Control System II - Lecture Notes

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Control Systems II (EEEN30041)

Lecturer: Zhengtao Ding

Email: [email protected]

Webpage: http://personalpages.manchester.ac.uk/staff/Zhengtao.Ding

EEEN30041, School of E&EE, University of Manchester, 2007 1

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Course Format and Assessment

The course will be delivered in 20 lectures with 4 tutorials and a computer-basedassignment. Assessment will be based on an assignment (10%) and examination(90%). Course material will include online lecture notes and handouts.

Leaning Outcomes

• Describe dynamic systems in continuous-time state space models, and dynamicmodels in discrete-time forms for industrial control applications.

• Design controllers in state space using state-feedback and output feedbackmethods including linear optimal control.

• Design and implement controllers in discrete-time using digital control includingimplementing PID in digital control.

• Appreciate important industrial control applications


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References

Dorf and Bishop, Modern Control Systems

Stefani, Shahian, Savant and Hostetter, Design of Feedback Control Systems

Kailath, Linear Systems

Ogata, Modern Control Engineering

Kuo, Automatic Control Systems

Nise, Control Systems Engineering


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Course Structure

Part 1. State space model and control design

Part 2. Digital control


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1. State Space Variables

The state variables describe the future response of a system, given the presentstate, the excitation inputs, and the equations describing the dynamics.

Example 1. Mass-spring-damper system

Md2y

dt2+ b

dy

dt+ ky = u(t) (1)

where M is the mass, b the friction coefficient, and k the spring constant.

Taking the position and velocity as the state variables, ie:

x1 = y

x2 =dy

dt


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the dynamics can then be written as

dx1

dt= x2

dx2

dt=

−b

Mx2 −

k

Mx1 +

1M

u

Question: Why y and dydt can be the state variables?

Example 2. RLC circuit

ic = Cdvc

dt= −iL + u(t) (2)

LdiLdt

= −RiL + vc (3)

with the voltage across R as the output, vo = RiL


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By taking x1 = vc and x2 = iL, we have

dx1

dt= − 1

Cx2 +

1C

u(t) (4)

dx2

dt=

1L

x1 −R

Lx2 (5)

y = Rx2 (6)

Question: Is the set of state variable unique?

Alternative choice can be x1 = vc and x2 = vL.


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2. State Differential Equations

State Space Equation

A set of first order differential equations can be written as

x1 = a11x1 + a12x2 + . . . + a1nxn + b11u1 + . . . + b1mum

x2 = a21x1 + a22x2 + . . . + a2nxn + b21u1 + . . . + b2mum

...

xn = an1x1 + an2x2 + . . . + annxn + bn1u1 + . . . + bnmum

(7)


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We can put the above in the matrix form

d

dt

x1

x2...

xn

=

a11 a12 . . . a1n

a21 a22 . . . a2n... ... . . . ...

an1 an2 . . . ann

x1

x2...

xn

+

b11 . . . b1m... . . . ...

bn1 . . . bnm

u1...

um

(8)

The column vector consisting of the state variables is called state vector and it iswritten

x =

x1

x2...

xn

(9)

We may also write x ∈ Rn, as there are n state variables. Similarly we have theinput vector. Using the state vector and input vector, we have the compact


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notation of the state differential equation

x = Ax + Bu

where A is an n× n matrix or A ∈ Rn×n, and B is an n×m matrix orB ∈ Rn×m. We can also write the output in the compact form

y = Cx + Du

The dimensions of C and D depend on the dimension of the column vector y.Puting the above two equations together, we have the most common state spaceequation

x = Ax + Bu (10)

y = Cx + Du (11)

Therefore a dynamic system is characterised by the four matrices {A,B, C, D}.


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Example 3. Write down the state space equation for RLC system shown inExample 2.

x =[

0 − 1C

1L −R

L

]x +

[1C0

]u(t) (12)

y = [0 R]x (13)

For this example, we have

A =[

0 − 1C

1L −R

L

], B =

[1C0

], C = [0 R], D = 0

Question: Determine the state space matrices A,B, C, D for Example 1.


