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<math>\,x[n]=6cos(4\pi n)+2cos(\pi n)</math>
 
<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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and <math>\,N_2</math> is the lowest usable period(ie the highest) and <math>\,N=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>:
 
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>:
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<math>\,x[3]=4</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 13:27, 25 September 2008

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 $

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