Revision as of 06:17, 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)