ME 375 System Modeling and Analysis Section 4 – Solution ... Handouts... · 4.5 frequency domain; in other words, short events translate into a “broadband” of energies that

1

ME 375 System Modeling and Analysis

Section 4 – Solution Methods and Transfer Function Analysis

System(w/ I.C.s)

Input Output

Spring 2009School of Mechanical Engineering

Douglas E. AdamsAssociate Professor

2

Motivational ExampleDifferential equations vs. algebra – choose wisely

4.1

Consider the vibration of the f ll i i t d

INCHES POWERED TETRASILICIDE AND BOROSILICATE GLASS

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following ceramic-coated silica tile during launch and reentry of the orbiter.

TILE DENSITY HIGH PURITY SILICA 99.8% AMORPHOUS FIBERS

1

.008

SIP -- NEEDLED STRUCTURE

.016 to .018

RTV BOND

.09/.115/.16

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SIP NEEDLE FIBERS AND RTV BOND .125

NOMEX FELT INSULATION – RESISTANT UP TO 800F

RTV BOND – CURES AT RM TEMP & VACUUM BAG PRESSWhich of the following problems would you

RTV BOND -- CURES AT ROOM TEMP UNDER VACUUM BAG .008

VEHICLE SURFACE TILES NOMINAL SIZE IS 1”(to 5”) x 6” x 6”

rather solve?

2 0as bs c+ + =OR

We can find solutions usingalgebra if we transform intothe frequency domain

Find s

© 2009 D. E. AdamsME 375 – Solutions and Transfer Function Analysis

OR

0ay by cy+ + =&& &OR

We can find solutions usingdifferential equations if westay in the time domain

Find y

3

Frequency-Domain SolutionsPartial fraction expansions

When the Laplace transform, This limit ofintegration is

4.2

[ ] ∫+∞

−==0

)()(L)( dttyetysY stThis limit ofintegration isassociated with

integration isassociated withsteady-state response

is applied to a differential equation, eachterm is transformed into the frequency domain where we can use algebra to solve simultaneously for both responses

transient response

ji dd ( ) Initial Condition

Y s = +j

j

jj

ii

i

i dtudb

dtyda ∑∑ =

( )

Y sTermsForcing Function

Terms

= +


4

Important Examples of TransformsFinding Laplace transforms of common functions

Some simple examplesSteps

dttyesY st )()( =+∞

−∫

4.3

StepsExponentialsRampsTrigonometricImpulses

t

y

s

es

dte stst

1

1

00

0

=

−==∞

−∞+

−∫

∫

y s

t

yα=

∞

+∞−−∫ dteesY tst

11

)(0

t

y

)( tdtesY st= ∫+∞

−

ααα

+=

+−= +−

se

sts 11

0

)(

2

0

1s

=

∫


5

Some simple examplesTrigonometric ∫

+∞t

Important Examples of TransformsFinding Laplace transforms of common functions

4.4

TrigonometricImpulses

22

0

sin)(

ωω

ω

+=

= ∫ −

s

tdtesY st

t

y

( ))(0

=+∞

−∫ st dttesY δt

y 0

2 2

( ) cosstY s e tdt

ss

ω

ω

+∞−=

=+

∫

10== −ste

Impulses transform into constantswith spectral energy at all frequencies


6

An impulse in the time domain produces a constant in the frequency domain; in other words short events translate

What Is the Frequency Domain?The special case of an impulse

4.5

frequency domain; in other words, short events translate into a “broadband” of energies that excite systems throughout their frequency ranges (e.g., TPS impacts)

Typical Example of Original Input Time

History vs Estimated

5

10

15

20

25

Forc

e - l

b f

OriginalEstimated

History vs. Estimated

0.008 0.01 0.012 0.014 0.016 0.018 0.02-5

0

Time - sec


7

Going Back to Time DomainTransform inversion

You do not really need to use the inversion formula,which should make you happy considering

4.6

which should make you happy considering…

∫∞+

∞−

=jc

jc

st dtesFj

ty )(21)(π

Usually have to usecontour integration(in the complex plane)

Instead, you can just use the transform formulas you already know to get the inverse time functions

∞

)(ty )(sY1

∫ −=0

)()( dtetysY st )(tus s1


8

Laplace PropertiesFor using Laplace to solve for system responses

If you forget a property, go back and derive it again using th f l t d

4.7

these formulae, e.g., steps and ramps

t

y

[ ] )()(L saYtay = [ ] )()(L sYdsdtty −=

t

y

t

y

t

y[ ] )0()()(L yssYty −=&[ ] )()()()(L sZsYtzty +=+

Superposition

We use these a lot!

Differentiator

t

y

[ ] )()(L asYtye at +=−

Modulation t

y[ ]

ssYdtty )()(L =∫

Integrator


9

More Useful PropertiesUsing Laplace transforms to solve for system responses

[ ] )()()(L sYeatuaty sa−

y

4.8

[ ] )()()(L sYeatuaty s =−−

[ ] [ ]tztytx += )()(L)(L

Superposition

taSome examples:

=y

+z

a

x[ ] [ ]

( )sa

sa

es

se

s

zy

−

−

−=

−=

11

1)()()(=

t

Step

+t

Time-shiftedstep

at

Unit pulse

a

[ ] [ ]( ) )0()(L)(L ytysty &&&& −=[ ] [ ]( )( )

)0()0()()0()0()(

)0()(L)(L

2 ysysYsyyssYs

ytysty

&

&

−−=

−−==

Differentiation


10

The poles determine the nature of the dynamic responseThey do the same kinds of things here!

