1. Chapter 7 Steady State Error <<<4.1>>>
###Continuous Time Signals###

2. Test waveforms for evaluating steady-state errors of position control systems
Steady-state error is the difference between the input and the output for a prescribed test input as t →∞

3. Test inputs for steady-state error analysis and design vary with target type

4. Steady-state error: a. step input; b. ramp input

5. Closed-loop control system error: a. general representation; b
Closed-loop control system error: a. general representation; b. representation for unity feedback systems

6. System with: a. finite steady-state error for a step input; b
System with: a. finite steady-state error for a step input; b. zero steady-state error for step input
For pure gain K as in (a), Css= K ess or ess =1/K Css there will always be error
But if the pure gain is replaced by an integrator as in (b) the steady state error will be zero.

7. Feedback control system for Example 7.1
Problem Find the steady-state error for the system in(a) if T(s) = 5/(s2+7s+10) and the input is a unit step.
Solution:
Using E(s) = R(s)[1-T(s)]
We can apply the final value theorem
To yield e(∞) =1/2

8. Steady-state error in terms of G(s)
E(s) =R(s) – C(s),
C(s) =E(s)G(s) so
E(s) = R(s)/1+G(s)
Using final value theorem

9. Steady-state error in terms of G(s)
For step input, the steady state error will be zero if there is at least one pure integration in the forward path.
For ramp input, the steady state error will be zero if there is at least 2 pure integration in the forward path.
For parabolic input, the steady state error will be zero if there is at least 3 pure integration in the forward path.

10. Steady state error for systems with no integrations
Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the system. u(t) is a step function.
Solution First verify the closed loop system is stable.
For input 5u(t)
For input 5tu(t)
For input 5t2u(t)

11. Steady state error for systems with one integration
Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the system. u(t) is a step function.
Solution First verify the closed loop system is stable.
For input 5u(t)
For input 5tu(t)
For input 5t2u(t)

12. Static error constants
Position constant step input
Velocity constant ramp input
Acceleration constant parabolic input

13. Steady state error via static error constants
Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs..
Solution First verify the closed loop system is stable.
Thus for step input,
For ramp input
For parabolic input

14. Steady state error via static error constants
Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs.
Solution First verify the closed loop system is stable.
Thus for step input,
For ramp input
For parabolic input

15. Steady state error via static error constants
Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs.
Solution First verify the closed loop system is stable.
Thus for step input,
For ramp input
For parabolic input

16. Feedback control system for defining system type
System type is the value of n in the denominator, or the number of pure integrations in the forward path.

17. Relationships between input, system type, static error constants, and steady-state errors

18. Feedback control system for Example 7.6
Problem Find the value of K so that there is 10% error in the steady state..
Solution:
Since the system is Type 1, the error must apply to a ramp input, thus
and
and K = 672
Applying Routh-Hurwitz we see that the system is stable at this gain.

19. Feedback control system showing disturbance
Applying final value theorem we obtain

20. Previous system rearranged to show disturbance as input and error as output, with R(s) = 0
Assume step disturbance D(s) = 1/s
We get
This shows that the steady-state error produced by a step disturbance can be reduced by increasing the dc gain of G1(s) or decreasing the dc gain of G2(s)

21. Feedback control system for Example 7.7
Problem Find the steady state error due to a step disturbance for the shown system..
Solution: The system is stable, we find

22. G(s) = G1(s)G2(s) H(s) = H1(s)/G1(s)
Forming an equivalent unity feedback system from a general non-unity feedback system
G(s) = G1(s)G2(s)
H(s) = H1(s)/G1(s)

23. Non-unity feedback control system for Example 7.8
Problem Find the system type, the appropriate error constant, and the steady state error for a unit step input..
Solution: The system is stable, we convert the system into an equivalent unity feedback system
The system is Type 0, and Kp= -5/4. The steady-state error is

24. Steady-State error for systems in State Space
The Laplace transform of the error is
E(s) = R(s) –Y(s), and since
Y(s) = R(s)T(s).
Then
E(s) = R(s)[1-T(s)]
Using T(s) =Y(s)/U(s) = C(sI-A)-1B + D
We have
E(s) = R(s)[1-C(sI-A)-1B]
Applying final value theorem, we have

25. Steady-State error for systems in State Space
Example: Evaluate the ss error for the system described by
Solution:
Substitute A,B, and C in
For unit step, R(s)=1/s and e(∞)=4/5, for unit ramp, R(s) = 1/s2 and e(∞)= ∞