complex analysis, differential equations, and laplace transform

1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – Mathematical Topics

Complex Analysis, Differential Equations, and Laplace Transform

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell


iyxjyxz +=+=

1ji −==x)zRe( =

Where:

Complex Analysis

y)zIm( =Real part of the functionImaginary part of the function

111 jyxz +=

222 jyxz +=( ) ( )212121 yyjxxzz +++=+

Addition


Complex Numbers – Multiplication and Division

)y x y j(x) y – yx (x · zz 12212121 21 ++=2

12

111 y x * · zz +=onjugate complex c* where: z1 =

jy x zif: x – jy * and: z

1

1

+==

Multiplication

Division

( ) ( )2

22

2

12212121

22

22

22

11

22

11

yxxyxyjyyxx

jyxjyx

jyxjyx

jyxjyx

+

−++=

−−

∗++

=++


Example – Complex Number Multiplication

Perform the following multiplication and express the result in rectangular form.

( )( )2j35j2 −+−

Solution: Treating the two complex numbers as binomials, the product is obtained as

10j15j4j6 2−++−=( )( )2j35j2 −+−

1019j6 ++−= 19j4+=


Example – Complex Number DivisionPerform the following division of complex numbers and express the result in rectangular form.

5j23j1

++−

Solution:

2911j

2913

2911j13

5j25j2

5j23j1

+=+

=−−

⋅++−


Example – Complex Conjugate

Given: Evaluate: j21z +−= *zz

Solution:

( )( )j21j21*zz −−+−=

( ) ( ) 521 22 =+−=


Location of Complex Conjugates in the Complex Plane

jyx ±−

Complex Conjugates


jyxzsinrycosrx

+=θ=θ=

Complex Numbers in Polar Form


( ) θ

θ

=θ+θ=

θ+θ=+=θ+θ=

j

j

ersinjcosrz

sinrjcosrjyxzsinjcose

Polar Form - Euler’s Formula

{ } { }

( )directionccwinpositivexytan

yxzImzRezr

1

2222

θ

=θ

+=+==

−


Example

Express the complex number

in polar form.

3j3z +=

3212z ==63

3tan 1 π=

=θ −

Solution:

π

= 6j

e32z


ExampleDetermine the location and the phase angle of the complex number:

j12+−

Solution: Express this number in standard rectangular form by multiplying its numerator and denominator by the conjugate of the denominator.

j111

j22j1j1

j12

−−=+−−

=−−−−

⋅+−

Location = third quadrant Phase Angle = +225º or -135º


Multiplication and Division (Polar Form)

Multiplication

( )21

21

j2121

j22

j11

errzz

erz;erzθ+θ

θθ

=

==

Division( )21j

2

1

2

1 err

zz θ−θ=

Complex Conjugate

2jj

j1

rrere*zz

re*z11

1

=⋅=⋅

=θ−θ

θ−


( )nmwhere)s(D)s(N

bsbsbsasasasK)s(G

js

n1n1n

1n

m1m1m

1m

<=

++++++++

=

ω+σ=

−−

−−

K

K

Complex FunctionsComplex Variable, s


Complex FunctionThe complex function can be expressed in POLE-ZERO form as:

( )( ) ( )( )( ) ( )pnspsps

zszszsK)s(G21

m21−−−−−−

=K

K

Often this can be written is partial fraction form as:

n

n

2

2

1

1ps

aps

aps

a)s(G−

+−

+−

= K

The roots of the numerator are referred to as ZEROS.The roots of the denominator are referred to as POLES.


ExampleExpress the given complex function in pole-zero form. Identify the zeros and the poles, as well as the multiplicity of each.

( ) ( )3s102ss1s2)s(G 2 ++

+=

Solution: G(s) can be written in pole-zero form as:

( )( ) ( )3.0s2ss10

5.0s2)s(G 2 ++

+=

Simple zero: s=-0.5Two simple poles: s=0 and s=-0.3Double pole: s=-2

( ) ( )3.0s2ss5.0s

51

2 ++

+

=


( )0for)t(fx1x >τ=τ

+&

Differential Equations1st Order ODE

τ is the time constant

ζ - damping ratiocc – critical dampingωn – natural frequency

)t(fxx2x 2nn =ω+ςω+ &&&

mk;m2c;

cc 2

nncc

=ωω==ζ

2nd Order ODE


2nd order – 2 integrals/derivatives – 2 constants

Initial Value Problem

Constant Coefficients – time independent

x is dependent variable – t is independent

If coefficients do not depend on x, then the equations are linear (linear superposition possible)

Has Xh and Xp – homogeneous & particular

2nd Order ODE


Example – Types of ODEs

( )( )

( ) 0x9x1xxtxx2

t3sinxtsinx0x2x1t2xtsin2x9x3x

0xx2

2

=+++=+

=+=+−+

=++=+

&&&

&

&

&&&

&&&

&

If the coefficients are constants or functions of t, the ODE is linear. Otherwise it is non-linear.