Poles of the Laplace TransformThe “fingerprints” of dynamic response

4.9

They do the same kinds of things here!

t

y

t

y

Im

X

y

t X XX

y

t X

y y

XX t

yX

X

O

X

ORl

t tX

poleszeros

sDsNsY ==)()()(


11

Simple Inversion TechniquesSome requirements – classifications of Y(s)

When Y(s) is the ratio of two polynomials in s, we call it rational

4.10

on

nn

om

m

asasbsb

sDsNsY

+++++

== −− L

L1

1)()()(

Do we always havethis form? What about time delays?

When n≥m, we say Y(s) is properWhen n>m, we say it is strictly proper (more denominator terms)

Partial fraction expansions can be used to find inversetransforms as long as Y(s) is proper

There are two general cases to consider:Distinct poles (roots)Repeated poles (roots)


12

Partial Fraction Expansions – Distinct PolesSuperposition of modes (in the frequency domain)

When Y(s) is strictly proper with distinct roots…

4.11

∑∑==

=−

=n

r

tsr

n

r r

r reAtyss

AsY11

)()( so Each one of theseterms representsa “mode” of thesolution/function

How do we find the “residues” (Ar )

( ) ( )[ ]q

q

ssqss

n

r r

rqq sYss

ssAssA

===

−=⎥⎦

⎤⎢⎣

⎡−

−= ∑ )(1

We could justdo thisusing algebra

We can use this technique for real and complex (complex conjugates) poles – if we are careful with the algebra!


13

Inversion ExampleTwo distinct poles (roots)

( )( )ssY 5)( +

= 1 2

1.4

4.12

( )( )

ssA

sA

sA

ss

1

21

34

45

41

41)(

=++

=

++

+=

++

where0

0.2

0.4

0.6

0.8

1

1.2

y(t)

Mode #1

Mode #1 + Mode #2

s

s

sssY

ssA

s

42

1

43/1

13/4)(

31

15

34

−=

−=

+−

+=

−=++

=

+

so

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.4

-0.2

Time [s]

Mode #2

Note that Mode #1 dominatesthe response for long times

tt eety 4

31

34)( −− −= and

p g

What would happen if there was azero at s=-1 (pole/zero cancellation)?


14

Partial Fraction Expansions – Distinct C.C. PolesSuperposition of modes (in the frequency domain)

When Y(s) is strictly proper with distinct complex roots…

4.13

How do we find the “residues” (Ar ) ?

∑∑==

=−

=n

r

tsr

n

r r

r reAtyss

AsY11

)()( so A pair of theseterms representsan oscillatingmode in thesolution

N t th t th id f l j t i

( ) ( )[ ]q

q

ssqss

n

r r

rqq sYss

ssAssA

===

−=⎥⎦

⎤⎢⎣

⎡−

−= ∑ 1

)(

Note that the residues form a complex conjugate pair

( )**12 bjaAbjaA +==−=


15

Inversion ExampleTwo complex conjugate poles (roots)

4 8( ) sY +

4.14

2

1 2

0.461

4 8( )2 5

( 1 2) ( 1 2)

4 8where 2 5 j

sY ss s

A As j s j

sA j e−

+=

+ +

= +− − + − − −

+= = − =1

1 2

* 0.462 1

0.46 0.46

where 2 5( 1 2)

2 5

5 5so ( )( 1 2) ( 1 2)

s j

j

j j

A j es j

A A j e

e eY ss j s j

=− +

+

− +

− − −

= = + =

= +− − + − − −

( )and ( ) 2 5 cos 2 0.46ty t e t−= −Euler’s formulaThis function should be REAL!!!!


16

Multiple Second Order TermsGreater than two complex conjugate poles (roots)

In the majority of systems, there are more than two complex conjugate poles; however the procedure for

4.15

complex conjugate poles; however, the procedure for finding solutions is identical in higher order systems

( )Y s Pairs= ∑ 1st pair

2nd pair


17

Inversion ExampleA repeated pole of multiplicity two

s + 5 1.8

2

4.16

( )

( )sA

sA

sA

sssY

=+=

++

+=

++

=

45

11

15)(

1

22

1

2

where 0 4

0.6

0.8

1

1.2

1.4

1.6

y(t)

e-t

4te-t

(4t+1)e-t

( )

( )

s

s

sssY

sdsdA

s

−=

−=

++

+=

=+=

11

14)(

15

5

2

12

11

so

w e e

Note the effect of the repeated pole.Completing the square also works.

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

Time [s]

( )( ) tettys

−+= 14)( and

p g q

What would happen if there was amultiplicity of three? Can you doit when there are many different roots?


