(New page: == Fourier series coefficients for DT signal == =DT Signal= :<math>x[n]=sin(3 \pi n)\,</math> =Fourier series coefficients= To find N: :<math>N=\frac{2\pi}{3\pi} \times k </math> where k ...) |
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| − | + | = Fourier series coefficients for DT signal = | |
| − | =DT Signal= | + | |
| + | ==DT Signal== | ||
:<math>x[n]=sin(3 \pi n)\,</math> | :<math>x[n]=sin(3 \pi n)\,</math> | ||
| − | =Fourier series coefficients= | + | ==Fourier series coefficients== |
To find N: | To find N: | ||
:<math>N=\frac{2\pi}{3\pi} \times k </math> where k is the smallest integer that makes this be an integer. | :<math>N=\frac{2\pi}{3\pi} \times k </math> where k is the smallest integer that makes this be an integer. | ||
Latest revision as of 17:14, 26 September 2008
Fourier series coefficients for DT signal
DT Signal
- $ x[n]=sin(3 \pi n)\, $
Fourier series coefficients
To find N:
- $ N=\frac{2\pi}{3\pi} \times k $ where k is the smallest integer that makes this be an integer.
- $ \therefore k=3\ and\ N=2 $
- $ \, x[0]= 1 $
- $ \, x[1]=-1 $
- $ \, x[2]=1 $
- $ \, x[3]=-1 $
- $ \, a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} \rightarrow a_k=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-jk\pi n} $
- $ \, a_0=\frac{1+(-1)}{2}=0 $
- $ \, a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-jk\pi n} $
- $ \, a_1=\frac{1}{2}(x[0]e^{0}+x[1]e^{-j\pi})=\frac{1}{2}(1 \cdot 1+(-1)(-1)=1 $
- $ \, \therefore \ a_o=0, a_1 =1 $
