(New page: == Guess the signal == 1. DT signal x[n] is even. 2. X[n] has a period of 2. 3. <math>\sum_{n=0}^{1}x[n]=3</math> 4. <math>\sum_{n=0}^{1}(-1)^nx[n]=5</math> == Answer == From 2. w...) |
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== Guess the signal == | == Guess the signal == | ||
| − | 1. DT signal x[n] | + | 1. DT signal x[n] has a period of 2. |
| − | 2. | + | 2. <math>a_k = 0</math> for |k|>1 |
3. <math>\sum_{n=0}^{1}x[n]=3</math> | 3. <math>\sum_{n=0}^{1}x[n]=3</math> | ||
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== Answer == | == Answer == | ||
| − | From | + | From 1. we know the period = 2, therefore: |
<math>x[n]=\sum_{n=0}^{1}a_ke^{jk(2\pi/2)n}</math> | <math>x[n]=\sum_{n=0}^{1}a_ke^{jk(2\pi/2)n}</math> | ||
Latest revision as of 10:39, 25 September 2008
Guess the signal
1. DT signal x[n] has a period of 2.
2. $ a_k = 0 $ for |k|>1
3. $ \sum_{n=0}^{1}x[n]=3 $
4. $ \sum_{n=0}^{1}(-1)^nx[n]=5 $
Answer
From 1. we know the period = 2, therefore:
$ x[n]=\sum_{n=0}^{1}a_ke^{jk(2\pi/2)n} $
From 3. we know that:
$ a_0=\frac{1}{2}\sum_{n=0}^{1}x[n]=\frac{1}{2}3 = \frac{3}{2} $
From 4. we know:
$ a_1=\frac{1}{2}\sum_{n=0}^{1}x[n](-1)^n= \frac{1}{2}\sum_{n=0}^{1}x[n]e^{-jn\pi} = \frac{5}{2} $
Therefore our function x[n]:
$ x[n]=\frac{3}{2}+\frac{5}{2}e^{jn\pi} $