18

What if Y(s) Is Only Proper?Long division and partial fractions

When Y(s) is only proper with distinct roots…

4.17

∑∑==

+=−

+=n

r

tsr

n

r r

r reAtCtyss

ACsY11

)()()( δ so

Proper Y(s)’s are rare because real systems cannot Respond instantaneously regardless of the input

ss2 23872 ++Example:

tt eettyssss

sssY

2

2

23)(2)(2

21

3223872)(

−− −+=+

−+

+=++++

=

δ


19

Solving ODEs with LaplaceFirst order system (overdamped) and transfer functions

We can solve for both parts of the solution at the same time

4.18

MgtfKyyCtfKyyC dd

+=+

=+

)()(

&

&E.O.M.(time)

K

f

C

y

( ) CysFsYKCs dd +=+ )0()()(E.O.M.(freq.)

Solutions

kCsCy

KCssFsY d

d ++

+=

)0()()(

( )

( )s

MgCysFsYKCs

dd

++=+ )0()()(

( q )

)()()( sH

sFsY

≡Roots of

( )KCssMg

kCsCy

KCssFsY

kCsKCs

++

++

+=

++)0()()(

)(sFRoots ofdenominatorpolynomialare poles


20

Solving ODEs with LaplaceApplying the principle of superposition to identify solution modes

Before spending a great deal of time and energy transforming the solution back into the time domain we

4.19

transforming the solution back into the time domain, we should use the principle of superposition to speculate:

K

f

CSolutions

( )( ) (0)( ) F s Cy MgY s

Cs K Cs k s Cs K= + +

+ + +f y ( )Cs K Cs k s Cs K+ + +

y1(t)Depends onForce !

y2(t) y3(t)

Time

Force !

Time Time


21

Solving ODEs with LaplaceFirst order system (overdamped)

How do the locations of the poles determine the type of response?

4.20

response?

+⎥⎦⎤

⎢⎣⎡

+=

−− tCK

dd eyKCs

sFty )0()(L)( 1

K

f

C

y

⎟⎟⎞

⎜⎜⎛−+

+⎥⎦⎤

⎢⎣⎡

+=

⎦⎣ +

−

−−

tCK

tCK

eMg

eyKCs

sFty

KCs

1

)0()(L)( 1

Real

Imag

XCK

−

⎟⎟⎠

⎜⎜⎝

+ eK

1

Complex plane (s plane)Zeros of the CE


22

What Happens for Excitations?Superimposing canonical excitations to create general case

For linear systems, the principle of superposition works so we solve for the forced response simultaneously with

4.21

so we solve for the forced response simultaneously with the free response by adding up the contributions for all of the forces that are applied to our system

Many different kinds of inputs to consider…

t

u

STEP(on-off)

t

u

IMPULSE(shock)

t

u

SINUSOID(cyclic)

t

u

RAMP(change)

Generalinput = + + +


23

Multiple Types of ExcitationSuperposition applies in this case (rivet gun)

When a rivet gun is used to manually install rivets, the forces may contain impulses, sinusoids and steps depending on the

4.22

y p , p p gquality of the rivet produced.

ZXRivet

Frequency [Hz]

Y

Part

Bucking bar

.


24

Solving ODE with LaplaceFirst order system – Step input

For a forced response problem, just substitute form of F(s)

4.23

Solutionsfor stepinput

( )

( ) ( )KCssMg

kCsCy

KCsssY

kCsCy

KCsssY d

d

++

++

+=

++

+=

)0(1)(

)0(1)(Does this makesense?

Step

K

f

C

y

⎟⎞

⎜⎛

⎟⎞

⎜⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−−−

−−

tCKt

CKt

CK

tCK

d

tCK

d

Mg

eyeK

ty

1)0(11)(

)0(11)(

( )⎟⎠⎞

⎜⎝⎛ +

=+

CKssCKCss

111

⎟⎟⎠

⎜⎜⎝−++⎟⎟

⎠⎜⎜⎝−= CCC e

Kgeye

Kty 1)0(1)( ⎠⎝ C


25

SDOF System – Step Response1st order system (overdamped)

We will need to understand how the transient response changes as a function of the key system parameters

4.24

changes as a function of the key system parameters.

K

f

B

yg

0.7

0.8

0.9

1

0.2

0.3

0.4

0.5

0.6

y d(t)

τ=1 sτ=0.5 sτ=0.4 sτ=0.3 s

Do you remember this?

yd = c1e−tτ +FoK

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

Time [s]

Do you remember this?98% of the way there fort=5τ.


26

Solving ODE with LaplaceFirst order system - Impulse

We can repeat the previous steps for any kind of excitation

4.25

Solutionsfor impulseinput

( )

( ) ( )KCssMg

kCsCy

KCssY

kCsCy

KCssY d

d

++

++

+=

++

+=

)0(1)(

)0(1)(

Impulse

K

f

C

y

⎟⎟⎞

⎜⎜⎛

++

+=

−−−

−−

tCKt

CKt

CK

tCK

d

tCK

d

eMgeyety

eyeC

ty

1)0(1)(

)0(1)(

⎟⎟⎠

⎜⎜⎝−++= CCC e

Keye

Cty 1)0()(


27

Solving ODEs with LaplaceSecond order system (underdamped)

When the order of the ODE is higher, there is more algebra

4.26

( ) ( ) yMyCMssFsYKCsMs +++=++ )0()0()()(2 &E.O.M.(freq )

K

f y

g

M

C E.O.M.(time) MgtfKyyCyM

tfKyyCyM ddd

+=++

=++

)()(

&&&

&&&

( ) ( )

( ) ( )s

MgyMyCMssFsYKCsMs

yMyCMssFsYKCsMs ddd

++++=++

+++=++

)0()0()()(

)0()0()()(

2 &

(freq.)

