Robot Control (4th)
(A Basic Study of Dynamics & Control)
Autumn semester
School of Robotics
BICAR(Biologically-inspired Control and Robot) Lab.
Prof. Woosung Yang
1
Outline
1. Introduction of Control System
2. Basic Study of Electric Circuits
• Example: Control unit used in mechatronics
• Components of Electric Circuits
• Circuit Equations
3. Basic Study of Dynamics
• Mass, Force and Acceleration
• Dynamic Equation
4. Basic Study of Mathematics
• Differential Equation
• Linear Calculation
5. Understanding of Feedback System
Bio-inspired Control & Robot (BICAR) Lab.
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1. Introduction of Control System
(1st or 2nd order systems)
(1st or 2nd order systems)
◆
◆
◆
The control action is independent of the output.
The ability to perform accurately is determined
by the calibration.
Not usually troubled with problems of
instability.
◆
◆
◆
◆
◆
◆
◆
The control action is somehow dependent on the
output.
Increased accuracy
Tendency toward oscillation or instability
Reduced sensitivity of the ratio of output to input
to variations in system parameters and other
characteristics
Reduced effects of nonlinearities
Reduced effects of external disturbances or noise
Increased bandwidth
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1. Introduction of Control System
◆
◆
◆
Requires the identification of all possible disturbances and their direct
measurement
Any changes in the parameters of a process cannot be compensated
Requires a good model of the process
◆
Feedback Control
◆
Does not require knowledge or measurement of disturbances
Not very sensitive to changes in process parameters or to errors
in the process model
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1. Introduction of Control System
◆
◆
◆
◆
◆
◆
◆
Establish the control goals
Control System Design Process
Identify the variables to be controlled
Write the specifications
Establish the system configuration
Obtain a model for the process, the actuator, and the sensor
Describe a controller and select key parameters to be adjusted
Optimize the parameters and analyze the performance
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1. Introduction of Control System
Basic Representations (Models)
◆
◆
◆
◆
Mathematical models; differential equations, difference equations,
and/or other mathematical relations, e.g., Laplace- and z-transforms
Block diagrams
Signal flow graphs
Bond graphs
A Dynamic System
Its instantaneous behavior is dependent on its past
history, so that the behavior of the system at time t >
t0 can be determined given
(1) the forcing function (that is, the input), and
(2) the state of the system at t = t0.
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1. Basic Study of Electric Circuits
▪ Example: Control unit
스위치
저항
코일
캐패시터
확장커넥터
전원
직류모터/엔코더
통신포트
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2. Basic Study of Electric Circuits
▪ Components of electric circuits
- Active(IC, Transistor, Power, etc.) or Passive(Resistor, Inductor, Conductor, etc.)
▪ R, L, C
vR = Ri
Controlling current/voltage
Transfer voltage to current
1
vC = ∫
idt
C
d
vL = L i
dt
Charge the electricity
Cutoff DC currents
Pass AC currents
Stabilizing currents
Cutoff noise (AC currents)
Pass DC currents
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2. Basic Study of Electric Circuits
▪ Circuit equation → RLC circuit
Kirchhoff’ law
d
1
Ri + L i + ∫
idt = v
dt C
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3. Basic Study of Dynamics
▪ Newton's first law: Inertia law
“The first law states that if the net force (the vector sum of all forces acting on
an object) is zero, then the velocity of the object is constant”
F =0
d
v=0
dt
Consequently,
An object that is at rest will stay at rest unless an external force acts upon it.
An object that is in motion will not change its velocity unless an external force
acts upon it.
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3. Basic Study of Dynamics
▪ Newton's second law: Acceleration law
“The second law states that the net force on an object is equal to the rate of
change (that is, the derivative) of its linear momentum p in an inertial reference
frame”:
d
d
F =
p=
(mv ) = ma
dt
dt
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3. Basic Study of Dynamics
▪ Newton's third law: Action-reaction law
“The third law states that all forces exist in pairs”: if one object A exerts a force
FA on a second object B, then B simultaneously exerts a force FB on A, and the two
forces are equal and opposite: FA = −FB
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.)
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d
1
i+ ∫
idt = v
dt
C
v = 100cos(2π × 120t)
Ri + L
R = 100, L = 1mH , C = 10F
kx
mx + cx + kx = F
mx
m = 10kg , c = 10 N sec/ m, k = 5 N / m
cx
F
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4. Basic Study of Mathematics (D.E.)
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The natural response of a system is when there is no input to the system forcing the
variable to change but it is just changing naturally. The forced response of a system is
when there is an input to the system forcing it to change.
A first-order system with no forcing input has a differential equation of the form
dx
a1 + a0 x = 0
dt
and this has the solution x = e − a0t / a1
When there is a forcing function the differential equation is of the form
dx
a1 + a0 x = b0 y
dt
(
and the solution is x = steady-state value  1 − e − a0t / a1
)
The time constant  is the time the output takes to rise to 0.63 of its steady-state
value and is (a1 / a0 )
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.)
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A second-order system with no forcing input has a differential equation of the form
d 2x
dx
m 2 + c + kx = 0
dt
dt
The natural angular frequency is given by n 2 = k / m and the damping constant by  2 = c 2 / 4mk .
The system is over-damped when we have  >1 and the general solution for x n is
x n = Ae s1t + Be s2t
with s = −n  n  2 − 1
When  =1 the system is critically damped and
xn = ( At + B ) e −nt
and with  <1 the system is under-damped and
xn = e −nt ( P cos t + Q sin t )
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When we have a forcing input F the second-order differential equation becomes
d 2x
dx
m 2 + c + kx = F
dt
dt
and for the over-damped system
F
x = Ae s1t + Be s2t +
k
for the critically damped system
F
x = ( At + B ) e −nt +
k
and for the under-damped system
x = e −nt ( P cos t + Q sin t ) +
F
k
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The rise time 𝒕𝒓 is the time taken for the response 𝒙 to rise from 0 to the steady-state value 𝒙𝒔𝒔 and
is a measure of how fast a system responds to the input and is given by 𝝎𝒕𝒓 = 𝟏Τ𝟐 𝝅. The peak time
𝒕𝒑 is the time taken for the response to rise from 0 to the first peak value and is given by 𝝎𝒕𝒑 = 𝝅.
The overshoot is the maximum amount by which the response overshoots the steady-state value
and is
 − 

overshoot = xss exp 
 1−  2 


The subsidence ratio or decrement is the amplitude of the second overshoot divided by that of
the first overshoot and is
 −2
subsidence ratio = exp 
 1−  2