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Solution of state space equation

Consider the first order differential equation

x = ax + bu (14)

The solution is given by

x(t) = eatx(0) +∫ t

0

ea(t−τ)bu(τ)dτ (15)

With the matrix exponential function defined by

eAt = I + At +A2t2

2!+ . . . +

Aktk

k!+ . . . (16)


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we have the solution of the state space equation given by

x(t) = eAtx(0) +∫ t

0

eA(t−τ)Bu(τ)dτ (17)

Sometimes we denote Φ(t) = eAt and Φ(t) is referred to as the state transitionmatrix. Hence the above equation can be written as

x(t) = Φ(t)x(0) +∫ t

0

Φ(t− τ)Bu(τ)dτ (18)

It can be shown that Φ(t) equals the inverse Laplace transform of [sI −A]−1.


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3. Transfer Functions from State Equations

We can obtain the transfer function for a single-input-single-output (SISO) systemfrom its state space equation. Taking the Laplace transform of the state spaceequation

x = Ax + Bu (19)

y = Cx (20)

we have

sX(s) = AX(s) + BU(s) (21)

Y (s) = CX(s) (22)

From the first equation, we have

X(s)(sI −A) = BU(s) (23)


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and

X(s) = (sI −A)−1BU(s) (24)

Hence we have

Y (s) = C(sI −A)−1BU(s) (25)

Therefore the transfer function G(s) = Y (s)U(s) is given by

G(s) = C(sI −A)−1B (26)

Note that we can write G(s) = CΦ(s)B, as Φ(s) = (sI −A)−1

Example 4. Determine the transfer function of the RLC system in Example 3.


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Answer:

G(s) =R

LC

s2 + RLs + 1

LC

(27)

Φ(s) := (sI −A)−1 is a matrix with elements of Laplace transformed functions.Therefore we can use inverse Laplace transform to obtain Φ(t) which can be usedto evaluate the time response.

Example 5. Obtain the time respnse of the RLC system shown in Example 2,assuming R = 3, L = 1, C = 1/2, u(t) = 0 and x(0) = [1 1]T .

Solution: When u(t) = 0, we have

x(t) = Φ(t)x(0) (28)


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In this case, we have

A =[

0 − 1C

1L −R

L

]=

[0 −21 −3

](29)

and

Φ(s) =1

s2 + 3s + 2

[s + 3 −2

1 s

](30)

From the inverse Laplace transform, we obtain

Φ(t) =[

(2e−t − e−2t) (−2e−t + 2e−2t)(e−t − e−2t) (−e−t + 2e−2t)

](31)

and subsequently

x(t) = Φ(t)[

11

]=

[e−2t

e−2t

](32)


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Stability of State Variable Systems

When we convert the state space model to the transfer function, we observe thatthe denominator of the transfer function is the determinant of (sI −A), ie,

d(s) = |sI −A| (33)

Therefore the stability of the state space model depends on the characteristicpolynomial of the state matrix A. The system is stable if all the poles are in theleft half plane, ie, the solutions of d(s) = 0 are all with negative real parts(equivalent to the eigenvalues of A are all with negative real parts).


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4. State Space Realization and Canonical Forms

Given a transfer function, how to write its state space equations? For example, ifthe transfer function is given by

G(s) =6

s2 + 3s + 2(34)

we know that its state space realization is, based on the previous RLC example,

x =[

0 −21 −3

]x +

[20

]u(t) (35)

y = [0 3]x (36)

Consider a general second transfer function

G(s) =b1s + b2

s2 + a1s + a2(37)


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One realization is given by

x =[

0 1−a2 −a1

]x +

[01

]u(t) (38)

y = [b2 b1]x (39)

The state space realization is not unique. For the same transfer function, we canhave another realization as

x =[−a1 1−a2 0

]x +

[b1

b2

]u(t) (40)

y = [1 0]x (41)

Note that the state space variables in the two realizations are different. The firstrealization is referred to as the controller canonical form, and the second as the


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observer canonical form, as one is convenient for controller design and the other isconvenient for observer design.

Controller Canonical Form

For a general transfer function given by (assuming the numerator and thedenominator are coprime)

G(s) =b1s

n−1 + b2sn−2 + . . . + bn

sn + a1sn−1 + . . . + an(42)

the controller canonical form is given by

x =

0 1 . . . 0... ... . . . ...0 0 . . . 1−an −an−1 . . . −a1

x +

0...01

u(t) (43)

y = [bn bn−1 . . . b1]x (44)


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Observer Canonical Form

For the above transfer function, the observer canonical form is given by

x =

−a1 1 . . . 0

... ... . . . ...−an−1 0 . . . 1−an 0 . . . 0

x +

b1...

bn−2

bn−1

u(t) (45)

y = [1 0 . . . 0]x (46)

Example 5. Put the state space equation shown for the RLC system into thecontroller and observer canonical forms.