( )KCM

yMKCM

yCMsKCM

sFsY ddd +

++= 222

)0()0()()(& Solutions

Transfer Function

( )( )KCsMss

MgKCsMs

yMKCsMs

yCMsKCsMs

sFsY

KCsMsKCsMsKCsMsd

+++

+++

+++

+++

=

++++++

2222

222

)0()0()()(

)(

&


28

( )+⎟⎟⎞

⎜⎜⎛

−+⎥⎦⎤

⎢⎣⎡= Ω−Ω−− ωζω ζζ

12121 sin)0(arctancos)0()(L)( 1111 teytey

KCMsFty tdt

dd&


4.27

( )

( ) ( )

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−

−−

Ω+

++⎥⎦⎤

⎢⎣⎡

+=

⎟⎠

⎜⎝ −⎥⎦⎢⎣ ++

Ω−

Ω−Ω−−

ζζ

ωζ

ωω

ω

ωζ

ζ

ζζ

2

12

11

11

11

212

1arctansin

111

sin)0(cos)0()(L)(

1)()(

11

1111

teg

teyteyKCs

sFty

yKCsMs

y

t

tt

dd

&

( ) yMyCMssFsY dd ++

+=)0()0()()(

& Solutions

⎠⎝ ⎠⎝ ζζ1

( )( )KCsMss

MgKCsMs

yMKCsMs

yCMsKCsMs

sFsY

KCsMsKCsMsKCsMssYd

+++

+++

+++

+++

=

+++

+++

++=

2222

222

)0()0()()(

)(

&


29

( )+⎟⎟⎞

⎜⎜⎛

−+⎥⎦⎤



KCMsFty tdt

dd&


4.28

( )

( ) ( )

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−

−−

Ω+

++⎥⎦⎤

⎢⎣⎡

+=

⎟⎠

⎜⎝ −⎥⎦⎢⎣ ++

Ω−

Ω−Ω−−

ζζ

ωζ

ωω

ω

ωζ

ζ

ζζ

2

12

11

11

11

212

1arctansin

111

sin)0(cos)0()(L)(

1)()(

11

1111

teg

teyteyKCs

sFty

yKCsMs

y

t

tt

dd

&

2111

22 21)0()0(

ζ Ω+Ω+=

++ ssy

KCsMsyM

dd &&

( )KCM

yMKCM

yCMsKCM

sFsY ddd ++

+++

++

++= 222


⎠⎝ ⎠⎝ ζζ1

( ) 21

211

1

1

)0(ωζ

ωω +Ω+

=s

yd&

( )( )KCsMss

MgKCsMs

yMKCsMs

yCMsKCsMs

sFsY

KCsMsKCsMsKCsMs

+++

+++

+++

+++

=

++++++

2222

222

)0()0()()(&


30

No matter what the order is, problems are solved the same wayLook at our SDOF oscillator (MKC or LCR) ( )+

⎟⎟⎞

⎜⎜⎛

−+⎥⎦⎤



KCMsFty tdt

dd&


4.29

Look at our SDOF oscillator (MKC or LCR)We have free oscillations now, so we must havecomplex conjugate poles (why?)

E O M

K

f yg

M

C E.O.M.(time) MgtfKyyCyM

tfKyyCyM ddd

+=++

=++

)()(

&&&

&&&Does this makesense to you?Inertia accel.Damping veloc.

( )

( ) ( )

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−

−−

Ω+

++⎥⎦⎤

⎢⎣⎡

+=

⎟⎠

⎜⎝ −⎥⎦⎢⎣ ++

Ω−

Ω−Ω−−

ζζ

ωζ

ωω

ω

ωζ

ζ

ζζ

2

12

11

11

11

212

1arctansin

111

sin)0(cos)0()(L)(

1)()(

11

1111

teg

teyteyKCs

sFty

yKCsMs

y

t

tt

dd

&

( ) ( ) ( )( )

( ) 111

21

211

11112111

211

2

sinsincoscoscos2

2)0()0(

ϕωϕωϕωωζ

ζζζζ

+=−+Ω+Ω+Ω+

=Ω+Ω+

Ω+=

+++

tttss

sssy

KCsMsyCMs

dd

( ) ( )

( ) ( )s

MgyMyCMssFsYKCsMs

yMyCMssFsYKCsMs ddd

++++=++

+++=++

)0()0()()(

)0()0()()(

2

2

&

&E.O.M.(freq.)

02 =++ KCsMs

( ) yMyCMssFsY ddd +

++= 222


⎠⎝ ⎠⎝ ζζ1( )

211

111

1sin1cos

ζζωω−

⋅+⋅= tt

( )( )KCsMss

MgKCsMs

yMKCsMs

yCMsKCsMs

sFsY

KCsMsKCsMsKCsMsd

+++

+++

+++

+++

=

++++++

2222

222

)0()0()()(

)(

&


31

SDOF System – Step Response2nd order system (underdamped)

We will need to understand how the transient response changes as a function of the key system parameters

4.30

changes as a function of the key system parameters.

