Exam

1
Exam 2 Review

• Nov. 14, 1110-12
• Format same as Exam 1.
• Chapters covered are 5,6 and 7. (Note You need
  to know what a transfer function is!)
• Important topics
• - Performance of 1st and 2nd-order systems
• (System output parameters such as
  overshoot,
• Settling time, etc.)
• - Root Locations and the Transient Response
• - Stability Analysis
• - Root Locus Analysis and Design
• - P, PI and PD Controllers

2
Time (or Transient) Response

• Key Concepts
• 1. (System) Characteristic Equation
• 2. Poles and Zeros

3
Time (or Transient) Response

• Key Concepts
• 1. System Characteristic Equation
• 2. Poles and Zeros
• Let the system transfer function be
  G(s)p(s)/q(s).
• gt
• The (System) Characteristic Eqn. is q(s)0.
• And, Poles are the roots of q(s)o.
• Zeros are the roots of p(s)0.

4
Transfer Function of First-Order Systems

• Ex. G(s)V2(s)/V1(s)1/(RCs1) where RC is called
  
• the time constant ?.
• ?In general,
• G(s) K/(?s1), where Ksystem DC gain
• (i.e.,
  KG(0)).
• R(s)
  Y(s)

G(s)
5
Performance of First-Order Systems

• G(s) K/(?s1), where Ksystem DC gain (G(0)),
• and ? time
  constant
• Step response R(s)1/s Y(s)
• gt
• Y(s)K/s K/(s1/?)
• gt
• y(t) K(1-e t/? )

G(s)
?
6
Performance of First-Order Systems

• G(s) K/(? s1), where Ksystem DC gain (i.e.,
  KG(0)).
• 1. Step response U(s)1/s gt
• Y(s)K/s K/(s1/? gt y(t)
  K(1-e -t/? )
• 2. Ramp response R(s)1/s2 gt
• Y(s) K/s2 K? /s K? /(s1/?)
• gt
• y(t) Kt - K? K? e t/?

7
Performance of a 2nd-Order System

• Standard Form
• Y(s)G(s)/(1G(s))R(s)
• ?n2/(s22? ?ns ?n2)R(s) (5.7)
• where,
• ? damping ratio
• ?nnatural frequency
• gt Thus, the standard 2nd-order system
  characteristic equation is
• q(s) s22? ?ns ?n2 0.

8
Performance of a 2nd-Order System

• Unit Step Response
• Y(s)G(s)R(s), R(s)1/s
• ?n2/s(s22? ?ns ?n2) (5.8)
• gt y(t) 1 (1/ß)e - ? ?nt sin(?n ß t
  ?) (5.9)

9
Step Response of a 2nd-Order System

• Time (or Transient) Response

Overshoot
Peak Time
Settling Time
Rise Time
10
Step Response of a 2nd-Order System

• Standard form G(s) ?n2/s(s2? ?ns ?n2)
• For complex poles, the unit step response is
• y(t) 1 (1/ß)e - ? ?nt sin(?n ß t
  ?)
• Key Response Parameters (used in Design)
• -Tr (Rise time)
• -Tp (Peak time)
• -Mpt(Peak Value)
• -P.O.( Overshoot)
• ((Mpt-fv)/fv)x100
• where fvthe steady-state,
• or final, value of y(t)

11
Effect of Pole Locations of 2nd-Order Systems

• Standard transfer function
• G(s) ?n2/(s22? ?ns ?n2)
• Poles of G(s) for ? lt1 are
• p1,2 - ? ?n j?n 1-?2

Complex Conjugate pair
12
Effect of Pole Locations of 2nd-Order Systems

• Standard transfer function
• G(s) ?n2/(s22? ?ns ?n2)
• gt Poles are p1,2 - ? ?n j?n 1-?2
• Unit Step Responce
• Y(s)G(s)R(s), R(s)1/s
• ?n2/s(s22? ?ns ?n2) (5.8)
• gt
• y(t) 1 (1/ß)e - ? ?nt sin(?n ß t ?)
  (5.9)
• gt

13
Steady-State Error

• System Error E(s) R(s)-Y(s)
• R(s)-
• 
  (G(s)/(1GH(s))R(s)
• gt
• E(s)G(s)R(s), G(s)(1GH(s)-G(s))/
• (1GH(s))
• If H(s)1,
• E(s)R(s)/(1G(s))
• gt
• The steady state error is
• lim e(t) ess lim sR(s)/(1G(s))
• t-gt s-gt0

14
Definition of System Stability

• 1. A stable system is a dynamic system with a
  bounded response to a bounded input.
• This is called a BIBO stability.
• Ex. Step response of a 2nd-order system

BIBO Stability
15
Effect of Pole Locations on Impulse Responses
(Figure 5.17)
Marginally stable
Unstable
Stable
16
Concept of Root Locus

• The root locus method is a graphical technique to
  analyze/design system stability. A graph or locus
  of roots of the characteristic equation as one
  system parameter (e.g., the controller gain K)
  varies is known as a root locus plot.

