Upload
others
View
14
Download
0
Embed Size (px)
Citation preview
1
Amme 3500 : System Dynamics and Control
Frequency Domain Modelling
Dr. Stefan B. Williams
Slide 2 Dr. Stefan B. Williams Amme 3500 : Introduction
Course Outline Week Date Content Assignment Notes 1 1 Mar Introduction 2 8 Mar Frequency Domain Modelling 3 15 Mar Transient Performance and the s-plane 4 22 Mar Block Diagrams Assign 1 Due 5 29 Mar Feedback System Characteristics 6 5 Apr Root Locus Assign 2 Due 7 12 Apr Root Locus 2 8 19 Apr Bode Plots No Tutorials 26 Apr BREAK 9 3 May Bode Plots 2 Assign 3 Due 10 10 May State Space Modeling 11 17 May State Space Design Techniques 12 24 May Advanced Control Topics 13 31 May Review Assign 4 Due 14 Spare
Slide 3 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
A Familiar Mechanical Example • In the last lecture, we
considered this mechanical system
• We derived a differential equation defining this system
• How do we solve for y(t)? M
f(t)
y(t) F =m!!y!m!!y = f (t)"Ky(t)"Kd !y(t)m!!y+Kd !y(t)+Ky(t) = f (t)
Slide 4 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
How do we find y(t)?
• As we saw, the input-output relationship is usually expressed in terms of a differential equation
h(t) u(t) y(t)
• If we can describe the characteristics of the plant using a general function, h(t), we can compute the output y(t) given some arbitrary input u(t)
1
1 0 01
( ) ( ) ( )( ) ( )n n m
n mn n m
d y t d y t d u ta a y t b b u tdt dt dt
−
− −+ + + = + +L L
2
Slide 5 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
LTI Systems
• In this course, we will consider Linear Time Invariant Systems – Linear : the output of
the system is equal to the sum of the input responses
– Time Invariant : the system’s dynamic characteristics are time independent
1
1( )
n
iin
ii
y y
f u
=
=
=
=
∑
∑
y
u1
u2
u3
Plant
Slide 6 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
The Unit Impulse • Recall the impulse function, d
(t) • In the limit, we can represent
an arbitrary function as a sum of impulses
• By the principal of superposition, if we can find the response of the system to an impulse, we will be able to find the response to an arbitrary input
( ) ( ) ( )f t d f tτ δ τ τ∞
−∞− =∫
f(t)
t
Slide 7 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Impulse Response
• The impulse response represents the system response to an impulse input
• We will examine techniques for calculating the impulse response shortly
• We usually denote the impulse response of a system as h(t)
1
1 01
( ) ( ) ( ) ( )n n
nn n
d y t d y ta a y t tdt dt
δ−
− −+ + + =L
Slide 8 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Convolution Integral • If we consider the input
to a system as a sum of impulses, we can solve for the system output by integrating the signal with its impulse response over time
• This yields the system response to an arbitrary input signal u(t)
( ) ( ) ( )
( ) ( )* ( )
y t u t h d
ory t u t h t
τ τ τ∞
−∞= −
=
∫
3
Slide 9 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Solving in time
• Assume we have a system described by the following differential equation
• The impulse response for this system is
• We’ll see how we can find the impulse response shortly
!y (t )+ ky (t ) = u (t )
( ) kth t e−=
Slide 10 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
• If we want to find the response of the system to a sinusoidal input, sin(wt)
Example : Solving in time
( ) ( ) ( )
sin( ( ))k
y t h u t d
e t dτ
τ τ τ
ω τ τ
∞
−∞∞ −
−∞
= −
= −
∫∫
Slide 11 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Cascaded systems
• Now what happens if we have multiple components in the system?
• It clearly becomes difficult to manipulate these integrals beyond the simplest cases
h (t) u(t) y1(t)
h1(t) y(t)
1
1 1
1
( ) ( ) ( )
( ) ( ) ( )
( ) ( ( )* ( ))* ( )
y t u t h d
y t y t h d
y t u t h t h t
τ τ τ
τ τ τ
∞
−∞∞
−∞
= −
= −
=
∫∫
Slide 12 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
The input signal as an Exponential
• If we further assume that the input signal can be represented by an exponential, est
• The constant s may be complex, expressed as
( )
( ) ( ) ( )
( )
( )
s t
st s
y t u t h d
e h d
e e h d
τ
τ
τ τ τ
τ τ
τ τ
∞
−∞
∞ −
−∞
∞ −
−∞
= −
=
=
∫∫∫
!
s = " + j#
4
Slide 13 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
The input signal as an Exponential
• This provides us with a rich basis function for describing functions
• Through Euler’s formula, we find
• Fourier analysis tells us that this is sufficient for representing any signal
( )
(cos sin )
st j t
t j t
t
e ee ee t j t
σ ω
σ ω
σ ω ω
+=== +
Slide 14 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
The input signal as an Exponential
• Now let us replace the integral term by the constant H(s)
( ) ( )
( )
( ) ( )
st s
st
s
y t e e h d
H s e
where H s e h d
τ
τ
τ τ
τ τ
∞ −
−∞
∞ −
−∞
=
=
=
∫
∫
Slide 15 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Laplace Transform
• In general
0( ) ( ) stF s f t e dt
−
∞ −= ∫
1( ) ( )2
j st
jf t F s e ds
jσ
σπ+ ∞
− ∞= ∫
Where
!
s = " + j#
Slide 16 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Table of Laplace Transforms • What about that nasty
integral in the Laplace operation?