1 2

1.4

1.6

1.8

K

f y

g

M

B

( )KFtcey ot

d ++= ϕωσ11 cos1

0.4

0.6

0.8

1

1.2

y d(t)

ς=0.1ς=0.2ς=0.9

0 1 2 3 4 5 6 7 8 9 100

0.2

Time [s]Transient S.S.


32

The denominator of the transfer function has roots, called the l ith l d i i t

Poles in the Complex Plane(Continued) The “fingerprints” of dynamic systems

4.31

poles, with real and imaginary partsReal part (damping) and imaginary part (frequency)

Imag

Real

gX

nωdω

nζω−2

2 2

2 ( )

2 0

n ny y y u t

s s

ζω ω

ζω ω

+ + =

⇓

+ + =

&& &

Complex plane (s plane)1

2

2 0

1

1

n n

n n

n n

s s

s j

s j

ζω ω

ζω ζ ω

ζω ζ ω

+ + =

= − + −

= − − −


33

Solving ODEs with LaplaceSecond order system – Impulse

We haven’t computed the second order impulse response yetLaplace transforms provide an easy way to do this2

11111L)( 2221

ssMKCMKCsMstyd Ω+Ω+

==⎥⎦⎤

⎢⎣⎡

++= −

ζ

4.32

Laplace transforms provide an easy way to do thisRemember – the impulse response has zero initial conditions

K

f yg

M

C ( )

( )1

11

2

221

21

21

21

211

1112

M

sM

ssMMKs

MCsMKCsMs

+Ω+=

Ω−Ω+Ω+=

Ω+Ω+++⎥⎦⎢⎣ ++

ωζω

ω

ζζ

ζ

( )KCsMs

yMKCsMs

yCMsKCsMs

sY ddd ++

+++

++

++= 222

)0()0(1)(& Solutions

( )

( ) )(sin1)( 11

21111

11 thteM

ty

sM

td ≡=

+Ω+

Ω− ωω

ωζω

ζ

( )( )KCsMss

MgKCsMs

yMKCsMs

yCMsKCsMs

sY

KCsMsKCsMsKCsMs

+++

+++

+++

+++

=

++++++

2222

)0()0(1)(&


34

Time Data from Frequency DomainInitial value and final value theorems

Sometimes we move into the frequency domain for certain information, and don’t want to go back (for stable Y(s))

4.33

, g ( ( ))Initial values (allow ‘s’ to approach infinity)Final values (allow ‘s’ to approach zero)

lim)0(

)(lim)0(

22

2

+=+

=+

∞→

∞→

ωssy

ssYy

s

sy

22)(ω+

=s

ssY[ ]y(t)L

1=+∞→ ωsst

! Order of numerator < order of denominator !

lim)(

)(lim)(0

=∞

=∞→

say

ssYysy

( )asY =)(

[ ]y(t)L

( )1

lim)(0

=+

=∞→ ass

ys

No poles in RHP or on imaginary axis andonly one pole at the origin

t( )ass +

)(


35

System StabilityStability, CE, the system poles, and T.F.s

Systems determine their own poles through the characteristic ti d l d t i th t bilit f t

4.34

equation and poles determine the stability of systemsThe roots of the denominator of the T.F. determine stability

Models are imperfect!Operating environments can affect marginally stable systems.

Imag

Real

g

Relative 22 ( )ζ&& &

AbsoluteStability

Unstable

Complex (s-plane)

Marginallystable

RelativeStability(gain/phase margin)

22 ( )n ny y y u tζω ω+ + =&& &


36


Passive linear systems are usually stableAny initial energy in the system is usually

In linearsystems it

4.35

y gy y ydissipated in real-world systems (poles in LHP)If there are no dissipation mechanisms, then therewill be poles on the imaginary axis (jω axis)If any coefficients of the denominator polynomialare zero, then there will be poles with zero real part

systems, itis alrightto say thata “system”is stable;however,in nonlinearsystems

Active systems can be unstableAny initial energy in the system can be amplifiedby internal sources of energy (feedback)If all coefficients of the denominator polynomial

t th i th t i t bl

systems,we cannotsay thisbecause nonlinearsystems actdifferentlyaround are not the same sign, the system is unstable

Even if all the coefficients of the denominatorpolynomial are the same sign, instability can occur(complex conjugate in RHP)

aroundvariousoperatingpoints


37


Consider a SDOF vibrating system without any force

4.36

jsss ±−=⇒=++ 2,12

23

2101

Stable

K

f yg

M

C

asas ±=⇒=−

22

2,122 0

Unstable

Marginally “Unstable”

K

f yg

M

C

jasas ±=⇒=+ 2,122 0 K

f yg

M


38

Time Domain Analysis Frequency Domain Analysis( )sy ay u t+ =& ( )sy ay u t+ =&

1 Set the right hand side to zero:0y ay+ =&

[ ] 1( ) (0)s a Y s y+ − =

1 Take the Laplace transform:

2 Assume exponential yc(t): [ ] ( ) ( )ys

p yc( )

