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Jodi.Hodge at 04:58, 19 December 2009

2009-12-19T04:58:48Z

← Older revision		Revision as of 21:58, 18 December 2009
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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'''

Jodi.Hodge at 04:57, 19 December 2009

2009-12-19T04:57:08Z

← Older revision		Revision as of 21:57, 18 December 2009
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	b)If <math>S(f_0)\neq 0</math>		b)If <math>S(f_0)\neq 0</math>
	Then <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\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>
			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>

Jodi.Hodge at 04:52, 19 December 2009

2009-12-19T04:52:53Z

← Older revision		Revision as of 21:52, 18 December 2009
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	and <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f)- S(f_0) \right] \,d\lambda \! </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>


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

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






	~~In my notations <math>S(f_0)=S(f)\|_{f=0~~ \!</math>

	~~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>~~

	<math>\~~mbox~~{ if } S(0) ~~= 0~~\,\,\, \int_{-\infty}^{\infty} s(t) ~~dt =0~~ \!</math>		b)If <math>S(f_0)\neq 0</math>
			Then <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\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>

Jodi.Hodge at 04:42, 19 December 2009

2009-12-19T04:42:40Z

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

	<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>

Jodi.Hodge at 04:40, 19 December 2009

2009-12-19T04:40:30Z

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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>

	~~This means~~ <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!</math>		Then <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!</math>

	~~In the problemstatement it says to make <math>S(f_0)=0 \!</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>~~ 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> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </math>		<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </math>

Jodi.Hodge at 04:36, 19 December 2009

2009-12-19T04:36:33Z

← Older revision		Revision as of 21:36, 18 December 2009
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	'''Answer'''		'''Answer'''

	a)		a)Remember that dummy variable <math> \lambda \!</math> was used in substitution such that <math> \lambda= t-t_0 \! </math>

	~~Remember that dummy variable <math> \lambda \!</math> was used in substitution such that~~ <math> \lambda= t-t_0 \! </math>		This means <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!</math>

	~~This means <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!</math>~~ In ~~my notations~~ <math>S(f_0)~~=S(f)\|_{f~~=0 \!</math>		In the problemstatement it says to make <math>S(f_0)=0 \!</math>

	~~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>~~



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






			In my notations <math>S(f_0)=S(f)\|_{f=0 \!</math>

			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>

			<math>\mbox{ if } S(0) = 0\,\,\, \int_{-\infty}^{\infty} s(t) dt =0 \!</math>

Jodi.Hodge at 04:15, 19 December 2009

2009-12-19T04:15:47Z

← Older revision		Revision as of 21:15, 18 December 2009
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	a)		a)

	Remember that dummy variable<math> \lambda</math> was used ~~as a~~ substitution such that <math> \lambda= t-t_0 \! </math>		Remember that dummy variable <math> \lambda \!</math> was used in substitution such that <math> \lambda= t-t_0 \! </math>

	~~See then that~~ <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!		This means <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!</math> In my notations <math>S(f_0)=S(f)\|_{f=0 \!</math>

	<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>		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>



	<math>\mbox{ if } S(0) = 0\,\,\, \int_{-\infty}^{\infty} s(t) dt =0 \!</math>		<math>\mbox{ if } S(0) = 0\,\,\, \int_{-\infty}^{\infty} s(t) dt =0 \!</math>

Jodi.Hodge at 04:02, 19 December 2009

2009-12-19T04:02:29Z

← Older revision		Revision as of 21:02, 18 December 2009
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	a)		a)

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

			See then that <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \!

	<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>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>\mbox{ if } S(0) = 0\,\,\, \int_{-\infty}^{\infty} s(t) dt =0 \!</math>


			</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>

Jodi.Hodge: New page: Back to my home page '''Problem Statement''' 6(a) Show $\mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S(0...$

2009-12-19T03:58:09Z

New page: Back to my home page '''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...

New page

[[Jodi Hodge|Back to my home page]]

'''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(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>?

'''Answer'''

a)

Remember dummy variable <math> \lambda= t-t_0 \! </math> Then <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \! </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>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>

<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>

<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>