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5. Controllability and Observability

Controllability: A system is completely controllable if there exists a control inputu(t) that can transfer any initial state x(0) to any other desired location x in afinite time.

We can check the controllability of the system

x = Ax + Bu (47)

y = Cx (48)

by examining the algebraic condition

rank[B AB . . . An−1B] = n (49)

For a SISO system, we have B as a vector, and therefore if we define

Pc = [B AB . . . An−1B] (50)


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then Pc is an n× n matrix. In this case, to check the controllability is equivalentto check if the determinant of Pc is nonzero.

Example 6. Check the controllability of the system

x =[−2 0d −3

]x +

[10

]u(t) (51)

y = [0 1]x (52)

where d is a constant.

Observability: A system is completely observable if and only if there exists a finitetime T such that the initial state x(0) can be determined from the observationhistory y(t) given the control u(t).

For a SISO system

x = Ax + Bu (53)


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y = Cx (54)

the system is completely observable is the determinant of the observability matrix

Po =

C

CA...

CAn−1

(55)

is nonzero.

Example 7. Check the observability of the system

x =[

2 0−1 1

]x +

[1−1

]u(t) (56)

y = [1 1]x (57)


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6. Full State Feedback Design

In the first step in the state variable design, we assume all the state variables areavailable for feedback control. In this case, we can design the control input as

u = −Kx (58)

where K is the gain matrix. The full state feedback design is to decide a suitablefeedback gain matrix K.

For the closed-loop control system, we have

x = Ax + B(−Kx) = (A−BK)x (59)

The stability of the closed-loop system depends on the characteristic polynomial of(A−BK)

Pole Assignment. To assign the closed loop poles at given locations via full statefeedback.


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Pole assignment can be achieved if the system is completely controllable.

Example 8. Design a full state feedback control of the system described by

d3y

dt3+ 5

d2y

dt2+ 3

dy

dt+ 2y = u (60)

such that the closed-loop poles are at {−1,−2,−3} respectively.

Solution: The transfer function of the system is

G(s) =1

s3 + 5s2 + 3s + 2(61)

If we realize the system in the controller canonical form, we have

x =

0 1 00 0 1−2 −3 −5

x +

001

u(t) (62)


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y = [1 0 0]x (63)

If the control input is designed as

u = [k1 k2 k3]x (64)

The closed-loop system is given by

x =

0 1 00 0 1−2 −3 −5

x +

001

(−[k1 k2 k3]x) (65)

=

0 1 00 0 1

−(k1 + 2) −(k2 + 3) −(k3 + 5)

x := (A−BK)x (66)

The closed-loop characteristic polynomial is

|sI − (A−BK)| = s3 + (k3 + 5)s2 + (k2 + 3)s + (k1 + 2) (67)


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Comparing it with the desired characteristic polynomial

(s + 1)(s + 2)(s + 3) = s3 + 6s2 + 11s + 6 (68)

we have k3 = 1, k2 = 8, k1 = 4, ie,

K = [4 8 1] (69)

Therefore the control input is designed as

u = −[4 8 3]x = −4x1 − 8x2 − x3 (70)

Note that for this realization we have x1 = y, x2 = dydt and x3 = d2y

dt2. The control

input is then expressed as

u = −4y − 8dy

dt− d2y

dt2(71)


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Question: How to implement the feedback control if only the output y is available?

From Example 8 we have seen the convenience of using the controller canonicalform for the full state feedback design. In fact, for a given state space system, wecan transform the system to the controller canonical form if the system iscontrollable. To avoid the state transform, we have Ackermann’s formula for thefull state feedback control design.

Ackermann’s Formula. For a system {A,B}, the state feedback control gain K forthe closed loop control u = −Kx to achieve the desired closed-loop characteristicpolynomial

d(s) = sn + α1sn−1 + . . . + αn (72)

is given by

K = [0 0 . . . 1]P−1c d(A) (73)


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where

d(A) = An + α1An−1 + . . . + αnI (74)

and Pc is the controllability matrix.


