me 3253 linear systems theoryccao/teaching_fall08_me3253-s2/review2.pdf · me 3253 linear systems...

ME 3253

Linear Systems Theory

Midterm Review

1. System modeling (Mechanical system,

Electrical systems, water level system, etc)

2. Transient performance of first and second 2. Transient performance of first and second

order system (curve shape, overshot,

settling time)

3. Routh’s Stability Criterion

4. Frequency domain analysis: Bode plots

Mechanical Systems

Friction

Static friction: not moving

Sliding friction: translational motion (sliding friction

coefficient smaller than static friction coefficient)

Rolling friction: friction force magnitude unknown.

Rolling without sliding:

x Rθθθθ====

Example of Mechanical

Systems

1. Dynamic equation (ODE)

2. Laplace domain

2.1 Input, output signal

2.2 Transfer function

3. Laplace transformation of input signal

4. Laplace transformation of output signal

5. Inverse Laplace transformation ���� Output

signal

Properties:

1. Stability (real part of components)

2. Natural frequency (imaginary part of

components)

3. Initial/End values

Laplace Transform

Differentiation Theorem

[ ]df

Ldt

( ) (0)sF s f−−−−

2( ) (0) (0)s F s sf f− −− −− −− − &&&&

2

2[ ]d f

Ldt2

[ ]Ldt

Final Value Theorem: if exists

0lim ( ) lim ( )t s

f t sF s→∞ →→∞ →→∞ →→∞ →

====

lim ( )t

f t→∞→∞→∞→∞

Initial Value Theorem:

(0) lim ( )s

f sF s→∞→∞→∞→∞

====

Electrical systems

Basic Laws (continue):

For single loop circuit, loop law is enough;

For circuits with multiple loops, need to use both loop

law and current law

Dynamic modeling of circuits:

combination of circuit laws and basic element

definitions (what does this mean?? Read Example

Figure 6.10 very very carefully 9)

Examples 6-1 and 6-2. pp 261-263 of Ogata

Electrical Systems

Z R====

Z Ls====

Complex impedances: directly write system equations

in Laplace domain

( ) ( ) ( )E s Z s I s====

e iR==== E IR====di

e L==== Z Ls====

1e idtC

==== ∫∫∫∫Advantage: can be used in parallel and series circuit

modeling, just like resistances

Only suitable for Transfer Function Derivation (why??)

die L

dt==== E LsI====

1Z

Cs====1

E ICs

====

How to Solve Electrical systems

1. Loop law (Voltage equations)

2. Node law (Current equations)

3. Basic element definitions

In Laplace domain: a set of linear equations:

Laplace transformation of input signals.

Solve to get Laplace transformation of output signal.

Inverse-Laplace transformation.

Initial values?

Basic Elements

1. Resister

2. Capacitor

3. Inductor

4. Source (Voltage and Current)

Sign conventionSign convention

+_

i

e+

_

i

e+

_

i

e

Consider Initial Conditions

Initial Conditions

1∫∫∫∫

e iR====di

e Ldt

====

RsIsE )()( =

))0()(()( issILsE −=

( ))0()()( essECsI −=1e idtC

==== ∫∫∫∫ ( ))0()()( essECsI −=

Current Law

Kirchhoff’s current law: (node law) the sum of all

currents entering and leaving a node is zero (sign

convention: entering + leaving -)

Loop Law

Basic Laws (continue):

Kirchhoff’s voltage law: (loop law) the sum of voltage

around any loop is zero. Sign convention:

1. Choose a loop direction

2. A rise in voltage is positive2. A rise in voltage is positive

going through an active element from negative to

positive terminal – a rise in voltage

Example

Example problem of Electrical System – Review

Prob B-6-5

E.O.M. – Loop law and node law

+ − =+ − =+ − =+ − =1 1 2 1 2

( ) 0R i R i i

+ − =+ − =+ − =+ − =∫∫∫∫ 2 2 2 1

1( ) 0i dt R i i

C

Solving for the Integral Equation using Laplace

Transform (Integration theorem: page 27)

Plug in initial condition:

And solve9.

+ − =+ − =+ − =+ − =∫∫∫∫ 2 2 2 1( ) 0i dt R i i

C

+ − =+ − =+ − =+ − =1 1 2 1 2( ) [ ( ) ( )] 0R I s R I s I s

−−−−

+ + − =+ + − =+ + − =+ + − =1

2 2

2 2 1

( ) (0)1[ ] [ ( ) ( )] 0I s i

R I s I sC s s

−−−−===== = == = == = == = =∫∫∫∫1

2 2 0 0(0) ( ) | (0)

ti i t dt q e C

System Modeling

First order examples of water level system, hydraulic

system, thermal system.

Concept of linearization covered in class.

For a nonlinear system, how to perform linearization

around equilibrium point to get a LTI system

Transient Performance

Step response of first and second order systems

• Steady state value

• Settling time

• Time constant• Time constant

• Overshot

• Damping rate of second order system

• Natural frequency of second order system

Given transfer function

Stable: All roots of A(s) in left half plane.

