Homework Four: Difference between revisions

Latest revision as of 08:36, 8 November 2009

Nick Christman

1. Find ${\mathcal {F}}\left[g(t)e^{j2\pi f_{0}t}\right]$

This is a fairly straightforward property and is known as complex modulation

${\mathcal {F}}\left[g(t)e^{j2\pi f_{0}t}\right]=\int _{-\infty }^{\infty }\left[g(t)e^{j2\pi f_{0}t}\right]e^{-j2\pi ft}\,dt$

Combining terms, we get:

$\int _{-\infty }^{\infty }\left[g(t)e^{j2\pi f_{0}t}\right]e^{-j2\pi ft}\,dt=\int _{-\infty }^{\infty }g(t)e^{-j2\pi (f-f_{0})t}\,dt$

Now let's make the following substitution $\displaystyle \theta =f-f_{0}$

This now gives us a surprisingly familiar function:

$\int _{-\infty }^{\infty }g(t)e^{-j2\pi (f-f_{0})t}\,dt=\int _{-\infty }^{\infty }g(t)e^{-j2\pi \theta t}\,dt$

This looks just like $\displaystyle G(\theta )$ !

We can now conclude that:

${\mathcal {F}}\left[g(t)e^{j2\pi f_{0}t}\right]=G(\theta )=G(f-f_{0})$

Looks good - Kevin

@@ Line 4: / Line 4: @@
 [[Nick Christman|<b><u>Nick Christman</u></b>]]<br><br>
-. '''Find <math>\mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_{0}} \,dt \right] </math><br/>'''
+. '''Find <math>\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] </math><br/>'''
+This is a fairly straightforward property and is known as ''complex modulation''<br/>
-To begin, we know that<br/>
 <math>
-\mathcal{F} \left[\int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right]
+\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = \int_{- \infty}^{\infty} \left[ g(t)e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt
-= \int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right) e^{-j2 \pi ft}\,dt
 </math>
-<br/>
-After some factoring and combinting of like terms we get:
+Combining terms, we get:
-<br/>
 <math>
-\mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right]
+\int_{- \infty}^{\infty} \left[ g(t)e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt = \int_{- \infty}^{\infty} g(t)e^{-j2 \pi (f-f_{0})t} \,dt
-= \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} 10^{t}g(t) \,dt \right) e^{j2 \pi f(t_0-t)}\,dt
 </math>
 <br/>
+Now let's make the following substitution <math> \displaystyle \theta = f-f_{0}</math>
-But recall that
-<math>\int_{- \infty}^{\infty}e^{j2 \pi f(t_{0}-t)} \,dt \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>
+This now gives us a surprisingly familiar function:
-Because of this definition (and some "math magic") our problem has been simplified significantly: <br/>
 <math>
-\mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right]
+\int_{- \infty}^{\infty} g(t)e^{-j2 \pi (f-f_{0})t} \,dt = \int_{- \infty}^{\infty} g(t)e^{-j2 \pi \theta t} \,dt
+</math>
-= \int_{- \infty}^{\infty}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)
-</math>
 <br/>
+This looks just like <math> \displaystyle G(\theta )</math>!
-Therefore,
+We can now conclude that:
+<br/>
+<math>
+\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = G(\theta ) = G(f-f_{0})
+</math>
+<br>
+Looks good - Kevin
-<math> \mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right] = 10^{t_0}g(t_0) </math>

Homework Four: Difference between revisions

Latest revision as of 08:36, 8 November 2009

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