[ ]( ) 0st sty t Ae s a Ae= ⇒ + =

( )1 (0)( ) yY s

s s a s a= +

+ +

2 Solve for Y(s):3 Solve the characteristic equation:0s a s a+ = ⇒ = −

3 E d i ti l f ti4 Plug roots back into guess:

( ) aty t Ae−

( )1 1/ 1/A B a a

s s a s s a s s a= + = −

+ + +

3 Expand using partial fractions:( )cy t Ae=

4 Take the inverse Laplace transform:

5 Assume yp(t) with same form as input: ( ) ( )sy t Bu t=

6 Solve for particular solution:10 ( ) ( ) ( ) ( )s s p saBu t u t y t u t+ = ⇒ =

( )

( )

1 1( ) ( ) 0

1 10 for 0

at ats

at

y t u t e y ea a

y e ta a

− −

−

= − +

⎛ ⎞= + − >⎜ ⎟⎝ ⎠

( ) ( ) ( ) ( )s s p sya

7 Write the solution as :c py y+1( ) ( )at

sy t Ae u ta

−= +8 Apply I.C.s:

( )1 1( ) 0 for 0aty t y e ta a

−⎛ ⎞= + − >⎜ ⎟⎝ ⎠

39

Frequency Response FunctionsAn important class of transfer functions

We are very often interested in how linear systems respond to sinusoidal or periodic excitations because these

4.37

excitations are common.

To apply frequency response function (FRF) methods, we must always remember these things:

1) FRFs work for linear systems (FRF does not change with amplitude)) y ( g p )

2) FRFs work for time-invariant systems (does not change with time)

3) FRFs work for stable systems4) FRFs apply for a simple harmonic excitation5) FRFs apply in the steady-state only6) FRFs relate the input amplitude/phase to the output

amplitude/phase at one frequencyamplitude/phase at one frequency7) FRFs contain Rl/Im parts with amplitude/phase info7) FRFs do not change when the input changes


40

Consider a general linear system with transfer function H(s)

Frequency Response – Sinusoidal ResponseDerivation and importance of FRFs in system analysis

4.38

( ) [ ][ ] EOM intoit Plugthen

, If

⇒=

==tj

tjoo

Yety

eFtFtfω

ωω

Rl)(

Rlcos)( Could use the method of undetermined coefficients

AAAsF n

Or we can use the transfer function concept PFENo I C s

)(

)(

)()()()(

21

321

3

2122 ωωω

ωω

ωω

tyeAeA

eAeAeAty

ssA

jsA

jsA

ssFsHsFsHsY

sstjtj

n

k

tsk

tjtjt

n

k k

ko

k

≡+=

++=

−+

−+

+=

+==

−

=

−∞→

=

∑

∑

... stab. Abs.

No I.C.s

Cosine input

Steady state is reachedfor a stable system

( ) ( )

( ))(cos)()(

2)( *

2221

ωωω

ωωω ω

jHtjHFty

AjHFjss

sFsHA

oss

o

js

o

∠+=

==⎥⎦⎤

⎢⎣⎡ +

+=

−=

Thus

where


41

Sinusoidal Responses from T.F.s1st order system (overdamped)

Evaluate the T.F. along the imaginary axis

4.39

)(tfKyyC dd =+&E.O.M. ( )oo tFtf ϕω += cos)(

Thi d t ff t th S S

K

f

C

y

( )

[ ]eeFtfKCj

HjsKsCsF

sYsH

CysFsYKCs

jtjo

d

dd

ωωω

ϕωRl)(

1)(1)()()(

)0()()(

=

+=⇒=

+=≡

+=+

that us tellsFRF then the, Since

let Now

Frequency response functionThis does not affect the S.S.

KCty

CKFty ssd

ossd

ωϕω

arctan)(,)(222

−=∠+

= −−

t

y


42

Sinusoidal InputsAnd how linear systems respond to them (1st order)

FRFs are useful because they are complex and contain twopieces of information (relative magnitude and relative phase)

4.40

pieces of information (relative magnitude and relative phase)Everything is relative to the sinusoidal (or cosinusoidal) input

Linear Simulation Results

Am

plitu

de

0

0.05

0.1

Linear S imulation Results

Am

plitu

de

0

0.05

0.1


Am

plitu

de

0

0.05

0.1

ssd

Fty )(−

Time (sec.)

0 1 2 3 4 5 6 7 8-0.05

Time (sec.)

0 1 2 3 4 5 6 7 8-0.05

Time (sec.)

0 1 2 3 4 5 6 7 8-0.05


mpl

itude

0.05

0.1

)()( tfty ssd ∠−∠ −ωoF

ω Time (sec.)

Am

0 1 2 3 4 5 6 7 8-0.05

0

ω

These are Bode Diagrams© 2009 D. E. AdamsME 375 – Solutions and Transfer Function Analysis

43

Sinusoidal Response of 2nd Order SystemsSecond order system (underdamped)

No matter what the order is, problems are solved the same way using the FRF to amplify and modulate the excitation

4.41

way using the FRF to amplify and modulate the excitation

( )E O M

K

f yg

M

CE.O.M.(time) MgtfKyyCyM

tfKyyCyM ddd

+=++

=++

)()(

&&&

&&& ( )ϕω += tFtf o cos)(

( )+=⇒=++

= 2

cos)(

1)()(

ϕωω tFtfjsKCsMssF

sYd

thenWhenSet

conditions initial zerofor

( ) ( )

( ) ( )s

MgyMyCMssFsYKCsMs

yMyCMssFsYKCsMs ddd

++++=++

+++=++

)0()0()()(

)0()0()()(

2

2

&

&E.O.M.(freq.)