*Highest derivative identifies the order

First-order, linear with constant coefficientsSecond-order, linear with constant coefficientsSecond-order, linear

First-order, linear

First-order, nonlinear

Second-order, nonlinear


Example – 1st Order ODE

Consider the single-tank, liquid-level system shown in the figure below. The mathematical model of this system is given by the following first-order, linear ODE with constant coefficients.

)t(qgRhh

gRA

i=+&

FlowrateIN

FlowrateOUTLevel


Example – 1st Order ODE

The ODE can easily be expressed in the standard form as:

)t(qA1h

RAgh i=+&

As a result, the system’s time constant is identified as:

gRA

=τ


Free Response – First Order

Homogeneous Solution

00x1x >τ=τ

+&

Characteristic Equation

τ−=λ⇒=

τ+λ

101

The solution becomes

τ−=

tce)t(x


Step Response – First Order

Consider a 1st order system described as:

)t(fBxxA =+&

And subjected to a step input.

Bf(t)x

dtdx

=+τ

In standard form:

Af(t)x

τ1

dtdx

=+



First Order homogeneous response is in the form of an exponential function

th e)t(y λ−= where

τ=λ

1

The solution is

where C is determined from initial condition for a particular solution and yp indicates the particular solution.

)t(yCe)t(y pt+= τ−



If the system is subjected to a step change

>=≤=

µ=0t10t0

)t()t(f s

The particular solution can be found to be

1Ce)t(yt+= τ−

Given the initial conditions when the system is initially at rest (t=0, ys(0)=0 requires that C=-1 which then gives

τ−−=

te1)t(y


Step Response - First Order

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Time (sec)

Perc

enta

ge o

f Res

pons

e

τ 2τ 3τ

63.2

86.595.0



Step Response – First OrderResponse is broken up into two regions:1. Transient region where system is still responding

dynamically2. Steady-state region where system has reached its final

valueNote: There is no clear break point; often, 3τ, 4τ, or 5τ ischosen based on desired accuracy

The initial slope of the response may be found by differentiating y(t) to give

τ=

1dtdy



Using the following equation

HOMEWORK ASSIGNMENT -----

)t(fBxxA =+&

Calculate the response by hand and plot by hand. Let A = month and B = day of your birthday

Use MATLAB to confirm your results


Transfer Function of Step Response

)t(fA1x1x

)t(fB1xx

BA

)t(fBxxA

BA =

+

=+

=+

&

&

&

( )ABs

A1

f(s)x(s)

+

=

Diff. Eq.

reduces to

or

Transfer Function is

Transferring to the LaPlace domain:



Using MATLAB and the equivalent Laplace form, the system transfer function is described as

+

=

+

−

ABS

A1

ABS

A1 1

› NUM = [0 1/A];› DEN = [1 B/A];› step(NUM,DEN)

Step Response - First Order

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Time (sec)

Perc

enta

ge o

f Res

pons

e

τ 2τ 3τ

63.2

86.595.0


Free Response – Second Order

Homogeneous Solution

0xx2x 2nn =ω+ζω+ &&&

Characteristic Solution

02 2nn

2 =ω+λζω+λ

( )

12nn

2n

2nn2,1

−ζω±ζω−=

ω−ζω±ζω−=λ

mk;m2c;

cc 2

nncc

=ωω==ζ



For purposes of development of these general equations a simple mass, spring, dashpot system will be used.

Equation of motion is obtained from Newton’s second law

)t(fkxdtdxc

dtxdm 2

2=++

With I.C.’s

ox)0(x = 0x)0(x && =and

k

c

m

x(t)

f(t)



A solution of the form fitstex λ=

( ) 0ekcm t2 =+λ+λ λ

0kcm 2 =+λ+λ

Characteristic equation

m2

mk4cc 2

−±−

=λ⇒

Solution has 2 roots and 3 possible solutions depending on the term under the √.


