(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...)
 
(→‎Computing the Fourier series coefficients for a Discrete Time signal x[n])
 
(One intermediate revision by the same user not shown)
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<math>x[n] = 7sin(7\pi n + \frac{\pi}{8})</math>
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== Computing the Fourier series coefficients for a Discrete Time signal x[n] ==
  
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<math>x[n] = \frac{7}{2j}(e^{j(7\pi n + \frac{\pi}{8})} - e^{-j(7\pi n + \frac{\pi}{8})})</math>
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<math>x[n] = 7sin(7\pi n + \frac{\pi}{8})\,</math>
  
−
<math>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}}</math>
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<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\frac{\pi}{8}} - \frac{7}{2j}e^{-j3\pi n}e^{-j\frac{\pi}{8}}\,</math>
  
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Using the fact that <math>e^{j\frac{\pi}{8}}</math> and <math>e^{-j\frac{\pi}{8}}</math> are equal to j...
 
  
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<math>x[n] = \frac{7}{2}e^{j3\pi n} + \frac{7}{2}e^{-j3\pi n}</math>
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Using the fact that <math>e^{j\frac{\pi}{8}}</math> and <math>e^{-j\frac{\pi}{8}}\,</math> are equal to j...
  
 +
<math>x[n] = \frac{7}{2}e^{j3\pi n} + \frac{7}{2}e^{-j3\pi n}\,</math>
  
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Also, <math>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}</math>
 
  
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<math>x[n] = \frac{7}{2}e^{j\pi n} + \frac{7}{2}e^{j\pi n} = 7e^{j\pi n}</math>
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Also, <math>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}\,</math>
  
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<math>x[n] = a_0e^{0j\omega_0 n} + a_1e^{1j\omega_0 n} </math>
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<math>x[n] = \frac{7}{2}e^{j\pi n} + \frac{7}{2}e^{j\pi n} = 7e^{j\pi n}\,</math>
  
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<math>/rightarrow k= 7, a_1 = 7, a_0 = 0</math>
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<math>x[n] = a_0e^{0j\omega_0 n} + a_1e^{1j\omega_0 n}\,</math>
 +
 
 +
<math>\rightarrow a_1 = 7, a_0 = 0\,</math>
 +
 
 +
<math>x[n] = 0e^{0j\omega_0 n} + 7e^{j\omega_0 n}\,</math>

Latest revision as of 11:47, 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}\, $

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