EE362-Ch2.pdf - Feedback and Control Systems

System Modeling in the Time Domain
Assoc. Prof. Enver Tatlicioglu
Department of Electrical & Electronics Engineering
Izmir Institute of Technology
Chapter 2
Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control Systems Chapter 2 1 / 64

Mathematical modeling
Q. What is a model?
A. The term model, as it is used and understood by control engineers,
primarily means a set of differential equations that describe the
dynamic behaviour of the system.
The first step in designing a controller is to develop a mathematical
description (also called the dynamic model) of the process to be
controlled.
Developing the model is the 80%–90% of the effort in designing a
controller.
Mathematical model is required to:
– Understand system behavior (analysis).
– Design a controller (synthesis).

System identification
When we generate models of system dynamics, we are performing
system identifications.
Q. How is a model obtained?
A. When we use known properties from physics and knowledge of the
system’s structure we are performing white–box system identification
(analytical system modeling).
A. If the system is very complex, or if the underlying physics is not well
understood, we need to use input/output data to generate a system
model, so we perform black–box system identification (empirical
system identification).
System identification is a topic for a whole course!
No model is exact!
Inaccuracies are due to:
– Unknown parameter values.
– Unmodeled dynamics (to obtain simpler models).

Linear time–invariant systems
In this course, we mainly focus on control of causal, linear,
time–invariant (LTI) systems for t ≥ 0.
Suppose a system maps
x (t) → y (t) as y (t) = T [x (t)] .
– This system is linear iff
T [ax1 (t) + bx2 (t)] = aT [x1 (t)] + bT [x2 (t)] .
– This system is time–invariant iff
y (t − t0) = T [x (t − t0)] ∀t0.
A system is causal if the output at any time depends on values of the
input at only the present and past times.

A resistor model
Consider the following resistor model
+
i (t)
R
+
−
v (t)
−
From Ohm’s Law, we know that
v (t) = i (t) R
where v (t) is the voltage, i (t) is the current, and R is the resistance.
Is the resistor model LTI?

A resistor model
Consider a 1 ohm, 2 watt resistor.
Apply v (t) = 1 volt:
? ⇒ 1 ampere of current is predicted to flow
? ⇒ Power dissipated = V 2/R = 1 watt.
Apply v (t) = 10 volts:
? ⇒ 10 amperes of current is predicted to flow
? ⇒ Power dissipated = V 2/R = 100 watts.
Model will no longer be accurate.
True behavior of the model depends on input signal level which is
nonlinear.
Model is accurate in certain range of input signal values.

Important components of translational mechanical systems
The fundamental law governing mechanical systems is Newton’s
second law:
X
F = ma
Vector sum of forces = mass of object × inertial acceleration
Figure: Important components of translational mechanical systems

Cruise control model
Write the equations of motion for the speed and forward motion of a
car assuming the engine imparts a forward force of u (t).
Figure: Simplified cruise control model
Assume that friction is proportional to car’s speed.
Assume rotational inertia of wheels is negligible.

Cruise control model
Dynamic model of the cruise control system is as follows
X
F = ma
⇒ u (t) − bẋ (t) = mẍ (t)
⇒ ẍ (t) +
b
m
ẋ (t) =
1
m
u (t) .
If the variable of interest is speed (v (t) = ẋ (t)), then
v̇ (t) +
b
m
v (t) =
1
m
u (t) .
Is this model accurate?
What are some of the unmodeled dynamics?
What parameters could be uncertain?

A two–mass system
A system consisting of one quarter of the car mass is referred to as
quarter–car model.
Figure: Quarter–car model

A two–mass system
The force from the spring is proportional to its stretch.
The force from the damper (i.e., shock absorber) is proportional to
the rate–of–change of its stretch.
Figure: Free–body diagrams for the suspension system

A two–mass system
Dynamic model of the quarter–car model is
X
F = ma
b (ẏ − ẋ) + ks (y − x) − kw (x − r) = m1ẍ
−ks (y − x) − b (ẏ − ẋ) = m2ÿ.
After rearranging, we obtain
ẍ +
b
m1
(ẋ − ẏ) +
ks
m1
(x − y) +
kw
m1
x =
kw
m1
r
ÿ +
b
m2
(ẏ − ẋ) +
ks
m2
(y − x) = 0.

