Signals and Systems Laplace Transform Analysis of
Transcript of Signals and Systems Laplace Transform Analysis of
Laplace Transform Analysis ofSignals and Systems
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Transfer Functions
• Transfer functions of CT systems can befound from analysis of– Differential Equations
– Block Diagrams
– Circuit Diagrams
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A circuit can be described by a system of differentialequations
R t Lddt
tddt
t tg1 1 1 2i i i v( )+ ( )( ) − ( )( )
= ( )
Lddt
tddt
tC
d R tc
t
i i i v i2 1 2 2 2
0
10 0( )( ) − ( )( )
+ ( ) + ( ) + ( ) =−
−∫ λ λ
Transfer Functions
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Using the Laplace transform, a circuit can be described bya system of algebraic equations
R s sL s sL s sg1 1 1 2I I I V( )+ ( )− ( ) = ( )
sL s sL ssC
s R sI I I I2 1 2 2 2
10( )− ( )+ ( )+ ( ) =
Transfer Functions
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A circuit can even be described by a block diagram.
Transfer Functions
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A mechanical system canbe described by a system ofdifferential equations
or a system of algebraic equations.
f x x x xt K t K t t m td s( )− ′ ( )− ( )− ( )[ ] = ′′( )1 1 1 2 1 1
K t t K t m ts s1 1 2 2 2 2 2x x x x( )− ( )[ ] − ( ) = ′′( )
F X X X X
X X X X
s K s s K s s m s s
K s s K s m s s
d s
s s
( )− ( )− ( )− ( )[ ] = ( )( )− ( )[ ] − ( ) = ( )
1 1 1 2 12
1
1 1 2 2 2 22
2
Transfer Functions
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The mechanical system can also be described by a blockdiagram.
Time Domain Frequency Domain
Transfer Functions
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System Stability
• System stability is very important
• A continuous-time LTI system is stable if itsimpulse response is absolutely integrable
• This translates into the frequency domain asthe requirement that all the poles of thesystem transfer function must lie in the openleft half-plane of the s plane (pp. 675-676)
• “Open left half-plane” means not includingthe ω axis
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System InterconnectionsCascade
Parallel
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Feedback
E X H Ys s s s( ) = ( )− ( ) ( )2
Y H Es s s( ) = ( ) ( )1
HYX
HH H
sss
ss s
( ) = ( )( ) = ( )
+ ( ) ( )1
1 21
E(s) Error signal
Forward path transfer function or the“plant”
Feedback pathtransfer function or the“sensor”.
Loop transfer function
H1 s( )
H2 s( )
T H Hs s s( ) = ( ) ( )1 2
HH
Ts
ss
( ) = ( )+ ( )
1
1
System Interconnections
T H Hs s s( ) = ( ) ( )1 2
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Analysis of Feedback SystemsBeneficial Effects
HH
sK
K s( ) =
+ ( )1 2
If K is large enough that then . This
means that the overall system is the approximate inverse of the system in the feedback path. This kind of system can be usefulfor reversing the effects of another system.
K sH2 1( ) >> HH
ss
( ) ≈ ( )1
2
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Analysis of Feedback Systems
A very important example of feedback systems is anelectronic amplifier based on an operational amplifier
Let the operational amplifier gain be
HVV1
0
1s
ss
Asp
o
e
( ) = ( )( ) = −
−
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Analysis of Feedback Systems
The amplifier can be modeled as a feedback system with thisblock diagram.
The overall gain can be written as
VV
Z
Z Z
0 0
01 1
ss
A s
sp
A ssp
si
f
i f
( )( ) =
− ( )
− +
( )+ −
( )
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If the operational amplifier low-frequency gain, , is verylarge (which it usually is) then the overall amplifier gainreduces at low-frequencies to
the gain formula based on an ideal operational amplifier.
VV
Z
Z0 s
s
s
si
f
i
( )( ) ≅ −
( )( )
A0
Analysis of Feedback Systems
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Analysis of Feedback Systems
The change in overall system gain is about 0.001% for a changein open-loop gain of a factor of 10.
The half-power bandwidth of the operational amplifier itself is15.9 Hz (100/2π). The half-power bandwidth of the overall
amplifier is approximately 14.5 MHz, an increase in bandwidthof a factor of approximately 910,000.
If and thenA p j j0710 100 100 9 999989 0 000011= = − −( ) = − +H . .
