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^{j \dfrac{\pi}{2} n}) \, $
- $ N=\dfrac{2\pi}{\pi/2}K $, where K is the smallest integer, that makes N an integer.
- $ K = 1,N = 4\, $
- $ a0=average of signal = 0 \, $
- $ a1=1/N\sum_{n=0}^{N-1} -5e^{j \dfrac{\pi}{2} n}e^{- j k \dfrac{2\pi}{N} n} $
- $ a1=1/4\sum_{n=0}^{4-1} -5e^{j \dfrac{\pi}{2} n}e^{- j \dfrac{2\pi}{4} n} $
- $ a1=1/4\sum_{n=0}^{3} -5e^{j \dfrac{\pi}{2} n}e^{- j \dfrac{\pi}{2} n} $
- $ a1=-5/4\sum_{n=0}^{3} 1 $
- $ a1=-5/4(1+1+1+1)=-5 \, $
- $ a2=1/4\sum_{n=0}^{4-1} -5e^{j \dfrac{\pi}{2} n}e^{- j 2 \dfrac{2\pi}{4} n} $
- $ a2=1/4\sum_{n=0}^{3} -5e^{j \dfrac{\pi}{2} n}e^{- j \pi n} $
- $ a2=1/4\sum_{n=0}^{3} -5e^{-j \dfrac{\pi}{2} n}\ $
- $ a2=-5/4(1+-j+-1+j=0) \, $
- $ a3=1/4\sum_{n=0}^{4-1} -5e^{j \dfrac{\pi}{2} n}e^{- j 3 \dfrac{2\pi}{4} n} $
- $ a3=1/4\sum_{n=0}^{3} -5e^{j \dfrac{\pi}{2} n}e^{- j \dfrac{3\pi}{2} n} $
- $ a3=-5/4\sum_{n=0}^{3} e^{-j n} $
- $ a3=-5/4(1+-1+1+-1=0) \, $
You can also note that you do not have to use the formula for this problem. x[n] looks like a Fourier series. wo=pi/2, so a1=-5 from the first term and since there are no other terms a2=a3=0. Then a0= average of signal.