The settling time 𝒕𝒔 is the time taken for the response to fall within and remain within some
specified percentage, e.g. 2%, of the steady-state value, this being given by
ts =
4
n
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 1st order systems
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▪ 4.1 First-order systems
- Consider a first-order system with y(t) as the input to the system
and x(t) the output and which has a forcing input b0y and can be
described as the following
a1
dx
+ a0 x = b0 y
dt
where a1, a0 and b0 are constants.
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4. Basic Study of Mathematics (D.E.) – 1st order systems
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▪ 4.1.1 Natural Response
- The input y(t) can make many forms.
- Consider first the situation when the input is 0.
- This system shows natural response.
dx
a1 + a0 x = 0
dt
Using the technique called separation of variables.
a
dx
= − 0 dt
x
a1
Integrating this between the initial value of x=1 at t=0, i.e. a unit
step input, and x at t gives
a0
ln x = − t
a1
Bio-inspired Control & Robot (BICAR) Lab.
3. Basic Study of Mathematics (D.E.) – 1st order systems
so we have
19
x = e − a0t / a1
Assume the differential eq. would have a solution x=Aest(A and s
are constants.). Differentiating this, dx/dt=sAest and so when these
values are substituted in the differential eq. we obtain
a1sAe st + a0 Ae st = 0
and so a1s+a0=0 and s=-a0/a1.
x = Ae− a0t / a1
This is termed the natural response since there is no forcing
function. We can determine the value of the cons. A given some
initial (boundary) condition. Thus if x=1 when t=0 then A=1.
x = e − a0t / a1
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▪ 4.1.2 Response with a forcing input
- Consider the differential eq. when there is a forcing function, i.e.
a1
dx
+ a0 x = b0 y
dt
Consider the solution to this eq. to be made up of two parts, i.e. x = u + v.
One part represents the transient part of the solution and the other the
steady-state part. Substituting this into the differential eq. gives
d (u + v )
a1
+ a0 (u + v ) = b0 y
dt
Rearranging this gives
 du
  dv

a
+
a
u
+
a
+
a
v
 1
0   1
0  = b0 y
 dt
  dt

dv
du
+ a0 v = b0 y then we must have a1
If we let, a1
+ a0 u = 0
dt
dt
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 1st order systems
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Two differential equations, one of which contains a forcing
function and one which is just the natural response equation.
The solution of the natural response eq.
u = Ae− a0t / a1
-
The other differential eq. contains y
This solution depends on the form of the input y.
1) Step input y (const. and > 0), y = k.
→ The solution v = A. (A is const.)
2) Input y = a+bt+ct2+⋯(a, b and c are const.),
→ Assume that the output v = A+Bt+Ct2
3) For a sinusoidal signal
→ The output response, v = Acosωt + Bsinωt
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4. Basic Study of Mathematics (D.E.) – 1st order systems
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-
Assume there is a step input at a time of t=0 with the size of the
step being k.
- Then v=A. Differentiating a const. gives 0.
If so a0A=b0k, then v=(b0/a0)k.
Hence, (y=u+v)
y = Ae
− a0t / a1
b0
+ k
a0
We can determine the value of the constant A given some initial
(boundary) conditions. Thus if the output y=0 when t=0 then.
b0
0 = A+ k
a0
Thus A=-(b0/a0)k. The solution then becomes.
(
b0
x = k 1 − e −a0t / a1
a0
)
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4. Basic Study of Mathematics (D.E.) – 1st order systems
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When t ⟶ ∞ the exponential term tends to 0. The exponential
term thus gives that part of the response which is the transient
solution. The steady-state response is the value of x when t ⟶ ∞
and , so is (b0/a0)k. Thus the eq. can be written as.
x = steady − state value  (1 − e − a0t / a1 )
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – review(C.S.)
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• Response of a first order circuit
complete response= transient response + steady-state response
• In general,
the response at steady state
the response that dies out
complete response= natural response + forced response
• Natural response: the general solution of the differential equation
representing the first order circuit when the input is set to zero.
-For a first-order circuit, natural response:
K depends on the init. cond., e.g., capacitor voltage at Ke− ( t −t0 ) 
• Forced response: a particular solution of the differential equation t = t0 .
representing the first order circuit.
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
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• Solving the first order differential equation to a Const. Input
d
v(t ) Voc
v(t ) +
=
dt
Rt C Rt C
R
R
d
i (t ) + t i (t ) = t I sc
dt
L
L
d
x(t )
x(t ) +
=K
dt

x(t ) = x() + [ x(0) − x()]e −t 
• Time constant: how fast the exponential decays
x() − x(0)
Time constant:  =
d
x(t )
dt
t =0
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
• RC circuit
d
v(t ) Voc
v(t ) +
=
dt
Rt C Rt C
v(t ) = Voc + (v(0) − Voc )e−t ( Rt C )
 = Rt C
26
• RL circuit
R
R
d
i (t ) + t i (t ) = t I sc
dt
L
L
i (t ) = I sc + (i (0) − I sc )e−( Rt
L
=
Rt
L )t
forced response
natural response
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
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• Example 4.1: Find the capacitor voltage after the switch opens. What is the
value of the capacitor voltage 50ms after the switch opens?
d
v(t ) Voc
v(t ) +
=
dt
Rt C Rt C
v(t ) = Voc + (v(0) − Voc )e −t ( Rt C )
 v(0) = 2V, Rt = 10k and Voc = 8V
 v(t ) = 8 − 6e − t / 20 V
let t =50. Then v(50) = 8 − e −50/ 20 = 7.51V
v(t ) = 8 − 6e −t 20 V
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• Example 4.2: Find the capacitor voltage after the switch opens. What is the
value of the capacitor voltage 50ms after the switch opens?
d
v(t ) Voc
d
v(t ) +
=
 Rt C v(t ) + v(t ) = Voc
dt
Rt C Rt C
dt
v(t ) = 8 − 6e −t 20 V
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
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• Example 4.3: Find the inductor current after the switch closes. How long
will it take for the inductor current to reach 2mA?
Rt
Rt
d
i (t ) + i (t ) =
I sc
dt
L
L
i (t ) = 4 − 4e−t 5 mA
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
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• The case when the input is a time varying function.
d
v(t ) Voc
v(t ) +
=
dt
Rt C Rt C
R
R
d
i (t ) + t i (t ) = t I sc
dt
L
L
d
v(t ) vs (t )
v(t ) +
=
dt
Rt C
Rt C
R
R
d
i (t ) + t i (t ) = t is (t )
dt
L
L
dx(t )
+ ax(t ) = y (t )
dt
• Integrating factor method: multiply e at to the D.E.
dx(t )
+ ax(t ) = y (t )
dt
x(t ) = e − at  y (t )e at dt + Ke − at
=
x f (t )
+ xn (t )
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
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• The case when the input is a time varying function.
d
v(t ) Voc
v(t ) +
=
dt
Rt C Rt C
R
R
d
i (t ) + t i (t ) = t I sc
dt
L
L
d
v(t ) vs (t )
v(t ) +
=
dt
Rt C
Rt C
R
R
d
i (t ) + t i (t ) = t is (t )
dt
L
L
dx(t )
+ ax(t ) = y (t )
dt
• Integrating factor method: multiply e at to the D.E.
dx(t )
+ ax(t ) = y (t )
dt
at


