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3. <math>\sum_{n=0}^{3}x[n](-j)^n = -20j</math> | 3. <math>\sum_{n=0}^{3}x[n](-j)^n = -20j</math> | ||
| − | 4. <font "size"=4> <math>a_{ | + | 4. <font "size"=4> <math>a_{k} \; = \; -a_{k-2}</math> </font> |
==Solution== | ==Solution== | ||
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Recall that <math>a_{k}=\frac{1}{N} \sum_{n=0}^{N-1} x[n]e^{-jk \frac{2\pi}{N}n} = \frac{1}{4} \sum_{n=0}^{3} x[n]e^{-jk \frac{\pi}{2}n}</math> | Recall that <math>a_{k}=\frac{1}{N} \sum_{n=0}^{N-1} x[n]e^{-jk \frac{2\pi}{N}n} = \frac{1}{4} \sum_{n=0}^{3} x[n]e^{-jk \frac{\pi}{2}n}</math> | ||
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| + | If k = 1, then <math>a_{1}=\frac{1}{4} \sum_{n=0}^{3} x[n]e^{-j \frac{\pi}{2}n}</math>, and, realizing that <math>e^{-j \frac{\pi}{2}n} = (-j)^n</math>, <math>a_{1}=\frac{1}{4} \sum_{n=0}^{3} x[n](-j)^n</math>. | ||
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| + | By 3., <math>\sum_{n=0}^{3} x[n](-j)^n = -20j</math>, so <math>a_{1}=\frac{1}{4}(-20j) = -5j</math>. | ||
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| + | By 4. <math>a_{3} = -a_{1}</math>, so <font "size"=5><math>a_{3} = 5j</math> </font>. Similarly, <math>a_{2} = -a_{0}</math>, so <font "size"=5><math>a_{2} = 0</math> </font>. | ||
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| + | We now have <math>a_{0}</math>, <math>a_{1}</math>, <math>a_{2}</math>, and <math>a_{3}</math>, which is all we need since <math>x[n] = \sum_{k=0}^{N-1} a_{k}e^{jk \frac{2\pi}{N}n} = \sum_{k=0}^{3} a_{k}e^{jk \frac{\pi}{2}n}</math>. | ||
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| + | <math>x[n] = \sum_{k=0}^{3} a_{k}e^{jk \frac{\pi}{2}n} = -5je^{j \frac{\pi}{2}n} + 5je^{j \frac{3\pi}{2}n} = -5je^{j \frac{\pi}{2}n} + 5je^{j \frac{4\pi}{2}n}e^{-j \frac{\pi}{2}n} = -5je^{j \frac{\pi}{2}n} + 5je^{-j \frac{\pi}{2}n}</math> | ||
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| + | <math>=-5j(e^{j \frac{\pi}{2}n} - e^{-j \frac{\pi}{2}n}) = \frac{5}{j}(e^{j \frac{\pi}{2}n} - e^{-j \frac{\pi}{2}n}) = \frac{10}{2j}(e^{j \frac{\pi}{2}n} - e^{-j \frac{\pi}{2}n}) = 10\frac{e^{j \frac{\pi}{2}n} - e^{-j \frac{\pi}{2}n}}{2j} = 10sin[\frac{\pi}{2}n]</math> | ||
Latest revision as of 20:45, 24 September 2008
Determine the Periodic Signal
A periodic signal x[n] has the following characteristics:
1. N = 4
2. The average value of the signal over the interval $ 0 \leq n \leq 7 $ is 0.
3. $ \sum_{n=0}^{3}x[n](-j)^n = -20j $
4. $ a_{k} \; = \; -a_{k-2} $
Solution
Coupling 1. and 2. together, we see that the interval is twice the fundamental period. The coefficient $ a_{0} $ is equal to the average value of the signal over one period, so $ a_{0} = \frac{0}{2} = 0 $.
Recall that $ a_{k}=\frac{1}{N} \sum_{n=0}^{N-1} x[n]e^{-jk \frac{2\pi}{N}n} = \frac{1}{4} \sum_{n=0}^{3} x[n]e^{-jk \frac{\pi}{2}n} $
If k = 1, then $ a_{1}=\frac{1}{4} \sum_{n=0}^{3} x[n]e^{-j \frac{\pi}{2}n} $, and, realizing that $ e^{-j \frac{\pi}{2}n} = (-j)^n $, $ a_{1}=\frac{1}{4} \sum_{n=0}^{3} x[n](-j)^n $.
By 3., $ \sum_{n=0}^{3} x[n](-j)^n = -20j $, so $ a_{1}=\frac{1}{4}(-20j) = -5j $.
By 4. $ a_{3} = -a_{1} $, so $ a_{3} = 5j $ . Similarly, $ a_{2} = -a_{0} $, so $ a_{2} = 0 $ .
We now have $ a_{0} $, $ a_{1} $, $ a_{2} $, and $ a_{3} $, which is all we need since $ x[n] = \sum_{k=0}^{N-1} a_{k}e^{jk \frac{2\pi}{N}n} = \sum_{k=0}^{3} a_{k}e^{jk \frac{\pi}{2}n} $.
$ x[n] = \sum_{k=0}^{3} a_{k}e^{jk \frac{\pi}{2}n} = -5je^{j \frac{\pi}{2}n} + 5je^{j \frac{3\pi}{2}n} = -5je^{j \frac{\pi}{2}n} + 5je^{j \frac{4\pi}{2}n}e^{-j \frac{\pi}{2}n} = -5je^{j \frac{\pi}{2}n} + 5je^{-j \frac{\pi}{2}n} $
$ =-5j(e^{j \frac{\pi}{2}n} - e^{-j \frac{\pi}{2}n}) = \frac{5}{j}(e^{j \frac{\pi}{2}n} - e^{-j \frac{\pi}{2}n}) = \frac{10}{2j}(e^{j \frac{\pi}{2}n} - e^{-j \frac{\pi}{2}n}) = 10\frac{e^{j \frac{\pi}{2}n} - e^{-j \frac{\pi}{2}n}}{2j} = 10sin[\frac{\pi}{2}n] $
