Dynamic Systems System Response 031906 -...

1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.451 Dynamic Systems – System Response

System Response

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell

TRANSFER FUNCTION

)S(F)S(H)S(Y •=

LAPLACE DOMAIN

IMPULSE

)t(f)t(h)t(y ⊗=

∫ ττ−τ=t

0d)t(f)(h)t(y

TIME DOMAIN

FRF

)j(F)j(H)j(Y ω•ω=ω

FREQUENCY DOMAIN

RESPONSEMODELS


Transient Response – First-order System

oy)0(y.C.I0)t(fyy =>τ=+τ&

Free Response No forcing function – Only I.C.Laplace Transform

[ ] 0)s(Yy)s(sY o =+−ττ+

=+τ

τ=⇒ 1

oo

sy

1sy)s(Y

Inverse Laplace( )τ−

=t

oey)t(ystart at and exponentially decay to zero as t ∞

oy

Source: Dynamic Systems – Vu & Esfandiari



0otherwise0tifA)t(f =>=Forced Response – Step Response

[ ]sA)s(Yy)s(sY o =+−τ

( )1ssAsy)s(Y o

+τ+τ

=⇒

Laplace Transform

τ+

+=

τ+

τ+

= 1s

csc

1ss

Asy)s(Y 21

o

break up using partial fraction expansion



A1sAsy)s(sYc 0s

o0s1 =

+τ+τ

== ==

where

( )

τ+

−+=∴ 1s

1AysA)s(Y o

Ays

Asy)s(Y1sc o1s

o1s2 −=τ

+=

τ+=

τ−=τ−=



τ−−+=

toSTEP e)Ay(A)t(y

Inverse Laplace

−=→= τ− t

STEPo e1A)t(y0yif

A632.0)(y =τA865.0)2(y =τA950.0)3(y =τA982.0)4(y =τA995.0)5(y =τ

Scan Fig. 7.2 p. 337



Transient Response – First-order SystemForced Response – Ramp Response

0otherwise0tifAt)t(f =>=Laplace Transform

[ ] 2o sA)s(Yy)s(sY =+−τ

τ+

++=

τ+

+τ= 1s

Cs

BsB

1ss

syA

)s(Y 122

2

2o

where AyC;AB;AB o12 τ+=τ−==


Transient Response – First-order SystemInverse Laplace

Note that the SS response is . After the transient portion decays, the response will track the ramp but with an error of .

τ− AAt

τ−τ+τ−=

tRAMP eAAAt)t(y

τA



Transient Response – Second-order System

Free Response – No forcing function – Only I.C.

oo2

nn x;x.C.I)t(fxx2x &&&& =ω+ζω+

Laplace Transform

[ ] [ ] 0)s(X)0(x)s(sX2)0(x)0(sx)s(Xs 2nn

2 =ω+−ζω+−− &

( )2

nn2

00n

ωs2ζsxx2ζsX(s)

++−+

=•

ωω

0s2s 2nn

2 =ω+ζω+Solution form depends on the poles of the characteristic equation

( ) 2n

2nn2,1s ω−ζω±ζω−=∴

12nn −ζω±ζω−=


Transient Response – Second-order SystemThree cases of damping•Case 1 – Overdamped (ζ>1)•Case 2 – Critically damped (ζ=1)•Case 3 – Underdamped (0<ζ<1)•Case 4 – Undamped (ζ=0)

Scan Fig. 7.5 p. 343



Transient Response – Second-order SystemCase 1 – Overdamped (ζ>1) – Two Real Roots

( ) ( )( )( )21

00n2

nn2

00nssss

vx2ss2s

vx2s)s(X−−+ζω+

=ω+ζω+

+ζω+=

Partial Fraction Expansion

;1s 2nn1 −ζω+ζω−= 1s 2

nn2 −ζω−ζω−=

2

2

1

1ss

Ass

A)s(X−

+−

=12

vx1A

2n

002

n1

−ζω

+

−ζ+ζω

=

12

vx1A

2n

002

n2

−ζω

+

−ζ−ζω−

=


Transient Response – Second-order SystemCase 1 – Overdamped (ζ>1) – Two Real Roots