Important components of rotational mechanical systems
Newton’s second law is modified as
X
M = Iα
Vector sum of moments = moment of inertia × angular acceleration
Figure: Important components of rotational mechanical systems

Satellite attitute control model
Consider the below simple model for a satellite.

Satellite attitute control model
Satellites usually require attitude control so that antennas, sensors,
and solar panels are properly oriented.
Antennas are usually pointed toward a particular location on earth,
while solar panels need to be oriented toward the sun for maximum
power generation.
For simplicity reasons, planar motion is considered
Fc (t) d = Iθ̈ (t)
⇒ θ̈ (t) =
d
I
Fc (t) .
The output of the system θ (t) integrates torque twice which is called
as double integrator plant.

Summary of developing models for rigid bodies
Assign variables such as x (t) and θ (t) that are both necessary and
sufficient to describe any arbitrary position of the object.
Draw a free–body diagram of each component, and indicate all forces
acting on each body and the accelerations of the center of mass with
respect to an inertial reference.
Apply Newton’s laws:
?
P
F = ma
?
P
M = Iα
Combine the equations to eliminate internal forces.
The final form must be in terms of only the input to the system and
its derivatives, and the output of the system and its derivatives.

Dynamics of electrical circuits
Electrical circuits are frequently used in control systems due to ease
of manipulation and processing electric signals.
Analog circuits are still used because they are faster and less
expensive when compared to the digital ones.
For some control systems, analog circuits are required (i.e., a power
amplifier for electromechanical control).

Dynamics of electrical circuits
The basic equations of electric circuits are Kirchoff’s laws.
– Kirchoff’s current law (KCL):
The algebraic sum of currents leaving a junction or a node equals the
algebraic sum of currents entering that node.
– Kirchoff’s voltage law (KVL):
The algebraic sum of all voltages taken around a closed path in a
circuit is zero.
Node analysis is a powerful tool to analyse complex circuits with
many elements. Briefly, one node is selected as a reference node and
the voltages of the other nodes are assumed to be unknown. The
equations for the unknowns are written using KCL at each node. If
the circuit contains voltage sources, KVL is used for such sources.

Important elements of electric circuits

Bridged tee circuit
Determine the differential equations for the bridged tee circuit where
v1 = vi is the input and v3 = vo is the output.
−
+
vi (t)
R1
C1
R2
+
−
vo (t)
C2
1 2 3
4

Bridged tee circuit
Select node 4 as the reference and the voltages v1, v2 and v3 at nodes
1, 2 and 3 as the unknowns.
– KVL at node 1:
v1 = vi .
– KCL at node 2:
v2 − v1
R1
+
v2 − v3
R2
+ C1
dv2
dt
= 0.
– KCL at node 3:
v3 − v2
R2
+ C2
d (v3 − v1)
dt
= 0.
– After rearranging the above expression, we obtain
v3 + R2C2 (v̇3 − v̇1) = v2.

Bridged tee circuit
Substituting the above expression in the expression for the KCL at
node 2 and then rearranging results in
0 =
(v3 + R2C2 (v̇3 − v̇1)) − v1
R1
+
(v3 + R2C2 (v̇3 − v̇1)) − v3
R2
+C1 (v̇3 + R2C2 (v̈3 − v̈1)) .
After rearranging the above expression, we obtain
0 = [(v3 + R2C2 (v̇3 − v̇1)) − v1] + R1C2 (v̇3 − v̇1)
+R1C1 (v̇3 + R2C2 (v̈3 − v̈1)) .
After rearranging the above expression, we obtain
R1R2C1C2v̈3 + (R2C2 + R1C2 + R1C1) v̇3 + v3
= R1R2C1C2v̈1 + (R2C2 + R1C2) v̇1 + v1.

Op–amp
Op–amp is a feedback system and it is very important for
electromechanical systems.
−
+
R1
+
−
vi (t)
C
R2
+
−
vo (t)

Op–amp
The equations of the (ideal) op–amp circuit above are
i (t) =
vi (t)
R1
v0 (t) = −R2i (t) − vc (t) .
Taking the time derivative of the second expression results in
dv0
dt
= −R2
di
dt
−
dvc
dt
= −
R2
R1
dvi
dt
−
1
C
i
= −
R2
R1
dvi
dt
−
1
R1C
vi .
Rearranging and then changing the notation results in
R1Cv̇0 = −R2Cv̇i − vi .

