signals and systems laplace transform analysis of

Laplace Transform Analysis ofSignals and Systems

5/10/04 M. J. Roberts - All Rights Reserved 2

Transfer Functions

• Transfer functions of CT systems can befound from analysis of– Differential Equations

– Block Diagrams

– Circuit Diagrams


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


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


A circuit can even be described by a block diagram.

Transfer Functions


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


The mechanical system can also be described by a blockdiagram.

Time Domain Frequency Domain

Transfer Functions


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


System InterconnectionsCascade

Parallel


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


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


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

( ) = ( )( ) = −

−



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

( )( ) =

− ( )

− +

( )+ −

( )


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




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 . .



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



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.



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).


Analysis of Feedback SystemsStable Oscillation Using Feedback

Prototype Feedback

System

Feedback System Without

Excitation


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).



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



Laser action begins when a photon is spontaneously emitted fromthe pumped medium in a direction normal to the mirrors.




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.




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.



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



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


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 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



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



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



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



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

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



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


.

.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


σ = −( )∑ ∑1m

finite poles finite zeros


Analysis of Feedback SystemsRoot Locus Examples


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



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.


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.



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



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



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.



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)


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


Block Diagram ReductionMoving a Summer


Block Diagram Reduction

Combining Two Summers



Move Pick-Off Point



Move Summer



Combine Summers



Combine Parallel Blocks



Combine Cascaded Blocks



Reduce Feedback Loop



Combine Cascaded Blocks


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:


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( )


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


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



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


Let . Then the unit step response is

Unit Step Response

H sA

sp

( ) =−1

y ut A e tpt( ) = −( ) ( )1




Let

Η sA

s s( ) =

+ +ω

ζω ω02

20 0

22

(pp. 710-712)

Unit Step Response



Let Η sA

s s( ) =

+ +ω

ζω ω02

20 0

22

Unit Step Response


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


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



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{



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



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


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


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.


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


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


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


Standard Realizations of SystemsCanonical Form

The system can then be realized in this form,


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


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,


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


Standard Realizations of SystemsParallel Form


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


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



• 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


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


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( ) = ( )



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



A block diagram for the system can be drawn directly from the state and output equations.



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


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



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( )



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

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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


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

. .

. .



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


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=



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=

signals and systems laplace transform analysis of

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