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'''Problem Statement'''
'''Problem Statement'''


6(a) Show <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S(0) = 0 </math>.
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>.


6(b) If <math> S(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>?
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>?


'''Answer'''
'''Answer'''


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 as a substitution such that <math> \lambda= t-t_0 \! </math>
Then <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!</math>


See then that <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!
and <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f)- S(f_0) \right] \,d\lambda \! </math>


The problem statement says to make <math>S(f_0)=0 \!</math> that makes the above equation simplify to
<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>


<math>\mbox{ if } S(0) = 0\,\,\, \int_{-\infty}^{\infty} s(t) dt =0 \!</math>
<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </math>


Taking the inverse Fourier Transform and changing the order of intgration


</math> and <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f)- S(f_0) \right] \,d\lambda \! </math>
<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>


Then


<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}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </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>


<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>


b)If <math>S(f_0)\neq 0</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>


Therefore <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \!</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>

Latest revision as of 20:58, 18 December 2009

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Problem Statement

6(a) Show .

6(b) If can you find in terms of ?

Answer

a)Remember that dummy variable was used in substitution such that

Then

and

The problem statement says to make that makes the above equation simplify to

Taking the inverse Fourier Transform and changing the order of intgration

Then

Therefore it is demonstrated that


b)If

Then