Difference between revisions of "HW5.1 Allen Humphreys ECE301Fall2008mboutin" - Rhea

Latest revision as of 17:59, 8 October 2008

Allen Humphreys_ECE301Fall2008mboutin
Homework 5_ECE301Fall2008mboutin | .1 | .2 | .3 | .4

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}) $

This problem is done strictly from the properties in the tables 4.1 and 4.2.

First, find the time shift of the sine function by factoring out a $ 2\pi $

$ f(t) = sin\Big(2 \pi (t + \frac{1}{8})\Big) $

Then we have a new function:

$ g(t) = sin\big(2\pi t) $

From table 4.2

$ G(\omega) = \frac{\pi}{j}\Big[ \delta(\omega - 2\pi ) - \delta(\omega + 2\pi )\Big] $

From table 4.1, the time shift property

$ F(\omega) = e^{-j\omega t_0} G\big(\omega)\,\,\,\,,t_0 = -\frac{1}{8} $

Plug in -1/8 and the negatives cancel:

$ F(\omega) = \frac{\pi}{j}\Big[ \delta(\omega - 2\pi )e^{j\omega \frac{1}{8}} - \delta(\omega + 2\pi )e^{j\omega \frac{1}{8}}\Big] $

The final final answer comes when we realize the the delta functions multiplied by each of the exponentials are only valid when $ \omega $ is such that it is delta(0):

$ F(\omega) = \frac{\pi}{j}\Big[ \delta(\omega - 2\pi )e^{2\pi \frac{1}{8}j} - \delta(\omega + 2\pi )e^{-2\pi \frac{1}{8}j}\Big] = \frac{\pi}{j}\Big[ \delta(\omega - 2\pi )e^{\frac{\pi}{4}j} - \delta(\omega + 2\pi )e^{-\frac{\pi}{4}j}\Big] $

b

$ f(t) = 1 + cos(6 \pi t + \frac{\pi}{8}) $

This is similar to the previous problem so I won't go through it in so many steps

$ F(w) = \mathcal{F}(1) + \pi\Big[ \delta(\omega - 6\pi )e^{\omega \frac{1}{48}j} - \delta(\omega + 6\pi )e^{\omega \frac{1}{48}j}\Big] $

$ = 2\pi\delta(\omega) + \pi\Big[ \delta(\omega - 6\pi )e^{\frac{\pi}{8}j} - \delta(\omega + 6\pi )e^{-\frac{\pi}{8}j}\Big] $

Problem 4.4

a

$ X_1\big(j\omega) = 2\pi \delta(\omega) + \pi \delta(\omega - 4\pi) + \pi \delta(\omega + 4 \pi) $

Apply the inverse fourier transform integral:

$ x_1(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\Big(2\pi \delta(\omega)e^{-j\omega t}\Big)\,d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\Big(\pi \delta(\omega - 4\pi)e^{-j\omega t} + \pi \delta(\omega + 4 \pi)e^{-j\omega t}\Big)\,d\omega $

Cancel the pi:

$ x_1(t) = \int_{-\infty}^{\infty}\Big(\delta(\omega)e^{-j\omega t}\Big)\,d\omega + \frac{1}{2}\int_{-\infty}^{\infty}\Big( \delta(\omega - 4\pi)e^{-j\omega t} + \delta(\omega + 4 \pi)e^{-j\omega t}\Big)\,d\omega $

Apply the sifting property:

$ x_1(t) = e^{0} + \frac{1}{2}\Big( e^{-4\pi j t} + e^{4\pi j t}\Big) $

Simplify using euler's formula

$ x_1(t) = 1 + cos\big(4\pi t) $

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{3}{2} \omega + \pi $

The entire integral:

$ \frac{2}{2\pi}\int_{-\infty}^{\infty}\Bigg(e^{-\frac{3}{2} j \omega + \pi j + j\omega t} u(\omega + 3) - e^{-\frac{3}{2} j \omega + \pi j + j\omega t} u(\omega - 3)\Bigg)\,d\omega $

Change the limits:

$ \frac{1}{\pi}e^{\pi j}\Bigg\{ \int_{3}^{\infty}\Bigg(e^{j\omega(t-\frac{3}{2})}\Bigg)\,d\omega - \int_{-3}^{\infty}\Bigg(e^{j\omega(t-\frac{3}{2})} \Bigg)\,d\omega \Bigg\} $

Integrate:

$ \frac{1}{\pi}e^{\pi j}\Big\{ \Big(\frac{e^{j\omega(t-\frac{3}{2})}}{ jt- j\frac{3}{2} }\Big)\Bigg|_{3}^{\infty} - \Big(\frac{e^{j\omega(t-\frac{3}{2})}}{ jt- j\frac{3}{2} } \Big)\Bigg|_{-3}^{\infty} \Big\} $

The infinite terms cancel out:

$ \frac{1}{\pi}e^{\pi j}\Big\{ \Big( \frac{e^{3j(t-\frac{3}{2})}}{ jt- j\frac{3}{2} } \Big) - \Big( \frac{e^{-3j(t-\frac{3}{2})}}{jt - j\frac{3}{2}} \Big) \Big\} $

Combine terms:

$ \frac{1}{\pi}e^{\pi j}\Big\{ \frac{e^{3j(t-\frac{3}{2})} - e^{-3j(t-\frac{3}{2})}} {j(t-\frac{3}{2})} \Big\} $

Simplify using euler's crap:

$ -\frac{2}{\pi}\Bigg\{ \frac{sin(3(t-\frac{3}{2}))} {t-\frac{3}{2}} \Bigg\} $

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 $

b

c

d

e

f

g

h

i

j

Problem 4.22