(New page: == Guess the signal == Guess <math>x[n]\,</math> == Properties== 1. period of the function is 3 2. <math>\sum_{n=0}^{2}x[n]=4</math> 3. <math>\,a_k=a_{k+2}</math> 4. <math>x[n]\,</math>...) |
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== Final Answer == | == Final Answer == | ||
We still have to compute the sum, so | We still have to compute the sum, so | ||
| + | we have <math>\,x[0]</math> and <math>\,x[1]</math>, now get <math>\,x[2]</math> | ||
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| + | <math>x[2]=\frac{4}{3}e^{-jk*2*\frac{2\pi }{3}}</math> | ||
| + | |||
| + | So summing them makes: | ||
| + | |||
| + | <math>x[n]=\frac{4}{3}+\frac{4}{3}e^{-jk\frac{4\pi }{3} n}</math> | ||
Latest revision as of 15:00, 25 September 2008
Guess the signal
Guess $ x[n]\, $
Properties
1. period of the function is 3
2. $ \sum_{n=0}^{2}x[n]=4 $
3. $ \,a_k=a_{k+2} $
4. $ x[n]\, $ has a minimum power among all signals that satisfy rules 1-3
Derivations
1.
Given that the definition of a signal defined by fourier transforms is shown like:
$ x[n]=\sum_{N=0}^{N-1}a_k e^{-jk \frac {2\pi}{\omega_0} n} $
We know that our answer must be in the form of:
$ x[n]=\sum_{N=0}^{2}a_k e^{-jk \frac {2\pi}{3} n} $
2.
We know that our $ a_k\, $ values are defined as
$ a_k=\frac{1}{N}\sum_{N=0}^{N-1}x[n] e^{-jk \frac {2\pi}{\omega_0} n} $
so:
$ a_0=\frac{1}{3}\sum_{N=0}^{2}x[n] e^{0}=\frac{4}{3} $
3.
If $ \,a_k=a_{k+2} $, then
$ \,a_0=a_2 $
4.
We minimize the other values, so $ \,a_1=0 $
Final Answer
We still have to compute the sum, so we have $ \,x[0] $ and $ \,x[1] $, now get $ \,x[2] $
$ x[2]=\frac{4}{3}e^{-jk*2*\frac{2\pi }{3}} $
So summing them makes:
$ x[n]=\frac{4}{3}+\frac{4}{3}e^{-jk\frac{4\pi }{3} n} $
