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 $
