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<math>x[n]=-\frac{1}{4}e^{j3\pi n}-\frac{1}{4}e^{-j3\pi n}+\frac{1}{j2}e^{3\pi n}-\frac{1}{j2}e^{-j3\pi n}</math> | <math>x[n]=-\frac{1}{4}e^{j3\pi n}-\frac{1}{4}e^{-j3\pi n}+\frac{1}{j2}e^{3\pi n}-\frac{1}{j2}e^{-j3\pi n}</math> | ||
| − | <math>a_{-1} = -\frac{1}{4}-j\frac{1}{2}</math> | + | <math>a_{-1} = -\frac{1}{4}-j\frac{1}{2}</math> |
| − | <math>a_{0} = 0\!</math> | + | <math>a_{0} = 0\!</math> |
| − | <math>a_{1} = -\frac{1}{4}+j\frac{1}{2}</math> | + | <math>a_{1} = -\frac{1}{4}+j\frac{1}{2}</math> |
Latest revision as of 15:43, 26 September 2008
Periodic DT Signal
$ x[n]=-\frac{1}{2}cos(3\pi n)+sin(3\pi n)\! $.
Fourier Series Coefficients
$ x[n]=-\frac{1}{2}[\frac{e^{j3\pi n}+e^{-j3\pi n}}{2}]+\frac{e^{j3\pi n}-e^{-3\pi n}}{j2} $
$ x[n]=-\frac{1}{4}e^{j3\pi n}-\frac{1}{4}e^{-j3\pi n}+\frac{1}{j2}e^{3\pi n}-\frac{1}{j2}e^{-j3\pi n} $
$ a_{-1} = -\frac{1}{4}-j\frac{1}{2} $
$ a_{0} = 0\! $
$ a_{1} = -\frac{1}{4}+j\frac{1}{2} $
