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<math>x[6] = 2\,</math>
 
<math>x[6] = 2\,</math>
  
<math>x[7] = 2\,</math>
+
<math>x[7] = 2\,</math>,  which continue to repeat in this way every 4 integers.
  
:<center><math>.\,</math></center>
+
<math>a_k = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-jk\frac{\pi}{2} n}\,</math>
 +
 
 +
<math>a_0 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^0\,</math>
 +
 
 +
<math> = \frac{1}{4}\sum^{3}_{n = 0} x[n]\,</math>
 +
 
 +
<math> = \frac{1}{4}(2 - 6 + 2 + 2)\,</math>
 +
 
 +
 
 +
<math>a_0 = 0\,</math>
 +
 
 +
 
 +
<math>a_1 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-j\frac{\pi}{2} n}\,</math>
 +
 
 +
<math> = \frac{1}{4}(x[0] + x[1]e^{-j\frac{\pi}{2}} + x[2]e^{-j\pi} + x[3]e^{-j\frac{3\pi}{2}})\,</math>
 +
 
 +
<math> = \frac{1}{4}(2 - 6(-j) + 2(-1) + 2(j))\,</math>
 +
 
 +
<math> = \frac{1}{4}(8j)\,</math>
 +
 
 +
 
 +
<math>a_1 = 2j\,</math>
 +
 
 +
 
 +
<math>a_2 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-j\pi n}\,</math>
 +
 
 +
<math> = \frac{1}{4}(x[0] + x[1]e^{-j\pi} + x[2]e^{-j2\pi} + x[3]e^{-j3\pi})\,</math>
 +
 
 +
<math> = \frac{1}{4}(2 - 6(-1) + 2(1) + 2(-1))\,</math>
 +
 
 +
<math> = \frac{1}{4}(8)\,</math>
 +
 
 +
 
 +
<math>a_2 = 2\,</math>
 +
 
 +
 
 +
<math>a_3 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-j\frac{3\pi}{2} n}\,</math>
 +
 
 +
<math> = \frac{1}{4}(x[0] + x[1]e^{-j\frac{3\pi}{2}} + x[2]e^{-j3\pi} + x[3]e^{-j\frac{9\pi}{2}})\,</math>
 +
 
 +
<math> = \frac{1}{4}(2 - 6(j) + 2(-1) + 2(-j))\,</math>
 +
 
 +
<math>a_3 = \frac{1}{4}(-8j)\,</math>
 +
 
 +
 
 +
<math>a_3 = -2j\,</math>
 +
 
 +
So, the function is
 +
 
 +
<math>x[n] = \sum^{3}_{k = 0} a_k e^{jk\frac{\pi}{2} n}\,</math>, with coefficients which repeat every four integers,
 +
 
 +
 
 +
<math>a_0 = 0 , k = ..., -8, -4, 0, 4, 8, ... \,</math>
 +
 
 +
<math>a_1 = 2j , k = ..., -7, -3, 1, 5, 9, ... \,</math>
 +
 
 +
<math>a_2 = 2 , k = ..., -6, -2, 2, 6, 10, ... \,</math>
 +
 
 +
<math>a_3 = -2j , k = ..., -5, -1, 3, 7, 11, ... \,</math>

Latest revision as of 12:42, 25 September 2008

Define a Periodic DT Signal and Compute the Fourier Series Coefficients

I am going to choose a sine signal, since there have been many cosines done already.

DT signal: $ x[n] = 2\sin(\pi n + \frac{\pi}{2}) + 4\sin(\frac{\pi}{2} n + \pi)\, $


Now, each sine has its own period, and the fundamental period of the function is the greater of the separate periods.

$ N_2sin = \frac{2\pi}{\pi} k = \frac{2}{1} k $


$ N_4sin = \frac{2\pi}{\frac{\pi}{2}} k = \frac{2}{\frac{1}{2}} k $


Take $ k = 1\, $,

$ N_2sin = 2\, $

$ N_4sin = 4\, $, so the overall fundamental period is


$ N = 4\, $

In order to find the coefficients, we must first calculate the values of $ x[n]\, $ for four consecutive integer values of $ n\, $. By plugging values of $ n\, $ into the given signal, we find that

$ x[0] = 2\, $

$ x[1] = -6\, $

$ x[2] = 2\, $

$ x[3] = 2\, $

$ x[4] = 2\, $

$ x[5] = -6\, $

$ x[6] = 2\, $

$ x[7] = 2\, $, which continue to repeat in this way every 4 integers.

$ a_k = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-jk\frac{\pi}{2} n}\, $

$ a_0 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^0\, $

$ = \frac{1}{4}\sum^{3}_{n = 0} x[n]\, $

$ = \frac{1}{4}(2 - 6 + 2 + 2)\, $


$ a_0 = 0\, $


$ a_1 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-j\frac{\pi}{2} n}\, $

$ = \frac{1}{4}(x[0] + x[1]e^{-j\frac{\pi}{2}} + x[2]e^{-j\pi} + x[3]e^{-j\frac{3\pi}{2}})\, $

$ = \frac{1}{4}(2 - 6(-j) + 2(-1) + 2(j))\, $

$ = \frac{1}{4}(8j)\, $


$ a_1 = 2j\, $


$ a_2 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-j\pi n}\, $

$ = \frac{1}{4}(x[0] + x[1]e^{-j\pi} + x[2]e^{-j2\pi} + x[3]e^{-j3\pi})\, $

$ = \frac{1}{4}(2 - 6(-1) + 2(1) + 2(-1))\, $

$ = \frac{1}{4}(8)\, $


$ a_2 = 2\, $


$ a_3 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-j\frac{3\pi}{2} n}\, $

$ = \frac{1}{4}(x[0] + x[1]e^{-j\frac{3\pi}{2}} + x[2]e^{-j3\pi} + x[3]e^{-j\frac{9\pi}{2}})\, $

$ = \frac{1}{4}(2 - 6(j) + 2(-1) + 2(-j))\, $

$ a_3 = \frac{1}{4}(-8j)\, $


$ a_3 = -2j\, $

So, the function is

$ x[n] = \sum^{3}_{k = 0} a_k e^{jk\frac{\pi}{2} n}\, $, with coefficients which repeat every four integers,


$ a_0 = 0 , k = ..., -8, -4, 0, 4, 8, ... \, $

$ a_1 = 2j , k = ..., -7, -3, 1, 5, 9, ... \, $

$ a_2 = 2 , k = ..., -6, -2, 2, 6, 10, ... \, $

$ a_3 = -2j , k = ..., -5, -1, 3, 7, 11, ... \, $

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett