Allen Humphreys_ECE301Fall2008mboutin
Homework 5_ECE301Fall2008mboutin
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Contents
Problem 4.1
Using Calculus
Part a
Part b
Using the Table
Part a
$ f(t) = e^{-2(t-1)} \times u(t-1) $
remove the time shift where $ t_o = 1 $
- $ g(t) = e^{-2t} \times u(t) $
the time shift property in table 4.1 says:
- $ F(j\omega) = e^{-j\omega t_o} G(j\omega) $
from table 4.2 the FT of $ g(t) $ can be found
- $ G(j\omega) = \frac{1}{2 + j\omega} $
then the final answer can be found substituting 1 for $ t_o $
- $ F(j\omega) = e^{-j\omega} G(j\omega) = \frac{e^{-j\omega}}{2 + j\omega} $
Part 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} $
Problem 4.2
a
$ f(t) = \delta\big(t+1) + \delta(t-1) $
- $ F(j\omega) = \int_{-\infty}^{\infty}\delta(t+1)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty}\delta(t-1)e^{-j\omega t}\,dt $
by the sifting property of the delta function:
- $ F\big(j\omega) = e^{j \omega} + e^{-j \omega} = 2cos\big(\omega ) $
b
$ f(t) = \frac{d\lbrace u(-2-t) + u(t-2)\rbrace }{dt} = \delta(-2-t) + \delta(t-2) $ because $ \frac{d\{u\big(t)\} }{dt} = \delta(t) $
So very similar to part a we can take the integral and use the sifting property of the delta function
- $ F(j\omega) = \int_{-\infty}^{\infty}\delta(-2-t)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty}\delta(t-2)e^{-j\omega t}\,dt $
Paying special attention to the first integral, the resulting exponential is negative because the delta function is time reversed
- $ F\big(j\omega) = -e^{-2j \omega} + e^{2j \omega} = -2jsin\big(2\omega ) $
Problem 4.3
a
$ f(t) = sin(2 \pi t + \frac{\pi}{4}) $
b
$ f(t) = 1 + cos(6 \pi t + \frac{\pi}{8}) $
Problem 4.4
a
$ X_1\big(j\omega) = 2\pi \delta(\omega) + \pi \delta(\omega - 4\pi) + \pi \delta(\omega + 4 \pi) $
b
$ X_2\big(j\omega) = \begin{cases} 2, \,\,\,\,\,\,\,\, 0 \le \omega \le 2 \\ -2,\,\, -2 \le \omega < 0 \\ 0,\,\,\,\,\,\, |\omega| > 2 \end{cases} $
Problem 4.5
Find the inverse Fourier transform of:
- $ X(j\omega) = |X(j\omega)|e^{j \sphericalangle X(j\omega)} $
Given that:
- $ \big|X(j\omega)| = 2\lbrace u(\omega +3) - u(\omega - 3)\rbrace $
- $ \sphericalangle X(j \omega) = -\frac{2}{3} \omega + \pi $
Problem 4.21
Compute the Fourier transform of each of the following signals:
a
$ f\big(t) = [e^{-\alpha t}cos(\omega_0 t)]u(t), \alpha > 0 $
