chapter 11 digital control systems

Chapter 11 Digital Control Systems

Topics to be covered: - Discrete-time systems - z-Transforms - State Equation models of discrete systems

(continuous differential equations)(discrete difference equations)

ECE 4923/6923 Chapter 11 Part 2 2

Digital Control Systems

Simulation Diagrams and Flow Graphs: (How to make an electrical simulation of a system.)

For a discrete-time simulation, the main operation is the delay operator. (Recall that for a continuous-time system, we work with integrators. A delay operator is simpler, and, as a result, cheaper.)

We can also represent the delay as a z-1 operator.



Example:Draw a simulation diagram for the difference equation: m(k) = e(k) – e(k-1) – m(k-1)

Components required:2 delay elements and 1 summer(This is a special purpose computer, which can only solve the given

equation.) Alternate representation (in SFG form):

Note that the z-1 and the -1 paths can be combined in a “self-loop”



• Consider a general nth order difference equation of the following form:

• By using the real translation theorem, we get:

• This leads to the transfer function:

• The following two pages show a simulation diagram and a flow graph for this system.

m k a m k a m k n b e k b e k b e k nn n n( ) ( ) ( ) ( ) ( ) ( ) 1 0 1 01 1

M z a z M z a z M z b E z b z E z b z E znn

n nn( ) ( ) ( ) ( ) ( ) ( )

1

10 1

10

G z M zE z

b b z b za z a z

n nn

nn( ) ( )

( )

11

0

11

01



• For a general nth order difference equation, we can draw the following simulation diagram:



• As we have seen before, the SFG form is generally much more compact and is easier to draw...



• State Variables - Consider the following TF and the corresponding simulation diagram and SFG:Y zU z

a a z a zb z b z

n nn

nn

( )( )

11

0

11

01

a z a z a zz b z b zn

nn

n

nn

n1

10

0

11

00



Alternativerepresentations:


Digital Control Systems• State Variables – Let the outputs of each delay be a state, and we

can write state variable equations as:

and

(Phase Variable form of the State Equations)

x k x kx k x kx k x k

x k b x k b x k b x k u kn n n n n

1 2

2 3

3 4

1 2 1 0 1

111

1

( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( ) ( )

y k a x k a x k a x k a b x k b x k b x k u kn n n n n n n( ) ( ) ( ) ( ) ( ( ) ( ) ( ) ( )) 0 1 1 2 1 1 2 1 0 1

x1xn-1xn



• Note that these state equations can be more compactly written as:

(Note that these give a linear, time-invariant discrete system description.)A somewhat more general form is the linear time-varying form of the state equations:

x k A x k B u ky k C x k D u k

( ) ( ) ( )( ) ( ) ( )

1

x k A k x k B k u ky k C k x k D k u k

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

1



Example: Write State Equations for a system with the TF

(Technique: generate a SFG, label outputs of delays as states, and then write equations.)

(see board for details…)

Y zU z

a a z a zb z b z

( )( )

2 1

10

2

11

021



• Solution of State Equations:– Plug in each value of k:x(1) = A x(0) + B u(0)x(2) = A x(1) + B u(1) = A2 x(0) + AB u(0) + B u(1)x(3) = A3 x(0) + A2B u(0) + AB u(1) + B u(2). . . x(n) = An x(0) + An-1B u(0) + An-2B u(1) + . . .

+ AB u(n-2) + B u(n-1)So,

x n A x A B u kn n k

k

n

( ) ( ) ( )

0 1

0

1



• Solution of State Equations:– By z-Transforms:x(n+1) = A x(n) + B u(n)

The term multiplying the initial condition is the z-transform of the state transition matrix and

z X z x A X z BU zzI A X z z x BU z

X z z zI A x zI A BU z

( ) ( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

00

01 1

( )k z Z I A A k Z 1 1



Example: Draw a simulation diagram:

And assign states (from right to left):

G z Y z

U zz

z zz

z zzz z

( ) ( )( )

2

1

1 23 2 1 2 1 3 2

Y zU z

zz z

X zX z

Y z z X zU z X z z X z z X z

X z U z z X z z X z

( )( )

( )( )

( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1

1 2

1

1 2

1 2

1 3 2

3 23 2



• Example, continuedWith states assigned, we can then write out the

specific state equations:

In matrix form, theyend up as:

x k x kx k u k x k x k

y k x k

1 2

2 1 2

2

11 2 3

( ) ( )( ) ( ) ( ) ( )

( ) ( )

x k x k u k

y k x k

10 12 3

01

0 1



• Example, continuedTo solve, we can use

the sequential technique:(assume that x(0)=0 and u(k)=1, for all k.

x k x k u k

y k x k

10 12 3

01

0 1

x and y( ) ,10 12 3

00

01

101

1 0 101

1

x and y( ) ,20 12 3

01

01

114

2 0 114

4

x and y( ) ,30 12 3

14

01

14

113 0 1

411

11



• Example, continued. . .or, by z-transforms:

Recall that:

So, with x(0)=0, we get

and

Since we get

and

x k x k u k

y k x k

10 12 3

01

0 1

zI Az

zzI A z z

12 3

3 22,

X z z zI A x zI A BU z( ) ( ) ( ) 1 10

X z zI A BU zz z

zz

U zz z z

U z( ) ( ) ( ) ( )

12 2

13 2

3 12

01

13 2

1

U z u nT zz

( ) ( ) ,

Z1

Y z C X zz z

zzzz

zz z z

zz z

p fe zz

zz

zz

( ) ( )

( )

0 1 13 2

1

11 3 2

1 2 12

12

2

2 2

2

2

2

2 2

y k k u kk( ) ( ) ( ) 2 2 2



• Transfer Functions (TF’s) and State Variables For the SISO case, we have

So, with zero IC’s (since we are looking at TF’s):and

So, we end up with

x k A x k Bu ky k C x k D u k

( ) ( ) ( )( ) ( ) ( )

1

z X z A X z BU zzI A X z BU z

X z zI A BU z

( ) ( ) ( )( ) ( )

( ) ( )

1

Y z C X z DU zC zI A BU z DU z

C zI A B D U z

( ) ( ) ( )( ) ( )

( )

1

1

Y zU z

G z C zI A B D( )( )

( ) 1



Example:

Using a previous example setup,

we get

Since D = 0, the transfer function is:

Y zU z

G z C zI A B D( )( )

( ) 1

x k x k u k

y k x k

10 12 3

01

0 1

zI Az

z

zz

z z

1

1

2

12 3

3 123 2

G z Y zU z

C zI A B

zz

z zz

z z( ) ( )

( )

1

2 20 1

3 12

01

3 2 3 2



Note: If you don’t really like computing matrix inverses, you can alternatively use the following:

This method allows us to compute the TF without the need to determine a matrix inverse.

For our previous example,

G z Y zU z

zI A bc zI AzI A

d( ) ( )( )

d e t d e td e t

d e t de t

d e t de t de t

( )

zI Az

zz z

zI A bcz

zz

zz z

G zz z z z

z zz

z z

12 3

3 2

12 3

0 00 1

12 2

2 2

2 2 3 2

3 2 3 2

2

2

2 2

2 2


Digital Control SystemsFinally, we can use MATLAB to automate much of the work:

Example:

num=[0 1 0.95];den=[1 -1.9 0.93];printsys(num,den,’z’)[A,B,C,D]=tf2ss(num,den)

Results:

Then, we can use:dstep(A,B,C,D) (MATLAB chooses # of points)dimpulse(A,B,C,D)

dstep(A,B,C,D,IU,50) (we specify 50 points)dimpulse(A,B,C,D,IU,50) (IU is # of input, only one in this case)

G z zz z

( ) .. .

0 95

1 9 0 9 32

A B C D

1 9 0 9 31 0

10

1 0 9 5 0. .

, , . , [ ]



More MATLAB:Still using:

But note that if we specify:

which is different from before, we find that:[num,den]=ss2tf(A,B,C,D)gives

num=[0 1 0.95]den = [1 -1.9 0.93]

which is the SAME TF as before! What gives????

Answer: These are two different forms of the state equations, but are equivalent. Note that there are an infinite number of valid state equations for any given system.

G z zz z

( ) .. .

0 9 5

1 9 0 932

A B C D

0 10 9 3 1 9

01

0 9 5 1 0. .

, , . , [ ]

chapter 11 digital control systems

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