d
e


d
x
(
)
(
)
dx

at 
at
at
at
at


e 
+
ae
x
=
e
y
→
e
+
x
=
e
y
(
)



 dt 
 dt   dt 
x(t ) = e − at  y (t )e at dt + Ke − at
=
x f (t )
+ xn (t )
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
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• The case when the input is a time varying function.
d
v(t ) Voc
v(t ) +
=
dt
Rt C Rt C
R
R
d
i (t ) + t i (t ) = t I sc
dt
L
L
d
v(t ) vs (t )
v(t ) +
=
dt
Rt C
Rt C
R
R
d
i (t ) + t i (t ) = t is (t )
dt
L
L
dx(t )
+ ax(t ) = y (t )
dt
• Integrating factor method: multiply e at to the D.E.
d at
at
dx(t )
e
x
=
e
y
(
)
+ ax(t ) = y (t )
dt
dt
at


d
e


d
x
(
)
(
)
dx

at 
at
at
at
at


e 
+
ae
x
=
e
y
→
e
+
x
=
e
y
(
)



 dt 
 dt   dt 
x(t ) = e − at  y (t )e at dt + Ke − at
=
x f (t )
+ xn (t )
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• Forced response to a forcing function, first-order circuit
Forcing Function,
y (t )
Forced Response, x f (t )
Constant,
y (t ) = M
x f (t ) = N , constant
Exponential
y (t ) = Me− bt
x f (t ) = Ne − bt
Sinusoid
y (t ) = M sin(t +  )
x f (t ) = A sin t + B cos t
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – review(C.S.)
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• Example 4.4: Find the current for
t  0 when vs (t ) = 10e−2t u (t ) V.
Assume the circuit is in steady state at t = 0−.
vs
di R
di
+ i=

+ 4i = 10e −2t , i (0) = 2A
dt L
L
dt
i (t ) = (−3e −4t + 5e −2t ) A, t  0.
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – review(C.S.)
• Example 4.5: Find the response
35
v(t ) for t  0 when v(0) = 0 and
is (t ) = (10sin 2t )u (t ) A.
C
di 1
dv v
+ i = is  0.5
+ = 10 sin 2t , v(0) = 0V
dt R
dt 4
160
 160 −t 2 40

v(t ) = 
e +
sin 2t −
cos 2t  V.
17
17
 17

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4. Basic Study of Mathematics (D.E.) – 1st order systems
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▪ 4.1.3 The time constant
- For a first-order system subject to a step input of size k.
x=
(
b0
k 1 − e −a0t / a1
a0
or
)
(
x = steady − state value  1 − e − a0t / a1
)
When t=(a1/a0), then e-1=0.37.
x = steady − state value  (1 − 0.37 )
In this time the output has risen to 0.63 of its steady-state value.
This time is called the “time constant” τ.
=
a1
a0
In a time of 2(a1/a0)=2τ, e-2=0.14 and so
x = steady − state value  (1 − 0.14)
In this time the output has risen to 0.86 of its steady-state value.
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 1st order systems
-
In terms of the time const. τ, we can write the eq.
describing the response of a first-order system as
(
x = steady − state value  1 − e −t /
The time const. τ is (a1/a0), thus we can write our
general form.
37
)
dx
a1 + a0 x = b0 y
dt
as

b
dx
+x= 0 y
dt
a0
But b0/a0 is the factor by which the input y is multiplied to give the
steady-state value. We can term it the “steady-state gain”.
Thus we denote this by Gss .
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 1st order systems
38
dx
 + x = Gss y
dt
-
v0 of a first-order system varies with time when
subject to a step input of 5V. τ is the time taken for
the output to change from 0 to 0.63 of its final steady-state value.
In this case this time is about 3s.
The steady-state output is 10V. Thus Gss is (steady-state
output/input)=10/5=2.
The differential eq. for a first-order system can be written as.

dx
+ x = Gss y
dt
Thus, for this system, we have
3
dvo
+ vo = 2vi
dt
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 2nd order systems
39
▪ 4.2 Second-order systems
d 2x
dx
m 2 + c + kx = F
dt
dt
where m is the mass, c the damping const. and k the spring const.
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 2nd order systems
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▪ 4.2.1 Natural response
- Consider a mass on the end of a spring with no damping without
being force. The output of the 2nd-order system is a continuous
oscillation(simple harmonic motion). Thus, suppose we describe
this oscillation by the eq.
x = A sin nt
where x is the displ. at a time t, A the ampl. and ωn the angular freq.
of the free undamped oscillations. Differentiating this gives
dx
= n A cos nt
dt
Differentiating a second time gives
d 2x
2
2
=
−

A
sin

t
=
−

n
n
nx
2
dt
This can be reorganized to give the differential eq.
d 2x
2
+

nx =0
2
dt
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4. Basic Study of Mathematics (D.E.) – 2nd order systems
41
If considering a restoring force of kx and thus
d 2x
m 2 = −kx
dt
This can be written as.
d 2x k
+ x=0
2
dt
m
Thus, comparing the two differential eqs., we must have
n2 =
k
m
and x = Asinωnt is the solution to the differential eq.
-
Now consider when we gave damping. The motion of the mass is
then described by
2
d x
dx
m 2 + c + kx = 0
dt
dt
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 2nd order systems
42
To solve this eq. xn=Aest. ⇒ dx/dt = Asest and dx/dt=As2est
Then,
mAs2e st + cAsest + kAest = 0
ms 2 + cs + k = 0
Thus xn=Aest can only be a solution provided the above eq. equals 0. This eq.
is called the “auxiliary equation”.
By using the formula for the roots of a quadratic eq.
Thus
2
2
s=
− c  c − 4mk
c
k
 c 
=−
 