Free Response - Case 1 ( )1mk4c2 <ζ<

( )m2

cmk4c 2

−−±−

=λ

jm2

cmk4m2c 2

2,1−

±−=λ

jβ±α−= (Two complex conjugate roots)

The solution is

( ) ( )tjtjh BeAex β−α−β+α− +=


Factor out: te α−

( )tjtj ββ −−+=⇒ BeAeex αt

h

Recall that:

θ−θ=θ+θ= θ−θ sinjcose;sinjcose jj

Then,

( ) ( )[ ][ ]tjctce

tjtBtjtAext

th

ββ

ββββα

α

sincos

sincossincos

21 +=

−++=−

−

Free Response - Case 1



Using

equation can be written as

ysinxcosycosxsin)yxsin( +=+

=α

m2c( )φβα += − tcex t

h sin

φ+

−=

−t

m2cmk4sincex

2tm2c

h



Now if we divide through by m, then

0mfxx2x 2

nn ==ω+ζω+ &&& (homogeneous)

whereζ = damping ratio =

ωn = natural frequency =

km2c

cc

c=

mk

( )φ+ω=⇒ ζω− tsinex dtn

where 2nd 1 ζ−ω=ω



1. For small ζ ---› ωn ≈ ωd2. ωn is independent of damping3. If c = 0, then ωn ≡ ωd4. Solution response is always of the form of a

damped exponentially decaying sinusoidal formFree Response - Damped Exponential Decay

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 0.5 1 1.5 2 2.5 3

Time (sec)

Mag

nitu

de


Free Response – Case 2 ( )1mk4c2 >ζ>

m2mk4c

m2c 2 −

±=λ Two REAL Roots

t2

t1

21 ececx λλ +=Recall

θ−θ=θ+θ= θ−θ coshsinhe;coshsinheThen

tcoshctsinhcx 2413 λ+λ=

Solution of this type will always be of the form of a lag in the system.


Free Response – Case 2

Response will have an exponential envelope but will not have oscillatory motion about steady-state.

Damping ζ > 1

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Time (sec)

Am

plitu

de


Free Response – Case 3 ( )14mkc2 == ζ

m2c

11 −=λ=λ Two REAL REPEATED Roots

t2

t1

21 tececx λλ +=

Solution of this type will also be in the form of a lag to the system, but this system will return to steady state faster than any other damping without overshoot.

This is the break point between structural dynamics and controls problems.


Free Response - Summary

1

1

1

>ζ

=ζ

<ζ


S-plane Representation

dn1 jω+ζω−=λ

dn*

1 jω−ζω−=λ

n2,1 ζω−=λ1

1

1

>ζ

=ζ

<ζX – Conjugate Poles

– Repeated Roots

– Real Roots


jω

σ

UNSTABLESTABLE

TIME

FRF

FRF

TIME

FRF

TIME

FRF

TIME

TIME

TIME

> 1.0ζ

= 1.0ζ

= 0.7ζ

= 0.3ζ

= 0.1ζ = 0ζ

S-PLANE PLOTS FOR IMPULSE RESPONSE

OF A SINGLE DEGREE OF FREEDOM

MECHANICAL SYSTEM


Example – 1.12

Example from Dynamic Systems by Vu and Esfandiari


Example – 1.12 (cont)



Example – 1.13



Example – 1.14



Example – 1.16



Example – 1.17



Example – 1.18



Example –1.18 (cont)



Example – 1.19



Example 1.19 (cont)



Using Matlab to Solve Differential Equations

Matlab’s “dsolve” command is a common alternative to solving complicated differential equations by hand.

Example 1.17 will be solved again using Matlab


Using Matlab to Solve Differential Equations

The same solution can be obtained with Matlaband compared to the solution from Example 1.17.

( ) t4t2t e31ee

35tx −−− +−=


Differential Equations with force

or I.C. in time domain

Algebraic Equationin

Laplace domain

Rearrange terms intogood form

Closed-Form Solution or

Numerical Solution

Laplace Transform using I.C.

Inverse Laplace

Alternative: Convolution Integral

(very difficult)

Laplace Transform

Frequency domain X(s)Time domain x(t)


Laplace Transform Equation

{ }

{ }{ } )0()0()()(

)0()()(

)()()(

2

0

ggssGstggsGstg

dttgetgsG st

&&&

&

−−=

−=

⋅== ∫∞ −

L

L

L

Derivatives


Laplace Transform of several functions

ExponentialExponential

SINSIN

COS

Unit SinusoidUnit Sinusoid

1Unit Impulse (Dirac Unit Impulse (Dirac delta function)delta function)

Unit PulseUnit Pulse

Unit RampUnit Ramp

Unit StepUnit Step

Laplace TransformLaplace TransformFunction f(t)

ssµ

s12s1

22

22

ωssωs

ω

+

+

as1+


Laplace Transform

{ } ∫∞ −==

0)()()( dtetusutu st

sssL

su

sue

su sssts =

−−−=−= ∞− 0|0

•Unit Step

•Unit Ramp

{ } ∫∞ −==

0)()( dtetsutu st

rsL

202

stst

00

st

s1|

sedt

se|

set =

−=

−−

−= ∞

−−∞∞−

∫

Requires manipulation of Laplace domain equation to get in a suitable form to apply L -1.

•Inverse Laplace


Convolution Integral

{ } { } ))(()()()( 11 thgsHsGsF ⊗=•= −− LL

•If G(s) and H(s) have known inverses g(t) and h(t), then the product of GH can be obtained by the convolution integral.