( )( )

( ) ( )⎟⎠⎞

⎜⎝⎛

−−+

+−=

∠++=

+=⇒=

−

222arctancos1

)(cos)()(cos)(

ωωϕω

ωω

ωϕωω

ϕωω

MKCt

CMKF

HtHFtytFtfjs

o

ossd

o then, When Set

Does this make sense intuitively?…

The input informationThe system FRF information


44

Sinusoidal InputsAnd how linear systems respond to them (2nd order)

FRFs are useful because they are complex and contain twopieces of information (relative magnitude and relative phase)

4.42

pieces of information (relative magnitude and relative phase)Everything is relative to the sinusoidal input

Am

plitu

de


-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Am

plitu

de


-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Am

plitu

de


-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

ssd

Fty )(−

Am

plitu

deLinear Simulation Results

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Time (sec.)

0 5 10 15 20 25

-0.8

Time (sec.)

0 5 10 15 20 25

-0.8

Time (sec.)

0 5 10 15 20 25

-0.8

ude


0.2

0.4

0.6

0.8

)()(

tfty ssd

∠

−∠ −

ωoF

ω

Time (sec.)

0 5 10 15 20 25

-0.8

Time (sec.)

Am

plitu

0 5 10 15 20 25

-0.8

-0.6

-0.4

-0.2

0)(tf∠ ω

More Bode Diagrams…© 2009 D. E. AdamsME 375 – Solutions and Transfer Function Analysis

45

Multiple FrequenciesSuperposition applies in this case (large engine valve assembly)

When there are multiple frequencies in the excitation spectrum, then we simply add the results for all frequencies.

4.43

p y q

Rubbing between hammer andplunger causes sticking / material loss


46

Resonance and Why It OccursOptimum condition for energy transfer

Resonance occurs when the input is in phase with the derivative of the output:

4.44

pEnergy exchange is more efficient at certain frequencies?

o

ssd

Fty )(− K

H 1)( ≈ω

STIFFNESS dominates here1o

ω

)()(

tfty ssd

∠

−∠ −

ω

STIFFNESS dominates here

1)(H

( ) yfyyCyKyyMCj

H

&&&&&& +−=+

≈ω

ω 1)(

DAMPING dominates here

2nd order system frequency response

)(tf∠ ω2)(

ωω

MH

−≈

INERTIA dominates here


47

Multiple DOF SystemsMultiple inputs and multiple outputs

Example: Two DOF system (MDOF)- Can we draw the Bode diagrams by just observing the response?

4.45

M2

f 2x 2

M1

C2

K2

x 1 f 1

C1

K1


48

Multiple DOF SystemsMultiple transfer functions and multiple frequency response functions

Example: Two DOF system (MDOF)

We just transform each equation C2C1

x1 x2

4.46

j qSubstitute jω in the transfer functionInterpret the MIMO case

( )

⎫⎧⎤⎡⎫⎧⎤⎡ −+⎫⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡

++−−−−++++

)0(0)0()(

)()(

2

1

222

222

2221212

1

xMxCCsMsF

sXsX

KsCsMKsCKsCKKsCCsM

&

M1 M2

K1 K2f2f1

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡+

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡+−

−++

⎭⎬⎫

⎩⎨⎧

)0()0(

00

)0()0(

)()(

2

1

2

1

2

1

222

211

2

1

xx

MM

xx

CsMCCCsM

sFsF

&

( )( ) ⎪

⎪⎫

⎪⎪⎧

∠

+∠+

⎫⎧

⎭⎬⎫

⎩⎨⎧

++

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

)()(

)(cos)(

)(

)()()()()()()()(

)()(

)()()()(

)()(

11111

222121

212111

2

1

2221

1211

2

1

ωωω

HtHF

HtHF

t

sFsHsFsHsFsHsFsH

sFsF

sHsHsHsH

sXsX

o

Output 1 due toInputs 1 and 2( )

( )( ) ⎪

⎪

⎭

⎪⎬

⎪⎪

⎩

⎪⎨

∠+

+∠+∠+

=⎭⎬⎫

⎩⎨⎧

)(cos)(

)(cos)()(cos)(

)()(

22222

21211

12122

2

1

ωωω

ωωωωωω

HtHF

HtHFHtHF

txtx

o

o

oInputs 1 and 2

Output 2 due toInputs 1 and 2

How many amplitude/phasediagrams are there?


49

Multiple DOF SystemsMore inputs and outputs means more FRFs

Bode Diagrams

0

From: U1 From: U2

4.47

gnitu

de (

dB) -200

-100

0

300-200-100

0

To: Y

1

X1= *F1 + *F2 M2

C2

K2

f 2x 2

Pha

se (

deg)

; M

ag -300

-200

-100

0

-1000

Y2

X2= *F1 + *F2M

1

CC

1

K1

K

x 1 f 1

Frequency (rad/sec)

10-1

100

101

102

103

-300-200

To: Y

10-1

100

101

102

103

K


50

Multiple DOF Systems (Design)How do we design better systems using FRFs?

4.48

Video of sine sweep experiment of

By selecting the “right” mountstiffness, the resonances of the

Lumped-parameter model of vehiclechassis, suspension and powertrain

p pSSR prototype vehicle system

stiffness, the resonances of thepowertrain can be positioned toabsorb energy from the vehicleproviding a smoother ride.