( )t12

n

002

n 2ne

12

vx1)t(x −ζ−ζω−

−ζω

+

−ζ+ζω

=

( )t12

n

002

n 2ne

12

vx1−ζ+ζω−

−ζω

+

−ζ−ζω

−


Transient Response – Second-order SystemCase 2 – Critically damped (ζ=1) – Two Real Repeated Roots

( ) t00n

t0

nn tevxex)t(x ω−ω− +ω+=

roots) repeated real two(s n2,1 ω−=

( ) ( )( )2n

00n0n2

nn2

00n

svxxs

s2svx2s)s(X

ω+

+ω+ω+=

ω+ω+

+ω+=

( )2n

00n

n

0

svx

sx)s(X

ω+

+ω+

ω+=

Inverse Laplace


Transient Response – Second-order SystemCase 3 – Underdamped (0<ζ<1) – Complex Conjugate Pair

( ) 2n

2n

22n

2nn

2 ss2s ω+ωζ−ζω+=ω+ζω+

Define: nζω=σ2

nd 1 ζ−ω=ω

( ) ( )( ) 2

d2

0002

nn2

00n

svxx2s

s2svx2s)s(X

ω+σ+

++σ+=

ω+ζω+

+ζω+=

Then

( ) 2d

2s ω+σ+=

Then


Transient Response – Second-order Systemwhich can be rearranged into

( )( ) ( ) 2

d2

002

d2

0

svx

sxs)s(X

ω+σ+

+σ+

ω+σ+

σ+=

Inverse Laplace

( )( )

( )( ) 2

d2d

d

002

d2

0

svx

sxs

ω+σ+

ωω+σ

+ω+σ+

σ+=

tsinevxtcosex)t(x dt

d

00d

t0 ω

ω+σ

+ω= σ−σ−

OR

ω

ω+σ

+ω= σ− tsinvxtcosxe dd

00d0

t


Forced Response – Unit Impulse

with I.C.

Grouping terms

oo x;x &)t()t(fxx2x 2nn δ==ω+ζω+ &&&

Laplace Transform

[ ] [ ] )s(F)s(X)0(x)s(sX2)0(x)0(sx)s(Xs 2nn

2 =ω+−ζω+−− &

( )2

nn2

oon2

nn2 s2s

vx2ss2s)s(F)s(X

ω+ζω+

+ζω++

ω+ζω+=

If I.C. are zero, then the result is the system transfer function

2nn

2 s2s1

)s(F)s(X)s(H

ω+ζω+==

The inverse Laplace of this yields the impulse response of the system. (Note that I.C. are zero)


Forced Response – Unit ImpulseCase 1 – Overdamped

( ) ( )t12

n

t12

n

2n

2n e

12

1e12

1)t(x −ζ+ζω−−ζ−ζω−

−ζω−

−ζω=

( )t12

n

002

n 2ne

12

vx1−ζ−ζω−

−ζω

+

−ζ+ζω

+

( )t12

n

002

n 2ne

12

vx1−ζ+ζω−

−ζω

+

−ζ−ζω

−

I.C.



Case 2 – Critically Damped

( ) toon

to

t nnn tevxexte)t(x ω−ω−ω− +ω++=

I.C.



Case 4 – Undamped

Case 3 – Underdamped

( )

ω

ω+σ

+ω+ωω

= σ−σ− tsinvxtcosxetsine1)t(x dd

oodo

td

t

d

I.C.

tsinvtcosxtsin1)t(x nn

onon

nω

ω+ω+ω

ω=


Forced Response – Unit Step

with I.C. oo x;x &)t(u)t(fxx2x 2nn ==ω+ζω+ &&&



Frequency Response Function

For a 1st order system

The FRF can be obtained from the Fourier Transform of Input-Output Time Response (and is commonly done so in practice)The FRF can also be obtained from the evaluation of the system transfer function at s=jω.