Dynamics of electromechanical systems
Electromechanical systems convert energy from electrical to
mechanical or vice versa.
DC motor directly generates rotational motion and indirectly
generates translational motion.
Mechanical resistance of the load is translated into an electrical
resistance called the back emf (electro–motive force).
Figure: DC motor model

DC motor
The equations of the electrical circuit are
ea (t) = Raia (t) + La
dia (t)
dt
+ eb (t)
eb (t) = Ke
dθ (t)
dt
τ (t) = Kτ ia (t)
where Ke , Kφ and Kτ , K1φ with φ being the flux.
The equations of motion are
X
M = Jθ̈ (t)
⇒ τ (t) − bθ̇ (t) = Jθ̈ (t) .

DC motor
It is assumed that the electrical response is faster than the mechanical
response (i.e., La ≈ 0)
Jθ̈ (t) = τ (t) − bθ̇ (t)
= Kτ ia (t) − bθ̇ (t)
=
Kτ
Ra
(ea (t) − eb (t)) − bθ̇ (t)
=
Kτ
Ra

ea (t) − Ke
dθ (t)
dt

− bθ̇ (t) .
Rearranging results in
Jθ̈ (t) +

Kτ Ke
Ra
+ b

θ̇ (t) =
Kτ
Ra
ea (t) .

DC generator
Generator has field windings (input) and rotor/armature windings
(output).
+
−
ef (t)
Rf if (t)
Lf −
+ eg (t)
Ra La ia (t)
Zl
+
−
ea (t)
Field circuit Rotor circuit Load circuit
Let the generator be driven at a constant speed.

DC generator
The equation of the field circuit is
ef (t) = Rf if (t) + Lf
dif (t)
dt
where ef (t) is the input and if (t) is the output.
The relation between the field circuit and the rotor circuit can be
expressed as
eg (t) = Kφ
dθ (t)
dt
where K depends on the generator structure, and φ is the flux which
is proportional to if (t). Since dθ(t)
dt is the constant angular velocity,
we obtain
eg (t) = Kg if (t) .

DC generator
The equation of the rotor circuit is
eg (t) = Raia (t) + La
dia (t)
dt
+ Zl ia (t)
where eg (t) is the input and ia (t) is the output.
For the load circuit,
ea (t) = Zl ia (t)
where ia (t) is the input and ea (t) is the output.
This is a preview of a block diagram used to simplify our
understanding of the system dynamics.
Field circuit Kg Rotor circuit Zl
ef (t) if (t) eg (t) ia (t) ea (t)

Linearity of a model
We can convert a differential equation into a first order vector
differential equation
ẋ = f (x, u)
where x (t) is the state vector and u (t) is the input.
If the system is LTI then the right–hand side of the above expression
can be written as follows
ẋ = Ax + Bu
with A and B being constant matrices with appropriate dimensions.

Torsional pendulum
A torsional pendulum is used in clocks enclosed in glass domes. A
similar device is the read/write head on a hard–disk drive.
Figure: A torsional pendulum (left) and its simplified model (right)

Torsional pendulum
The equations of motion are
X
M = Jθ̈ (t)
⇒ τ (t) − bθ̇ (t) − kθ (t) = Jθ̈ (t) .
Rearranging results in
θ̈ (t) +
b
J
θ̇ (t) +
k
J
θ (t) =
1
J
τ (t) .
Let
x (t) =

x1 (t)
x2 (t)

=

θ (t)
θ̇ (t)

.

Torsional pendulum
So the dynamic model becomes
ẋ1 (t) = x2 (t)
ẋ2 (t) +
b
J
x2 (t) +
k
J
x1 (t) =
1
J
τ (t) .
Combining above equations in matrix form results in

ẋ1 (t)
ẋ2 (t)

=

0 1
−k
J −b
J

x1 (t)
x2 (t)

+

0
1
J

τ (t)
where A =

0 1
−k
J −b
J

and B =

0
1
J

are constant matrices thus
our model of the torsional pendulum is linear.

Pendulum
Consider the below pendulum model.
Figure: A pendulum model

Pendulum
Moment of inertia = I = ml2
X
M = Iθ̈ (t)
⇒ Iθ̈ (t) = τ (t) − mgl sin (θ (t))
⇒ θ̈ (t) +
g
l
sin (θ (t)) =
1
ml2
τ (t) .
The expression above is nonlinear due to the sin (θ (t)) term.
Why? Can you show it?

