DFS

Given the periodic sequence, \(x[n]=\{...,2,0,4,-5,2,0,4,-5,2,0,4,-5,...\}\), and that the first -5 is n=0 find the DFS coefficients \(X[k]\).
(Alternative prompt for DFT: Given a sequence defined below, find the DFT coefficients for the sequence)
\(\(x[n]=\Bigl\{ \begin{array}{l}{-5,2,0,4},\; \text{ for } 0\le n <N \\0,\; otherwise\end{array}\)\)
1. The pattern repeats every 4 n, so N=4.
2. DFS Coefficients Definition: \(\large X[k]=\sum\limits_{n=0}^{N-1}x[n]e^{-jk \frac{2\pi}{N}n}\)
3. Apply N=4: \(\large X[k]=\sum\limits_{n=0}^{3}x[n]e^{-jk \frac{2\pi}{4}n}\)

1. Evaluate Sum: \(\large =x[0]e^{-jk \frac{2\pi}{4}0} + x[1]e^{-jk \frac{2\pi}{4}1} + x[2]e^{-jk \frac{2\pi}{4}2} + x[3]e^{-jk \frac{2\pi}{4}3}\)

2. Simplify: \(=\large -5(1) + 2e^{-jk \frac{\pi}{2}} + 0e^{-jk\pi} + 4e^{-jk \frac{3\pi}{2}}\)

   1. \(\large X[k]=-5 + 2e^{-jk \frac{\pi}{2}} + 4e^{-jk \frac{3\pi}{2}}\)

3. Find \(\{X[k]\}\):

   1. \(\large X[0]=-5 + 2e^{-j0 \frac{\pi}{2}} + 4e^{-j0 \frac{3\pi}{2}} = -5+2+4=1\)
   2. \(\large X[1]=-5 + 2e^{-j\frac{\pi}{2}} + 4e^{-j\frac{3\pi}{2}} = -5+2(-j)+4(+j)=-5+2j\)
   3. \(\large X[2]=-5 + 2e^{-j2\frac{\pi}{2}} + 4e^{-j2\frac{3\pi}{2}} = -5+2(-1)+4(-1)=-11\)
   4. \(\large X[3]=-5 + 2e^{-j3\frac{\pi}{2}} + 4e^{-j3\frac{3\pi}{2}} = -5+2(+j)+4(-j)=-5-2j\)

4. Answer: \(X[k]=\{1,-5+2j,-11,-5-2j\}\)

• Let's use the previous example's coefficients: \(X[k]=\{1,-5+2j,-11,-5-2j\}\)
• The Synthesis equation: \(\(\LARGE x[n]=\frac{1}{N}\sum\limits_{k=0}^{N-1}X[k]e^{j2\pi \frac{k}{N}n}\)\)
• If we manually build the sinusoid form of those coefficients using Eulers Formula, we'll find that all of the imaginary parts of the our equation will either cancel out or be zero-valued for integer values of n, leaving us with just the real portions at our sampling points:
  \(\(\large x[n]=1-5cos(2\pi \frac{n}{4})-11cos(2\pi \frac{n}{2})-5cos(2\pi \frac{3n}{4})-2sin(2\pi \frac{n}{4})+2sin(2\pi \frac{3n}{4})\)\)
• Here is the graphical representation of the added real series (lined) and the original sequence (circles). Keep in mind that we could have also graphed the complex magnitude of the DFT, or graphed the imaginary part on it's own.
  
  

  clear all
  x=[-5 2 0 4];
  m=0:3;
  n=0:0.1:3;
  y=1-5*cos(2*pi*n/4)+-11*cos(2*pi*n/2)+-5*cos(2*pi*3*n/4)-2*sin(2*pi*n/4)+2*sin(2*pi*3*n/4);
  plot(m,x,'o',n,y/4,'x');
  title("Real Fourier Series vs Real Periodic Sequence")
  xlabel("Discrete Time, n");
  ylabel("Amplitude")
  ylim([-5.5,4.5]);
  xlim([-0.5,3.5]);