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7. Observer Design

Full state feedback control needs the values of all the state variables. In industrialsystems, it is common that not all the state are available, and in this case, anobserver can be designed to provide an estimate of the unknown state variables.

Luenberger Observer. For a dynamic system

x = Ax + Bu (75)

y = Cx (76)

a full-state observer is designed as

˙x = Ax + Bu + L(y − Cx) (77)

where x denotes the estimate of the state variable, and L is the observer gain.


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Define the observer error as

e = x− x (78)

we have the error dynamics

e = (Ax + Bu)− (Ax + Bu + L(y − Cx))

= (A− LC)e (79)

To ensure the error asymptotically converge to zero as t →∞, we need thecharacteristic equation |sI − (A−LC)| = 0 to have all the roots in the left half ofthe complex plan.

Example 9. Design a full state observer for the dynamic system

x =

−5 1 0−3 0 1−2 0 0

x +

001

u(t) (80)


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y = [1 0 0]x (81)

Solution: Note the system is in the observer canonical form. For L = [l1 l2 l3]T ,we have

A− LC =

−5 1 0−3 0 1−2 0 0

− l1

l2l3

[1 0 0] =

−(l1 + 5) 1 0−(l2 + 3) 0 1−(l3 + 2) 0 0

(82)

The characteristic polynomial is

|sI − (A− LC)| = s3 + (l1 + 5)s2 + (l2 + 3)s + (l3 + 2) (83)

If we place the poles for the observer at {−2,−2,−2}, the desired characteristicpolynomial is

(s + 2)3 = s3 + 6s2 + 12s + 8 (84)


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By comparing the polynomials we have l1 = 1, l2 = 9, l3 = 6, ie,

L = [1 9 6]T (85)

The full state observer is given by

˙x =

−5 1 0−3 0 1−2 0 0

x +

001

u(t) +

196

(y − [1 0 0]x) (86)

The observer canonical form makes the design of observer gain easier. For thesystem what is not in the observer canonical form, we can still evaluate thecharacteristic polynomial, and then by comparing the coefficients with the desiredone to obtain the observer gain matrix. For design observer gains, we also haveAckermann’s formula.

Ackermann’s Formula. For a system {A,B, C}, the observer gain L to achieve the


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desired closed-loop characteristic polynomial

d(s) = sn + α1sn−1 + . . . + αn (87)

is given by

L = d(A)P−1o [0 0 . . . 1]T (88)

where

d(A) = An + α1An−1 + . . . + αnI (89)

and Po is the observability matrix.


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8. Compensator Design

We aim at design dynamic feedback control from the system output. There arethree steps in the compensator design.

• Full state feedback design

• Full state observer design

• Compensator design using the state estimate to replace the state variable in thefull state control design

The final control design is given by

u = −Kx(t) (90)

Question: How to ensure the stability of the closed-loop system?


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The separation principle plays an important part.

Consider the system

x = Ax + Bu (91)

y = Cx (92)

with the observer

˙x = Ax + Bu + L(y − Cx) (93)

and the control law

u = −Kx(t) (94)

The closed-loop system with the observer can be written as

x = Ax−BKx


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˙x = Ax−BKx + LC(x− x)

In terms of the state variable x and observer error e = x− x, we have

x = (A−BK)x + BKe

e = (A− LC)e

Treating [xT eT ]T as the augmented state variable, we have

d

dt

[xe

]=

[A−BK BK

0 A− LC

] [xe

](95)

The characteristic polynomial of the augmented system is given by∣∣∣∣sI − [A−BK BK

0 A− LC

]∣∣∣∣ = |sI − (A−BK)||sI − (A− LC)| (96)


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Therefore if |sI − (A−BK)| = 0 and |sI − (A− LC)| = 0 have all the roots inthe left half of the complex plane, the augmented system is stable. This is theseparation principle.

Example 9. Design a dynamic output feedback control (compensator) for thesystem

x =[

0 11 1

]x +

[01

]u(t) (97)

y = [1 0]x (98)

The poles for the closed loop controller are at {−1,−1} and the poles for theobserver error dynamics are at {−2,−2}.