If and thenA p j j0610 100 100 9 99989 0 00011= = − −( ) = − +H . .
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Analysis of Feedback Systems
Feedback can stabilize an unstable system. Let a forward-pathtransfer function be
This system is unstable because it has a pole in the right half-plane. If we then connect feedback with a transfer function,K, a constant, the overall system gain becomes
and, if K > p, the overall system is now stable.
H ,1
10s
s pp( ) =
−>
H ss p K
( ) =− +
1
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Analysis of Feedback Systems
Feedback can make an unstable system stable but it can alsomake a stable system unstable. Even though all the polesof the forward and feedback systems may be in the open lefthalf-plane, the poles of the overall feedback system can bein the right half-plane.
A familiar example of this kind of instability caused byfeedback is a public address system. If the amplifier gainis set too high the system will go unstable and oscillate,usually with a very annoying high-pitched tone.
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Analysis of Feedback Systems
Public Address SystemAs the amplifier gain isincreased, any soundentering the microphonemakes a stronger soundfrom the speaker until, atsome gain level, thereturned sound from thespeaker is a large as theoriginating sound into themicrophone. At that pointthe system goes unstable(pp. 685-689).
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Analysis of Feedback SystemsStable Oscillation Using Feedback
Prototype Feedback
System
Feedback System Without
Excitation
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Analysis of Feedback SystemsStable Oscillation Using Feedback
Can the response be non-zero when theexcitation is zero? Yes, if the overallsystem gain is infinite. If the systemtransfer function has a pole pair on theωaxis, then the transfer function is infiniteat the frequency of that pole pair and there can be a responsewithout an excitation. In practical terms the trick is to be sure thepoles stay on the ω axis. If the poles move into the left half-plane
the response attenuates with time. If the poles move into the righthalf-plane the response grows with time (until the system goes non-linear).
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Analysis of Feedback Systems
A real example of a system that oscillates stably is a laser.In a laser the forward path is an optical amplifier.
The feedback action is provided by putting mirrors at eachend of the optical amplifier.
Stable Oscillation Using Feedback
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Analysis of Feedback Systems
Laser action begins when a photon is spontaneously emitted fromthe pumped medium in a direction normal to the mirrors.
Stable Oscillation Using Feedback
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Analysis of Feedback Systems
If the “round-trip” gain of thecombination of pumped lasermedium and mirrors is unity,sustained oscillation of light willoccur. For that to occur thewavelength of the light must fit intothe distance between mirrors aninteger number of times.
Stable Oscillation Using Feedback
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Analysis of Feedback Systems
A laser can be modeled by a block diagram in which the K’srepresent the gain of the pumped medium or the reflection ortransmission coefficient at a mirror, L is the distance betweenmirrors and c is the speed of light.
Stable Oscillation Using Feedback
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Analysis of Feedback SystemsThe Routh-Hurwitz Stability Test
The Routh-Hurwitz Stability Test is a method fordetermining the stability of a system if its transfer functionis expressed as a ratio of polynomials in s. Let thenumerator be N(s) and let the denominator be
D s a s a s a s aDD
DD( ) = + + + +−
−1
11 0L
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Analysis of Feedback Systems
The first step is to construct the “Routh array”.
The Routh-Hurwitz Stability Test
D a a a a
D a a a
D b b b
D c c c
d d
e
f
D D D
D D D
D D D
D D D
− −
− − −
− − −
− − −
−−−
2 4 0
1 3 5
2 4 6
3 5 7
2 0
1
0
1 0
2 0
3 0
2 0 0 0
1 0 0 0 0
0 0 0 0 0
L
L
L
L
M M M M M M
D even
D a a a a
D a a a a
D b b b
D c c c
d d
e
f
D D D
D D D
D D D
D D D
− −
− − −
− − −
− − −
−−−
2 4 1
1 3 5 0
2 4 6
3 5 7
2 0
1
0
1
2 0
3 0
2 0 0 0
1 0 0 0 0
0 0 0 0 0
L
L
L
L
M M M M M M
D odd
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The first two rows contain the coefficients of the denominator polynomial. The entries in the following row are found by the formulas,
The Routh-Hurwitz Stability Test
b
a a
a a
aD
D D
D D
D−
−
− −
−
= −2
2
1 3
1
b
a a
a a
aD
D D
D D
D−
−
− −
−
= −4
4
1 5
1
...
The entries on succeeding rows are computed by the same process based on previous row entries. If there are any zeros or sign changes in the column, the system is unstable. The number of sign changes in the column is the number of poles in the right half-plane (pp. 693-694).
aD
Analysis of Feedback Systems
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Analysis of Feedback SystemsRoot Locus
Common Type of Feedback System
System Transfer Function HH
H Hs
K sK s s
( ) = ( )+ ( ) ( )
1
1 21
Loop Transfer Function T H Hs K s s( ) = ( ) ( )1 2
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Analysis of Feedback SystemsRoot Locus
Poles of H(s) Zeros of 1 + T(s)
T is of the form TPQ
s Kss
( ) = ( )( )
Poles of H(s) Zeros of 1 + Kss
PQ
( )( )
Poles of H(s)
Q Ps K s( )+ ( ) = 0
orQ
Ps
Ks
( ) + ( ) = 0
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Analysis of Feedback SystemsRoot Locus
K can range from zero to infinity. For K approaching zero,using
the poles of H are the same as the zeros of whichare the poles of T. For K approaching infinity, using
the poles of H are the same as the zeros of whichare the zeros of T. So the poles of H start on the poles of Tand terminate on the zeros of T, some of which may be atinfinity. The curves traced by these pole locations as K isvaried are called the root locus.
Q Ps K s( )+ ( ) = 0
Q s( ) = 0
QP
sK
s( ) + ( ) = 0
P s( ) = 0
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Analysis of Feedback SystemsRoot Locus
Let and let .
Then
No matter how large K getsthis system is stable because the poles always lie in the left half-plane (although for large K the system may bevery underdamped).
H1 1 2s
Ks s
( ) =+( ) +( ) H2 1s( ) =
T sK
s s( ) =
+( ) +( )1 2 RootLocus
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Analysis of Feedback SystemsRoot Locus
RootLocus
Let
and let .
At some finite value of Kthe system becomesunstable because twopoles move into the righthalf-plane.
H2 1s( ) =
H1 1 2 3s
Ks s s
( ) =+( ) +( ) +( )
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Analysis of Feedback Systems
1. Each root-locus branch begins on a pole of T andterminates on a zero of T.
2. Any portion of the real axis for which the sum of thenumber of real poles and/or real zeros lying to its right onthe real axis is odd, is a part of the root locus.
3. The root locus is symmetrical about the real axis...
Four Rules for Drawing a Root Locus
Root Locus
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.
.4. If the number of finite poles of T exceeds the number of finitezeros of T by an integer, m, then m branches of the root locusterminate on zeros of T which lie at infinity. Each of these branchesapproaches a straight-line asymptote and the angles of theseasymptotes are at the angles,
with respect to the positive real axis. These asymptotes intersect onthe real axis at the location,
km
kπ
, , , ,...=1 3 5
Analysis of Feedback SystemsRoot Locus
σ = −( )∑ ∑1m
finite poles finite zeros
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Analysis of Feedback SystemsRoot Locus Examples
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Analysis of Feedback SystemsGain and Phase Margin
Real systems are usually designed with a margin of error toallow for small parameter variations and still be stable.
That “margin” can be viewed as a gain margin or a phasemargin.
System instability occurs if, for any real ω,
a number with a magnitude of one and a phase of -πradians.
T jω( ) = −1
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Analysis of Feedback SystemsGain and Phase Margin
So to be guaranteed stable, a system must have a T whosemagnitude, as a function of frequency, is less than one whenthe phase hits -π or, seen another way, T must have a phase, as
a function of frequency, more positive than - π for all |T|
greater than one.
The difference between the a magnitude of T of 0 dB and themagnitude of T when the phase hits - π is the gain margin.
The difference between the phase of T when the magnitudehits 0 dB and a phase of - π is the phase margin.
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Analysis of Feedback SystemsGain and Phase Margin
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Analysis of Feedback SystemsSteady-State Tracking Errors in Unity-Gain Feedback Systems
A very common type of feedback system is the unity-gainfeedback connection.
The aim of this type of system is to make the response“track” the excitation. When the error signal is zero, theexcitation and response are equal.
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Analysis of Feedback SystemsSteady-State Tracking Errors in Unity-Gain Feedback Systems
The Laplace transform of the error signal is
The steady-state value of this signal is (using the final-value theorem)
If the excitation is the unit step, , then the steady-state error is
EXH
ss
s( ) = ( )
+ ( )1 1
lim e lim E limXHt s s
t s s ss
s→∞ → →( ) = ( ) = ( )
+ ( )0 011
A tu( )
lim e limHt s
tA
s→∞ →( ) =
+ ( )011
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Analysis of Feedback SystemsSteady-State Tracking Errors in Unity-Gain Feedback Systems
If the forward transfer function is in the common form,
then
If and the steady-state error is zero and theforward transfer function can be written as
which has a pole at s = 0.
H11
12
21 0
11
22
1 0
sb s b s b s b s ba s a s a s a s a
NN
NN
DD
DD( ) = + + + +
+ + + +−
−
−−
L
L
lim e limt s
NN
NN
DD
DD
tb s b s b s b s ba s a s a s a s a
aa b→∞ → −
−
−−
( ) =+ + + + +
+ + + +
=+0
11
22
1 0
11
22
1 0
0
0 0
1
1L
L
a0 0= b0 0≠
H11
12
21 0
11
22 1
sb s b s b s b s b
s a s a s a s aN
NN
N
DD
DD( ) = + + + +
+ + +( )−
−
−−
−L
L
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Analysis of Feedback SystemsSteady-State Tracking Errors in Unity-Gain Feedback Systems
If the forward transfer function of a unity-gain feedbacksystem has a pole at zero and the system is stable, thesteady-state error with step excitation is zero. This typeof system is called a “type 1” system (one pole at s = 0 inthe forward transfer function). If there are no poles ats = 0, it is called a “type 0” system and the steady-stateerror with step excitation is non-zero.
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Analysis of Feedback SystemsSteady-State Tracking Errors in Unity-Gain Feedback Systems
The steady-state error with ramp excitation is
Infinite for a stable type 0 system
Finite and non-zero for a stable type 1 system
Zero for a stable type 2 system (2 poles at s = 0 inthe forward transfer function)
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Block Diagram ReductionIt is possible, by a series of operations, to reduce a complicated block diagram down to a single block.
Moving a Pick-Off Point
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Block Diagram ReductionMoving a Summer
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Block Diagram Reduction
Combining Two Summers
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Block Diagram Reduction
Move Pick-Off Point
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Block Diagram Reduction
Move Summer
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Block Diagram Reduction
Combine Summers
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Block Diagram Reduction
Combine Parallel Blocks
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Block Diagram Reduction
Combine Cascaded Blocks
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Block Diagram Reduction
Reduce Feedback Loop
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Block Diagram Reduction
Combine Cascaded Blocks
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Mason’s Theorem
Number of Paths from Input to Output - N p
Number of Feedback Loops - NL
Transfer Function of ith Path from Input to Output - Pi s( )
Loop transfer Function of ith Feedback Loop - Ti s( )
∆ s s s s s s sii
N
i jij
i j ki j k
L
( ) = + ( )+ ( ) ( )+ ( ) ( ) ( )+=∑ ∑ ∑1
1
T T T T T Tth loopandth loop not
sharing a signal
th, th, th loops not
sharing a signal
L
Definitions:
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Mason’s Theorem
The overall system transfer function is
HP
ss s
s
i ii
N p
( ) =( ) ( )
( )=∑ ∆
∆1
where is the same as except that all feedbackloops which share a signal with the ith path, , areexcluded.
∆ i s( ) ∆ s( )Pi s( )
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Mason’s Theorem
P1
10s
s( ) =
P2
13
ss
( ) =+
N p = 2 NL =1
∆ ss s s s
( ) = ++
= ++( )1
1 38
13
8∆ ∆1 2 1s s( ) = ( ) =
HP
ss s
ss s
s s
s s
s s s
i ii
N p
( ) =( ) ( )
( ) =+
++
+( )= +( ) +( )
+( ) + +( )=∑ ∆
∆1
2
10 13
13
8
8 11 303 8 3
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System Responses to Standard Signals
HND
sss
( ) = ( )( )Let be proper in s. Then the Laplace transform
of the unit step response is
Y HND
ND
s ss
s sss
Ks
( ) = ( ) = ( )( ) = ( )
( ) +−11
If the system is stable, the inverse Laplace transform ofis called the transient response and the steady-state
response is .
ND
1 ss( )( )
H 0( )s
K = ( )H 0
Unit Step Response
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System Responses to Standard Signals
HND
sss
( ) = ( )( )Let be proper in s. If the Laplace transform of the
excitation is some general excitation, X(s), then the Laplace transform of the response is
YND
XND
ND
ND
ND
sss
sss
ss
ss
ss
x
x
x
x
( ) = ( )( ) ( ) = ( )
( )( )( ) = ( )
( ) + ( )( )
1 1
same polesas system
same polesas excitation
123 123
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Let . Then the unit step response is
Unit Step Response
H sA
sp
( ) =−1
y ut A e tpt( ) = −( ) ( )1
System Responses to Standard Signals
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System Responses to Standard Signals
Let
Η sA
s s( ) =
+ +ω
ζω ω02
20 0
22
(pp. 710-712)
Unit Step Response
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System Responses to Standard Signals
Let Η sA
s s( ) =
+ +ω
ζω ω02
20 0
22
Unit Step Response
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System Responses to Standard SignalsH
ND
sss
( ) = ( )( )Let be proper in s. If the excitation is a suddenly-
applied, unit-amplitude cosine, the response is
which can be reduced and inverse Laplace transformed into(pp. 713-714)
If the system is stable, the steady-state response is a sinusoid ofsame frequency as the excitation but, generally, a differentmagnitude and phase.
YND
sss
ss
( ) = ( )( ) +2
02ω
y
ND
H cos H utss
j t j t( ) = ( )( )
+ ( ) + ∠ ( )( ) ( )−L 1 10 0 0ω ω ω
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Pole-Zero Diagrams andFrequency Response
If the transfer function of a system is H(s), the frequency response is H(jω). The most common type of transfer function
is of the form,
Therefore H(jω) is
H s As z s z s z
s p s p s pN
D
( ) =−( ) −( ) −( )−( ) −( ) −( )
1 2
1 2
L
L
H j Aj z j z j z
j p j p j pN
D
ω ω ω ωω ω ω
( ) =−( ) −( ) −( )−( ) −( ) −( )
1 2
1 2
L
L
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Pole-Zero Diagrams andFrequency Response
Let H ss
s( ) =
+3
3
H jj
jω ω
ω( ) =
+3
3
The numerator, jω, and the
denominator, jω + 3, can be
conceived as vectors in the s plane.
H jj
jω ω
ω( ) =
+3
3∠ ( ) = ∠ + ∠ − ∠ +( )
=H j j jω ω ω3 3
0{
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Pole-Zero Diagrams andFrequency Response
lim H limω ω
ω ωω→ →+ +
( ) =+
=0 0
33
0jj
jlim H lim
ω ωω ω
ω→ →− −( ) =
+=
0 03
30j
jj
lim H limω ω
ω ωω→+∞ →+∞
( ) =+
=jj
j3
33lim H lim
ω ωω ω
ω→−∞ →−∞( ) =
+=j
jj
33
3
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Pole-Zero Diagrams andFrequency Response
lim Hω
ω π π→ +
∠ ( ) = − =0 2
02
jlim Hω
ω π π→ −
∠ ( ) = − − = −0 2
02
j
lim Hω
ω π π→−∞
∠ ( ) = − − −
=j
2 20 lim H
ωω π π
→+∞∠ ( ) = − =j
2 20
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Butterworth FiltersThe squared magnitude of the transfer function of an nth order, unity-gain, lowpass Butterworth filter with a corner frequency of 1 radian/s is
This is called a normalizedButterworth filter becauseits gain is normalized to one and its corner frequencyis normalized to 1 radian/s.
H j nωω
( ) =+
2
2
11
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Butterworth FiltersA Butterworth filter transfer function has no finite zeros andthe poles all lie on a semicircle in the left-half plane whoseradius is the corner frequency in radians/s and the anglebetween the pole locations is always π/n radians.
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Butterworth FiltersFrequency Transformations
A normalized lowpass Butterworth filter can be transformedinto an unnormalized highpass, bandpass or bandstop Butterworthfilter through the following transformations (pp. 721-725).
Lowpass to Highpass
Lowpass to Bandpass
Lowpass to Bandstop
ss
c→ ω
sss
L H
H L
→ +−( )
2 ω ωω ω
ss
sH L
L H
→−( )
+ω ω
ω ω2
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Standard Realizations of Systems
There are multiple ways of drawing a system block diagramcorresponding to a given transfer function of the form,
HYX
,sss
b s
a s
b s b s b s bs a s a s a
ak
k
k
N
kk
k
NN
NN
N
NN
N N( ) = ( )( ) = = + + + +
+ + + +==
=
−−
−−
∑
∑0
0
11
1 0
11
1 0
1L
L
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Standard Realizations of SystemsCanonical Form
The transfer function can be conceived as the product of twotransfer functions,
and H
YX1
1
11
1 0
1s
ss s a s a s aN
NN( ) = ( )
( ) =+ + + +−
− L
HYY2
11
11 0s
ss
b s b s b s bNN
NN( ) = ( )
( ) = + + + +−− L
5/10/04 M. J. Roberts - All Rights Reserved 72
Standard Realizations of SystemsCanonical Form
The system can then be realized in this form,
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Standard Realizations of SystemsCascade Form
The transfer function can be factored into the form,
and each factor can be realized in a small canonical-formsubsystem of either of the two forms,
and these subsystems can then be cascade connected.
H s As zs p
s zs p
s zs p s p s p s p
N
N N N D
( ) = −−
−−
−− − − −+ +
1
1
2
2 1 2
1 1 1L L
5/10/04 M. J. Roberts - All Rights Reserved 74
Standard Realizations of SystemsCascade Form
A problem that arises in the cascade form is that some polesor zeros may be complex. In that case, a complex conjugatepair can be combined into one second-order subsystem of theform,
5/10/04 M. J. Roberts - All Rights Reserved 75
Standard Realizations of SystemsParallel Form
The transfer function can be expanded in partial fractions ofthe form,
Each of these terms describes a subsystem. When all thesubsystems are connected in parallel the overall system isrealized.
H sK
s pK
s pK
s pD
D
( ) =−
+−
+ +−
1
1
2
2
L
5/10/04 M. J. Roberts - All Rights Reserved 76
Standard Realizations of SystemsParallel Form
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State-Space Analysis• In larger systems it is important to keep the
analysis methods systematic to avoid errors
• One popular method for doing this is throughthe use of state variables
• State variables are signals in a system which,together with the excitations, completelycharacterize the state of the system
• As the system changes dynamically the statevariables change value and the system moveson a trajectory through state space
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State-Space Analysis
• State space is an N-dimensional space whereN is the order of the system
• The order of a system is the number of statevariables needed to characterize it
• State variables are not unique, there can bemultiple correct sets
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State-Space Analysis
• There are several advantages to state-spaceanalysis– Reduction of the probability of analysis errors
– Complete description of the system signals
– Insight into system dynamics
– Can be formulated using matrix methods and thesystem state can be expressed in two matrixequations
– Combined with transform methods it is verypowerful
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State-Space AnalysisTo illustrate state-space methods, let the system be this circuit
Let the state variables be the capacitor voltage and the inductorcurrent and let the output signals be v i .out Rt t( ) ( )and
5/10/04 M. J. Roberts - All Rights Reserved 81
State-Space AnalysisTwo differential equations in the two state variables, called the state equations, characterize the circuit,
Two more equations called the output equations define the responses in terms of the state variables,
′ ( ) = ( )i vL CtL
t1 ′ ( ) = − ( )− ( )+ ( )v i v iC L C int
Ct
GC
tC
t1 1
v vout Ct t( ) = ( ) i vR Ct G t( ) = ( )
5/10/04 M. J. Roberts - All Rights Reserved 82
State-Space Analysis
The state equations can be written in matrix form as
and the output equations can be written in matrix form as
′ ( )′ ( )
=− −
( )( )
+
( )[ ]i
v
i
viL
C
L
Cin
t
tL
CGC
t
t Ct
01
1
01
v
i
i
viout
R
L
Cin
t
t G
t
tt
( )( )
=
( )( )
+
( )[ ]0 1
0
0
0
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State-Space Analysis
A block diagram for the system can be drawn directly from the state and output equations.