Pendulum
Let
x (t) =

x1 (t)
x2 (t)

=

θ (t)
θ̇ (t)

.
So the dynamic model becomes
ẋ1 (t) = x2 (t)
ẋ2 (t) +
g
l
sin (x1 (t)) =
1
ml2
τ (t) .
Combining above equations in matrix form results in

ẋ1 (t)
ẋ2 (t)

=

x2 (t)
−g
l sin (x1 (t))

+

0
1
ml2

τ (t)
where A =

x2 (t)
−g
l sin (x1 (t))

and B =

0
1
ml2

.
Notice that A is not a constant matrix.

Pendulum
We can linearize the sin (θ (t)) term for small θ (t) values
sin (θ (t)) ≈ θ (t) .
So we obtain the following linearized model
θ̈ (t) +
g
l
θ (t) =
1
ml2
τ (t) .

Small signal linearization
Approximation using Taylor series expansion of the differential
equation (ẋ = f (x, u)) around some operating condition (equilibrium
value where ẋ0 = 0 = f (x0, u0) where x0 is the operating state and u0
is the nominal control value).
Let
x = x0 + δx
u = u0 + δu.
After Taylor series expansion, we obtain
ẋ = ẋ0 + δẋ ≈ f (x0, u0) + Aδx + Bδu + H.O.T..
Subtracting out the equilibrium solution ẋ0 = f (x0, u0) and after
neglecting the higher order terms result in
δẋ = Aδx + Bδu
which is linear.

Small signal linearization
This is how we linearized the rotational pendulum, with u0 = 0 and
θ0 = 0
sin (θ) = θ −
θ3
3!
+
θ5
5!
+ H.O.T..
≈ θ.

Feedback linearization (computed torque)
Recall the rotational pendulum model
ml2
θ̈ (t) + mgl sin (θ (t)) = τ (t) .
Design
τ (t) = u (t) + mgl sin (θ (t))
where u (t) is the auxiliary input and the other term cancels the
nonlinear term.
Substitute the above control into the dynamic model to obtain
ml2
θ̈ (t) = u (t) .
No matter how large θ (t) is, the final model is linear!

Analogous systems
Differential equations of many very different physical systems appear
identical. They behave in similar ways (dynamic response) and can be
controlled with similar controllers.
Mechanical Translational mẍ (t) + bẋ (t) = u (t)
Mechanical Rotational Jθ̈ (t) + bθ̇ (t) + kθ (t) = τ (t)
Satellite Iθ̈ (t) = Fc (t) d
Op-Amp R1Cv̇0 (t) = −R2Cv̇i (t) − vi (t)
DC motor (for La = 0) Jθ̈ (t) +

Kτ Ke
Ra
+ b

θ̇ (t) = Kτ
Ra
ea (t)
These are all of the form
a2ẍ (t) + a1ẋ (t) + a0x (t) = b2ü (t) + b1u̇ (t) + b0u (t)
which is called a second order form.
If we learn how to control the general class of systems, we can apply
this knowledge to specific systems.

Analogous systems
In this table, the force f and the current i are analogs and are
classified as through variables. There is a physical similarity between
the through variables in the sense that a measuring instrument must
be placed in series with the system in both cases.
Also, the velocity across a mechanical element is analogous to voltage
across an electrical element. Again, there is a physical similarity in
the sense that a measuring instrument must be placed across the
system in both cases. A voltmeter must be placed across a circuit to
measure voltage and it must have a point of reference. A velocity
indicator must also have a point of reference.
Symbol Quantity Symbol Quantity
f Force i Current
v Velocity V Voltage
M Mass C Capacitance
K Stiffness coefficient 1/L Reciprocal inductance
B Damping coefficient G = 1/R Conductance

Block diagrams
We have already seen block diagrams
Shows information/energy flow in a system.
When used with Laplace transform, can simplify complex system
dynamics.

Block diagrams
H (s)
U (s) Y (s)
Y (s) = H (s) U (s)

Block diagrams
H1 (s) H2 (s)
U (s) Y (s)
Y (s) = [H1 (s) H2 (s)] U (s)

Block diagrams
Example
Recall DC generator model
+
−
ef (t)
Rf if (t)
Lf −
+ eg (t)
Ra La ia (t)
Zl
+
−
ea (t)
Field circuit Rotor circuit Load circuit
with its block diagram obtained as

Block diagrams
Example
The equations of the field circuit are
ef (t) = Rf if (t) + Lf
dif (t)
dt
where ef (t) is the input and if (t) is the output.
Laplace transform of the above expression is
Ef (s) = Rf If (s) + Lf sIf (s)
and the transfer function is obtained as
If (s)
Ef (s)
=
1
Rf + Lf s
.
Note that, since we are obtaining a transfer function the initial
conditions are set to zero.

