Outline
Steady-State Error
for State-Space Models
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
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M. Sami Fadali
Professor of Electrical Engineering
University of Nevada, Reno
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Steady-state error for state-space
equations.
Error due to step.
Error due to disturbance.
Output tracking.
Perturbation model.
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State-Space Equations
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Block Diagram
State-space model
•
•
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Feedback control
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Closed-loop dynamics
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Square system:
,
Closed-loop system with state feedback.
Preamplifier with gain .
square matrix.
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4
Steady-State Error
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Error due to Step
(square)
Assume stable closed-loop dynamics
Final Value Theorem (limit exists)
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All inputs step functions:
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Internally stable system implies nonsingular
.
→
→
→
(square)
(square)
→
→
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Gain
Zero Steady-State Error: Step
Assume a square system (includes scalar)
→
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Assume stable closed-loop system.
For
(square: recall matrix is full rank)
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For
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→
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, can use a (right) pseudoinverse
for
For
: zero error by
adding a preamplifier with gain to make
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8
Transmission Zero
SISO Example
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Closed-loop transfer function (zero at 0)
Closed-loop transfer function (without preamp.)
Step Input:
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Step Response:
If the matrix is singular, there exists
→
i.e. the system has a transmission zero at zero.
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
→
Cannot follow a step.
Cannot track a step input.
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Evaluate Error Due to Step
Example: Step Input
1
0
0   x1 (t )  0
 x1 (t )   0
 x (t )   0
0
1
0   x 2 (t )   0 
 2 
    r (t )

0
0
1   x3 (t )  0
 x3 (t )   0
 x (t )    156  139  53  11  x (t )  1 
 4   
 4  
→
→
 x1 (t ) 
 x (t ) 
y (t )  100 0 4 2 2 
 x3 (t ) 
 x (t ) 
 4 
Can make steady-state error zero by adding a
preamplifier of gain
2
11
4
53
100
139
156
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Example: Error Due to Step
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State-space Model
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For zero steady-state error
Effect of Disturbance
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Linear system: use superposition.
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Transfer functions
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Example: Error Due to Step Disturbance
& Step Input
Total Steady-State Error
_
1
0
0   x1 (t )  0
 x1 (t )   0
 x (t )  0
0
1
0   x2 (t ) 0
 2 

 x3 (t )   0
0
0
1   x3 (t )  0
 
 


 x 4 (t )  156  139  53  11  x4 (t ) 1
 x1 (t ) 
 x (t )
y (t )  100 0 4 2 2 
 x3 (t ) 


 x4 (t )
4 s 3  62 s 2  178s  788
Td ( s )  4
s  11s 3  53s 2  139 s  156
Assume a square system
→
→
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→
0
0  r (t ) 
1 d (t )

0
→
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Steady-state Error
Output Regulation
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→
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Design a regulator for
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Equilibrium:
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→
in the steady state.
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→
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Example
Solve for the Reference Input
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Equilibrium:
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Find the equilibrium state and reference input
for a steady state output
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or
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For
, the solution exists if the matrix is
invertible (full rank)
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or
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Reference input
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Perturbations from Equilibrium
Perturbation Block Diagram
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Error Convergence to Zero
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Error converges to zero if the closed-loop
dynamics are stable.
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