Homework Four: Difference between revisions

Latest revision as of 09:36, 8 November 2009

Nick Christman

1. Find $ℱ [g (t) e^{j 2 π f_{0} t}]$

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

$ℱ [g (t) e^{j 2 π f_{0} t}] = \int_{- \infty}^{\infty} [g (t) e^{j 2 π f_{0} t}] e^{- j 2 π f t} d t$

Combining terms, we get:

$\int_{- \infty}^{\infty} [g (t) e^{j 2 π f_{0} t}] e^{- j 2 π f t} d t = \int_{- \infty}^{\infty} g (t) e^{- j 2 π (f - f_{0}) t} d t$

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

This now gives us a surprisingly familiar function:

$\int_{- \infty}^{\infty} g (t) e^{- j 2 π (f - f_{0}) t} d t = \int_{- \infty}^{\infty} g (t) e^{- j 2 π θ t} d t$

This looks just like $G (θ)$ !

We can now conclude that:

$ℱ [g (t) e^{j 2 π f_{0} t}] = G (θ) = G (f - f_{0})$

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/>'''
-To begin, we know that<br/>
+This is a fairly straightforward property and is known as ''complex modulation''<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 <math> \Longleftarrow </math> "combining-Kevin" 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/>
-But recall that
+Now let's make the following substitution <math> \displaystyle \theta = f-f_{0}</math>
-<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
-= \int_{- \infty}^{\infty}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)
+</math>
-</math>
 <br/>
-Therefore,
+This looks just like <math> \displaystyle G(\theta )</math>!
+We can now conclude that:
+<br/>
-<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>
+<math>
+\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = G(\theta ) = G(f-f_{0})
+</math>
 <br>
-Other than the one typo looks good - Kevin
+Looks good - Kevin

Homework Four: Difference between revisions

Latest revision as of 09:36, 8 November 2009

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