Link title: Difference between revisions
		
		
		
		Jump to navigation
		Jump to search
		
| Jodi.Hodge (talk | contribs) No edit summary | Jodi.Hodge (talk | contribs) No edit summary | ||
| Line 32: | Line 32: | ||
| 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> | |||
Revision as of 21:57, 18 December 2009
Problem Statement
6(a) Show .
6(b) If can you find in terms of ?
Answer
a)Remember that dummy variable was used in substitution such that
Then
and
The problem statement says to make that makes the above equation simplify to
Taking the inverse Fourier Transform and changing the order of intgration
Then
Therefore it is demonstrated that
b)If   
Then