ωτ+=ω=⇒

+τ= ω= j1

1)j(H)s(H1s

1)s(H js

For a 2nd order system

2nn

2 s2s1)s(H

ω+ζω+=

2nn

2jsj2

1)j(H)s(Hω+ωζω+ω−

=ω=⇒ ω=


Frequency Response FunctionThe FRF for a mechanical system

kjcm1)j(H

kcsms1)s(H 22 +ω+ω−

=ω⇒++

=

This is normally presented in a LOG-MAG and PHASE plot called a BODE DIAGRAM.


FRF – Bode Diagram – 1st Order



FRF – Bode Diagram – 2nd Order



Fourier Series

A periodic function is a function that satisfies the relationship

where is the period of the function( ) ( )Ttftf +=

T

The fundamental angular frequency is defined by

(rad/sec) T2

oπ

=ω


Fourier Series

Consider a family of waveforms

( )( ) ( )

( ) ( )

( ) ( )

( ) ( )t14sin71tf

t10sin51tf

t6sin31tf

t2sintf1tf

4

4

3

2

1

π=

π=

π=

π==

Summed as

( ) ( ) 5N1 tftgN

1nn ≤<= ∑

=


Fourier Series

The summed waveforms can be plotted in Matlab to approximate a square wave


Fourier Series

A Fourier Series representation of a real periodic function is based upon the summation of harmonically related sinusoidal components.

Note: These two are equivalent – one is written with sine and cosine terms – the other is written as a sin with amplitude and phase

If the period is , then the harmonics are sinusoids with frequencies that are integer multiples of ; This is writtenas

( )tf

Toω

( )non tnsinA φ+ω=

( ) ( ) ( )tnsinbtncosatf ononn ω+ω=


Fourier Series

The Fourier Series representation of an arbitrary periodic waveform is an infinite sum of harmonically related sinusoids and is commonly written as

( )tf

( )[ ]∑∞

=φ+ω+=

1nnono tnsinAa

21

( ) ( ) ( )[ ]∑∞

=ω+ω+=

1nonon0 tnsinbtncosaa

21tf

Or( ) ( ) ( )tjntjnntjntjnn oooo ee

j2bee

2atf ω−ωω−ω +++=

( ) ( ) tjnnn

tjnnn

oo ejba21ejba

21 ωω −++−=

[ ]∑∞

∞−

ω= tjnn

oeF


Fourier Coefficients

The derivation of the expression for computing the coefficients in a Fourier Series is beyond the scope of this course. Without proof, the coefficients are

( ) ( )∫+

ω=Tt

ton

1

1

dttnsintfT1b

( )∫+

ω−=Tt

t

tjnn

1

1

o dtetfT1F

which is evaluated over any periodic segment or in sinusoidal form

nF

( ) ( )∫+

ω=Tt

ton

1

1

dttncostfT1a


Response of Linear System Due to Periodic Inputs

Consider a linear SDOF system with a steady state periodic input (assume that the initial conditions and transients have decayed to zero)

( )∑∞

=φ+ω+=

1nnono tnsinAa

21

( ) ∑∞

∞−

ω= tjnn

oeFtf

The input excitation can be described by a Fourier Series representation as



The system output response is related to the input excitation through the system frequency response function as

( ) ( )φ+ω= tnsinAtf onn( )tynGenerates an output sinusoidal component with a

magnitude and phase determined by the nth component of

( ) ( ) ( )ω⋅ω=ω jFjHjY

The nth real harmonic input component of the Fourier Series

( )ωjH

( ) ( ) ( )[ ]ononon jnHnsinAjnHty ω∠+φ+ωω=


Response of Linear System Due to Periodic InputsFrom superposition, the total output is the sum of all then components as

