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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]] | ||
+ | ---- | ||
+ | |||
+ | |||
<math> x(t) = e^{-|t-1|} \,</math><br><br> | <math> x(t) = e^{-|t-1|} \,</math><br><br> | ||
<math> X(w) = \int_{-\infty}^{\infty}e^{-|t-1|}e^{-jwt}dt</math><br><br> | <math> X(w) = \int_{-\infty}^{\infty}e^{-|t-1|}e^{-jwt}dt</math><br><br> | ||
<math> X(w) = \int_{-\infty}^{1}e^{(t-1)}e^{-jwt}dt+\int_{1}^{\infty}e^{-(t-1)}e^{-jwt}dt</math><br><br> | <math> X(w) = \int_{-\infty}^{1}e^{(t-1)}e^{-jwt}dt+\int_{1}^{\infty}e^{-(t-1)}e^{-jwt}dt</math><br><br> | ||
− | <math> X(w) = \int_{-\infty}^{1}e^{-1}e^{(1-jw)t}dt+\int_{1}^{\infty}e^{1}e^{(1+jw)t}dt</math><br><br> | + | <math> X(w) = \int_{-\infty}^{1}e^{-1}e^{(1-jw)t}dt+\int_{1}^{\infty}e^{1}e^{-(1+jw)t}dt</math><br><br> |
− | <math> X(w) = {\left.\frac{e^{-1}e^{(1-jw)t}}{1-jw}\right] | + | <math> X(w) = {\left.\frac{e^{-1}e^{(1-jw)t}}{1-jw}\right]^1_{-\infty} }+{\left.\frac{-e^{1}e^{-(1+jw)t}}{1+jw}\right]^{\infty}_1 }</math><math> = e^{-1}\frac{e^{(1-jw)}}{1-jw}+e^{1}\frac{e^{-(1+jw)}}{1+jw}</math><br><br> |
+ | <math> X(w) = \frac{e^{-jw}}{1-jw}+\frac{e^{-jw}}{1+jw}</math><br><br> | ||
+ | <math> X(w) = \frac{2e^{-jw}}{1+w^2}</math><br><br> | ||
+ | ---- | ||
+ | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] |
Latest revision as of 11:33, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
$ x(t) = e^{-|t-1|} \, $
$ X(w) = \int_{-\infty}^{\infty}e^{-|t-1|}e^{-jwt}dt $
$ X(w) = \int_{-\infty}^{1}e^{(t-1)}e^{-jwt}dt+\int_{1}^{\infty}e^{-(t-1)}e^{-jwt}dt $
$ X(w) = \int_{-\infty}^{1}e^{-1}e^{(1-jw)t}dt+\int_{1}^{\infty}e^{1}e^{-(1+jw)t}dt $
$ X(w) = {\left.\frac{e^{-1}e^{(1-jw)t}}{1-jw}\right]^1_{-\infty} }+{\left.\frac{-e^{1}e^{-(1+jw)t}}{1+jw}\right]^{\infty}_1 } $$ = e^{-1}\frac{e^{(1-jw)}}{1-jw}+e^{1}\frac{e^{-(1+jw)}}{1+jw} $
$ X(w) = \frac{e^{-jw}}{1-jw}+\frac{e^{-jw}}{1+jw} $
$ X(w) = \frac{2e^{-jw}}{1+w^2} $