(New page: <math>x[n] = 7sin(7\pi n + \frac{\pi}{8})</math> <math>x[n] = \frac{7}{2j}(e^{j(7\pi n + \frac{\pi}{8})} - e^{-j(7\pi n + \frac{\pi}{8})})</math> <math>x[n] = \frac{7}{2j}e^{j3\pi n}e^{j...) |
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| + | == Computing the Fourier series coefficients for a Discrete Time signal x[n] == | ||
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<math>x[n] = 7sin(7\pi n + \frac{\pi}{8})</math> | <math>x[n] = 7sin(7\pi n + \frac{\pi}{8})</math> | ||
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<math>x[n] = a_0e^{0j\omega_0 n} + a_1e^{1j\omega_0 n} </math> | <math>x[n] = a_0e^{0j\omega_0 n} + a_1e^{1j\omega_0 n} </math> | ||
| − | <math> | + | <math>\rightarrow a_1 = 7, a_0 = 0</math> |
| + | |||
| + | <math>x[n] = 0e^{0j\omega_0 n} + 7e^{j\omega_0 n} </math> | ||
Revision as of 10:41, 25 September 2008
Computing the Fourier series coefficients for a Discrete Time signal x[n]
$ x[n] = 7sin(7\pi n + \frac{\pi}{8}) $
$ x[n] = \frac{7}{2j}(e^{j(7\pi n + \frac{\pi}{8})} - e^{-j(7\pi n + \frac{\pi}{8})}) $
$ x[n] = \frac{7}{2j}e^{j3\pi n}e^{j\frac{\pi}{8}} - \frac{7}{2j}e^{-j3\pi n}e^{-j\frac{\pi}{8}} $
Using the fact that $ e^{j\frac{\pi}{8}} $ and $ e^{-j\frac{\pi}{8}} $ are equal to j...
$ x[n] = \frac{7}{2}e^{j3\pi n} + \frac{7}{2}e^{-j3\pi n} $
Also, $ e^{j3\pi n} = e^{j\pi n} \rightarrow e^{-j3\pi n} \times 1 = e^{-j3\pi n}e^{2j2\pi n} = e^{j\pi n} $
$ x[n] = \frac{7}{2}e^{j\pi n} + \frac{7}{2}e^{j\pi n} = 7e^{j\pi n} $
$ x[n] = a_0e^{0j\omega_0 n} + a_1e^{1j\omega_0 n} $
$ \rightarrow a_1 = 7, a_0 = 0 $
$ x[n] = 0e^{0j\omega_0 n} + 7e^{j\omega_0 n} $