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9. Tracking and Internal Model Design

Consider

x = Ax + Bu (99)

y = Cx (100)

We need to design a control input such that the output asymptotically tracks agiven reference r, which can be written as the output for the reference state spacemodel

xr = Arxr (101)

r = drxr (102)

Consider the case of tracking a constant reference, we have the reference model as

xr = 0 (103)


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r = xr (104)

Define the tracking error e = y − r. Taking the derivative of the tracking errorgives

e = y = Cx (105)

Let us use the notation z = x and v = u. Take e and z as the state variable of anaugmented state space system

d

dt

[ez

]=

[0 C0 A

] [ez

]+

[0B

]v (106)

If the augmented system is controllable, we can design a state feedback controllerin the form

v = −k1e− k2z (107)


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where k2 is a matrix (vector) in general. The state feedback ensures the stabilityof the augmented system which implies the asymptotic tracking with e convergingto zero. The control is input is given by integrating v as

u(t) = −k1

∫ t

0

e(τ)dτ − k2x(t) (108)

The integration in the above equation reflects the dynamics of the tracking signal.It is clear to see in a block diagram that the controller acts as an internal model.

Example 10. Internal model design for a unit step input for the system

x =[

0 1−2 −2

]x +

[01

]u(t) (109)

y = [1 0]x (110)

by placing the pools of the augmented system at {−1± j,−10}.


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The same control design can be extended to the case of tracking a polynomial oftime.

Question: How to design an internal model based controller to track a sinusoidalsignal?


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10. Optimal Control

The performance of a control system can be represented by a performance indexsuch as

J =∫ tf

0

g(x, u, t)dt (111)

Optimal control is concerning with the control design to minimize a performanceindex. Consider a particular performance index

J =∫ tf

0

xTQxdt (112)

where Q is an n× n positive definite matrix. To simplify the problem, we let tftend to infinity, that is, the index is given by

J =∫ ∞

0

xTQxdt (113)


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How to design a full state feedback law to minimise the index?

Consider the control design u = −Kx for x = Ax + Bu. The closed-loop system is

x = (A−BK)x := Hx (114)

If we have a positive definite matrix P such that

d

dt(xTPx) = −xTQx (115)

then substituting the above into the performance index, we have

J = −∫ ∞

0

d

dt(xTPx)dt = −xTPx |∞0 = xT (0)Px(0) (116)

where we assume that the closed-loop system is stable (x(∞) = 0).


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A direct evaluation gives

d

dt(xTPx) = xTPx + xTPx

= xTHTPx + xTPHx

= xT (HTP + PH)x (117)

From the differential equation ddt(x

TPx) = −xTQx we have

HTP + PH = −Q (118)

Therefore the optimal control design with the performance index J =∫∞0

xTQxdtcan be carried out in two steps:

• Solve the matrix equation

HTP + PH = −Q (119)


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to obtain matrix P that depends on the control gain K.

• Minimize the index

J = xT (0)Px(0) (120)

to determine the control gain or other parameters in the system.

Example 11. Design the state feedback control K = [k1 k2] by restricting k1 = 1with the optimal control index of Q = I and x(0) = [1 1]T for the system

x =[

0 10 0

]x +

[01

]u(t) (121)

(122)

Example 12. Continue from Example 11 by restricting the control gain asK = [k k] (ie., k1 = k2) and x(0) = [1 0]T .


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Solution: Solving the matrix equation HTP + PH = −I gives

p11 =1 + 2k

2k(123)

It is easy to see the control index is given by

J = xT (0)Px(0) = p11 = 1 +12k

(124)

The minimum value of J is obtained when k tends to infinity.


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11. Linear Quadratic Regulator

Consider the solution of Example 12. The bigger the controller gain, the smallerthe performance index; there is no optimal solution of the control gain. This is dueto the fact that we did not consider the control effort in the performance index.

In engineering systems, the bigger control input efforts often mean the biggerenergy consumption. To consider the input in the performance index, we oftendefine

J =∫ ∞

0

[xTQx + uTRu]dt (125)

where R is a positive definite matrix. For SISO case, we have R as a constantscaler.