−

2m
2m
2
m

 m
c
k  c2  k

 −
=−

2m
m  4mk  m
But ωn2=k/m and so, if we let ζ2=c2/4mk.
s = − n  n  2 − 1
(ζ is termed the damping factor)
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 2nd order systems
43
1. Over-damped
- ζ>1: two different real roots s1 and s2
s1 = − n + n  2 − 1
s2 = − n − n  2 − 1
xn = Ae s1t + Be s2t
2. Critically-damped
- ζ=1: two equal roots s1=s2=-ωn
xn = ( At + B )e −nt
Bio-inspired Control & Robot (BICAR) Lab.
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44
3. Under-damped
- ζ<1: two complex roots involving the square root of (-1)
s = − n  n  2 − 1 = − n  n − 1 1 −  2
s = − n  jn 1 −  2
if we let
 = n 1 −  2
then we can write s=-ζωn ± jω and so the two roots are
s1 = −n + j , s2 = −n − j
the solution is
(
xn = Ae(−n + j )t + Be(−n − j )t = e −nt Ae jt + Be− jt
)
e jt = cos t + j sin t , e − jt = cos t − j sin t
xn = e −nt ( A cos t + jA sin t + B cos t − jB sin t )
= e −nt ( A + B ) cos t + j ( A − B )sin t 
 xn = e − nt (P cos t + Q sin t )
Bio-inspired Control & Robot (BICAR) Lab.
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45
▪ 4.2.2 Response with a forcing input
- With an input F the differential eq. becomes
d 2x
dx
m 2 + c + kx = F
dt
dt
Let x = xn+xf
m
if we let
d 2 (xn + x f )
dt
2
+c
d (xn + x f )
dt
+ k (xn + x f ) = F
d 2 xn
dx
m 2 + c n + kxn = 0
dt
dt
then we must have
m
d 2xf
dt
2
+c
dx f
dt
+ kx f = F
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – 2nd order systems
46
To solve the forcing eq.,
m
d 2xf
dt
2
+c
dx f
dt
+ kx f = F
we need to consider a particular form of input signal.
Thus for a step input of size F at t=0, xf=A. (A is const.)
Here dxf/dt=0 and d2xf/dt2=0
Thus, when these are substituted in the differential eq.,
0 + 0 + kA = F and so A = F/k and xf=F/k.
The complete solution is
F
st
s t
x = Ae 1 + Be 2 +
for the critically damped system
x = ( At + B )e
for the under-damped system
x=e
−n t
−n t
k
F
+
k
F
(P cos t + Q sin t ) +
k
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4. Basic Study of Mathematics (D.E.) – 2nd order systems
47
When t ⟶ ∞, the above eq. all lead to the solution x=F/k.
This is the steady-state condition.
Thus a 2nd-order differential eq. in the form
d 2x
dx
a2 2 + a1 + a0 x = b0 y
dt
dt
has a natural frequency given by
n2 =
a0
a2
and a damping factor given by
a12
 =
4a2 a0
2
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4. Basic Study of Mathematics (D.E.) – 2nd order systems
48
▪ 4.2.3 Examples of 2nd–order systems
- Consider a series RLC circuit with R=100Ω, L=2.0H, C=20μF이다.
and a step input V, the current i in the circuit is given by
d 2i R di 1
V
+
+
i
=
dt 2 L dt LC
LC
Comparing the eq. with the general 2nd-order differential eq. of
d 2x
dx
a2 2 + a1 + a0 x = b0 y
dt
dt
then the natural angular frequency is given by
n2 =
1
1
=
LC 2.0  20 10−6
and so ωn = 158Hz. Comparison with the general 2nd-order eq. also
gives
2
2
2
−6
(
)
R
/
L
R
C
100

20

10
2=
=
=
4  (1 / LC ) 4 L
4  2.0
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49
Thus ζ = 0.16. Since ζ < 1, the system is under-damped. The
damped oscillation frequency ω is
 = n 1 −  2 = 158 1 − 0.162 = 156Hz
the sol. will be of the same form as
x=e
and so
−n t
F
(P cos t + Q sin t ) +
k
i = e −0.16158 t (P cos 156t + Q sin 156t ) + V
Since i=0 when t=0, then 0=1(P+0)+V. Thus P=-V. Since di/dt = 0
when t=0, then differentiating the above eq. and equating it to 0
gives
di
= e −nt (P sin t − Q cos t ) −  n e −nt (P cos t + Q cos t )
dt
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50
Thus 0=1(0-ωQ)-ζωn(P+0) and so
Q=
 n P
 V
0.16 158V
=− n =−
 −0.16V