17
The Routh-Hurwitz Stability Criterion

• The Routh-Hurwitz stability criterion states that
  the number of roots of q(s) with positive real
  parts is equal to the number of changes in sign
  of the first column of the Routh array.
• Ex.
• gt
• System is unstable.

18
Special Case

• Zeros in the first column (p.319)
• gt Replace the zero with a small positive number
  e.
• Ex.
• q(s)s52s42s34s211s10 (6.10)

19
Special Case

• Zeros in the first column
• Replace the zero with
• a small positive number e.
• Ex. q(s)
• s52s42s34s211s10
• (6.10)

20
6.4 Stability of State Variable Systems

• Ex. 6.8 Consider the following 2nd-order system
• dx1/dt -3x1 x2
• dx1/dt -Kx1 x2 Ku
• gt
• dx/dt A x Bu
• ?

21
Stability of State Variable Systems

• Solution of State Equation (Ch.3)
• dx/dt Ax Bu
• gt
• x(s) F(s)x0 F(s)Bu(s)ds
• where
• F(s) (sI-A) -1 State Transition
  Matrix
• and
• det(sI-A) 0 System
  Characteristic
• 
  Equation

22
Stability of State Variable Systems

• Ex. 6.8 Consider the following 2nd-order system
• dx1/dt -3x1 x2
• dx2/dt -Kx1 x2 Ku
• gt
• dx/dt A x Bu
• and
• the system characteristic equation is given
  by
• q(s) det(sI A ) 0
• gt
• q(s) s2 2s (K-3) 0.
• gt
• From the Routh array, the system is stable
  if
• Kgt3

23
Note System Poles and Eigenvalues

• dx/dt Ax Bu and yCx
• x(s) F(s)x0 F(s)BU(s), x0x(0)
• where F(s) (sI-A)-1 State
  Transition Matrix
• and det(sI-A) 0 System
  Characteristic Equation
• det(sI-A) 0 gt system poles
• Note Eigenvalue of a matrix A is defined as
• the solution of
• det(?I-A) 0
• ?
• System poles Eigenvalues of A

24
Note System Poles and Eigenvalues

• dx/dt Ax Bu and yCx
• ?
• System poles Eigenvalues of A
• Ex.
• dx/dt Ax , where A0 -12 3
• Then, det(sI-A)s 1-2 s-3
• s2-3s2(s-1)(s-2)
• ?
• eigenvalues of A 1 and 2 ? poles

25
6.5 Design ExamplesEx. 6.10 Automatic vehicle
turning control

• Figure 6.8 (Note A vehicles powertrain consists
  of all the components that generate power and
  deliver it. This includes the engine,
  transmission, drivershafts and wheels.)

26
Stability Region

• System Ch. Eqn. is
• q(s)s(s1)(s2)(s5)K(sa)
• s48s317s2(K10)sKa0 (6.28)
• The Routh array ?
• b3(126-K)/8, c3b3(K10)-8Ka/b3
• gt
• Klt126, Kagt0, (K10)(126-K)-64Kagt0
• gt
• Region of Stability is shown in Figure 6.9
  (next slide)

27
Stability Region on the K-a space

• Figure 6.9 for Kgt0 and agt0 (Exercise)

28
Pole locations using pole(sys)
29
5.6 System Error

• System Error E(s) R(s)-Y(s)
• R(s)-
• 
  (G(s)/(1GH(s))R(s)
• gt For a unity feedback,
• E(s)R(s)/(1G(s))
• gt
• The steady state error then be obtained
• using the final value theorem
• lim e(t) ess lim sR(s)/(1G(s))
• t-gt s-gt0

30
Example First-Order Systems

• G(s) K/(?s1), where Ksystem DC gain (G(0)),
• and ? time
  constant
• gtStep response R(s)1/s
  Y(s)
• gt
• Y(s)K/s K/(s1/?) gt
  y(t) K(1-e t/? )
• Now, the system error is defines as
• E(s)R(s)-Y(s)
• R(s)/(1G(s))
• So, the steady state error is
• ess lim s(1/s)/(1G(s)), s-gt0
• 1/(1G(0) )
• 1/(1K)

G(s)
?
31
Design Examples

• 1. Radar Tracking System
• 2. Hubble Space Telescope Position Control

32
Example Radar Tracking System

• System Block Diagram
• System Transfer Function
• T(s)K/(s22sK)
• ?n2/(s22? ?ns ?n2)
• gt
• Fastest unit response with no overshoot gt
  ?1 (below)
• gt
• 2? ?n 21?n 2 gt ?n 1
• gt
• K1 Ans.

