1. Transient and Steady-State Response Analysis (4)
Chapter 5
Transient and Steady-State Response Analysis (4)

2. 5-7 The steady state error
Any physical control system inherently suffers steady-state error in response to certain types of inputs. A system may have no steady-state error to a step input, but the same system may exhibit nonzero steady-state error to ramp input.
In this section, we shall introduce the definitions of system’s error and the method to evaluate the steady-state error.

3. 1. Two definitions of system error
G(s)
H(s)
C(s)
R(s)
B(s) or
The definitions reduce to the same if the system is unity-feedback.
The steady state error of the control system is then defined as

4. 2. Final Value Theorem and its applying condition:
Theorem: Suppose f(t) has the Laplace transform F(s) and the limits limtf(t) and lims0sF(s) exist. Then the final value is
Let
and assume E(s) is a rational proper function. Then, the Final Value Theorem can be applied provided all the poles of sE(s) lie in the left-half s-plane!

5. Example. Consider the following control system:
Controller
Plant
R(s)
C(s)
Let r(t)=1(t). Find its steady-state error ess.
Solution:
The coefficient of the power s is zero, which implies that the system has poles in right-half s-plane.

6. Example. Consider the following control system:
Controller
Plant
R(s)
C(s)
Let r(t)=sin(t). Find its steady-state error ess.
Solution: Since
It does not agree with the applying condition of the FV theorem since sE(s) has two poles on the imaginary axis.

7. 3. Classification of control systems
G(s)
H(s)
C(s)
R(s)
B(s)
E(s)
Define:
Let
How can we deal with the case of second-order system?
Then

8. 4. Static position error constant Kp (r=1(t))
Let
Then
Why is Kp called static position error constant? How to explain position error?
where

9. (1) For type 0 system (N=0):
The static position error constant Kp:
(2) For type 1 system (N1):

10. Therefore,
Summary
For a unit-step input, the steady-state error

11. 5. Static velocity error constant Kv (r=t1(t))
Let
G(s)
H(s)
C(s)
R(s)
B(s)
E(s)
Then
where

12. Define the velocity error constant:
Summary

13. 6. Static acceleration error constant Ka (r=t2/2)
G(s)
H(s)
C(s)
R(s)
B(s)
E(s)
Let
Then
where

14. Define the acceleration error constant:
Summary

15. The steady state error in terms of Gain K
Number of
Integrations
in G(s)H(s)
Input
A1(t)
At1(t)
At2/2  1 2

16. 7. The steady state error with disturbance n(t)
C(s)
R(s)
E(s)
N(s)
By the principle of superposition, we have
When

17. Determine its steady-state error.
Example.
C(s)
R(s)=0
E(s)
N(s)=1/s
Determine its steady-state error.
Solution: sE(s) has a stable pole. In fact,
In general, we should consider the case of H =\1.

18. Since sE(s) has imaginary poles, ess does not exist.
Note that attention must be paid to the location of the poles of sE(s). For instance,
C(s)
R(s)=0
E(s)
N(s)=1/s
Since sE(s) has imaginary poles, ess does not exist.

19. Nevertheless, if G1(s) is changed as
C(s)
R(s)=0
E(s)
N(s)=1/s
the poles of sE(s) lie in the left-half s-plane,

20. Example. Consider the following two systems:-
y(t)
(a)
r(t)=0 -
y(t)
(b)

21. Determine their steady-state errors.

22. However, if we use a PI controller, ess is able to converge to zero:
r(t)=0 -
y(t)
(c)

23. 8. Effects of integral and derivative control actions on system performance
In the proportional control of a plant whose transfer function does not possess an integrator 1/s, there is a steady-state error, or offset, in the response to a step input. Such an offset can be eliminated if an integral control action is included in the controller.

24. Proportional control of systems
Consider the following control system:
Controller
Plant
R(s)=1/s
C(s)
Since the system is type 0 system,

25. A higher K leads to a smaller offset.
c(t) 1
offset
A higher K can improve the steady-state error, but may cause saturation of the amplifier. The derivation of an integral control action may eliminate the steady-state error. t
c(t) 1
A higher K leads to a smaller offset.

26. Integral control of systems
Consider the following control system:
Controller
Plant
R(s)=1/s
C(s)
If we want the system to track
Since the system is type 1 stable system,
This is an important improvement over the proportional control alone, which gives offset.

