Problem: Find the Fourier series coefficients of x[n], where x[n] is a square wave with fundamental period N = 4 with the following values:
- x[0] = 1
- x[1] = 1
- x[2] = 0
- x[3] = 0
(Then, x[4] = x[0] = 1, etc.)
Solution: A periodic DT signal can be expressed as a Fourier series in the following manner:
$ x[n] = \sum_{k=0}^{N-1} a_k e^{jk \frac{2 \pi}{N} n} $, where $ a_k = \frac{1}{N} \sum_{r=0}^{N-1} x[r] e^{-jk \frac{2 \pi}{N} r} $
So, since N = 4 in this problem,
$ x[n] = \sum_{k=0}^{3} a_k e^{jk \frac{\pi}{2} n} $, where $ a_k = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-jk \frac{\pi}{2} r} $
($ \frac{2 \pi}{n} = \frac{\pi}{2} $ is also known as the fundamental frequency of the signal.)
To determine the Fourier series coefficients (and, consequently, the Fourier series), all we need to do is plug in values:
- $ a_0 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(0) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{0} = \frac{1}{4} (x[0] + x[1] + x[2] + x[3]) = \frac{1}{4} (1 + 1 + 0 + 0) = \frac{1}{2} $
- $ a_1 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(1) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \frac{\pi}{2} r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \frac{\pi}{2}} + x[2] e^{-j \pi}+ x[3] e^{-j \frac{3 \pi}{2} }) $
$ = \frac{1}{4} (1(1) + 1(-j) + 0 + 0) = \frac{1-j}{4} $
- $ a_2 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(2) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \pi r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \pi} + x[2] ee^{-j 2 \pi}+ x[3] e^{-j 3 \pi}) $
$ = \frac{1}{4} (1(1) + 1(-1) + 0 + 0) = 0 $
- $ a_3 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(3) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \frac{3 \pi}{2} r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \frac{3 \pi}{2}} + x[2] e^{-j 3 \pi}+ x[3] e^{-j \frac{9 \pi}{2} }) $
$ = \frac{1}{4} (1(1) + 1(j) + 0 + 0) = \frac{1+j}{4} $