( ) ( ) ( )[ ]∑∞

=ω∠+φ+ωω+=

1nonoono jnHtnsinjnHAjoHa

21

( )tyn

Note: This is also a Fourier Series with the same fundamental and harmonic frequencies as the input

( ) ( )∑∞

==

0nn tyty

The complete output Fourier Series is

( ) ( ) ( )∑∑∞=

−∞=

ω∞=

−∞=ω==

n

n

tjnno

n

nn

oeFjnHtyty

( ) ( ) tjnnon

oeFjnHty ωω=In complex form, this is written as


Response of Linear System Due to Periodic InputsConsider a simple mass spring dashpot system that is under steady state sinusoidal excitation (but not a of a SDOF system

nω

( ) ( )tsintf ω= ( )th ( ) ( )φ+ω= tsinAty

( )ωjf ( )ωjh ( )ωjy

SystemTime Domain

Frequency Domain

FFT IFT


Response of Linear System Due to Periodic InputsNow consider excitation at the natural frequency, ,of the SDOF system

nω


Response of Linear System Due to Periodic InputsNow consider a periodic input which is not sinusoidal. However, the Fourier Series representation is nothing more than the sum of different amplitude sine waves.



The first-order electric network shown in Fig. 15.7 is excited with the sawtooth function. Find an expression for the series representing the output Vout(t).

Example 15.7



( )1RCs

1sH+

=

Example 15.7 (cont.)The electric network has a transfer function

and therefore has a frequency response function

( )( ) 1RC

1jH2 +ω

=ω

( ) ( ) ( )∑∞

=

+ω

π−

=1n

o

1ntnsin

n12tu

( ) ( )RCtanjH 1 ω−=ω∠ −

the input function u(t) may be represented by the Fourier Series



( ) ( ) ( ) ( )[ ]∑∞

=

+ω∠+ω

π−

ω=1n

oo

1njnHtnsin

n12|jH|ty

( ) ( )( )

( )[ ]∑∞

=

−+

ω−+ω

+ωπ

−=

1no

1o2

o

1nRCntantnsin

1RCnn

12ty

RCT π=

Example 15.7 (cont.)At the output the series representation is

As an example consider the response if the period of the input is chosen to be so that ,then

RC2

o =ω

( ) ( )( )

( )

−+

+π

−= −

∞

=

+

∑ n2tantRC

n2sin1n2n

12ty 1

1n2

1n



RCT π=

Example 15.7 (cont.)

RC2

o =ω

Figure 15.8 shows the computer response found by summing the first 100 terms of the Fourier Series.



( )π<≤π−

π<≤=

2t s/m 10t0 s/m 10

tVin

Example 15.8A cart shown in Figure 15.9a, with a mass m=1.0 kg is supported on low friction bearings that exhibit a viscous drag B = 0.2 N-s/m and is coupled through a spring with stiffness K=25 N/m to a velocity source with a magnitude of 10 m/s but which switches direction every π seconds as shown in Figure 15.9b.

( )tvmThe task is to find the resulting velocity of the mass


Response of Linear System Due to Periodic InputsExample 15.8 (cont.)

( ) ( ) ( )

+ω+ω+ω

π= ...t5sin

51t3sin

31tsin20

ooo

π= 2T( )tΩThe input has a period of and a fundamental frequency of . The Fourier Series for the input contains only odd harmonics:

rad/s1T/2o =π=ω

( ) ( )[ ]∑∞

=ω−

−π=

1not1n2sin

1n2120tu



( ) ( )[ ]n2.0jn2525jnH 2o+−

=ω

The frequency response of the system is

( ) ( ) ω+ω−=ω

2.0j2525jH 2

radians/s nn o =ω

which when evaluated at the harmonic frequencies of the inputis



The following table summarizes the first five odd spectral components at the system input and output.



Figure 15-10a shows the computed frequency response magnitude for the system and the relative gains and phase shifts (rad) associated with the first five terms in the series.

Dynamic Systems System Response 031906 -...

Documents

Transcript of Dynamic Systems System Response 031906 -...