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Control Engineering |
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| Apr21-09, 12:38 PM | #1 |
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Control Engineering
1. The problem statement, all variables and given/known data
If the steady-state error of the proportional-plus-derivative-plus acceleration system is to be less than 10% determine a suitable value for Kp. 2. Relevant equations The input of the system is: v The output of the system is: y The transfer function of the system including the controller is: (23(Kp+4s+s^2)) / (s^3+19s^2+48s+96+23Kp) 3. The attempt at a solution With steady state: s = 0 Therefore: y steady state = v x (23Kp / (96 + 23Kp)) The error of the system is: v-y Therefore: error = v - (23Kp / (96+23Kp))v error = 96v / (96 + 23Kp) How can I correctly substitute the error of 10% now in this equation. I think (v-y)/y = 0.1 [error] But this will leave 3 variables in the equation... Thanks in advance! |
| Apr23-09, 05:05 PM | #2 |
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Hi,
The error is a ratio. So when you get to error = ( 96 / ( 96 + 23 Kp ) ) v, The error in percentage is: %error = ( 96 / ( 96 + 23 Kp ) ) < 10% There you can then solve for the minimum Kp. |
| Apr26-09, 01:41 PM | #3 |
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Cheers.
So that means that: error = v - y error [%] = ((v-y)/v)*100% So: error = ( 96 / ( 96 + 23 Kp ) ) v error [%] = ( 96 / ( 96 + 23 Kp ) ) * 100% ( 96 / ( 96 + 23 Kp ) ) * 100% < 10% ( 96 / ( 96 + 23 Kp ) ) < 0.1 Then I get: Kp > 37.565 Is that correct? Thanks in advance! |
| Apr26-09, 01:48 PM | #4 |
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Control Engineering
Hi, you must have done something wrong in the algebra, because Kp should be positive.
( 96 / ( 96 + 23 Kp ) ) < 0.1 Conceptually, when the denominator is 960, you will have exactly 0.1 error. As Kp increases, the error decreases further. When you solve it using algebra you can verify that Kp to give you a denominator that is greater than 960. |
| Apr27-09, 11:55 AM | #5 |
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I see, I just edited my reply before your reply.
Thanks a lot! |
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