• We normally use tables of Laplace transforms rather than solving the preceding equations directly
• This greatly simplifies the transformation process
5
Slide 17 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Laplace Transform Theorems
• The Laplace Transform is a linear transformation between functions in the t domain and s domain
Slide 18 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
How does this help us? • It turns out that
convolution in time is equivalent to multiplication in the Laplace domain
• This means that the convolution operation can be replaced by algebraic manipulation
{ } { }( ) ( )* ( )( ) ( ) ( )y t u t h tY s U s H s
==
L L
H(s) U(s) Y1(s)
H1(s) Y(s)
1( ) ( ) ( ) ( )Y s U s H s H s=
Slide 19 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
How does this help us?
• If the Laplace transform of the input can also be found, then the Laplace transform of the output is simply the product of these two fractions
• We can then use partial fraction expansion to expand the response as a sum of simpler terms whose inverse LT can be found in the tables
Slide 20 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Comparing Solution Methods
• Starting with an impulse response, h(t), and an input, u(t), find y(t)
u(t), h(t)
U(s), H(s)
convolution y(t)
L L-1
Multiplication, algebraic manipulation
Y(s)
6
Slide 21 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Transfer Function
• The transfer function H(s) of a system is defined as the ratio of the output of the system and input with zero initial conditions
• This is effectively the Laplace transform of the impulse response for the system
Slide 22 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Transfer Function
• Recall the system we examined earlier • To find the transfer function for this
system, we perform the following steps
!y (t )+ ky (t ) = u (t )
!h (t )+ kh (t ) = !(t )sH (s )+ kH (s ) =1
H (s ) = 1s + k ( ) kth t e−=or
Slide 23 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Forced vs. Natural Response
• The output response of a system is the sum of two responses – The natural, or homogeneous, response
describes the way the system dissipates or acquires energy. The form of this response is dependent only on the system, not the input
– The forced, or particular, response represents the system response to a forcing function
• Together these elements determine the overall response of the system
Slide 24 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
A Familiar Mechanical Example
M!!y (t )+ Kd !y (t )+ Ky (t ) = f (t )
• Earlier, we considered this mechanical system
• Now we can attempt to find y(t)
M
f(t)
y(t)
M s 2Y (s )! sy (0)! !y (0)( )+ Kd sY (s )! y (0)( )+ KY (s ) = F (s )
7
Slide 25 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example: Natural Response
• The natural response of the system can be found by assuming no input force
• Taking inverse Laplace transform will give us the system response
!
Ms2 + Kd s + K( )Y (s) = Ms + Kd( )y(0) + M˙ y (0)
Y (s) =Ms + Kd( )y(0) + M˙ y (0)
Ms2 + Kd s + K( )
Slide 26 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example: Natural Response
• Suppose we wish to find the natural response of the system to an initial displacement of 0.5m
• Assume the mass of the system is 1kg and the systems constant kd is 4 Nsec/m and k is 3 N/m
Slide 27 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example: Natural Response
• By Partial Fraction expansion we find ( )
( )( )
( )( )
( ) ( )
2
3
0.5 4( )
4 3
0.5 41 3
3 14 41 3
( ) 0.75 0.25t t
sY s
s s
ss s
s s
y t e e− −
+=
+ +
+=
+ +
= −+ +
= −0 1 2 3 4 5 6 7 8 9 10
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Time(s)
y(m
)
Natural Response
Slide 28 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example: Natural Response • What if we change the spring constant to
20 N/m? ( )
( )( )
( )( )
( ) ( )( )
2
2
2 2
2
0.5 4( )
4 20
0.5 4( )
2 16
2 20.52 16 2 16
( ) 0.5 cos 4 0.5sin 4t
sY s
s s
scompleting squares
s
ss s
y t e t t−
+=
+ +
+=
+ +
⎛ ⎞+= +⎜ ⎟⎜ ⎟+ + + +⎝ ⎠= +
0 1 2 3 4 5 6 7 8 9 10-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
Time(s)
y(m
)
Natural Response
8
Slide 29 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Forced Response
• Assuming zero initial conditions, the forced response will be
• This is the transfer function for this system
( )2
2
( ) ( )
( ) 1( )
d
d
Ms K s K Y s F s
Y sF s Ms K s K
+ + =
=+ +
Slide 30 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Forced Response • Suppose we wish to find the response of
the system to a unit step input force f(t) • Assume the mass of the system is 1kg
