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===Answer 1=== | ===Answer 1=== | ||
There are a few things that you need to know to accomplish this problem. The two main formulas that you need are | There are a few things that you need to know to accomplish this problem. The two main formulas that you need are | ||
| − | <math> \omega = 2 \pi f </math> and <math>\delta(cx)= \frac{1}{c} \delta(x) | + | <math> \omega = 2 \pi f </math> and <math>\delta(cx)= \frac{1}{c} \delta(x)</math> for c>0. |
PROOF | PROOF | ||
Revision as of 04:14, 1 September 2011
Contents
Continuous-time Fourier transform: from omega to f
In ECE301, you learned that the Fourier transform of a step function $ x(t)=u(t) $ is the following:
$ {\mathcal X} (\omega) = \frac{1}{j \omega} + \pi \delta (\omega ). $
Use this fact to obtain an expression for the Fourier transform $ X(f) $ (in terms of frequency in hertz) of the step function. (Your answer should agree with the one given in this table.) Justify all your steps.
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Answer 1
There are a few things that you need to know to accomplish this problem. The two main formulas that you need are $ \omega = 2 \pi f $ and $ \delta(cx)= \frac{1}{c} \delta(x) $ for c>0.
PROOF
$ \int_{-\infty}^{\infty}\delta(x)dx = 1 $
$ y=cx => \frac{dy}{c}=dx $
$ \int_{-\infty}^\infty \delta(y)\frac{dy}{c}=\frac{1}{c} $
THEREFORE
$ \delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f) $
and
$ {\mathcal X} (\omega) = \frac{1}{j \omega} + \pi \delta (\omega ) = \frac{1}{2}(\frac{1}{j\pi f} + \delta(f)) $
-my
Answer 2
Write it here.
Answer 3
write it here.
