Lecture 7 (SS)
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Transcript of Lecture 7 (SS)
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8/13/2019 Lecture 7 (SS)
1/22
Lecture 7: State Variables Analysis II
1
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8/13/2019 Lecture 7 (SS)
2/22
Course Outline Control Elements: Transducer, Switches, Actuators, Valves,
Motors
Control Fundamentals: Open loop and closed loop systems,transfer function, signal flow graph, gain formula
Modeling: Mathematical modeling of linear electrical andmechanical systems, state variables, state equations andstate diagrams
Analysis and Design: Stability, controllability andobservability of systems, state variables, state transitionmatrix, transient and steady state response, root locusmethod. Nyquist criterion, PID controllers, lead lagcompensators, pole-zero cancellations.
Practical systems: Analog and microprocessor basedcontrol systems, design examples
2
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8/13/2019 Lecture 7 (SS)
3/22
Recap -Example 1
3
The motion of any finite dynamic system can be expressed as a set of first-order ordinary
differential equations. This is often referred to as the state-variable representation. For
example, Newtonslaw for a single mass Mmoving in one dimensionx under force Fis:
We define one state variable as the position and the other state variable as the velocity
, this equation can be written as:
Furthermore, first-order linear differential equations can be concisely expressed using matrix
notation. If we collect the state into a column vector x, and the coefficients of the state
equations into a square matrix F, and the coefficients of the input into the column vector G,
these equations can be written in matrix form as:
Where Fis the system matrix and Gis the input matrix. If we take the input to be
FxM
xx 1
xx 2
21
xx
M
Fx 2
M
F
1
0
x
x
00
10
x
x
2
1
2
1
GuFxx
1xy
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8/13/2019 Lecture 7 (SS)
4/22
Recap
4
This to can be expressed in matrix form as:
Where His a row vector, referred to as the output matrix. Collecting these matrix notations, we
have an extremely compact notation..
Similarly, the differential-equation models of more complex systems, such as those developed in
earlier lectures on mechanical, electrical, and electromechanical systems, can be described by
state variables through selection of positions, velocities, capacitor voltages, and inductor
currents as state variables.
2
1
01 x
xy
Hxy
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8/13/2019 Lecture 7 (SS)
5/22
Recap -Example 2
5
A robot-arm drive system for one joint can be represented by the following differential equation:
where v(t) = velocity, y(t) = position and i(t) is the control-motor current. Put the equations in state variable form
and set up the matrix form for
We know that the velocity is the derivative of the position, therefore we have
and from the problem statement
This can be written in matrix form as
),()()()(
321 tiktyktvk
dt
tdv
121 kk
vdt
dy
)()()( 321 tiktyktvkdt
dv
ikv
y
kkv
y
dt
dv
312
010
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8/13/2019 Lecture 7 (SS)
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Recap - Example 3
6
As an example of the formulation of equations in state-variable form, consider the two-mass
system below:
If we take the state as the position and velocity of each mass as follows:
dx
dx
yx
yx
4
3
2
1
udykdybyM )()(
0)()( ydkydbdm
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8/13/2019 Lecture 7 (SS)
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Recap
7
the equations in state-variable form are:
The state-space matrices are:
43214
43
43212
21
xM
bx
M
kx
M
bx
M
kx
xx
M
ux
M
bx
M
kx
M
bx
M
kx
xx
M
b
M
k
M
b
M
k
1000
M
b
M
k
M
b-
M
k-
0010
F
0
0
1
0
MG 0001H
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8/13/2019 Lecture 7 (SS)
8/22
Recap - Example 4
8
Consider the equation of motion for the simple pendulum shown below:
where . If we choose as the state variables and , then the equations are:
These equations are nonlinear but are in state-variable form. We can linearize the equations and
hence for smallx.
2
12
2
21
sinml
Txx
xx
c
2
2 sin
ml
Tc
lg2
1x
2x
sin
22
21
sinml
Tx
xx
c
0
0F
2
10
ml
G
0
1
H
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8/13/2019 Lecture 7 (SS)
9/22
State Diagrams
9
Phase-variable format
Input feedforward format
Controllability Canonical form
Observability Canonical form
Diagonal Canonical form
Jordan Canonical form
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8/13/2019 Lecture 7 (SS)
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State Diagrams
10
To illustrate the derivation of the state diagram, we consider the fourth-order transfer function:
Note that the system, is fourth order and hence we identify 4 state variables, hence we will use
four integrators. The state diagram will be:
The numerator terms represent forward-path factors in Masonsgain formula.
43
1
2
2
1
3
4
1
2
2
3
3
4
1
)(
sasasasa
sb
asasasas
b
sG
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8/13/2019 Lecture 7 (SS)
11/22
State Diagrams II
11
To illustrate the derivation of the state diagram, we consider the fourth-order transfer function:
The state diagram will be:
The numerator terms represent forward-path factors in Masonsgain formula. The general form
shown above is called thephase variable format.
