Homework Four: Difference between revisions

Revision as of 15:10, 19 October 2009

Find $ℱ [1 0^{t} g (t) e^{j 2 π f t_{0}}]$

To begin, we know that

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

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

The following needs to be fixed, because the previous thing (just above this) which we just fixed wasn't an identity. Hint: $1 0^{t}$ is related to $e^{t}$

Because of this definition, our problem has now been simplified significantly:

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

Therefore,

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

@@ Line 9: / Line 9: @@
 <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>
-But recall that <math>e^{j2 \pi f(t_{0}-t)} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>
+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>
+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>
 Because of this definition, our problem has now been simplified significantly: <br/>

Homework Four: Difference between revisions

Revision as of 15:10, 19 October 2009

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