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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...
 
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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>





Revision as of 21:02, 18 December 2009

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Problem Statement

6(a) Show ℱ[∫−∞ts(λ)dλ]=S(f)j2πf if S(0)=0.

6(b) If S(0)≠0 can you find ℱ[∫−∞ts(λ)dλ] in terms of S(0)?

Answer

a)

Remember that dummy variableλ was used as a substitution such that λ=t−t0

See then that s(λ)=s(t−t0)=ℱ[S(f)−S(f0)]<math>f0=0 where S(0)=S(f)|f=0=∫−∞∞s(t)e−j2πftdt=∫−∞∞s(t)dt

 if S(0)=0∫−∞∞s(t)dt=0


</math> and ∫−∞ts(λ)dλ=∫−∞tℱ[S(f)−S(f0)]dλ


∫−∞ts(λ)dλ=∫−∞tℱ[S(f)]dt

ℱ−1[S(f)−S(f0)]=∫−∞tej2πftdt∫−∞∞S(f)df=ej2πftj2πf∫−∞∞S(f)df=

∫−∞ts(λ)dλ=∫∞∞S(f)ej2πftj2πfdf=ℱ−1[S(f)j2πf]

Therefore ℱ[∫−∞ts(λ)dλ]=S(f)j2πf