51

Decomposing Bode DiagramsThe various parts of Bode Diagrams

The frequency response function contains coupled termsWe will decouple them by applying logarithms

4.49

We will decouple them by applying logarithms.Then we only need to understand a few terms.Constants, integrator/differentiator, first and second order.

Constants: 1I t/Diff

K

j

We are interestedin magnitude and phase

2

Int/Diff: ,

1First order: ,11

1S d d 1 2

jj

jωjω

ω ωj ζ

ωω

ττ

++

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟2Second order: ,1 2

1 2 n n

n n

- j ζω ω- j ζ

ω ω

ω ω

+⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎝ ⎠ ⎝ ⎠

+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


52

Example of How Bode Can Be UsedSpotting potential resonance issues with an exhaust system

Consider the exhaust system shown below and its Bode magnitude diagram; what information does this provide?

4.50

100

101

102

/lb]

FRF H21 in x-direction

H21lev1H21lev2H21lev3

FRFs H21 in the X-dir

10-3

10-2

10-1

10

Mag

. of F

RF

H21

( ω) [

g/

Mag

. H

pq[g

/lb]

0 50 100 150 200 250 300 350 40010-4

Frequency[Hz]

Frequency [Hz]


53

The Pieces of Bode diagramsConstants, 1st order, and 2nd order terms

1,1,1 >>≈<<For ωωω20

40

50

100

4.51

10log20log20

10

10

==

⇒=

KKdBKdBKK

for 10

-210

-110

010

110

2-40

-20

0

20

Mag

nitu

de

0

0.5

1

ase

ej

j

ωτωττω

π2

101log201,1

≈<<−

For

10-2

10-1

100

101

102

-100

-50

0

Mag

nitu

de

100

-50

0

ase

10-2

10-1

100

101

102

-1

-0.5

0

Freq [rad/s]

Pha

dB03.32log20 10 =dB0262log20

dB

ej

dB

j

j

ωττω

τω

π

π

2

2

1010

1l2011

0

11,1

log20log20

⇒

==

−−⇒

−

−

F

For 10-2

10-1

100

101

102

-150

-100

Ph

Freq [rad/s]

dB02.62log20 10 =

dB02.62log2021log20 1010 −=−=dB

ej

ωωτωττ

ω

10

210

log20

log20,

−⇒

≈>>For

Do you understand why n>m?© 2009 D. E. AdamsME 375 – Solutions and Transfer Function Analysis

54


dBj

For ω 10 01log2011

1,1=⇒≈

+<< 0

20

4.52

ej

dB

ej

j

For

For

ωτωττω

ωττω

ωττ

π

π

2

10

4

11

1,1

3707.0log20

707.01

1,1

1

⇒≈+

>>

−=

⇒≈+

=

+

−

−

10-2

10-1

100

101

102

-80

-60

-40

-20

Mag

nitu

de

-40

-20

0

hase

dB03.32log20 10 =

dB ωτωτ 1010 log201log20 −=

10-2

10-1

100

101

102

-100

-80

-60

Freq [rad/s]

Ph

dB02.62log20 10 =

dB02.62log2021log20 1010 −=−=


55

20

40

50

1002

1For , 1

1 2nω ω

ω ωζ

<< ≈⎛ ⎞ 0

20

0

5050


4.53

10-2

10-1

100

101

102

-40

-20

0

20

Mag

nitu

de

0

0.5

1

ase

10-2

10-1

100

101

102

-100

-50

0

Mag

nitu

de

100

-50

0

ase

22

2

2 2

1 2

1 1For ,2

1 2

1For ,

n n

n

n n

nn

j

e

j

e

π

π

ω ωζω ω

ω ωζω ωζ

ω ω

ωω ω

−

−

⎛ ⎞+ − ⎜ ⎟

⎝ ⎠

= =⎛ ⎞

+ − ⎜ ⎟⎝ ⎠

>> ≈ ⇒

10-2

10-1

100

101

102

-80

-60

-40

-20

Mag

nitu

de

-40

-20

0

ase

10-2

10-1

100

101

102

-150

-100

-50

Mag

nitu

de

-100

-50

0

hase

10-2

10-1

100

101

102

-100

-50

0

Mag

nitu

de

-2

-1

0

ase

10-2

10-1

100

101

102

-1

-0.5

0

Freq [rad/s]

Pha

dB03.32log20 10 =

10-2

10-1

100

101

102

-150

-100

Ph

Freq [rad/s]

2 2

2

10 10 102

,

1 2

20log 40log 40log

n

n n

nn

j

dB

ωω ωζω ω

ω ω ωω

⎛ ⎞+ − ⎜ ⎟

⎝ ⎠

= −10

-210

-110

010

110

2-100

-80

-60

Freq [rad/s]

Pha

10-2

10-1

100

101

102

-200

-150

Freq [rad/s]

Ph

10-2

10-1

100

101

102

-4

-3

Ph

Freq [rad/s]

g10

dB02.62log20 10 =

dB02.62log2021log20 1010 −=−=


Documents

ME 375 System Modeling and Analysis Section 4 – Solution ... Handouts... · 4.5 frequency domain; in other words, short events translate into a “broadband” of energies that