27. Tracking error e(t) of the integral control: ess=0

28. Application example: Response to torque disturbance
Proportional control
C(s)
R(s)=0
E(s)
D(s)=Td/s

29. sE(s) has two stable poles and therefore,
Proportional + Integral (PI) control
Let G1(s) be replaced by a proportional+integral controller:
Then

30. sE(s) is stable if and only if
On this premise, we have
Remark: It is important to point out that if the controller were an integral controller alone, the closed-loop characteristic equation would be
which is unstable.

31. Derivative control action
An advantage of using derivative control action is that it responds to the rate of change of the actuating error and can produce a significant correction before the magnitude of the actuating error becomes too large. Derivative control thus
anticipates the actuating error, initiates an early corrective action, and tends to increase the stability of the system.
Anticipate, 感知，预知；initiate，开始，启动

32. Consider the following proportional control of an inertia load:
Controller
Inertia Load
R(s)=1/s
C(s) 1
The response to a unit-step input oscillates indefinitely.

33. Proportional + Derivative (PD) control of a system with inertia load
Consider the following PD control of an inertia load:
Controller
Inertia Load
R(s)=1/s
C(s)
u=up+ud u 1

34. Proportional + Derivative (PD) control of Second-order systems
Consider the following proportional+derivative control of an inertia load:
Controller
Plant
R(s)=1/s
C(s)

35. The steady-state error for a unit-ramp input is
The characteristic equation is
it is possible to make both the steady-state error ess for a ramp input and the maximum overshoot for a step input small by making B small, Kp large, and Kd large enough so that d is between 0.4 and 0.7.

36. Summary of Chapter 5
In this chapter, time domain analysis is investigated.
1. We studied the responses of the following systems:
First-order system:

37. Main properties of the unit-step response are:
Second-order system:

38. Performance specifications for unit-step response of a second-order system with 0<<1:tp ts 1 Mp
Performance specifications for unit-step response of a second-order system with 1: No oscillation and

39. Unit-step response for higher-order systems:
which is the sum of a series of first-order and second-order systems.
The performance analysis for higher-order systems is based on dominant poles, where to determine dominant poles, dipole must be excluded.

40. 2. We studied the relationship between the impulse, the step and the ramp responses:
from which the following equations also hold:
The same conclusion can be applied to the parabolic response.

41. 1) The system is stable if and only if
3. We studied the stability criterion, Routh’s stability criterion. For a given closed-loop characteristic equation
1) The system is stable if and only if
All the coefficients of D(s) are positive and
All the terms in the first column of the array have positive signs.
2) The number of roots of the D(s) with positive real parts is equal to the number of changes in sign of the coefficients of the first column of the array.
Some special cases are also discussed.
More specifically, two special cases are discussed. Both cases indicate that some poles lie either in the right-half s-plane or on the imaginary axis.

42. By utilizing Routh’s stability criterion, we are able to investigate the relative stability and conditional stability problems: j Re  
G(s)

43. We studied the steady-state error.
C(s)
R(s)
E(s)
N(s)
B(s)
Let
then

44. To apply the Final Value Theorem, it is required that all the poles of both sER(s) and sEN(s) lie in the left-half s-plane:
G(s)
H(s)
C(s)
R(s)
B(s)
E(s)
Further, in case of no disturbance, for typical input signals 1(t), t·1(t) and t2·1(t)/2, the system

45. type, the number of the integrations of G(s)H(s), makes us determine ess conveniently. For
(1) When input signal r=1(t)
where static position error constant Kp:

46. (2) When input signal r=t·1(t)
where static velocity error constant Kv:

47. (3) When input signal r=t2·1(t)/2
where static acceleration error constant Ka:

48. We introduced some widely used controllers: Proportional controller
plant
R(s)
C(s)
Advantage: In many cases, simply adjusting Kp is able to improve transient and steady-state performance;
Disadvantage: Large K may cause system to become unstable and cause saturation of the amplifier. Moreover, for type 0 Gp(s), steady-state error is non-zero when R(s)=1/s.

49. (2) Integral Controller
Plant
R(s)
C(s)
Advantage: is able to eliminate steady-state error.
Disadvantage: may cause system to become unstable and therefore, is rarely used alone.

50. (4) Proportional +Integral (PI) controller
Plant
R(s)
C(s)
Advantage: is able to eliminate steady-state error while keeping system stability.

51. (5) Proportional + Derivative (PD) controller
Plant
R(s)
C(s)
Advantage: is able to improve system transient performance and stability;
Disadvantage: may amplify high frequency input disturbance.

52. (6) Servo system with velocity feedback
Advantage: is able to improve system transient performance and avoid high frequency output disturbance.