(New page: I would like to do some examples so my classmates can see a general way of solving problems from this chapter. All problems come from the textbook. ==Question 1== Lets start with a regio...) |
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==Question 1== | ==Question 1== | ||
| − | Lets start with a region of convergence problem. | + | Chapter 9 Problem 6 |
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| + | Lets start with a region of convergence problem. | ||
<math>\frac{(s-1)}{(s+2)(s+3)(s^2+s+1)}</math> | <math>\frac{(s-1)}{(s+2)(s+3)(s^2+s+1)}</math> | ||
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==Question 2== | ==Question 2== | ||
| + | Chapter 9 Question 7 | ||
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Find the Laplace Transform of the following equation... | Find the Laplace Transform of the following equation... | ||
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==Question 3== | ==Question 3== | ||
| + | Chapter 9 Question 16 | ||
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| + | Taking the Laplace transform of both sides of the equation... | ||
| + | |||
| + | <math>Y(s)(s^{3} + (1+\alpha)s^{2} + \alpha(\alpha+1)s + (\alpha)^{2}) = X(s)</math> | ||
| + | |||
| + | Transforming into the equation <math>H(s) = \frac{Y(s)}{X(s)}</math> we can turn out equation into... | ||
| + | |||
| + | <math>H(s) = \frac{Y(s)}{X(s)} = \frac{1}{(s^{3} + (1+\alpha)s^{2} + \alpha(\alpha+1)s + (\alpha)^{2})}</math> | ||
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| + | Taking the Laplace transform of both sides of the equations again, we can use | ||
| + | |||
| + | <math>G(s) = sH(s) + H(s)</math> | ||
| + | |||
| + | Plugging our H(s) into that equation we have.. | ||
| + | |||
| + | <math>G(s) = \frac{(s+1)}{(s^{3} + (1+\alpha)s^{2} + \alpha(\alpha+1)s + (\alpha)^{2})}</math> | ||
| + | |||
| + | Simplifying we get.. | ||
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| + | <math>\frac{1}{s^{2}+ (\alpha)s + \alpha^{2}}</math> | ||
| + | |||
| + | This equation tells us that we have 2 poles. | ||
Latest revision as of 17:20, 24 November 2008
I would like to do some examples so my classmates can see a general way of solving problems from this chapter. All problems come from the textbook.
Question 1
Chapter 9 Problem 6
Lets start with a region of convergence problem.
$ \frac{(s-1)}{(s+2)(s+3)(s^2+s+1)} $
Given the equation we know that there are 4 poles.
s0 = -2
s1 = -3
s3 = $ \frac{-1}{2}+\frac{3^{.5}}{2}j $
s3 = $ \frac{-1}{2}-\frac{3^{.5}}{2}j $
Given these poles, the regions of convergence are as follows...
Re{s} > $ \frac{-1}{2} $
-2 < Re{s} < $ \frac{-1}{2} $
-3 < Re{s} < -2
Re{s} < -3
Question 2
Chapter 9 Question 7
Find the Laplace Transform of the following equation...
$ X(s) = \frac{2(s+2)}{s^{2}+7s+12} $, (Re{s} > -3)
Using partial fraction expansion, we get...
$ X(s) = \frac{4}{s+4} - \frac{2}{s+3} $
Using the inverse Laplace transform we conclude that...
$ x(t) = 4(e^{-4t})u(t) - 2(e^{-3t})u(t) $
Question 3
Chapter 9 Question 16
Taking the Laplace transform of both sides of the equation...
$ Y(s)(s^{3} + (1+\alpha)s^{2} + \alpha(\alpha+1)s + (\alpha)^{2}) = X(s) $
Transforming into the equation $ H(s) = \frac{Y(s)}{X(s)} $ we can turn out equation into...
$ H(s) = \frac{Y(s)}{X(s)} = \frac{1}{(s^{3} + (1+\alpha)s^{2} + \alpha(\alpha+1)s + (\alpha)^{2})} $
Taking the Laplace transform of both sides of the equations again, we can use
$ G(s) = sH(s) + H(s) $
Plugging our H(s) into that equation we have..
$ G(s) = \frac{(s+1)}{(s^{3} + (1+\alpha)s^{2} + \alpha(\alpha+1)s + (\alpha)^{2})} $
Simplifying we get..
$ \frac{1}{s^{2}+ (\alpha)s + \alpha^{2}} $
This equation tells us that we have 2 poles.
