ASN6 a,b- fixing: Difference between revisions
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<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> \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> |
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<math>\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> |
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Therefore <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \!</math> |
Latest revision as of 19:55, 18 December 2009
Problem Statement
6(a) Show . Hint:
6(b) If can you find in terms of ?
Answer
a)
Remember dummy variable Then and
where
Therefore