156
Thus the sol. of the differential eq. is
i = V − Ve −25.3t (cos 156t + 0.16 sin 156t )
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – review(C.S.)
51
• A second order differential equation
d 2x
dx
a2 2 + a1
+ a0 x = f (t ), ai are constants and f (t ) is specified
dt
dt
• The complete response:
x(t ) = xn (t ) + x f (t )
natural response
forced response
• Natural response: the response when f (t ) = 0.
• Forced response: the response w.r.t. the forcing function.
• Natural response: exponential xn (t ) = Ae
Background: The exponential is the only function that is proportional to all
of its derivatives and integrals.
st
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52
• Natural response of the differential equation satisfies
d 2 xn
dxn
a2
+
a
+ a0 xn = 0
1
2
dt
dt
st
• Substituting xn (t ) = Ae
→ a2 As 2 e st + a1 Ase st + a0 Ae st = 0
→ (a2 s 2 + a1s + a0 ) Ae st = 0
→ (a2 s + a1s + a0 ) xn = 0
2
→ excluding trivial solution xn = 0
→ (a2 s 2 + a1s + a0 ) = 0
Alternatively, using operator
dn
s = n
dt
n
Characteristic equation
The characteristic equation is derived from the governing differential
equation for a circuit by setting all independent sources to zeros value and
assuming an exponential solution
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
53
• Natural response of the differential equation satisfies
d 2 xn
dxn
a2
+
a
+ a0 xn = 0
1
2
dt
dt
2
• Roots to the characteristic equation a2 s + a1s + a0 = 0
s1 =
− a1 + a12 − 4a2 a0
2a2
, s2 =
− a1 − a12 − 4a2 a0
2a2
• When there are two distinct roots, the natural response is of the form
xn (t ) = A1e s1t + A2e s2t , A1 and A2 are unknowns.
The roots of the characteristic equation contain all the information
necessary for determining the character of the natural response.
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – review(C.S.)
▪ Example 4.6: Find the natural response of
54
i2
di1
mesh1: − vs + 8i1 + 2
+ 4(i1 − i2 ) = 0
dt
di
mesh 2 : − 4(i1 − i2 ) + 1 1 = 0
dt
Using the operators,
(12 + 2 s )i1 − 4i2 = vs
−4i1 + (4 + s )i2 = 0
 i2 =
2vs
s 2 + 10 s + 16
( s 2 + 10 s + 16)i2 = 2vs
The roots of the characteristic eq. are
s1 = −2 and s2 = −8
xn (t ) = A1e−2t + A2e−8t
Bio-inspired Control & Robot (BICAR) Lab.
4. Basic Study of Mathematics (D.E.) – review(C.S.)
55
• A circuit containing one capacitor and one inductor is represented by
d 2 x(t )
dx(t )
2
+
2

+

0 x (t ) = f (t )
2
dt
dt
 : damping coefficient, 0 : resonant frequency
• Homogeneous differential equation: differential equation with
f (t ) = 0
• Parallel RLC circuit is chosen just for illustration
- You should be able to derive similar results for any circuits with two
energy storage elements
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56
• Differential equation for unforced parallel RLC circuit:
- KCL at Node v:
- VI relation for the inductor:
d 2 v 1 dv 1
C 2 +
+ v=0
dt
R dt L
- Characteristic equation: 2
1
1
s +
s+
=0
RC
LC
• Roots of the characteristic equation:
12
 1 
1
1 
s1 = −
+ 

 −
2 RC  2 RC 
LC 
2
• When
12
 1 
1
1 
, s2 = −
− 

 −
2 RC  2 RC 
LC 
2
s1  s2 , vn (t ) = A1e s1t + A2e s2t
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57
• Three cases for the roots of the characteristic equation:
12
 1  2
1
1 
s1 = −
+ 

 −
2 RC  2 RC 
LC 
12
 1  2
1
1 
, s2 = −
− 

 −
2 RC  2 RC 
LC 
1
1
2
 s1 = − +  −  , s2 = − −  −  , where  =
, 0 =
2 RC
LC
2
2
0
2
Case
2
0
Condition
Overdamped
Two real and distinct roots,
Critically damped
A real root with multiplicity 2,
 2  02
s1 = − +  2 − 02 , s2 = − −  2 − 02
s1 = s2 = −
Underdamped
 2 = 02
Two complex roots,
 2  02
s1 = − + jd , s2 = − − jd , where d = 02 −  2
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58
• Natural response unforced parallel RLC circuit given initial conditions
- DE:
d 2 v 1 dv 1
C 2 +
+ v=0
dt
R dt L
- Characteristic equation:
s2 +
1
1
s+
=0
RC
LC
• Overdamped case with initial conditions vn (0) = v(0) and i(0).
• Two relations:
v(t ) = vn (t ) = A1e s1t + A2e s2t
vn (t ) = A1e s1t + A2e s2t → t = 0 → vn (0) = A1 + A2
KCL
at the top node:
v(0)
dv
v(t )
dv(t )
+
i
(0)
+
C
=0
+ i (t ) + C
=0→t =0→
R
dt t =0
R
dt
v(0) i (0)
dvn
−
= s1 A1 + s2 A2 → s1 A1 + s2 A2 = −
RC
C
dt t =0
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59
• Example 4.7: Find the natural response of v(t) when
R = 2 / 3, L = 1H, C = 1/ 2F, v(0) = 10V, i(0) = 2A.
vn (t ) = (−14e−t + 24e−2t )V.
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60
• Parallel RCL circuit
• Critically damped case: double root for the characteristic equation
 s1 = s2
• Recall:
12
 1 
1
1 
s1 , s2 = −
 

 −
2 RC  2 RC 
LC 
2
 s1 = − +  2 − 02 , s2 = − −  2 − 02 ,
where  =
• Natural response:
1
1
, 02 =
2RC
LC
vn (t ) = ( A1t + A2 )e s1t
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61
• Example 4.8: Find the natural response for a parallel RCL circuit with
R = 1, L = 1H, C = 1/ 4F, v(0) = 5V, i(0) = −6A.
vn (t ) = e −2t (14t + 5)V.
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62
• Parallel RCL circuit
• Underdamped case: two complex roots for the characteristic equation
12
• Recall:
 1  2
1
1 
s1 , s2 = −
 