Block diagrams
Example
The relation between the field circuit and the rotor circuit can be
expressed as
eg (t) = Kg if (t)
where if (t) is the input and eg (t) is the output.
Eg (s) = Kg If (s)
Eg (s)
If (s)
= Kg .

Block diagrams
Example
The equations of the field circuit are
eg (t) = Raia (t) + La
dia (t)
dt
+ ea (t)
where eg (t) is the input and ia (t) is the output.
Eg (s) = RaIa (s) + LasIa (s) + Ea (s)
= RaIa (s) + LasIa (s) + Zl Ia (s)
= (Ra + Las + Zl ) Ia (s)
Ia (s)
Eg (s)
=
1
Ra + Las + Zl
.

Block diagrams
Example
For the load circuit,
ea (t) = Zl ia (t)
where ia (t) is the input and ea (t) is the output.
Ea (s) = Zl Ia (s)
Ea (s)
Ia (s)
= Zl .

Block diagrams
Example
Put everything together
Ea (s)
Ef (s)
=
Ea (s)
Ia (s)
×
Ia (s)
Eg (s)
×
Eg (s)
If (s)
×
If (s)
Ef (s)
= Zl ×
1
Ra + Las + Zl
× Kg ×
1
Rf + Lf s
=
Zl Kg
(Ra + Las + Zl ) (Rf + Lf s)
.

Block diagrams
H1 (s)
H2 (s)
+
U (s) +
+
Y (s)
Y (s) = [H1 (s) + H2 (s)] U (s)

Block diagrams
H (s)
U (s) Y1 (s)
Y2 (s)
H (s)
1
H(s)
U (s) Y1 (s)
Y2 (s)

Block diagrams
+ H (s)
U1 (s) +
U2 (s)
+
Y (s)
H (s)
H (s)
+
U1 (s)
U2 (s)
+
+
Y (s)

Block diagrams
+ H1 (s)
H2 (s)
R (s) + U1 (s) Y (s)
U2 (s)
Y2 (s)
−

Block diagrams
The auxiliary signals U1 (s), Y2 (s) and U2 (s) can be obtained as
U1 (s) = R (s) − Y2 (s)
Y2 (s) = H2 (s) U2 (s)
U2 (s) = H1 (s) U1 (s) .
From the second and third expressions, we can obtain
Y2 (s) = H2 (s) H1 (s) U1 (s)
and substituting this into the first expression results in
U1 (s) = R (s) − H2 (s) H1 (s) U1 (s)
⇒ U1 (s) =
R (s)
1 + H2 (s) H1 (s)
.
Finally, we obtain
Y (s) = H1 (s) U1 (s) =
H1 (s)
1 + H2 (s) H1 (s)
R (s) .

Block diagrams
1
H2(s) + H2 (s) H1 (s)
R (s) + Y (s)
−

Block diagrams
Example
+ + H1 (s) H2 (s) H5 (s)
H3 (s)
H4 (s)
H6 (s)
+
R (s) + +
+
−
+
+ Y (s)

Block diagrams
Example
+ H1(s)
1−H1(s)H3(s) H2 (s) H5 (s)
H4 (s)
H6 (s)
+
R (s) +
−
+
+ Y (s)

Block diagrams
Example
+ H1(s)
1−H1(s)H3(s) H2 (s) H5 (s)
H4 (s)
H6(s)
H2(s)
+
R (s) +
−
+
+ Y (s)

Block diagrams
Example
+ H1(s)H2(s)
1−H1(s)H3(s) H5 (s)
H4 (s)
H6(s)
H2(s)
+
R (s) +
−
+
+ Y (s)

Block diagrams
Example
H1(s)H2(s)
1−H1(s)H3(s)+H1(s)H2(s)H4(s)
H6(s)+H2(s)H5(s)
H2(s)
R (s) Y (s)

Block diagrams
Example
H1(s)H2(s)H5(s)+H1(s)H6(s)
1−H1(s)H3(s)+H1(s)H2(s)H4(s)
R (s) Y (s)

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