))(()()(0

thgdtthgt

∫ ⊗=−= ττ

))(()()(0

tghdttght

∫ ⊗=−= ττ


Unit Impulse Response – First Order

)t(x1x δ=τ

+&

•The Equation

1)s(X1)s(Xs =τ

+

0x0 =

•Laplace

1)s(X)1s( =τ

+

τ+

= 1s

1)s(X τ−=

te)t(x∴

look up inverse laplace


Laplace – First Order ODE

•The basic first order ODE can be expressed in the Laplace Domain as for unit impulse and can be recast as

•This can be stated as follows:The basic value of is multiplied by A -- This

value is equal to 1 minus B times the integral of

•Normalize the equation so the coefficient on =1,

sAB

A1s −=∴ &

1BssA =+&Bs1sA −=&

s&s&

s&


Block Diagram – First Order ODE

s1 SCOPE

integrate

Multiply byB/A

Normalizeto A coefStep

= 1/A

1

-

+ s& s


MATLAB/SIMULINK – (see tutorial)

•Simulink•File New Model (workspace appears)Select the following items and place them on the worksheet

Unit Step (from Sources) – change amplitudeSum (from Math) – need + and –Gain (from Math) – change gain valueIntegrator (from Continuous)Scope (from Sinks)

Double click items to view or change property GAIN block can be rotated by format.T branch – mouse online/CTRL and right mouse button to extend line.


SIMULINK – First Order Step Response


SIMULINK – First Order Impulse Response


Unit Impulse Response – Second Order

•For unit impulse, 0x0,xδ(t),f(t) 00 === &

Then,

∴

δ(t)xωx2ζx 2nn =++ &&& ω

with 00 =x& 0x0 =and

•Laplace with I.C. = 0

1)s(X)s2s( 2nn

2 =ω+ζω+

2nn

2 s2s1)s(X

ω+ζω+=


Unit Impulse Response - Second Order

•Note that (assume ζ<1)

2d

2)s(1)s(X

ω+σ+=So that

2n

2n

2n

2nn

2 )()s(s2s ω+ζω−ζω+=ω+ζω+

)1()s( 22n

2n ζ−ω+ζω+=

2d

2)s( ω+σ+=


++⋅= −

221

)(1)(

d

d

d stx

ωσω

ωL

•The inverse Laplace

te dt

d

ωω

σ sin1 −=

te dt

d

ωω

ζω sin1 −=

If I.C. ≠ 0, then a more involved (but possible solution) exists.

Unit Impulse Response - Second Order


Unit Step Response – Second Order

•For unit step

)t(uxx2x 2nn =ω+ζω+ &&&

•Laplace with IC=0 gives

s1)s(x)s2s( 2

nn2 =ω+ζω+

∴

with

0x,0x),t(u)t(f 00 === &

0x,0x 00 ==&

2nn

2 s2s1

s1)s(X

ω+ζω+⋅=


Unit Step Response – Second Order

•Again assume ζ<1 - then

+++

−= 222 2211

nn

n

n ωsζωsζωs

sωX(s)

•Then the inverse Laplace

++

+−= −

2221

2211)(

nn

n

n ωsζωsζωs

sωtX L

−+−= − ttetX dd

t

n

n ωζ

ζωω

ζω sin1

cos11)(22


Example – 1.40



General Laplace Formulation

•Laplace

)t(fxx2x 2nn =ω+ζω+ &&&

[ ] F(s)x(s)ωx(0)sx(s)2ζ(0)xsx(0)x(s)s 2nn

2 =+−+−− ω&

)()()()(

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

sFtFsXx

xssxxxsxsxsx

==

−=−−=

LLLL

&

&&&


General Laplace Formulation

If initial conditions are zero, then the system transfer function is

)s(x)s2s( 2nn

2 ω+ζω+

2nn

2 s2s1)s(H

)s(F)s(x

ω+ζω+==

Applied Force and Initial Conditions

(0)x)x(0)ζω2(sF(s) n &+++=

2nn

2n

2nn

2 ωsζω2s(0)x)x(0)ζω2(s

ωsζω2sF(s)x(s)

++++

+++

=&

inputoutput

Many books use G(s)!

kcsms1aka 2 ++


SIMULINK – 2nd Order Impulse Response (or RAMP)

)t(Fx1000x40x100 =++ &&&

OR…



)t(Fx1000x40x100 =++ &&&

In alternate form (using the Transfer Function Block)



This example is

General Equation is

)t(Fkxxcxm =++ &&&

[ ]xckx)t(Fm1x &&& −−=

)t(Fx1000x40x100 =++ &&&

complex analysis, differential equations, and laplace transform

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