 −
2 RC  2 RC 
LC 
 s1 = − +  2 − 02 , s2 = − −  2 − 02 ,
where  =
1
1
, 02 =
2RC
LC
 when  2  02 , s1,2 = −  j 02 −  2
define d = 02 −  2  s1,2 = −  jd
d : damped resonant frequency
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63
• Underdamped case: two complex roots for the characteristic equation
• Natural response:
s1t
s2 t
2
2
vn (t ) = A1e + A2e
s1,2 = −  j 0 −  = −  jd
• For underdamped case,
Euler identity: e jt = cos t  j sin t
vn (t ) = A1e s1t + A2e s2t
= A1e − t e jd t + A2 e − t e − jd t
= e − t ( A1e jd t + A2 e − jd t )
= e − t ( B1 cos d t + B2 sin d t ), where B1 = A1 + A2 , B2 = j ( A1 − A2 )
• Natural response of underdamped case:
vn (t ) = e− t ( B1 cos d t + B2 sin d t )
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4. Basic Study of Mathematics (D.E.) – review(C.S.)
64
• Finding the coefficients in terms of initial conditions v(0), i (0) :
vn (t ) = e− t ( B1 cos d t + B2 sin d t )
At t = 0, vn (0) = B1
dvn (0)
= d B2 −  B1
dt
B1 = vn (0)
vn (0) = v(0),
v(t )
dv(t )
i (t ) +
+C
=0
R
dt
dv (0)
v(0) i (0)
 n
=−
−
dt
RC
C
B2 =
1
d
( B1 −
v(0) i (0)
−
)
RC
C
2
• Period of damped oscillation: Td =
sec
d
1
• Frequency in Hertz: f d =
Hz
Td
i (t ) +
v(t )
dv(t )
+C
=0
R
dt
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65
• Example 4.9: Find the natural response of the parallel RLC circuit when
R = 25 / 3, L = 0.1H, C = 1mF,
v(0) = 10V, i(0) = −0.6A.
vn (t ) = 10e−60t cos80t V
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66
• Forced response: a solution to the differential equation without any
arbitrary constants.
2
d
x
dx
• Differential equation:
+ a1
+ a0 x = f (t )
2
dt
dt
d 2xf
dx f
• Forced response
xf :
+ a1
+ a0 x f = f (t )
2
dt
dt
• Forcing functions and their associated assumed solution
Forcing Function
Assumed Solution
K
A
Kt
At + B
Kt 2
At 2 + Bt + C
K sin t
A sin t + B cos t
Ke− t
Ae− t
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67
• Example 4.10: Find the forced response for the inductor current i f when
is (t ) = 8e−2t A, R = 6, L = 7H, C = 1/ 42F.
I relation (I source after t = 0)
v
dv
KCL(top node): is = + i + C
,
R
dt
di
dv
d 2i
v=L 
=L 2
dt
dt
dt
i f (t ) = −12e −2t A.
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68
• Example 4.11: Find the forced response for the inductor current i fwhen
is (t ) = I 0 A, R = 6, L = 7H, C = 1/ 42F.
is
d 2i
1 di
1
d 2i
di
 2+
+
i=
 2 + 7 + 6i = 6 I 0
dt
RC dt LC
LC
dt
dt
i f (t ) = I 0 .
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69
• Special case: when the forcing function is of the same form as on the
components of the natural response.
d 2i
di
−6 t
+
7
+
6
i
=
6
i
,
i
(
t
)
=
3
e
.
s
s
2
dt
dt
• Example:
−t
−6 t
- in (t ) = A1e + A2 e
- Try with
i f (t ) = De −6t
- Try with
i f (t ) = Bte −6t
 0 = 18e −6t
B=−
18
18
,  i f = − te −6t
5
5
In general, if the forcing function is of the same form as one of the
p
components of the natural response xn1 , try x f (t ) = Bt xn1.
- Use the lowest power p that is not duplicated in the natural response.
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70
• Example: series RLC circuit
d 2v
dv
LC 2 + RC
+ v = vs
dt
dt
R = 5, L = 1H, C = 1/ 6F,
2e − t
dv
vs (t ) =
V, v(0) = 10V,
(0) = −2V/s.
3
dt
•
d 2v
dv
2
Characteristic equation:
+
5
+
6
v
=
6
v

s
+ 5s + 6 = 0
s
2
dt
dt
Natural response:
vn = A1e −2t + A2 e −3t
2
Forced response:
v f = Be − t  e − t ( B − 5B + 6 B ) = 6 e − t  B = 2
3
Complete response: v = vv + v f = A1e −2t + A2 e −3t + 2e − t
•
•
•
• Applying initial conditions:
v(0) = 10  10 = A1 + A2 + 2
dv dt = −2 (t = 0)  −2 A1 − 3 A2 − 2 = −2
 v = 24e −2t − 16e −3t + 2e − t V
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71
• Example 4.12 : series RLC circuit
dvs
d 2v
dv
+
7
+
10
v
=
+ 6vs
2
dt
dt
dt
v(0) = 6V, i (0)=1A 
dv
(0) = −4V/s ( KCL)
dt
• Characteristic equation:
•
•
•
•
Natural response:
Forced response:
Complete response:
Applying initial conditions:
v(0) = 6  6 = A1 + A2 − 9
dv dt = −4  −2 A1 − 5 A2 + 27 = −4
v =
44 −2t 1 −5t
e − e + 9e −3t V
3
3
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72
▪ 3.5 Performance measures for 2nd-order systems
- The response of under-damped 2nd-order system to a step input
-
The rise time tr: the time taken for the
response x to rise from 0 to xss, This is
the time for the oscillating response to
complete a ¼ of a cycle, i.e. ½π. Thus
1
2
t r = 
tr is specified as the time taken for
the response to rise from 10% to 90%.
- The peak time tp: the time taken for the response to rise from 0 to
the first peak value. This is the time for the oscillating response to
complete one half cycle, i.e. π. Thus
t p = 
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4. Basic Study of Mathematics (D.E.) – 2nd order systems
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-
73
The overshoot is max. amount by which the response overshoots the s-s value (often written as a percentage of the s-s
value).
For the under-damped oscillations,
x = e −nt (P cos t + Q sin t ) + steady − state value
Since x=0 when t=0, then 0=1(P +0) + xss. And so P=-xss. The
overshoot occurs at ωt=π and thus
x = e −n / ( −(− xss ) + 0 ) + xss
The overshoot is the difference between the output at that time
and the s-s value. Hence
(
)
overshoot = xss e −n / 
Since  = n 1 −  2 ,then
 −  
n
overshoot = xss exp 
  1−  2
 n

 − 
 = x exp 
ss

 1−  2






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74
Expressed as a percentage of xss
 − 
percentage overshoot = exp 
 1−  2

-

 100%


The subsidence ratio or decrement: how fast oscillation decay.
This is the amp. of the 2nd overshoot divided by that of the 1st
overshoot (1st overshoot: ωt=π, 2nd overshoot: ωt=3π).
 −  

first overshoot = xss exp 
 1−  2 


 − 3
second overshoot = xss exp 
 1−  2
and so


second overshoot
subsidence ratio =
= exp 

first overshoot





− 2 
1 −  2 
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4. Basic Study of Mathematics (D.E.) – 2nd order systems
-
75
The settling time ts: the time taken for the oscillations to die
away. The response to fall within and remain within some
specified percentage, e.g. 2% (amp. of the oscillation < 2% of xss).
x = e −nt (P cos t + Q sin t ) + steady − state value
as derived earlier, P=-xss. The amp. of the oscillation is (x-xss)
when x is a max. value (The max. value → ωt is kπ, cosωt=-1 and
sinωt=0). tss is when the max. amp. is 2% of xss, i.e. 0.02xss. Thus
0.02 xss = e −nts (xss 1 + 0)
Taking logarithms gives ln 0.02=-ζωnts and since ln 0.02=-3.9 or
approximately -4, then
ts =
4
 n
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76
If the percentage is 5%, the eq. becomes
ts =
3
 n
The period time is 1/f, where f is the frequency. Since ω=2πf, then
the time to complete on cycle is 2π/f.
settling time
number of oscillations =
periodic time
and thus for a ts defined for 2% of the s-s value.
4 /  n
number of oscillations =
2 / 
Since  = n (1 −  2 ) , then
2n 1 −  2 2
number of oscillations =
=
n

