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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>. | 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>. | ||
| − | 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 | + | 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> using the definition of a periodic signal: a signal <math>x(n)</math> is periodic with period <math>N</math> if <math>x(n)=x(n+N)</math>. |
We start by applying the definition | We start by applying the definition | ||
| − | <math>e^{j(\omega_0(n+N)}</math> | + | <math>e^{j\omega_0(n+N)}</math> |
| + | |||
| + | Using the properties of exponentials | ||
| + | |||
| + | <math>e^{j(\omega_0(n+N))}=e^{j\omega_0n}e^{j\omega_0N}</math> | ||
| + | |||
| + | To make this equation equal to the original signal, we must find an N to make | ||
| + | |||
| + | <math>e^{j\omega_0N}=1</math> | ||
| + | |||
| + | To do this we use the property of complex exponentials, that | ||
| + | |||
| + | <math>e^{j2\pi m}=1</math>, where m is any integer. | ||
| + | |||
| + | Therefore we set | ||
| + | |||
| + | <math>e^{j\omega_0N}=e^{j2\pi m}</math> | ||
| + | |||
| + | From this it is easy to see that | ||
| + | |||
| + | <math>j\omega_0N=j2\pi m</math>, or equivalently <math>\omega_0N=2\pi m</math> | ||
| + | |||
| + | -I WILL FINISH WRITING THIS IN 30 minutes please do not change!!! | ||
--[[User:Asiembid|Adam Siembida (asiembid)]] 10:09, 22 July 2009 (UTC) | --[[User:Asiembid|Adam Siembida (asiembid)]] 10:09, 22 July 2009 (UTC) | ||
Revision as of 05:27, 22 July 2009
Periodicity
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} $ using the definition of a periodic signal: a signal $ x(n) $ is periodic with period $ N $ if $ x(n)=x(n+N) $.
We start by applying the definition
$ 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 any integer.
Therefore 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 $
-I WILL FINISH WRITING THIS IN 30 minutes please do not change!!!
--Adam Siembida (asiembid) 10:09, 22 July 2009 (UTC)
