Homework Four: Difference between revisions

Revision as of 13:41, 31 October 2009

Nick Christman

1. Find $ℱ [\int_{- \infty}^{\infty} 1 0^{t} g (t) e^{j 2 π f t_{0}} d t]$

To begin, we know that

$ℱ [\int_{- \infty}^{\infty} 1 0^{t} g (t) e^{j 2 π f t_{0}} d t] = \int_{- \infty}^{\infty} (\int_{- \infty}^{\infty} 1 0^{t} g (t) e^{j 2 π f t_{0}} d t) e^{- j 2 π f t} d t$

After some factoring and combinting of like terms we get:

$ℱ [\int_{- \infty}^{\infty} 1 0^{t} g (t) e^{j 2 π f t_{0}} d t] = \int_{- \infty}^{\infty} (\int_{- \infty}^{\infty} 1 0^{t} g (t) d t) e^{j 2 π f (t_{0} - t)} d t$

But recall that $\int_{- \infty}^{\infty} e^{j 2 π f (t_{0} - t)} d t \equiv δ (t_{0} - t) or δ (t - t_{0})$

Because of this definition (and some "math magic") our problem has been simplified significantly:

$ℱ [\int_{- \infty}^{\infty} 1 0^{t} g (t) e^{j 2 π f t_{0}} d t] = \int_{- \infty}^{\infty} 1 0^{t} g (t) δ (t - t_{0}) d t = 1 0^{t_{0}} g (t_{0})$

Therefore,

$ℱ [\int_{- \infty}^{\infty} 1 0^{t} g (t) e^{j 2 π f t_{0}} d t] = 1 0^{t_{0}} g (t_{0})$

@@ Line 3: / Line 3: @@
 ----
-'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''
+[[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/>'''
 To begin, we know that<br/>
-<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>
+<math>
+\mathcal{F} \left[\int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right]
+= \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/>
-But recall that <math>\int_{-\infty}^{\infty}e^{j2 \pi f(t_{0}-t)}df \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>
+After some factoring and combinting of like terms we get:
-The following needs to be fixed, because the previous thing (just above this) which we just fixed wasn't an identity.  Hint:  <math>10^{t}</math> is related to <math>e^{t}</math>
+<br/>
+<math>
+\mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right]
+= \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} 10^{t}g(t) \,dt \right) e^{j2 \pi f(t_0-t)}\,dt
+</math>
+<br/>
-Because of this definition, our problem has now been simplified significantly: <br/>
+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>
-<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)</math> <br/>
+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}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)
+</math>
+<br/>
 Therefore,
-<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = 10^{t_0}g(t_0) </math>
+<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

Revision as of 13:41, 31 October 2009

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