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| + | |||
| + | ==4.1== | ||
| + | ===a=== | ||
| + | ===b=== | ||
| + | :<math> f(t) = e^{-2 |(t-1)|} </math> | ||
| + | :<math> f(t) = \begin{cases} | ||
| + | e^{-2 (t-1)}, t>1\\ | ||
| + | e^{-2 (1-t)}, t<1 | ||
| + | \end{cases} = | ||
| + | |||
| + | \begin{cases} | ||
| + | e^{-2 (t-1)}\times u(t-1) = h(t)\\ | ||
| + | e^{-2 (1-t)}\times u(1-t) = k(t) | ||
| + | \end{cases} | ||
| + | </math> | ||
| + | |||
| + | By the properties of integrating an absolute value and the linearity of the Fourier transform. | ||
| + | |||
| + | <math> F(j\times \omega) = H(j\times \omega) + K(j\times \omega)</math> | ||
| + | |||
| + | <math>H(j\times \omega) = \frac{e^{-j \omega}}{(2 + j \omega)}</math> from part a. | ||
| + | |||
| + | <math> | ||
| + | k(t) = e^{-2 (1-t)}\times u(1-t) | ||
| + | </math> | ||
| + | |||
| + | remove the time shift and time reversal | ||
| + | |||
| + | <math> m(t) = e^{-2(t)}\times u(t)</math> | ||
| + | |||
| + | from the table 4.2: | ||
| + | |||
| + | <math> M(j \omega) = \frac{1}{2 + j \omega} </math> | ||
| + | |||
| + | apply the time shift property from table 4.1: | ||
| + | |||
| + | <math> M(j \omega) = \frac{e^{-j\omega}}{2 + j \omega} </math> | ||
| + | |||
| + | apply the time reversal property from table 4.1 '''making sure to only apply it to the FT of the base function and not to the portion added by the time shift''': | ||
| + | |||
| + | <math> K(j \omega) = \frac{e^{-j\omega}}{2 - j \omega} </math> | ||
| + | |||
| + | <math> H(j \omega) + K(j \omega) = \frac{e^{-j \omega}}{2 + j \omega} + \frac{e^{-j\omega}}{2 - j \omega} </math> | ||
| + | |||
| + | finding common denominators: | ||
| + | |||
| + | <math> \frac{(2-j\omega)e^{-j \omega}}{2^2 + \omega^2} + \frac{(2+j\omega)e^{-j\omega}}{2^2 + \omega^2}</math> | ||
| + | |||
| + | in the numerator the <math> j\omega </math> terms will cancel when added yielding the final answer: | ||
| + | |||
| + | <math> F(j\omega) = \frac{4e^{-j \omega}}{4 + \omega^2} | ||
Revision as of 06:13, 8 October 2008
Allen Humphreys_ECE301Fall2008mboutin
Homework 4_ECE301Fall2008mboutin
| .1
| .2
| .3
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4.1
a
b
- $ f(t) = e^{-2 |(t-1)|} $
- $ f(t) = \begin{cases} e^{-2 (t-1)}, t>1\\ e^{-2 (1-t)}, t<1 \end{cases} = \begin{cases} e^{-2 (t-1)}\times u(t-1) = h(t)\\ e^{-2 (1-t)}\times u(1-t) = k(t) \end{cases} $
By the properties of integrating an absolute value and the linearity of the Fourier transform.
$ F(j\times \omega) = H(j\times \omega) + K(j\times \omega) $
$ H(j\times \omega) = \frac{e^{-j \omega}}{(2 + j \omega)} $ from part a.
$ k(t) = e^{-2 (1-t)}\times u(1-t) $
remove the time shift and time reversal
$ m(t) = e^{-2(t)}\times u(t) $
from the table 4.2:
$ M(j \omega) = \frac{1}{2 + j \omega} $
apply the time shift property from table 4.1:
$ M(j \omega) = \frac{e^{-j\omega}}{2 + j \omega} $
apply the time reversal property from table 4.1 making sure to only apply it to the FT of the base function and not to the portion added by the time shift:
$ K(j \omega) = \frac{e^{-j\omega}}{2 - j \omega} $
$ H(j \omega) + K(j \omega) = \frac{e^{-j \omega}}{2 + j \omega} + \frac{e^{-j\omega}}{2 - j \omega} $
finding common denominators:
$ \frac{(2-j\omega)e^{-j \omega}}{2^2 + \omega^2} + \frac{(2+j\omega)e^{-j\omega}}{2^2 + \omega^2} $
in the numerator the $ j\omega $ terms will cancel when added yielding the final answer:
$ F(j\omega) = \frac{4e^{-j \omega}}{4 + \omega^2} $
