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=3\pi $
$ \,x[n]=6sin(4\pi n)+2cos(2\pi n) $
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 }{2\pi}k=\frac{1}{1}k $
so $ \,k=2 $ makes them both integers:
$ \,N_1=1 $
$ \,N_2=2 $
and $ \,N_2 $ is the lowest usable period(ie the highest) and $ \,N=2 $
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] $ and $ \,x[1] $:
