Find Kp for 10% Error in Control Engineering Homework

[Pietair]

Homework Statement

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.

Homework 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)

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!

 

[saltine]

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.

 

[Pietair]

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!

 

[saltine]

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.

 

[Pietair]

I see, I just edited my reply before your reply.

Thanks a lot!