Homework Six

Perform the following tasks:

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Nick Christman

(a) Show ${ℱ}[\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda ]=\frac{S(F)}{j2\pi f}\text{ if }S(0)=0$

(b)If $S(0)\ne 0$ can you find ${ℱ}[\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda ]$ in terms of $S(0)$?

(c) Do another property on the Wiki and get it reviewed (i.e. review a second property) -- Fourier Transform Properties

Find ${ℱ}[g(t-t_0)e^{j2\pi f_{0}t}]$

-- Using the above definition of complex modulation and the definition from class of a time delay (a.k.a "the slacker function"), I will attempt to show a hybrid of the two...

By definition we know that:

${ℱ}[g(t-t_0)e^{j2\pi f_{0}t}]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[g(t-t_0)e^{j2\pi f_{0}t}]e^{-j2\pi ft}dt$

Rearranging terms we get:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[g(t-t_0)e^{j2\pi f_{0}t}]e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t-t_0)e^{-j2\pi (f-f_0)t}dt$

Now lets make the substitution $\lambda =t-t_0\to t=\lambda +t_0$.
This leads us to:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t-t_0)e^{-j2\pi (f-f_0)t}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)(\lambda +t_0)}dt$

After some simplification and rearranging terms, we get:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)(\lambda +t_0)}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }{e}^{-j2\pi (f-f_0)t_0}dt$

Rearranging the terms yet again, we get:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }{e}^{-j2\pi (f-f_0)t_0}dt=e^{-j2\pi (f-f_0)t_0}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }dt]$

We know that the exponential in terms of $t_0$ is simply a constant and because of the Fourier Property of complex modualtion, we finally get:

${ℱ}[g(t)e^{j2\pi f_{0}t}]=G(f-f_0)e^{-j2\pi (f-f_0)t_0}$