1

2
−1
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5. Understanding of Feedback System
78
• Transient and Steady State Response Analysis
In general, the input signals into a system are not known, and it would be difficult to
express them analytically in a form suitable for system analysis.
It is a frequent practice to apply a reference input into the
system and to study its response in the time domain.
The type of reference input applied does depend on the
purpose of the system:
for example,
if the system may be subjected to sudden disturbances,
a step reference input may be best;
if the system is used to follow a gradually changing situation,
a ramp input may be better.
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5. Understanding of Feedback System
79
• Feedback System (Closed-loop Control System)
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80
5. Understanding of Feedback System
Response of First Order Systems
A first order system has an associate differential equation of the form
ay + y = u, y(0) = 0.
Y (s) =
1
U ( s ).
as + 1
For transient response analysis, it is customary to use
a reference unit step function u(t) for which
1
1a
1
1
1
=
= −
.
U ( s) = . Y (s) =
(as + 1) s ( s + 1 a) s s s + 1 a
s
On taking the inverse Laplace transform, we obtain
y (t ) = 1 − e−t / a , (t  0).
the time constant of the system
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5. Understanding of Feedback System
81
−1
Notice that when t = a, then y (t ) = y (a) = 1 − e = 0.63.
The response is in two parts,
1) the transient part
e−t a ,
approaches 0 as t → .
2) the steady-state part 1
eventually (when t =  ) the output is 1
(that is equal to the input).
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5. Understanding of Feedback System
The Final Value Theorem
If all poles of sY(s) are in the left half of s-plane, then
lim y (t ) = lim sY ( s ).
t →
s →0
The Initial Value Theorem
It is always possible to determine the initial value of the time function f(t) from its
Laplace transform.
+
For any Laplace transform pair, lim sF ( s) = f (0 ).
s→
In contrast with the Final Value Theorem, the Initial Value Theorem can be applied to
any function F(s).
Homework #1
Prove the final and initial value theorems.
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83
5. Understanding of Feedback System
If the derivatives of the input are involved in the differential equation of the system,
ay + y = bu + u
(bs + 1)
(s + z)
Y (s) =
U (s) = K
U ( s ),
(as + 1)
( s + p)
When U(s) = 1/s,
K1
K2
Y ( s) =
−
,
s s+ p
where
where
K=
K1 = K
b
1
1
,z = , p = .
a
b
a
z
z− p
, K2 = K
.
p
p
y (t ) = K1 − K 2e− pt .
With the assumption that z > p > 0, this response is
shown as follows:
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5. Understanding of Feedback System
Response of Second Order Systems
Applying Newton’s law, we find
d2y
dy
M 2 = −  − ky + u (t ),
dt
dt
where k = spring constant,  = damping coefficient, y is the distance of the
system from its position of equilibrium, and it is assumed that y (0) = y (0) = 0.
d2y
dy
u (t ) = M 2 +  + ky
dt
dt
On taking Laplace transforms, we obtain
Y ( s) =
1
K
U
(
s
)
=
U ( s),
2
2
Ms + s + k
s + a1s + a2
where K =
1

k
, a1 = , a2 = .
M
M
M
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5. Understanding of Feedback System
Applying a unit step input, we obtain
K
Y (s) =
,
s( s + p1 )( s + p2 )
where
p1 , p2 : the poles of the transfer function G ( s) =
a1  (a12 − 4a2 )
p1 , p2 =
.
2
K
s 2 + a1s + a2
the zeros of the denominator of G(s)
Case 1.
a12  4a2
(overdamped system)
p1 , p2
are real and unequal.
K
K2
K3
Y ( s) = 1 +
+
,
s s + p1 s + p2
Y ( s) =
2
Case 2. a1 = 4a2
2
Case 3. 4a2  a1
(critically damped system)
(underdamped system)
p1 , p2
are real and equal.
p1 = p2 =
a1
= p( say)
2
are complex conjugate
p1 , p2 having the form
p1 , p2 =   i ,
where  =
a1
1
, =
(4a2 − a12 ) .
2
2
K
K1
K2
K3
K1
K2
K3
Y
(
s
)
=
+
+
,
=
+
+
,
s s + p1 s + p2
s( s + p)2
s s + p ( s + p)2
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86
The system error e(t) = y(t) - u(t)
The performance index