Consider the full state feedback control u = −Kx, the performance index can be


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written as

J =∫ ∞

0

[xTQx + xTKTRKx]dt :=∫ ∞

0

xTSxdt (126)

where

S = Q + KTRK (127)

Similar to the case with no control in the performance index, we need to solve amatrix equation

HTP + PH = −S (128)

and the performance index is then given by

J =∫ ∞

0

[xTQx + xTKTRKx]dt :=∫ ∞

0

xTSxdt = x(0)TPx(0) (129)


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Example 13. Repeat Example 12 with the new performance index

J =∫ ∞

0

[xTx + ru2]dt (130)

General Solution for Linear Quadratic Regulator: Consider a dynamic systemdescribed by

x = Ax + Bu (131)

The optimal control input for the performance index

J =∫ ∞

0

[xTQx + uTRu]dt (132)

is given by

u = −R−1BTPx (133)


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where P satisfies

ATP + PA− PBR−1BTP + Q = 0 (134)

This equation is often referred to as the matrix algeraic Riccati equation, and itcan be easily solved using Matlab.


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12. Digital Control and Sampled Data System

Question: Why digital control?

Due to the application of digital computers in industrial control systems. [Thereare also some inherently discrete-time systems.]

Sampled Data. Sampled data (or a discrete signal) are data obtained for systemvariables only at discrete time intervals.

We assume the sampling at the same fixed period, T , which is called the samplingperiod. The sampled data for a continuous time variable x(t) are denoted byx(kT ) with k taking the values of integers.

How to decide the sampling period?

It depends on the dynamics of the system, the required accuracy and hardwareconstraints.


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Sampler. An ideal sampler is a switch that closes for every T seconds for a instant.For an continuous time signal r(t), sampled at kT , the output from the samplerr∗(t) is an impulse signal,

r∗(t) = r(kT )δ(t− kT )

where δ is the impulse function.

Zero-order-hold. After the sampling, it is assumed that the value is kept at sameuntil next sampling, ie, the value of x(t) is assumed to be at the constant ofx(kT ) for kT ≤ t < (k + 1)T .

The impulse response of the zero-order-hold is given by

g0(t) ={

1, 0 ≤ t < T0, otherwise.

(135)


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and therefore its transfer function is given by

G0(s) =1s− 1

se−Ts =

1− e−Ts

s(136)

Quantization error. An error due to a computer’s finite word size.


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13. The z-Transform

The output from an ideal sampler is a sequence of impulses with values r(kT ),and we write

r∗(t) =∞∑

k=0

r(kT )δ(t− kT ) (137)

Taking the Laplace transform of the above equation, we have

L{r∗(t)} =∞∑

k=0

r(kT )e−kTs =∞∑

k=0

r(kT )(eTs)−k (138)

We define the z-transform as

R(z) = Z{r∗(t)} =∞∑

k=0

r(kT )z−k (139)


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with z = eTs. Similar to the notation for the Laplace transformed functions, acapital letter R denotes the z-transformed functions of r(t).

Example 14. Determine the z-transform for the unit step function u(t).

U(z) =∞∑

k=0

u(kT )z−k =∞∑

k=0

z−k =1

1− z−1=

z

z − 1(140)

Example 15. Determine the z-transform of e−at.

F (z) =∞∑

k=0

e−akTz−k =∞∑

k=0

(eaTz)−k

=1

1− (eaTz)−1=

z

z − e−aT


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The z-transform table

x(t) X(s) X(z)δ(t) 1 1δ(t− kT ) e−kTs z−k

u(t), unit step 1/s zz−1

t 1/s2 Tz(z−1)2

e−at 1s+a

zz−e−aT

sinωt ωs2+ω2

z sin ωTz2−2z cos ωT+1

cos ωt ss2+ω2

z(z−cos ωT )z2−2z cos ωT+1

e−at sinωt ω(s+a)2+ω2

ze−aT sin ωTz2−2ze−aT cos ωT+e−2aT

e−at cos ωt s+a(s+a)2+ω2

z2−ze−aT cos ωTz2−2ze−aT cos ωT+e−2aT


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Inverse z transform

• Partial fraction method (similar to inverse Laplace transform)

• Long division

Transfer function of an open loop system

We need to multiply the transfer function of the zero-order-hold to the systemtransfer function Gp(s).

G(z) = Z{G0(s)Gp(s)} (141)

Example 16. Determine the z-transfer function of Gp(s) = 1s(s+1).


