frequency domain modelling - university of...

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)


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


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


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


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


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


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−=


•  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

∞

−∞∞ −

−∞

= −

= −

∫∫


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

∞

−∞∞

−∞

= −

= −

=

∫∫


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



•  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

+=== +



•  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

∞ −

−∞

∞ −

−∞

=

=

=

∫

∫


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#


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


Laplace Transform Theorems

•  The Laplace Transform is a linear transformation between functions in the t domain and s domain


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=


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


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


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


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


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


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


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( )



•  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



•  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


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


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

+ + =

=+ +


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

=+ +

=+ +



•  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 =−

= =+



•  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



•  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− −= − +


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;!


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


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



•  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


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


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


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


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


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


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


Further Reading

•  Nise – Section 2.1-2.3 and 4.1-4.2

•  Franklin & Powell – Section 3.1

frequency domain modelling - university of...

Documents