Q2 2 Second_order

8/2/2019 Q2 2 Second_order

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AE 230 - Modeling and

Simulation Laboratory



Second order systems

iqb

dt

idq

bdt

iqd

bo

qoa

dt

odq

adt

oqd

a012

2

212

2

2++=++

iqboqoadt o

dq

adt

o

qd

a 012

2

2 =++ iq

a

b

oq

dt

odq

a

a

dt

oqd

a

a

0

0

0

1

2

2

0

2 =++

iKq

oq

n

D

n

D=

++ 12

2

2

ω

ζ

ω

ty)(sensitivigainstatesteadysystem

essdimensionlratiodamping

timerad frequencynaturalundamped

where

Δ

0a

0b

ΔK

Δ

0a

2a

1a

Δζ

Δ

2a0

a

Δn

ω

Most general form of second order system

Special case and most practical in nature



Cascaded Second order systems




( )

( )1

1

122222

1

1 iioq

D

K K qK q D

+

==+

τ

τ

Output of one system going to input of other;




( )

( )1

1

122222

1

1 iioq

D

K K qK q D

+

==+

τ

τ

1212]1)12(21[2)12)(11(

2

iqK K oq D Doq D D=+++=++

τ τ τ τ τ τ

Output of one system going to input of other; second ordersystem but not as a single system




022s02012

0110111i

xK)x-x(B

xMxB-f

=

=

Bottom damper and spring canbe assumed to be a velocitysensor. Interested in finding X

02




1

) /()(

12

21212

12

21

212

1

02

++

+=

D BK

B BK M D

BK

B M

K B B D

f

x

s

s

s

s

i

21

1

02 x)( TF TF D f

x

i=

1

/ 1

1

1

1

1+

= D

B

M

BTF

1

/

2

2

22

2+

= D

K

B

K BTF

s

s




02202012

011020120111

)(

)(

xK x x B

x M x x B x B f

s

i

=−

=−−−

02

2

221

02

2

2220222

2

11 1)(

x D BK D B M

x D B

K D B B xK D B

B

B f

s

ssi

+=

−

+−+−

1)()(

))( /()(

212

21212

212

21

2122

1

02

++

+

++

+=

D B BK

B BK M

D B BK

B M

B BK B D

f

x

s

s

s

s

i




1)()(

))( /()(

212

21212

212

21

2122

1

02

+++++

+=

D B BK B BK M D

B BK B M

B BK B D

f

x

s

s

s

s

i

1

) /()(

12

21212

12

21

212

1

02

++

+=

D BK

B BK M D

BK

B M K B B D

f x

s

s

s

s

i

Approximate (no interaction/no loading effect)

Exact (with interaction/loading effect)




1)()(

))( /()(

212

21212

212

21

2122

1

02

++++

+

+=

D B BK B BK M D

B BK B M

B BK B D

f

x

s

s

s

s

i

If B2 is small compared to B1 then individual transferfunction will be a good approximation. System 2 is notsignificantly loading system1. Assuming all parameter areequal to 1, except B2 = 0.05

105.105.0

05.0)(

21

02

++=

D D D

f

x

i

10.10476.0

0476.0)(

2

1

02

++=

D D D

f

x

i

No Loading

Exact




Two isolated first order system when joined togetherdraws power from one of the system, if it is significant,system equation cannot be developed using just two firstorder system




o x M x B

o xs

K W W i f =−+−+

0)(

i f

o xs

K x Bo

x M =++0

iKf

oq

n

D

n

D=

++ 12

2

2

ω

ζ

ω

Force balance

Newton

meter

sK

KMs

K

B

M

sK

n

1,

2,

time

rad ∆∆∆ ζ




iKf oqn

D

n

D

=

++ 12

2

2

ω

ζ

ω 12 and12 −−−−+− ζ ω ω ζ ζ ω ω ζ nnnn

)(

roots,0,0:Undamped

φ

ω

+=

±===

tn

CSinocx

niB

)21(

21roots ,0.10,20:dUnderdampe

φω

ζ

+−−

=

−±−=<<<<

tn

Sint

nCeocx

nin

Ms

KB

Two roots of the characteristic equation




Mass M = 1 kg; K = 1 N/m and B = 0.2 N/(m/sec)

sec / 1time

rad rad

M

sK

n=∆ω 1.0

2=∆

M s

K

Bζ

The system is subjected to

i) Initial displacement (1m)

ii) Initial velocity (1 m/sec)iii) Step force (1 N)



Second order systems - underdamped

Response for initial displacement, initial velocity, step force input



Second order systems – Step input

Step response of undamped second order system ζ =0

)cos1( t nisKf o x ω −=




Step response of critically second order system ζ =1

))1(1(t n

et nisKf o xω

ω −

+−=



Second order systems - Overdamped

21

21

2

/

2

/

1

1,11roots ,0.1,2:dampedOver

τ

τζ

teC

teC

ocx

nMs

KB

−+

−=

=−±−=>>

Mass M = 1 kg; K = 1 N/m and B = 20 N/(m/sec)

0.102

=∆ M s

K

Bζ sec / 1

time

rad rad

M

sK

n=∆ω

−

−

−++

−

−

−+−= 2

2

2

1

2

2 /

12

1 /

12

11

τ

ζ

ζ ζ τ

ζ

ζ ζ t e

t eKf

o x is





0.102

=∆ M s

K

Bζ sec / 1

time

rad rad

M

sK

n=∆ω

−+−−=

t e

t eKf

o x nn

is

ω ω 95.19002.0

05.0002.11

−

−

−−+

−

−

−+−= 2

2

2

1

2

2 /

12

1 /

12

11

τ

ζ

ζ ζ τ

ζ

ζ ζ t e

t eKf

o x is





0.102

=∆ M s

K

Bζ sec / 1

time

rad rad

M

sK

n=∆ω

−−≈

−+−−=

t eKf

t e

t eKf

o x

nis

nnis

ω

ω ω

05.0002.11

95.19002.0

05.0002.11

−

−

−−+

−

−

−+−= 2

2

2

1

2

2 /

12

1 /

12

11

τ

ζ

ζ ζ τ

ζ

ζ ζ t e

t eKf

o x is




Non-dimensional step response of second order system




Effect of damping on overshoot



Significance of K, ζ, ωn

Steady state gain is only dependent of K

ωn largely governs the speed of response due to product (ωnt).Doubling the natural frequency will half the response time. Tospeed up by a factor n, natural frequency has to be increased by

a factor of n (when ζ is constant).

For step response, when ζ < 1.0 overshooting. To control theovershoot ζ should be adjusted. when ζ = 1.0, least time toreach steady state without overshoot. when ζ > 1.0, no

overshoot, time to reach steady state is more than when ζ = 1.0



Lab testing of second order systems – Step inputs



Lab testing of second order systems – Step inputs

etcd nd nd n ,,,

6,

4,

2,0

ω π

ω π

ω π

d n,

2

ω

π

Peaks occur at

Time period

Amplitude ratio of two successive peak is constant for anunder damped system, this can be used for finding thedamping ratio.

21

2

,

1, ζ

πζ

−

−

+= e

x

x

n p

n p



Assignment

Write system equation and transfer function for any twomechanical and two electrical systems given insubsequent slides



Second order – Electrical systems

Q2 2 Second_order

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