33
2. Ex. 5.10 Hubble Space Telescope Position
Control (p. 316-319)

• Design Goal
• Choose the Amplifier gain K and the telescope
  system parameter K1 so that
• (1) overshoot lt 10 , and
• (2) minimum steady-state error for a unit
  step input
• are achieved. Assume that the disturbance
  Td(s)0.

34
Design Steps

• Overshoot lt 10
• gt From the two diagrams, we
• conclude that overshoot 9.5
• for ? 0.6.
• Steady state error for a
• step response
• esslim Y(s)-R(s)
• s -gt 0
• -1/K
• gt Choose a large gain K.

35
Step Response Curve(Figure 5.41, p. 330)

• Compute step response for a second-order system
• Duplicate Figure 5.5 (a) on p. 282
• For zeta0.1,0.2,0.4,0.7,1.0,2.0 and wn1
• t00.112num1
• zet0.1 0.2 0.4 0.7 1 2
• for k16
• zetazet(k)
• Gtf(num,1 2zeta 1)
• step(G)
• hold on
• end
• xlabel('\omega_nt'),ylabel('y(t)')
• title('Step Response for
• zeta0.1,0.2,0.4,0.7,1.0,2.0')
• grid on

36
Concept of Root Locus

• The root locus method is a graphical technique to
  analyze/design system stability.
• A graph or locus of roots of the characteristic
  equation as a system parameter (e.g., the
  controller gain K) changes is known as a root
  locus plot.

37
Definition of Root Locus

• The root locus is the path (or plot) of the roots
  of the system characteristic equation (i.e. the
  poles of the closed-loop transfer function)
  traced out in the s-plane as a system parameter
  (e.g., K) changes.
• Ex.
• Plot of root
• locations of
• q(s)s22sK
• 0 for K0-gt20.

• gtgtK00.510
• den1
• num1 2 0
• systf(num,den)
• rlocus(sys)
• xlabel('Real axis'),
• ylabel('Imaginary axis')

38
Plot of Root LocationsAnother example (Fig.
6.20)

• Matlab commands
• gtgtK00.520
• for i1length(K)
• q1 2 4 K(i)
• proots(q)
• plot(real(p),imag(p),x)
• hold on
• end
• grid on

39
Root Locus and the Stability Region covered in
CH. 6 are closely related.

• Example

40
Plot of Root Locations for 0ltKlt20 (Figure 6.20)
Roots on the j?-axis if K8
K00.520 for i1length(K)
q1 2 4 K(i) proots(q)
plot(real(p),imag(p),'x') hold on
end grid
41
7.6 Root-Locus Approach to Controller Design

• Often the root-locus plot of the original system
  may indicate that (a) it will become unstable as
  K increases, or (b) a desired performance
  criteria cannot be met just by changing the
  controller gain K.
• Then it is necessary to reshape the root locus
  plot to, for example, meet the performance
  specifications such as the damping ratio (?)
• or the steady-state error (ess).
• Solution
• gt Use a different controller.

42
7.6 Root-Locus Approach to Controller Design

• Often the root-locus plot of the original system
  may indicate that a desired performance criteria
  cannot be met just by changing the controller
  gain K.
• Then it is necessary to reshape the root locus
  plot to meet the performance specifications such
  as the damping ratio (?) or the
• steady-state error (ess).
• Use a different controller P-Controller
  PD-Controller
• See Table 7.7-
• 3 and 4

43
Basic Controller Types

• There are three basic building blocks for
  commonly used controllers. They are
• a) P (or Gain) Controller U(s) KpE(s)
• b) I (Integral) Controller U(s) KiE(s)/s
• c) D (Derivative) Controller U(s) KdE(s)s
• where e(t), error or actuating signal, is the
  input to the controller and u(t) is the
  controller output
• or command.

44
Features of P Controller

• Advantages
• Typical P Controllers are amplifiers. Thus,
• they are
• 1. Easy to implement,
• 2. Relatively inexpensive, and
• 3. Amplify the output.
• ? Widely used in industry

45
Features of P Controller

• Advantages
• 1. Easy to implement
• 2. Relatively inexpensive,
• and
• 3. Amplify the output.
• Disadvantage
• 1. Cannot eliminate the steady-state error,
  and
• 2. High gain leads to system instability as we
  have seen from root locus plots.

46
Example of P-Controller

• Basic Input-Output relation for proportional
  control is OutputKpError
• Ex. The Position Control System below where a
  robotic arm is powered by a motor and gearhead.

