Control System Gain design problem

AI Thread Summary
A Type 0 Open Loop Transfer function cannot achieve zero steady state error for step inputs, as it inherently has a constant steady state error. The discussed transfer function G(s)=3/(s^2 +4s+3) confirms this limitation, resulting in infinite steady state error for ramp and parabolic inputs. However, increasing the gain K in the forward path can reduce the steady state error for step inputs, approaching zero as K increases significantly. The discussion emphasizes the importance of gain placement in the control system design to meet steady state error specifications. Ultimately, achieving zero steady state error requires careful adjustment of system parameters.
koochiee
Messages
16
Reaction score
0
Is it possible for a Type 0 Open Loop Transfer function to have 0 steady state error?
Context - Control System Gain design to meet a certain steady state error specification.
The open loop T.F - G(s)=3/(s^2 +4s+3) (This is type 0 (n=0))
The closed loop T.F is Gc(s)=G(s)/(1+KG(s)), But my problem is will changing the value of K or adding gain would make a difference? Because the Open loop T.F. is Type 0, the steady state error for 1. step i/p - constant
2. ramp & parabolic i/p - infinity, Any help is much appreciated. :)
 

Attachments

  • ASK.png
    ASK.png
    3.8 KB · Views: 536
Last edited by a moderator:
Physics news on Phys.org
Typically 'H' is the feedback gain and 'K' is the amplifier before the plant, which is G in this case.

Are you sure you have your amplifier in the correct location?
 
  • Like
Likes 1 person
Thanks for the reply!

And yes it is, the closed loop T.F is Gc(s)=G(s)/(1+KG(s)). Even after when I convert it to a unity feedback gain system it still is a type 0 system. Which can't have zero steady state error. And they're asking us to show that a gain (possibly in the forward path) of K+1 would enable it to maintain a zero steady state error.
 
with your system, you will always have a steady state error of 1/4*step input.

Now if your K was in the forward path (as in directly before the plant) then the steady state error for a step input would be step_input*(1/(1+3*k))

so if you made K infinitely large, the steady state error would approximately equal zero.
 
  • Like
Likes 1 person
Thank you!
 

Similar threads

Back
Top