and the systems constant kd is 4 Nsec/m and k is 3 N/m
2
( ) 1( ) 4 3
1 1( )( 1)( 3)
Y sF s s s
Y ss s s
=+ +
=+ +
Slide 31 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Forced Response
• We have found • We wish to write Y(s)
in terms of its partial-fraction expansion
31 2( )1 3
CC CY ss s s
= + ++ +
1( )( 1)( 3)
Y ss s s
=+ +
10
1 1( 1)( 3) 3s
Cs s =
= =+ +
21
1 1( 3) 2s
Cs s =−
= = −+
33
1 1( 1) 6s
Cs s =−
= =+
Slide 32 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Forced Response
• So
• We now take inverse Laplace transforms of these simple transforms
1 1 1( )3 2( 1) 6( 3)
Y ss s s
= − ++ +
31 1 1( )3 2 6
t ty t e e− −= − +
9
Slide 33 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Forced Response
• This represents the step response for this system
• This plot shows the response y(t) in time
Step Response
Time (sec)
Ampl
itude
0 1 2 3 4 5 60
0.05
0.1
0.15
0.2
0.25
0.3
0.3531 1 1( )3 2 6
t ty t e e− −= − +
Slide 34 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Using Matlab to verify result %!% Example 1: Step response for !% Y(s)/F(s) = 1/(s^2 + 4s + 3)!%!sys = tf(1,[1,4,3]);!step(sys);!hold on;!pause;!!%!% Solving for time domain response yields!% y(t) = 1/3 - 1/2*exp(-t) + 1/6*exp(-3*t)!%!t = 0:0.1:6;!y_t = 1/3 - 1/2*exp(-t) + 1/6*exp(-3*t);!plot(t, y_t, 'r');!pause;!
Slide 35 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Poles, Zeros and System Response • The poles of a transfer function are the
values of the LT variables, s, that cause the transfer function to become infinite
• The zeros of a transfer function are the values of the LT variables, s, that cause the transfer function to become zero
• A qualitative understanding of the effect of poles and zeros on the system response can help us to quickly estimate performance
Slide 36 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
The s-plane
1( )( 1)( 3)
Y ss s s
=+ +
31 1 1( )3 2 6
t ty t e e− −= − +
Re{s}
Im{s}
x x x
• Recall
10
Slide 37 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Forced Response
• What if we change the spring constant k to 20 N/m?
2
2
( ) 1( ) 4 20
1 1 1 1( )4 20 ( 2 4 )( 2 4 )
Y sF s s s
Y ss s s s s j s j
=+ +
= =+ + + + + −
( )2
2 2
( 2) 16
1 4 20
1 1 1, ,20 20 5
A Bs Cs s
A s s Bs Cs
A B C
+= ++ +
= + + + +
= = − = −Slide 38 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
The s-plane
1( )( 2 4 )( 2 4 )
Y ss s j s j
=+ + + − ( )21 1( ) cos4 4sin 4
20 20ty t e t t−= − +
Re{s}
Im{s}
x
x
x
• Now
Slide 39 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example : Forced Reponse
• Notice that the step response has changed in response to the change in the modified parameter k
• The steady state response and transient behaviour are different
Step Response
Time (sec)
Ampl
itude
0 0.5 1 1.5 2 2.5 30
0.01
0.02
0.03
0.04
0.05
0.06
0.07
Slide 40 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Using Matlab to verify result %!% Example 1: Step response for !% Y(s)/F(s) = 1/(s^2 + 4s + 20)!%!sys = tf(1,[1,4,20]);!step(sys);!hold on;!pause;!!
11
Slide 41 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Qualitative Effect of Poles
• A first order system without zeros can be described by:
• The resulting impulse response is
• When – σ > 0, pole is located at s < 0,
exponential decays = stable – σ < 0, pole is at s > 0,
exponential grows = unstable
1( )( )
H ss σ
=+
( ) th t e σ−=
( )1( ) 1 ty t e σ
σ−= −
Step Response
Slide 42 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
The s-plane • From the preceding we
can conclude that in general, if the poles of the system are on the left hand side of the s-plane, the system is stable
• Poles in the right hand plane will introduce components that grow without bound
Re{s}
Im{s}
stable unstable
Slide 43 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Example: Automobile Suspension
• Consider the modelling of an automobile suspension system.
• Wheel can be modelled as a stiff spring with the suspension modelled as a spring and damper in parallel
• Interested in describing the motion of the mass in response to changing conditions of the road
M
y(t)
u(t)
Ks Kd
Kw
Slide 44 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Conclusions
• We have presented the motivation behind solving systems of differential equations
• We have reviewed the Laplace Transform method for solving differential equations describing Linear Time Invariant systems
• We have given a number of examples and have begun to investigate the effect of poles and zeros on the system response
12
Slide 45 Dr. Stefan B. Williams AMME 3500 : Frequency Domain Modelling
Further Reading
• Nise – Section 2.1-2.3 and 4.1-4.2
• Franklin & Powell – Section 3.1