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State-Space Analysis
The state and output equations can be written compactly as
where
′( ) = ( )+ ( )q Aq Bxt t t
y Cq Dxt t t( ) = ( )+ ( )
q tt
tL
C
( ) =( )( )
i
vA =
− −
01
1L
CGC
B =
01C
x t tin( ) = ( )[ ]i
y tt
tout
R
( ) =( )
( )
v
iC =
0 1
0 GD =
0
0
5/10/04 M. J. Roberts - All Rights Reserved 85
State-Space AnalysisThe solution of the state and output equations can be foundusing the Laplace transform.
or
Multiplying both sides by
The matrix, , is conventionally designated by thesymbol, . Then
s s s sQ q AQ BX( )− ( ) = ( )+ ( )+0
s s sI A Q BX q−[ ] ( ) = ( )+ ( )+0
sI A−[ ]−1
Q I A BX qs s s( ) = −[ ] ( )+ ( )[ ]− +1 0
sI A−[ ]−1
Φ s( )
Q BX q BX qs s s s s s( ) = ( ) ( )+ ( )[ ] = ( ) ( )+ ( ) ( )+
−
+
−
Φ Φ Φ0 0zero stateresponse
zero inputresponse
1 24 34 1 24 34
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State-Space Analysis
The time-domain solution is then
where and is called the state transition matrix.
q Bx qt t t t( ) = ( )∗ ( )+ ( ) ( )−
+
−
φ φzero stateresponse
zero inputresponse
1 24 34 1 24 340
φ t s( )← → ( )L Φ φ t( )
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State-Space Analysis
To make the example concrete, let the excitation and initialconditions be
and let
i ut A t( ) = ( ) q 00
0
0
1+
+
+( ) =( )( )
=
i
vL
C
R C L= = =13
1 1, ,
Φ s ss
L
Cs
GC
sGC L
Cs
sGC
sLC
( ) = −( ) =−
+
=
+
−
+ +
−
−
I A 1
1
2
1
1
1
1
1
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State-Space AnalysisSolving for the states in the Laplace domain,
Q ssLC s
GC
sLC
L sGC
sLC
C sGC
sLC
s
sGC
sLC
( ) =+ +
++ +
+ +
++ +
11
11
11 1
2 2
2 2
Substituting in numerical component values,
Q s s s s s s
s ss
s s
( ) = + +( ) ++ +
+ ++
+ +
13 1
13 1
13 1 3 1
2 2
2 2
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State-Space AnalysisInverse Laplace transforming, the state variables are
and the response is
or
q te e
e et
t t
t t( ) =
− −+
( )
− −
− −
1 0 277 0 723
0 723 0 277
2 62 0 382
0 382 2 62
. .
. .u
. .
. .
y q xtG
e e
e et
t t
t t( ) =
+
=
− −+
( )
− −
− −
0 1
0
0
0
0 1
0 3
1 0 277 0 723
0 723 0 277
2 62 0 382
0 382 2 62
. .
. .u
. .
. .
y te e
e et
t t
t t( ) =
++
( )
− −
− −
0 723 0 277
2 169 0 831
0 382 2 62
0 382 2 62
. .
. .u
. .
. .
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State-Space Analysis
In a system in which the initial conditions are zero (thezero-input response is zero), the matrix transfer functioncan be found from the state and output equations.
or
The response is
and the matrix transfer function is
s s s sQ q AQ BX( )− ( ) = ( )+ ( )+
=
0
0123
Q I A BX BXs s s s s( ) = −[ ] ( ) = ( ) ( )−1 Φ
Y CQ DX C BX DX C B D Xs s s s s s s s( ) = ( )+ ( ) = ( ) ( )+ ( ) = ( ) +[ ] ( )Φ Φ
H C B Ds s( ) = ( ) +Φ
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State-Space AnalysisAny set of state variables can be transformed into anothervalid set through a linear transformation. Let be theinitial set and let be the new set, related to by
Then
and using ,
where
q1 t( )q2 t( ) q1 t( )
q Tq2 1t t( ) = ( )
′ ( ) = ′ ( ) = ( )+ ( )( ) = ( )+ ( )q Tq T A q B x TA q TB x2 1 1 1 1 1 1 1t t t t t t
q T q11
2t t( ) = ( )−
′ ( ) = ( )+ ( ) = ( )+ ( )−q TA T q TB x A q B x2 11
2 1 2 2 2t t t t t
A TA T2 11= − B TB2 1=
5/10/04 M. J. Roberts - All Rights Reserved 92
State-Space Analysis
By a similar argument,
where
Transformation to a new set of state variables does not changethe eigenvalues of the system.
y C q D xt t t( ) = ( )+ ( )2 2 2
C C T2 11= − D D2 1=