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] \,dt \! </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}^{-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>

Latest revision as of 20:55, 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)]dt

1[S(f)S(f0)]=tej2πftdtS(f)df=ej2πftj2πfS(f)df=

ts(λ)dλ=S(f)ej2πftj2πfdf=1[S(f)j2πf]

Therefore [ts(λ)dλ]=S(f)j2πf