Routh’s Stability Criterion

)(

)()(

sA

sBsT =

How to check stability without solving A(s)=0.

Pg. 539 in the text book.

1. Write

2.

Routh’s Stability Criterion: Methodology

)(

)()(

sA

sBsT =

n

nn asasasA +++= − ....)( 1

10

.00 >aLet2. .00 >aLet

.,..,1,0 niai =>No

Yes

Not Stable.

End.

Step 3

3.

Routh’s Stability Criterion: Methodology

1

50412

1

30211

2

7531

1

6420

.......:

.....:

.....:

baabbaab

a

aaaab

a

aaaabs

aaaas

aaaas

n

n

n

−−

−=

−=−

−

Build pattern

1

0

1

31511

1

21311

3

:

:

:

gs

b

baabc

b

baabcsn

−=

−=−

Stable: First column >0

[Unstable Roots (with positive real parts) equal

To the number of sign changes in first column]

Frequency analysis: Bode Plots

Given transfer function )(sT

)()()( sInsTsOut =

Stable system: Sinusoidal Input -���� Sinusoidal output in steady state

ωω

φωω

))( of (Magnitude ,)(

)sin()(:stateSteady

)sin()(

jTjTM

tMtOut

ttIn

=

+=

=

Bode plots: X axis: frequency in log-scale

Magnitude:

Phase:

( )φφωωωφ

ωω

sincos)( :equivalent isIt

))( of (Angle )),((

))( of (Magnitude ,)(

jMjT

jTjTangle

jTjTM

+=

=

=

))((

)(log20log20

ωφ

ω

ω

jTangle

jTM

=

=

Example: Bode Plots of First Order System

as

asT

+=)(First order system:

( )( ) ( )( )( )+=+

−+

++=

+

−+

+++

=

+−

=+−+

+−=

+=

222222222222

22

22

sincos)(

)()(

)(

))((

)()(

ωφωφωω

ω

ωωω

ω

ωωω

ωω

ωωω

ωω

jM

aj

a

a

a

a

aj

a

a

a

aa

a

jaa

ajaj

aja

aj

ajT

( )( ) ( )( )( )

( )

+

−=

+=

+=

2222arcsin,)(

where

sincos)(

ω

ωωφ

ωω

ωφωφω

aa

aM

jM

Frequency response:

( ) ( )( )ωφωω

ω

+=

=

tMtOut

ttIn

sin)(

:responseoutput stateSteady

)sin()( :Input

Example: Bode Plots of First Order System

Bode plots:

Magnitude plot: Plot with respect to

Phase plot: Plot with respect to

( ))(log20 ωM ωlog

)(ωφ ωlog

ωωω 0log20log20))(log(20 ,0when

222=

≈

+=→

a

a

a

aM

Magnitude plot:

( )ωωω

ωω

ω

log20)log(20

log20log20))(log(20 ,when 222

−=

=

≈

+=∞→

+

a

a

a

aM

aa

Use these two lines to approximate bode plot:

Example: Bode Plots of First Order System

( ))(log20 ωM

0

20−

( )ωωω

log20)log(20))(log(20:2 Line

0))(log(20:1 Line

−=

=

aM

M

ωlogωω log10=

1− 0 1 2 31.0 1 10 100 1000

40−

60−a

Intersection point: a=ω

Reason: ( ) . when ,0log20)log(20 aa ==− ωω

Example: Bode Plots of First Order System

( ))(log20 ωM

0

20−

( )ωωω

log20)log(20))(log(20:2 Line

0))(log(20:1 Line

−=

=

aM

M

ωlogωω log10=

1− 0 1 2 31.0 1 10 100 1000

40−

60−a

Intersection point: a=ω

Reason: Line 1 is correct when is small. Line 2 is correct when is large ω ω

Red line: Approximated bode plot

Example: Bode Plots of First Order System

( ))(log20 ωM

0

20−

( )ωωω

log20)log(20))(log(20:2 Line

0))(log(20:1 Line

−=

=

aM

M

ωlogωω log10=

1− 0 1 2 31.0 1 10 100 1000

40−

60−a

Intersection point: a=ω

Red line: Approximated bode plot

Blue line: True bode plot

Example: Bode Plots of First Order System

Bode plots:

Magnitude plot: Plot with respect to

Phase plot: Plot with respect to

( ))(log20 ωM ωlog

)(ωφ ωlog

( ) 00

0arcsinarcsin ,0when

2222 ω

ωωφω =

+

−≈

+

−=→

aa

Phase plot:

( ) ( )

( )

42

2arcsin

arcsinarcsin ,when

21arcsinarcsin ,when

0

2222

22

2222

π

ω

ωωφω

π

ω

ωωφω

ω

−=

−=

=

+

−=

+

−==

−=−≈

+

−=∞→

+

+

aa

a

aa

a

aa

Example: Bode Plots of First Order System

ωlogωω log10=

1− 0 1 2 31.0 1 10 100 1000

)(ωφ0

4

π−

a

4

point: a=ω

2

π−