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<math>\,x[n]=a_0 + a_1e^{j\frac{\pi}{2}n} + a_2e^{j\pi n} + a_3e^{j\frac{3\pi}{2}n}\,</math>
 
<math>\,x[n]=a_0 + a_1e^{j\frac{\pi}{2}n} + a_2e^{j\pi n} + a_3e^{j\frac{3\pi}{2}n}\,</math>
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The second property can be used to calculate the average of the signal over one period, which is precisely what <math>\,a_0\,</math> is
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<math>\,a_0=\frac{4}{T}=1\,</math>

Revision as of 19:09, 25 September 2008

Guess the Periodic Signal Question

Suppose a DT singal satisfies the following properties:

1. x[n] is periodic with period 4 and has Fourier coefficients $ \,a_k\, $.

2. $ \,\sum_{n=0}^{3}x[n]=4\, $

3. The Fourier coefficients $ \,a_1\, $ and $ \,a_3\, $ are equal.

4. $ \,a_1\, $ is maximal while $ \,a_2\, $ is non-negative.

5. The value of the signal at $ \,n=0\, $ is zero.


Answer

The first property tells us that the signal is in the form of

$ \,x[n]=\sum_{k=0}^{3}a_ke^{jk\frac{2\pi}{4}n}\, $

$ \,x[n]=a_0 + a_1e^{j\frac{\pi}{2}n} + a_2e^{j\pi n} + a_3e^{j\frac{3\pi}{2}n}\, $


The second property can be used to calculate the average of the signal over one period, which is precisely what $ \,a_0\, $ is

$ \,a_0=\frac{4}{T}=1\, $

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