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G(s) =1− e−Ts

s

1s(s + 1)

= (1− e−Ts)(1s2− 1

s+

1s + 1

) (142)

It follows that

G(z) = (1− z−1)Z(1s2− 1

s+

1s + 1

)

= (1− z−1)(Tz

(z − 1)2− z

z − 1+

z

z − e−T)

=(ze−T − z + Tz) + (1− e−T − Te−T )

(z − 1)(z − e−T )


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14. Discrete-time Systems

Difference equation

Consider

y(k + n) + a1y(k + n− 1) + . . . + any(k) = b1u(k + n) + . . . + bnu(k) (143)

The solution of this difference equation can be obtained by iteration using

y(k + n) = −a1y(k + n− 1)− . . .− any(k) + b1u(k + n− 1) + . . . + bnu(k)(144)

if necessary initial values are unknown.

Transfer function

Taking z-transform of the difference equation

znY (z) + a1zn−1Y (z) + . . . + anY (z) = b1z

n−1U(z) + . . . + bnU(z) (145)


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and the transfer function is given by

G(z) =Y (z)U(z)

=b1z

n−1 + . . . + bn

zn + a1zn−1 + . . . + an(146)

and finally we have

G(z) =b1z

−1 + . . . + bnz−n

1 + a1z−1 + . . . + anz−n(147)

System response using difference equation

For a given transfer function, we can obtain the difference and then obtain thesystem responses using the difference equation.

Example 17. For the system considered in Example 16 with T = 1, determine thefirst four terms of the output subject to the impulse input, assumingy(−2) = y(−1) = 0.


64

For T = 1, we have the transfer function

G(z) =Y (z)U(z)

=0.3678z + 0.2644

z2 − 1.3678z + 0.3678(148)

The difference equation is obtained as

y(k + 2)− 1.3678y(k + 1) + 0.3678y(k) = 0.3678u(k + 1) + 0.2644u(k) (149)

or

y(k + 2) = 1.3678y(k + 1)− 0.3678y(k) + 0.3678u(k + 1) + 0.2644u(k) (150)

Note the {u(k)} = {1, 0, . . . , 0, . . .}. We have

y(0) = 1.3678y(−1)− 0.3678y(−2) + 0.3678u(−1) + 0.2644u(−2) = 0 (151)


65

y(1) = 1.3678y(0)− 0.3678y(−1) + 0.3678u(0) + 0.2644u(−1) = 0.3678 (152)

and y(2) = 0.7675 and y(3) = 0.9145.

The method shown in the above example can also be used for evaluation ofinverse z transform.


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15. Closed-loop Transfer Functions and Stability

Closed-loop transfer function

Consider a closed-loop system

Y (z) = G(z)U(z) (153)

U(z) = D(z)E(z) (154)

E(z) = R(z)− Y (z) (155)

where D(z) denotes the transfer function of a digital controller, and E(z) andR(z) denote the feedback error and the reference signal.

The closed-loop transfer function is obtained as

Y (z)R(z)

=G(z)D(z)

1 + G(z)D(z)(156)


67

The above result is the same for the closed-loop transfer function of thecontinuous-time system. In general, the block diagram manipulation follows in thesame way as the block diagram manipulation of the continuous-time systems.

Stability analysis

Conside the mapping between s-plane and z-plane. Let s = σ + jω and we have

z = eTs = eσ+jω (157)

It can be seen that for σ < 0 we have |z| < 1, that is, the left-half of the s planecorresponds to the area within the unit circle in z-plane. Therefore we have thefollowing statement.

A sampled (discrete-time) system is stable if all the poles of the closed-looptransfer function lie within the unit circle of the z-plane.

Example 18. The open-loop transfer function considered in Example 17 is under a


68

closed-loop control with the controller D(z) = K, a constant. Evaluate thestability for the closed-loop system when K = 1 and K = 10.

The closed-loop transfer function is given by

KG(z)1 + KG(z)

=K 0.3678z+0.2644

z2−1.3678z+0.3678

1 + K 0.3678z+0.2644z2−1.3678z+0.3678

(158)

and therefore the characteristic equation is

d(z) = z2 + (0.3678K − 1.3678)z + (0.2644K + 0.3678) (159)

For K = 1, we have

d(z) = z2 − z + 0.6832 = 0 (160)

and the poles, z1,2 = 0.50± j0.6182, are within the unit circle and the system is


69

stable. For K = 10, we have

d(z) = z2 + 2.310z + 3.012 = 0 (161)

and the poles, z1,2 = −1.155± j1.295, are outside the unit circle and the systemis unstable.

