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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier transform of a CT SIGNAL == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
==signal == | ==signal == | ||
assume | assume | ||
| − | <math>x(t) = e^{-5t}u(t) + e^{-4(t-1)}u(t-3)</math> | + | <math>x(t) = e^{-5t}u(t) + e^{-4(t-1)}u(t-3)\!</math> |
| Line 8: | Line 16: | ||
== answer == | == answer == | ||
| − | <math> | + | <math>X(w) = \int_{-\infty}^{\infty}x(t)e^{-jwt}dt</math> |
<math>=\int_{-\infty}^{\infty}(e^{-5t}u(t) + e{-4(t-1)})e^{-jwt}dt</math> | <math>=\int_{-\infty}^{\infty}(e^{-5t}u(t) + e{-4(t-1)})e^{-jwt}dt</math> | ||
| Line 18: | Line 26: | ||
<math>=\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4*\int_{\infty}^{3}e^{-(4+jw)t}dt</math> | <math>=\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4*\int_{\infty}^{3}e^{-(4+jw)t}dt</math> | ||
| − | <math>\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4 * \frac{e^{-(4+jw)t}}{-(4+jw)}\bigg]_3^\infty</math> | + | <math>=\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4 * \frac{e^{-(4+jw)t}}{-(4+jw)}\bigg]_3^\infty</math> |
| + | |||
| + | <math>=\frac{1}{(5+jw)}+ e^4 * \frac{e^{-(12+3jw)}}{(4+jw)}</math> | ||
| + | |||
| + | <math>=\frac{1}{(5+jw)} + \frac{e^{-(8+3jw)}}{(4+jw)}</math> | ||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:31, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
signal
assume
$ x(t) = e^{-5t}u(t) + e^{-4(t-1)}u(t-3)\! $
answer
$ X(w) = \int_{-\infty}^{\infty}x(t)e^{-jwt}dt $
$ =\int_{-\infty}^{\infty}(e^{-5t}u(t) + e{-4(t-1)})e^{-jwt}dt $
$ =\int_{0}^{\infty}e^{-5t} e^{-jwt}dt + \int_{3}^{\infty}e^{-4(t-1)}e^{-jwt}dt $
$ =\frac{e^{-(5+jw)t}}{-(5+jw)}\bigg]^{\infty}_{0}+ \int_{3}^{\infty}e^{-4t}e^{4}e^{-jwt}dt $
$ =\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4*\int_{\infty}^{3}e^{-(4+jw)t}dt $
$ =\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4 * \frac{e^{-(4+jw)t}}{-(4+jw)}\bigg]_3^\infty $
$ =\frac{1}{(5+jw)}+ e^4 * \frac{e^{-(12+3jw)}}{(4+jw)} $
$ =\frac{1}{(5+jw)} + \frac{e^{-(8+3jw)}}{(4+jw)} $
