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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\, $
