ASN6 a,b- fixing: Difference between revisions

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


<math>  \mathcal{F}\left[ S (f)- S(f_0) \right]  = \int_{- \infty}^{t}e^{-j2 \pi f t}\int_{- \infty}^{\infty} S(f)e^{-j2 \pi f t}\,df = \frac{ e^{-j2 \pi f t}} {-j2 \pi f }\int_{- \infty}^{\infty} S(f) \,df </math>
<math>  \mathcal{F}\left[ S (f)- S(f_0) \right]  = \int_{- \infty}^{\infty}e^{-j2 \pi f t}\int_{- \infty}^{t} S(f)e^{-j2 \pi f t}\,df = \frac{ e^{-j2 \pi f t}} {-j2 \pi f }\int_{- \infty}^{t} S(f) \,df </math>

Revision as of 20:30, 18 December 2009

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Problem Statement

6(a) Show [ts(λ)dλ]=S(f)j2πf if S(0)=0. Hint: S(0)=S(f)|f=0=s(t)ej2π(f0)tdt=s(t)dt

6(b) If S(0)0 can you find [ts(λ)dλ] in terms of S(0)?

Answer

a)

Remember dummy variable λ=tt0 Then s(λ)=s(tt0)=[S(f)S(f0)] and ts(λ)dλ=t[S(f)S(f0)]dλ

f0=0 where S(0)=S(f)|f=0=s(t)ej2πftdt=s(t)dt

 if S(0)=0s(t)dt=0


ts(λ)dλ=t[S(f)]dλ

[S(f)S(f0)]=ej2πfttS(f)ej2πftdf=ej2πftj2πftS(f)df