Example 19. Range of T for stability

Consider a unit feedback sampled-data system with the plant transfer functionGp = 10

s+1. Determine the range of the sampling interval to ensure the closed-loopstability.


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16. Discrete-time State Space Systems

State Difference Equation

Similar to the state differential equations, we have state difference equation

x(k + 1) = Ax(k + 1) + Bu(k) (162)

y(k) = Cx(k) + Du(k) (163)

where x(k) is the state vector at the step k.

System Response

The response can be evaluated repeatedly with the given initial values:

x(1) = Ax(0) + Bu(0)

x(2) = Ax(1) + Bu(1) = A2x(0) + ABu(0) + Bu(1)


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...

x(k) = Akx(0) + Ak−1Bu(0) + . . . + Gu(k − 1)

Controllability and Observability

The controllability and observability can be checked in the same ways as for thecontinuous-time systems. The system is controllable if the controllability matrix

Pc = [B AB . . . An−1B] (164)

has full rank, and the system is observable if the observability matrix

Po =

C

CA...

CAn−1

(165)


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has full rank.

Transfer functions from state difference equations

Taking z-transform of x(k + 1) = Ax(k + 1) + Bu(k), we have

(zI −A)X(z) = BU(z)

X(z) = (zI −A)−1BU(z)

(166)

and

Y (z) = CX(z) = C(zI −A)−1BU(z) (167)

Therefore the transfer function is given by

G(z) =Y (z)U(z)

= C(zI −A)−1B (168)


73

Note that the transfer function is in the same form as for the continuous-timecase, with the only difference that s is replaced by z. We would expect the samekind of state space realizations for discrete-time transfer functions, and this indeedis the case.

State space realization

The canonical forms corresponding to the transfer functions in s domain work inthe same way for z domain.

Example 20. Obtain the state difference equation for the transfer function

G(z) =0.3678z + 0.2644

z2 − 1.3678z + 0.3678(169)

The realization in the controller canonical form is given by

x(k + 1) =[

0 1−0.3768 1.3768

]x +

[01

]u(t) (170)


74

y = [0.2644 0.3768]x (171)

Stability

The characteristic equation for the transfer function in z domain is given by

d(z) = |zI −A| = 0 (172)

The system is stable if all the roots of the characteristic equation are within theunit circle.

Full State Feedback Control Design

Design the full state feedback control as

u(k) = −Kx(k) (173)


75

which gives the closed-loop system

x(k + 1) = (A−BK)x(k) (174)

The full-state feedback control can be used to place the poles of the closed-loopsystem at the desired positions, if the system is controllable.

Example 21. Design the full state feedback control law for the system in Example20 to place the poles at {−0.5,−0.5}.


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17. Implementation of Digital Controllers

There are two methods for design a digital controller.

• Emulation method, ie, design the controller in continuous-time and then convertto discrete-time.

• Direct digital design, including discrete-time pole placement via full state feed-back shown in the previous section.

Digital implementation of PID controllers

In s domain, we have the transfer function for a PID controller as

U(s)X(s)

= Gc(s) = k1 +k2

s+ k3s (175)

We need approximations for differentiation and integration in discrete-time.


77

Backward difference rule for differentiation

u(KT ) =dx

dt|t=kT =

1T

(x(kT )− x((k − 1)T ) (176)

The z-transform for this equation is given by

U(z) =1− z−1

TX(z) =

z − 1Tz

X(z) (177)

Forward-rectangular integration

u(kT ) = u((k − 1)T ) + Tx(kT ) (178)

and the z-transform gives

U(z)X(z)

=Tz

z − 1(179)


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The z domain transfer function for the PID controller is given by

U(z)X(z)

= k1 +k2Tz

z − 1+ k3

z − 1Tz

(180)

The difference equation for this transfer function is

u(kT ) = u((k − 1)T ) + (k1 + k2T +k3

T)x(kT )

−(k1 +2k3

T)x((k − 1)T ) +

k3

T)x((k − 2)T ) (181)

This is called the velocity form of PID implementation, as the control input at thecurrent step is calculated based on the control input in the previous step, and thecontribution in the current step only contributes to the change of the controlinput.


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Control System II - Lecture Notes