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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−
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