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| − | =Periodicity= | + | = Periodicity Page = |
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| − | + | == Definition == | |
| + | A CT signal <math>x(t)</math> is periodic with period <math>T</math> if <math>x(t)=x(t+T)</math>. | ||
| + | A DT signal <math>x(n)</math> is periodic with period <math>N</math> if <math>x(n)=x(n+N)</math>. | ||
| + | |||
| + | == Finding the Period == | ||
| + | The period of a periodic CT signal of the form <math>e^{j(\omega_0t+\phi)}</math> or <math>cos(\omega_0t+\phi)</math> is easy to find. This is due to the fact that every different value for the fundamental frequency <math>\omega_0</math> corresponds to a unique signal with period <math>T=\frac{2\pi}{\omega_0}</math>. | ||
| − | We start by applying the definition | + | Finding the period of a DT signal becomes more complicated. This is due to the fact that different values of <math>\omega_0</math> can in fact lead to identical equations. As an example I will show how to find the period of a DT complex exponential of the form <math>e^{j\omega_0n}</math>. We start by applying the definition of a periodic signal |
<math>e^{j\omega_0(n+N)}</math> | <math>e^{j\omega_0(n+N)}</math> | ||
Revision as of 07:15, 22 July 2009
Periodicity Page
Definition
A CT signal $ x(t) $ is periodic with period $ T $ if $ x(t)=x(t+T) $. A DT signal $ x(n) $ is periodic with period $ N $ if $ x(n)=x(n+N) $.
Finding the Period
The period of a periodic CT signal of the form $ e^{j(\omega_0t+\phi)} $ or $ cos(\omega_0t+\phi) $ is easy to find. This is due to the fact that every different value for the fundamental frequency $ \omega_0 $ corresponds to a unique signal with period $ T=\frac{2\pi}{\omega_0} $.
Finding the period of a DT signal becomes more complicated. This is due to the fact that different values of $ \omega_0 $ can in fact lead to identical equations. As an example I will show how to find the period of a DT complex exponential of the form $ e^{j\omega_0n} $. We start by applying the definition of a periodic signal
$ e^{j\omega_0(n+N)} $
Using the properties of exponentials
$ e^{j(\omega_0(n+N))}=e^{j\omega_0n}e^{j\omega_0N} $
To make this equation equal to the original signal, we must find an N to make
$ e^{j\omega_0N}=1 $
To do this we use the property of complex exponentials, that
$ e^{j2\pi m}=1 $, where m is an integer.
So we set
$ e^{j\omega_0N}=e^{j2\pi m} $
From this it is easy to see that
$ j\omega_0N=j2\pi m $, or equivalently $ \omega_0N=2\pi m $
This means that $ N=\frac{2\pi m}{\omega_0} $, where m is an integer that will also cause N to be an integer. For this to be possible, $ N=\frac{2\pi}{\omega_0} $ needs to be a rational number.
--Adam Siembida (asiembid) 10:09, 22 July 2009 (UTC)
