(New page: == Definition of fourier transform for DT signal == These are the fourier coefficients, which must be calculated from the function in this case, rather than vice versa in CT signals. <ma...) |
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The primary importance in the DT example, is making sure that a constant K exists, so that the signal can be forced to be periodic. | The primary importance in the DT example, is making sure that a constant K exists, so that the signal can be forced to be periodic. | ||
| − | We will make a relatively low period, say <math>\,N= | + | We will make a relatively low period, say <math>\,N=4</math> |
| − | <math>\,x[n]= | + | <math>\,x[n]=6cos(4\pi n)+2cos(\pi n)</math> |
| + | |||
| + | === Getting the Period === | ||
We have to make the periods make sense in DT, so we must make sure that <math>\,\frac{2\pi }{\omega_0}</math> is a whole number for both functions, so multiply it in this fashion: | We have to make the periods make sense in DT, so we must make sure that <math>\,\frac{2\pi }{\omega_0}</math> is a whole number for both functions, so multiply it in this fashion: | ||
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<math>N_1=\frac{2\pi }{4\pi}k=\frac{1}{2}k</math>, and | <math>N_1=\frac{2\pi }{4\pi}k=\frac{1}{2}k</math>, and | ||
| − | <math>N_2=\frac{2\pi }{ | + | <math>N_2=\frac{2\pi }{\pi}k=\frac{2}{1}k</math> |
so <math>\,k=2</math> makes them both integers: | so <math>\,k=2</math> makes them both integers: | ||
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<math>\,N_1=1</math> | <math>\,N_1=1</math> | ||
| − | <math>\,N_2=2</math> | + | <math>\,N_2=4</math> |
| + | |||
| + | and <math>\,N_2</math> is the lowest usable period(ie the highest) and <math>\,N=4</math> | ||
| + | |||
| + | |||
| + | === Finding the values of the function === | ||
| + | |||
| + | Now we have our period, and thus the limits of our sum. The limit goes to N-1, so we need to find <math>\,x[0]</math> through <math>\,x[3]</math>: | ||
| + | |||
| + | <math>\,x[0]=8</math> | ||
| + | |||
| + | <math>\,x[1]=4</math> | ||
| + | |||
| + | <math>\,x[2]=8</math> | ||
| + | |||
| + | <math>\,x[3]=4</math> | ||
| + | |||
| + | === Solving for the coefficients === | ||
| + | |||
| + | Now use definition of fourier coefficients: | ||
| + | |||
| + | <math>a_k = \frac{1}{4} \sum^{3}_{n=0} x[n] e^{-jk\frac{\pi}{2} n}</math> | ||
| + | |||
| + | Start finding <math>\,a_0</math> through <math>\,a_3</math>: | ||
| + | |||
| + | <math>a_0 = \frac{1}{4} \sum^{3}_{n=0} x[n] e^{0}</math> | ||
| + | |||
| + | <math>a_0 = \frac{1}{4}*(x[0]+x[1]+x[2]+x[3])</math> | ||
| + | |||
| + | <math>a_0 = \frac{1}{4}*(24)</math> | ||
| + | |||
| + | <math>\,a_0 = 6</math> | ||
| + | |||
| + | These results are periodic, as will be shown later. | ||
| + | |||
| + | <math>a_1 = \frac{1}{4} \sum^{3}_{n=0} x[n] e^{-j\frac{\pi}{2} n}</math> | ||
| + | |||
| + | Quickly done using matlab, | ||
| + | |||
| + | <math>\,a_1 = 0</math> | ||
| + | |||
| + | <math>\,a_2 = 2</math> | ||
| + | |||
| + | <math>\,a_3 = 2</math> | ||
| + | |||
| + | === Showing Periodicity === | ||
| + | |||
| + | These results are periodic, so | ||
| + | |||
| + | for <math>\,k=0,4,8,...4n</math>, <math>\,a_k=6</math> | ||
| + | |||
| + | for <math>\,k=1,5,9,...4n+1</math>, <math>\,a_k=0</math> | ||
| − | + | for <math>\,k=2,6,10,...4n+2</math>, <math>\,a_k=2</math> | |
| − | + | for <math>\,k=3,7,11,...4n+3</math>, <math>\,a_k=2</math> | |
Latest revision as of 14:27, 25 September 2008
Contents
Definition of fourier transform for DT signal
These are the fourier coefficients, which must be calculated from the function in this case, rather than vice versa in CT signals.
$ a_k = \frac{1}{N} \sum^{N-1}_{n=0} x[n] e^{-jk\frac{2\pi}{N} n} $
Where N is the period of the function.
Example of a periodic DT signal
The primary importance in the DT example, is making sure that a constant K exists, so that the signal can be forced to be periodic.
We will make a relatively low period, say $ \,N=4 $
$ \,x[n]=6cos(4\pi n)+2cos(\pi n) $
Getting the Period
We have to make the periods make sense in DT, so we must make sure that $ \,\frac{2\pi }{\omega_0} $ is a whole number for both functions, so multiply it in this fashion:
$ N_1=\frac{2\pi }{4\pi}k=\frac{1}{2}k $, and
$ N_2=\frac{2\pi }{\pi}k=\frac{2}{1}k $
so $ \,k=2 $ makes them both integers:
$ \,N_1=1 $
$ \,N_2=4 $
and $ \,N_2 $ is the lowest usable period(ie the highest) and $ \,N=4 $
Finding the values of the function
Now we have our period, and thus the limits of our sum. The limit goes to N-1, so we need to find $ \,x[0] $ through $ \,x[3] $:
$ \,x[0]=8 $
$ \,x[1]=4 $
$ \,x[2]=8 $
$ \,x[3]=4 $
Solving for the coefficients
Now use definition of fourier coefficients:
$ a_k = \frac{1}{4} \sum^{3}_{n=0} x[n] e^{-jk\frac{\pi}{2} n} $
Start finding $ \,a_0 $ through $ \,a_3 $:
$ a_0 = \frac{1}{4} \sum^{3}_{n=0} x[n] e^{0} $
$ a_0 = \frac{1}{4}*(x[0]+x[1]+x[2]+x[3]) $
$ a_0 = \frac{1}{4}*(24) $
$ \,a_0 = 6 $
These results are periodic, as will be shown later.
$ a_1 = \frac{1}{4} \sum^{3}_{n=0} x[n] e^{-j\frac{\pi}{2} n} $
Quickly done using matlab,
$ \,a_1 = 0 $
$ \,a_2 = 2 $
$ \,a_3 = 2 $
Showing Periodicity
These results are periodic, so
for $ \,k=0,4,8,...4n $, $ \,a_k=6 $
for $ \,k=1,5,9,...4n+1 $, $ \,a_k=0 $
for $ \,k=2,6,10,...4n+2 $, $ \,a_k=2 $
for $ \,k=3,7,11,...4n+3 $, $ \,a_k=2 $
