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== Guess the Periodic Signal ==
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Suppose a DT singal satisfies the following properties:
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1. x[n] is periodic with period 4 and has Fourier coefficients <math>\,a_k\,</math>.
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2. <math>\,\sum_{n=0}^{3}x[n]=4\,</math>
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3. The Fourier coefficients <math>\,a_1\,</math> and <math>\,a_3\,</math> are equal.
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4. <math>\,a_1\,</math> is the maximal negative value while <math>\,a_2\,</math> is non-positive.
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5. The value of the signal at <math>\,n=0\,</math> is zero.
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== Answer ==
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The first property tells us that the signal is in the form of
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<math>\,x[n]=\sum_{k=0}^{3}a_ke^{jk\frac{2\pi}{4}n}\,</math>
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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>
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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>
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The last three properties can be combined to figure out <math>\,a_1, a_2, a_3\,</math>.  The fifth property shows that
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<math>\,x[0]=0\,</math>
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<math>\,0=a_0 + a_1e^{0} + a_2e^{0} + a_3e^{0}\,</math>
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<math>\,0=1 + a_1 + a_2 + a_3\,</math>
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By the third property, we can rewrite this as
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<math>\,-1=2a_1+a_2\,</math>
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Property four is satisfied when <math>\,a_2=0\,</math>
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<math>\,-1=2a_1\,</math>
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<math>\,a_1=a_3=-\frac{1}{2}\,</math>
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Therefore, the signal described by the five properties is
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<math>\,x[n]=1-\frac{1}{2}e^{j\frac{\pi}{2}n}-\frac{1}{2}e^{j\frac{3\pi}{2}n}\,</math>
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[[Image:Jkubasci guess signal_ECE301Fall2008mboutin.png]]

Latest revision as of 19:40, 25 September 2008

Guess the Periodic Signal

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 the maximal negative value while $ \,a_2\, $ is non-positive.

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\, $


The last three properties can be combined to figure out $ \,a_1, a_2, a_3\, $. The fifth property shows that

$ \,x[0]=0\, $

$ \,0=a_0 + a_1e^{0} + a_2e^{0} + a_3e^{0}\, $

$ \,0=1 + a_1 + a_2 + a_3\, $

By the third property, we can rewrite this as

$ \,-1=2a_1+a_2\, $

Property four is satisfied when $ \,a_2=0\, $

$ \,-1=2a_1\, $

$ \,a_1=a_3=-\frac{1}{2}\, $


Therefore, the signal described by the five properties is

$ \,x[n]=1-\frac{1}{2}e^{j\frac{\pi}{2}n}-\frac{1}{2}e^{j\frac{3\pi}{2}n}\, $


Jkubasci guess signal ECE301Fall2008mboutin.png

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