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==Scaling of the Dirac Delta (Impulse Function)== | ==Scaling of the Dirac Delta (Impulse Function)== | ||
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x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha}</math> | x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha}</math> | ||
| − | == | + | ==Therefore...== |
| − | <math>\displaystyle\delta(\omega)=\delta(\frac{ | + | <math>\displaystyle \delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f)</math> |
| − | + | <math>\displaystyle 2\pi\delta(\omega)=\delta(f)</math> | |
| + | To convert <math>\delta(f)</math> to radians, simply replace <math>\delta(f)</math> with <math>2\pi\delta(\omega)</math> | ||
==Which also means that..== | ==Which also means that..== | ||
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<math>P_T(\omega)=\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s})\;\;\;\;\;\;\;w_s=\frac{2\pi}{T_s}</math> | <math>P_T(\omega)=\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s})\;\;\;\;\;\;\;w_s=\frac{2\pi}{T_s}</math> | ||
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| + | [[Hw3ECE438F09boutin|Return to previous page]] | ||
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| + | [[ECE438|Return to ECE438]] | ||
Latest revision as of 06:37, 25 August 2010
Contents
Scaling of the Dirac Delta (Impulse Function)
$ \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;\alpha>0 $
Mini Proof
$ \int_{-\infty}^{\infty}\delta(x)dx = 1 $
$ \displaystyle Let\;\;\;y=\alpha x\;\;\;\;\;\;\;\;\;\;\;\;\;dx=\frac{dy}{\alpha} $
$ \displaystyle\int_{-\infty}^{\infty}\delta(\alpha x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha} $
Therefore...
$ \displaystyle \delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f) $
$ \displaystyle 2\pi\delta(\omega)=\delta(f) $
To convert $ \delta(f) $ to radians, simply replace $ \delta(f) $ with $ 2\pi\delta(\omega) $
Which also means that..
$ P_T(f)=\frac{1}{T_s}\sum_{n=-\infty}^{\infty}\delta(f-\frac{n}{T_s})\;\;\;\;\;\;\;\;\;\;\;f_s=\frac{1}{T_s} $
$ P_T(\omega)=\frac{2\pi}{T_s}\sum_{n=-\infty}^{\infty}\delta(w-n\frac{2\pi}{T_s})\;\;\;\;\;\;\;w_s=\frac{2\pi}{T_s} $
