DT Periodic function
Find the Fourier Series coefficients of x[n]
- $ x[n]=5cos(5/2\pi n +\pi) \, $
- $ x[n]=5cos(5/2 \pi n +\pi) = \dfrac{5(e^{5/2 j \pi n+\pi}-e^{-5/2 j \pi n-\pi})}{2} \, $
- $ x[n]= \dfrac{5}{2}(e^{5/2 j \pi n}e^{\pi}-e^{-5/2 j \pi n}e^{-\pi}) \, $
- $ e^{\pi}=-1,e^{-\pi}=1 \, $
- $ x[n]= \dfrac{5}{2}(e^{-5/2 j \pi n}-e^{5/2 j \pi n}) \, $
- $ x[n]= \dfrac{5}{2}(e^{-2 j \pi n}e^{-1/2 j \pi n}-e^{2 j \pi n}e^{1/2 j \pi n}) \, $
- $ x[n]= \dfrac{5}{2}(e^{-1/2 j \pi n}-e^{1/2 j \pi n}) \, $
Note that,
- $ 1 = e^{2\pi} \, $
- $ e^{2\pi}*e^{-1/2\pi}=e^{3/2\pi}=e^{1/2\pi}e^(\pi)=-e^{1/2\pi} \, $
- $ x[n]= \dfrac{5}{2}(-e^{-1/2 j \pi n}-e^{1/2 j \pi n}) \, $
- $ x[n]= \dfrac{5}{2}(-2e^{1/2 j \pi n}) \, $
- $ x[n]= -5(e^{1/2 j \pi n}) \, $
- $ N=\dfrac{2\pi}{\pi/2}K $, where K is the smallest integer, that makes N an integer.
- $ K = 1,N = 4\, $
- $ a1=1/4\sum_{k=0}^{N-1} -5(e^{1/2 j \pi n})e^{- j 2\pi/N n} or :<math>x[0]=-1 \, $
- $ x[1]=0 \, $
- $ x[2]=1 \, $
- $ x[3]=0 \, $
- $ x[4]=-1 \, $
a0=average of signal = 0