0
0
0
e 2 (t )dt ,
integral of error squared (IES)
| e(t ) | dt ,
integral of absolute error (IAE)
t | e(t ) | dt ,
integral of time multiplied absolute error criterion (ITAE)
Having chosen an appropriate performance index, the system which minimizes the
integral is called optimal. The objective of modern control theory is to design a system so
that it is optimal with respect to a performance index.
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5. Understanding of Feedback System
Steady State Error
87
ess
Depends on both the transfer function of the system and on the reference input.
R(t) the reference input, c(t) the output,
A(s) an open loop transfer function.
We define the error function e(t ) = r (t ) − c(t ).
Hence ess = lim e(t ).
t →
Since E(s) = R(s) – A(s)E(s), it follows that
and by the final value theorem
E (s) =
R( s)
1 + A( s )
sR( s)
.
s →0 1 + A( s )
ess = lim sE ( s) = lim
s →0
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88
(1) Step input, r(t) = ku(t) (k is a constant)
sk s
k
=
.
s →0 1 + A( s )
1 + lim A( s )
ess = lim
The position error constant K p
s →0
If k and ess are specified, the value of the transfer function at s = 0 is determined.
(2) Ramp input, r(t) = ktu(t)
k
k
k
k
R( s ) = 2 → ess = lim
, ess =
, Kv = ,
s →0 s (1 + A( s ))
s
Kv
ess
K v = lim sA( s )
s →0
(3) Parabolic input, r (t ) =
R( s) =
The velocity error constant.
1 2
kt u (t )
2
k
k
k
,
e
=
,
K
=
,
ss
a
3
s
Ka
ess
K a = lim s 2 A( s )
s →0
The acceleration error constant.
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90
Homework #2
Solve the below problem.
Find the error coefficient for the system having the
open loop transfer function
8
A( s ) =
.
s (4s + 2)
K p , Kv , Ka ?
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91
A( s)
G ( s) =
1 + A( s) B( s)
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5. Understanding of Feedback System
Example
a first order system
A( s) =
K
, B( s) = c
as + 1
(a constant)
G(s) =
K
K a
=
.
as + Kc + 1 s + Kc + 1
a
We take the inverse Laplace transform, that is, obtain the impulse response of the
system, where
(1) c = 0 (response of the open loop system)
g (t ) =
K −t a
e ,
a
(2) c  0
K −
g (t ) = e
a
where
Kc +1
t
a
=
t
K −
= e ,
a
a
.
Kc + 1
a and  are respectively the time-constants of the open loop and the closed loop systems.
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93
A( s) B( s)
G ( s) =
.
1 + A( s) B( s)
(1) The on-off controller
if e(t) > 0 then q(t) = Q1
and if e(t) < 0 then q(t) = Q2 ,
where q(t) is the output signal from the controller and Q1 and Q2 are some
constants.
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5. Understanding of Feedback System
(2) Proportional Controller
q(t ) = K p e(t ),
where K p is a constant, called the gain of the controller. B( s ) =
(3) Integral Controller
Q( s)
= K p.
E ( s)
t
q(t ) = K  e(t )dt , B( s) = K s .
0
(4) Derivative Controller
q(t ) = K
de
, B( s) = Ks.
dt
(5) Proportional-plus-derivative (PD) Controller
q (t ) = K p e(t ) + K1
K
de
, B (t ) = K p (1 + 1 s ).
dt
Kp
(6) Proportional-plus-integral (PI) Controller
t
q(t ) = K p e(t ) + K1  e(t )dt ,
0
t
de
+
K
K1 K p
2  e(t ) dt ,
0
dt
B( s ) = K p (1 +
).
s
K K
K
(7) Proportional-plus-derivative-plus-integral (PDI) B ( s ) = K p (1 + 1 s + 2 p ).
Kp
s
Controller
q (t ) = K p e(t ) + K1
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95
• Properties of a PID control
- Proportional feedback control reduces the error response to disturbances, but
results in nonzero steady-state error to constant inputs.
- A term proportional to the integral of the error eliminates the steady-state error to
constant inputs
- A term proportional to the derivative of the error improve the dynamic response
(anticipatory term).
Structure of PID control: B ( s ) = k P +
kI
+ kD s
s
kI
k p : proportional term, : integral term, k D : derivative term
s
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96
• Proportional Control (P)
u = k pe ,
U (s)
E (s)
= Bcl ( s ) = k p
- For 2nd order plant G( s ) =
A
,
2
s + a1 s + a2
Closed-loop characteristic equation:
1 + kpG( s ) = 0
s 2 + a1 s + a2 + k p A = 0

a1 
Note:

=
a
+
k
A

,

=


n
2
p
2n 

- The system with proportional control usually has a steady-state error in
respose to a constant reference input or to a constant disturbance input.
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97
• Proportional plus Integral Control (PI)
t
u ( t ) =k pe(t)+k I  e ( τ )dτ
U (s)
E(s)
t0
=Bcl ( s ) =k p +
kI
s
- Integral term raises the type to Type 1.
- The system can reject completely constant bias disturbances.
Ex. PI control in a speed control
Effects of PI control on the steady-state error to a step disturbance:
A
Y=
(U + W )
s +1
R −Y
PI control: U = k p ( R − Y ) + kI
s
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98
• Example: Speed control
Effects of PI control on the steady-state error to a step disturbance:
A
Y=
(U + W )
 s +1
R −Y
PI control: U = k p ( R − Y ) + k I
s
k
( s + 1) Y = A  k p + I  ( R − Y ) + AW
s 

( s + ( Ak + 1) s + Ak ) Y = A ( k s + k ) R + sAW
2
p
I
p
I
- Characteristic equation:
 s 2 + ( Ak p + 1) s + Ak I = 0
→ n = Ak I /  ,  = ( Ak p + 1) / 2n
→ May result in an unsatisfactory lightly damped response.
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5. Understanding of Feedback System
- For 2nd order plant G ( s ) =
Characteristic equation: 1 +
99
A
,
2
s + a1s + a2
k p s + kI
s
A
=0
2
s + a1s + a2
s 3 + a1s 2 + a2 s + Ak p s + Ak I = s 3 + a1s 2 + (a2 + Ak p )s + Ak I = 0
→ Controller parameters can be used to set only two of the coefficients.
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100
• Proportional-Integral-Derivative Control (PID)
de(t )
dt
kI
kP s + kI + kD s 2
U (s)
transfer function: Bcl ( s ) =
= kP + + kD s =
E (s)
s
s
t
u (t ) = k P e(t ) + k I  e( )d + k D
B term in feedback
B term in forward path
▶ With derivative in the feedback, the reference is not differentiated: can avoid
undesirable response to sudden change
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101
• Example: speed control with the second order plant
- For 2nd order plant G ( s ) =
A
,
2
s + a1s + a2
Characteristic equation: 1 + (k p +
kI
A
+ kD s) 2
=0
s
s + a1s + a2
s 3 + a1s 2 + a2 s + A(k p s + k I + k D s 2 ) = 0
→ s 3 + (a1 + Ak D ) s 2 + (a2 + Ak p ) s + Ak I = 0
→ Controller parameters can be used to set all coefficients.
→ The roots can be uniquely, and in theory, arbitrarily determined.
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5. Understanding of Feedback System
• PID Control of Motor Speed
(( L s + R )( J
a
a
m
s + b) + K t K e   m = K tVa + K wW )
Plant Parameters: see textbook.
Controller parameters: k p = 3, k I = 15 sec −1 , k D = 0.3 sec
Step disturbance
Step reference
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Homework #3
Solve the below problem.
Design a controller for a plant having the transfer function
A( s) =
1
s( s + 2)
so that the resulting closed loop system has a zero steady state error to a reference
ramp input.
What kind of controller are you going to build?
Explain your reason for choosing that controller.
Bio-inspired Control & Robot (BICAR) Lab.