says to make <math>S(f_0)=0 \!</math> that makes the above equation simplify to |
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<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </math> |
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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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Then |
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<math> \mathcal{F}^{-1}\left[ S (f)- S(f_0) \right] = \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}^{\infty} S(f)\frac{ e^{j2 \pi f t}} {j2 \pi f }\,df = \mathcal{F }^{-1}\left[ \frac{S(f)}{j2 \pi f} \right] \! </math> |
<math>\int_{- \infty}^{t} s(\lambda ) \,d\lambda = \int_{\infty}^{\infty} S(f)\frac{ e^{j2 \pi f t}} {j2 \pi f }\,df = \mathcal{F }^{-1}\left[ \frac{S(f)}{j2 \pi f} \right] \! </math> |
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Therefore <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \!</math> |
Therefore it is demonstrated that <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \!</math> |
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| ⚫ | |||
| ⚫ | |||
The problem statement says let <math>f_0=0 \!</math> where <math>S(0)= S(f)|_{f=0} = \int_{-\infty}^{\infty} s(t)e^{- j 2 \pi f t} dt = \int_{-\infty}^{\infty} s(t) dt \! </math> |
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<math>\ |
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 \! </math> |
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Latest revision as of 20:58, 18 December 2009
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 in terms of ?
![{\displaystyle {\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0af3b7173a594a013131920839a34842568e41f3)
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 \!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8c048e3de24dd4d756b826e0a82184bc19a1a33)