47
Example of P-Controller

• Basic Input-Output relation for proportional
  control is OutputKpError
• If there is no friction, overall system
  input-output relation, then, is shown below
• input output

48
Another Example of P-Controller

• Below is the diagram of a flow control system.
  This control system consists of a flow valve, a
  flow sensor and a P-controller. The controllers
  job is to maintain the flow of water through a
  pipe at 6 gal/min.
• The flow valve is operated with a signal of 0-5
  volts, where 0 volt corresponds to completely
  closed and 5 volts is all the way open. The flow
  sensor provides an output signal of 0-5 volts,
  which corresponds to 0-10 gal/min. The control
  system is designed so that a sensor voltage swing
  of 2.5 volts (50 of its range) will cause the
  flow valve to swing from full off to full on
  (i.e. 6 gal/min). For this reason, this system is
  called a 50 proportional control system.

49
Another Example of P-Controller

• The flow valve is operated with a signal of 0-5
  volts, where 0 volt corresponds to completely
  closed and 5 volts is all the way open. The flow
  sensor provides an output signal of 0-5 volts,
  which corresponds to 0-10 gal/min. The control
  system is designed so that a sensor voltage swing
  of 2.5 volts (50 of its range) will cause the
  flow valve to swing from full off to full on
  (i.e. 6 gal/min). For this reason, this system is
  called a 50 proportional control system.
• Output of P-ControllerKpE(sensor)
• gt
  KpOutput/E100/502
• The flow sensor provides an output signal of 0-5
  V, which corresponds to 0-10 gal/min. Thus, to
  maintain a flow rate of 6 gal/min, we must set
• SP6 (gal/min)x5V/10 gal/min))3V.

50
From P to PI Controller

• Disadvantage
• Cannot reduce the steady-sate error to zero.
• Q-How could we modify to reduce the error?
• Remember (from Ch. 5) that
• ess lim sE(s) 1/1K1G(0) ? 0 (5.36)
• s?0
• Add an integrator, i.e. ess lim 1/1K1G(0)/s
  ?0
• s?0

51
Example of an I Controller

• The Input-Output relation for integral control
  for the robot arm position control system below
  is
• Output Ki?(Error)dt

Controller Ki?(Error)dt
52
Example of The Robot Arm Control using an I
Controller

• The Input-Output relation for integral control
  for the robot arm position control system below
  is
• Output
  Ki?(Error)dt
• Consider the case where the robot arm has a
  steady- state error of 2 degrees due to friction.
  This error is shown below (a). As time elapses,
  the error remains at 2 degrees.

53
Example of The Robot Arm Control using an I
Controller

• The Input-Output relation for integral control
  for the robot arm position control system below
  is
• Output
  Ki?(Error)dt
• Consider the case where the robot arm has a
  steady- state error of 2 degrees due to friction.
  As time elapses, the error remains at 2 degrees.
• Figure (b) shows , if Ki1, how the restoring
  torque due to integral control increases with
  time and eventually brings the error to zero

54
PI Controlleru(t) K1e(t) K2?e(t)dt
55
Summary P-I Controller Response

• Proportional plus Integral (PI)-compensated
  system response
• vs.
• uncompensated systemresponse
• Note
• The steady state error is
• lim e(t) ess
• lim sR(s)/(1Gc(s)G(s))
• s-gt0
• 0

56
Features of PI Controller

• Advantage
• It can eliminate the steady-state error.
• Disadvantage
• It could make the dynamic response slower due
  to the integral part!

57
PD Controller

• PD controller is sometimes called a Phase-Lead
  Compensator.
• GPD(s) K1 K2s
• K(s a)

58
PD Controller

• Advantages
• PD Controller can improve the transient response.
• PD compensation could make unstable system
  stable.
• Disadvantage
• It wont reduce the steady-sate error to zero.
• Example
• English Channel Boring Machine

59
Channel Tunnel Boring Machine

• 580 tons

60
System Block Diagram

• A Boring Machine System with
• a PD-Controller

61
Ex. 4.6 Unit Step Response of the English
Channel Boring Machine using a PD controller (Ch.
4)

• For K100,
• gtgtnumg1deng1 1 0
• gtgtsysgtf(numg,deng)
• gtgtK100
• gtgtnumc11 Kdenc0 1
• gtgtsysctf(numc,denc)
• gtgtsysoseries(sysc,sysg)
• gtgtsysclfeedback(syso,1)
• gtgtt00.012.0
• gtgty,tstep(syscl,t)
• gtgt grid

62
Proportional-Integral-Derivative (PID) Feedback
Control

• Q-How do we improve both Steady-Sate Error and
  Transient Response?
• gt PID Controller
• The PID controller is widely used in industry due
  partly to their robust performance in a wide
  range of operating conditions and partly to their
  functional simplicity.
• ? 258 Class

63
?

• Any Questions?