What is
The output is simply:
43
1
2
2
1
3
43
1
2
2
1
3
1
2
2
3
3
4
1
2
2
3
3
1
)(
sasasasa
sbsbsbsb
asasasas
bsbsbsb
sG
uxaxaxaxa 43322114x
4332211)( xbxbxbxbtc
433221 ,, xxxxxx
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8/13/2019 Lecture 7 (SS)
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State Diagrams II
12
State equations
Then in matrix form we have:
and the output is:
uxaxaxaxa 43322114x
4332211)( xbxbxbxbtc
433221 ,, xxxxxx
ubAxx
)(
1
0
0
0
1000
0100
0010
4
3
2
1
32104
3
2
1
tu
x
x
x
x
aaaax
x
x
x
dt
d
4
3
2
1
3210 ,,,Dx)(
x
x
x
x
bbbbtc
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8/13/2019 Lecture 7 (SS)
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State Diagrams III
13
The state diagram in the previous example is not a unique representation of the given transfer
function and other equally useful structures can be designed. An alternative form known as the
input feedforward format is illustrated below.
Using the state diagram we can obtain the following set of first-order differential equations:
43
1
2
2
1
3
43
1
2
2
1
3
1
2
2
3
3
4
1
2
2
3
3
1
)(
sasasasa
sbsbsbsb
asasasas
bsbsbsbsG
ubxxax 32131 ubxxax 23122
ubxxax 14113 ubxax 14
)(
000
100
010
001
0
1
2
3
4
3
2
1
0
1
2
3
4
3
2
1
tu
b
b
b
b
x
x
x
x
a
a
a
a
x
x
x
x
dt
d
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8/13/2019 Lecture 7 (SS)
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State Diagrams IV
14
Consider the system represented by the equation:
The state variables, the outputs of the integrators are . In these terms,
the equations of motion are:
uyyyy 66116
uxxxx
xx
xx
661163213
32
21
yxandyxyx 321 ,
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8/13/2019 Lecture 7 (SS)
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State Diagrams V
15
The closed-loop transfer function of a system is:
The first model is the phase variable state diagram as illustrated below:
Recalling Masonsgain formula, the denominator can be considered to be one minus the sum ofthe loop gains whereas the numerator of the transfer function is equal to the forward-path.
and the output is:
321
321
23
2
61681
682
6168
682
)(
)()(
sss
sss
sss
ss
sR
sCsT
)(
1
0
0
8166
100
010
3
2
1
3
2
1
tu
x
x
x
x
x
x
dt
d
3
2
1
286)(
x
x
x
tc
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8/13/2019 Lecture 7 (SS)
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State Diagrams VI
16
The closed-loop transfer function of a system is:
The second model is the phase variable state diagram as illustrated below:
Recalling Masonsgain formula, the denominator can be considered to be one minus the sum ofthe loop gains whereas the numerator of the transfer function is equal to the forward-path.
and the output is:
321
321
23
2
61681
682
6168
682
)(
)()(
sss
sss
sss
ss
sR
sCsT
)(
1
0
0
8166
100
010
3
2
1
3
2
1
tu
x
x
x
x
x
x
dt
d
3
2
1
286)(
x
x
x
tc
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8/13/2019 Lecture 7 (SS)
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State Diagrams VII
17
The closed-loop transfer function of a system is:
The first model is the input feedforward format as illustrated below:
Recalling Masonsgain formula, the denominator can be considered to be one minus the sum ofthe loop gains whereas the numerator of the transfer function is equal to the forward-path.
and the output is:
321
321
23
2
61681
682
6168
682
)(
)()(
sss
sss
sss
ss
sR
sCsT
)(
6
8
2
006
1016
018
3
2
1
3
2
1
tu
x
x
x
x
x
x
dt
d
)()( 1 txtc
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8/13/2019 Lecture 7 (SS)
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System Transfer Functions
18
In order to relate the state-variable equations to our earlier consideration of poles and zeros,
we take the Laplace transform of:
We obtain:
which is now an algebraic equation. We collect the terms involving X(s) on the left-hand side,
we get:
If we multiply both sides by the inverse of , then:
The output of the system is:
Which means
GuFxx
)(G)FX(x(0))( sUsssX
x(0))(G)(X)FI( sUss
)FI( s
x(0))()(G)()(X -1-1 FsIsUFsIs
)()(H)( sJUsXsY
)(x(0))(H)(G)(H)( -1-1
sJUFsIsUFsIsY
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8/13/2019 Lecture 7 (SS)
19/22
System Transfer Functions
19
If we assume that the initial conditions are zero, the input-output relationship is:
JG)(H)(
)(
)( 1-
FsIsU
sY
sG
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8/13/2019 Lecture 7 (SS)
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Example 1
20
Consider the system with state-variable description:
Find the corresponding state diagram and compute the transfer function.
0165F
01G
01H 0J
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8/13/2019 Lecture 7 (SS)
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Example 1 (cont.)
21
Compute the transfer function form:
and compute:
Therefore:
ssFsI
165
6)5(51
6
)( 1
ss
s
s
FsI
6)5(
0
151
6)5(
0
1
51
610
)(
ss
s
ss
s
s
sG
)3)(2(
1)(
ss
sG
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8/13/2019 Lecture 7 (SS)
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State Diagram Example
22
The phase variable form of a system is given by:
Draw the flow graph model.
2