(New page: ==CT Signal Fourier Coefficients== An easy signal to compute is a sine or cosine based function. I'm trying to look around and find something less trivial but all i can find is functions...) |
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| − | == | + | ==DT Signal Fourier Coefficients== |
| − | + | Let's make up a signal. | |
| + | |||
| + | <math> x[0] = 0 </math> | ||
| + | |||
| + | <math> x[1] = 1 </math> | ||
| + | |||
| + | <math> x[2] = 1 </math> | ||
| + | |||
| + | <math> x[3] = 0 </math> | ||
| + | |||
| + | <math> x[4] = x[0] </math> etc, the function is periodic with period 4 | ||
| + | |||
| + | Using the formula | ||
| + | |||
| + | <math>x[n] = \sum_{k=0}^{N-1} a_k e^{jk \frac{2 \pi}{N} n}</math>, where <math>a_k = \frac{1}{N} \sum_{r=0}^{N-1} x[r] e^{-jk \frac{2 \pi}{N} r}</math> | ||
| + | |||
| + | Since the period is 4, N=4. | ||
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| + | <math>x[n] = \sum_{k=0}^{3} a_k e^{jk \frac{2 \pi}{4} n}</math>, where <math>a_k = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-jk \frac{2 \pi}{4} r}</math> | ||
| + | |||
| + | Now to find the fourier series coefficients: | ||
| + | |||
| + | <math>a_0 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(0) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{0} = \frac{1}{4} (x[0] + x[1] + x[2] + x[3]) = \frac{1}{4} (0 + 1 + 1 + 0) = \frac{1}{2}</math> | ||
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| + | <math>a_1 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(1) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \frac{\pi}{2} r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \frac{\pi}{2}} + x[2] e^{-j \pi}+ x[3] e^{-j \frac{3 \pi}{2} }) </math> | ||
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| + | <math>= \frac{1}{4} (0(0) + 1(-j) + 1*-1 + 0*i) = \frac{-1-j}{4}</math> | ||
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| + | <math>a_2 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(2) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \pi r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \pi} + x[2] ee^{-j 2 \pi}+ x[3] e^{-j 3 \pi}) </math> | ||
| + | |||
| + | <math>= \frac{1}{4} (0 + 1 * -1 + 1 * 1 + 0 ) = 0 </math> | ||
| + | |||
| + | <math>a_3 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(3) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \frac{3 \pi}{2} r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \frac{3 \pi}{2}} + x[2] e^{-j 3 \pi}+ x[3] e^{-j \frac{9 \pi}{2} }) </math> | ||
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| + | <math> = \frac{1}{4} (0 + 1(j) + 1 (-1) + 0) = \frac{j-1}{4} </math> | ||
Latest revision as of 16:36, 25 September 2008
DT Signal Fourier Coefficients
Let's make up a signal.
$ x[0] = 0 $
$ x[1] = 1 $
$ x[2] = 1 $
$ x[3] = 0 $
$ x[4] = x[0] $ etc, the function is periodic with period 4
Using the formula
$ x[n] = \sum_{k=0}^{N-1} a_k e^{jk \frac{2 \pi}{N} n} $, where $ a_k = \frac{1}{N} \sum_{r=0}^{N-1} x[r] e^{-jk \frac{2 \pi}{N} r} $
Since the period is 4, N=4.
$ x[n] = \sum_{k=0}^{3} a_k e^{jk \frac{2 \pi}{4} n} $, where $ a_k = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-jk \frac{2 \pi}{4} r} $
Now to find the fourier series coefficients:
$ a_0 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(0) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{0} = \frac{1}{4} (x[0] + x[1] + x[2] + x[3]) = \frac{1}{4} (0 + 1 + 1 + 0) = \frac{1}{2} $
$ a_1 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(1) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \frac{\pi}{2} r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \frac{\pi}{2}} + x[2] e^{-j \pi}+ x[3] e^{-j \frac{3 \pi}{2} }) $
$ = \frac{1}{4} (0(0) + 1(-j) + 1*-1 + 0*i) = \frac{-1-j}{4} $
$ a_2 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(2) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \pi r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \pi} + x[2] ee^{-j 2 \pi}+ x[3] e^{-j 3 \pi}) $
$ = \frac{1}{4} (0 + 1 * -1 + 1 * 1 + 0 ) = 0 $
$ a_3 = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j(3) \frac{\pi}{2} r} = \frac{1}{4} \sum_{r=0}^{3} x[r] e^{-j \frac{3 \pi}{2} r} = \frac{1}{4} (x[0] e^{0} + x[1] e^{-j \frac{3 \pi}{2}} + x[2] e^{-j 3 \pi}+ x[3] e^{-j \frac{9 \pi}{2} }) $
$ = \frac{1}{4} (0 + 1(j) + 1 (-1) + 0) = \frac{j-1}{4} $
