Second-Order Dynamic Systems KCC 2011

7/30/2019 Second-Order Dynamic Systems KCC 2011

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2nd-Order Dynamic System Response K. Craig 1

Time Response & Frequency Response

of a

2nd

-Order Dynamic System

I P

2

p I nc

2 22 P Ic n n

p p

KK KD 1

K D 1G GCKK 1 KK R 1 G G D 2 D

D D

Closed-Loop Transfer Function: 2nd-Order Dynamic System With Numerator Dynamics


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2

0 02 1 0 0 0 i2

2

0 00 i2 2

n n

d q dqa a a q b q

dt dt

d q dq1 2q Kq

dt dt

0n

2

1

2 0

0

0

aundamped natural frequency

a

a

damping ratio2 a a

bK steady-state gain

a

Step Response

of a

2nd-Order System

2nd-Order Dynamic

System Model


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2

0 00 i2 2

n n

d q dq1 2q Kq

dt dt

n t 2 1 2o is n2

1q Kq 1 e sin 1 t sin 1 1

1

Step Response

of a

2nd-Order System

2n

2n

21 t

2

o is2

1 t

2

11 e

2 1q Kq 1

1e

2 1

n to is nq Kq 1 1 t e 1

Over-damped

Critically Damped

Underdamped


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Frequency Response

of a

2nd

-Order System

o 2i

2

n n

Q KD

D 2 DQ1

1o

22i 2 2 n

2 n

n n

Q K 2i tanQ

41

Operational Transfer Function

Sinusoidal Transfer Function


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Frequency Response

of a2nd-Order System


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Frequency Response

of a2nd-Order System

-40 dB per decade slope


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Some Observations When a physical system exhibits a natural oscillatory

behavior, a 1st-order model (or even a cascade of

several 1st-order models) cannot provide the desired

response. The simplest model that does possess that

possibility is the 2nd-order dynamic system model.

This system is very important in control design.

System specifications are often given assuming thatthe system is 2nd order.

For higher-order systems, we can often usedominant pole techniques to approximate the system

with a 2nd-order transfer function.


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Damping ratio clearly controls oscillation; < 1 is requiredfor oscillatory behavior.

The undamped case ( = 0) is not physically realizable (totalabsence of energy loss effects) but gives us, mathematically, a

sustained oscillation at frequency n.

Natural oscillations of damped systems are at the dampednatural frequency d, and not at n.

In hardware design, an optimum value of = 0.64 is oftenused to give maximum response speed without excessive

oscillation.

Undamped natural frequency n is the major factor in responsespeed. For a given response speed is directly proportional to

n.

2d n 1


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Thus, when 2nd-order components are used in feedbacksystem design, large values ofn (small lags) are desirable

since they allow the use of larger loop gain before stabilitylimits are encountered.

For frequency response, a resonant peak occurs for 1.0), no oscillations exist,

and the determination of and nbecomes more difficult.Usually it is easier to express the system response in terms

of two time constants.


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For the over-damped step response:

where

2n

2n

1 2

21 t

2

o is2

1 t

2

t t

o 1 2

is 2 1 2 1

1

1 e2 1q Kq 1

1e

2 1

qe e 1

Kq

1 2

2 2

n n

1 11 1


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To find 1 and 2 from a step-function response curve, wemay proceed as follows:

Define the percent incomplete response Rpi as:

Plot Rpi on a logarithmic scale versus time ton a linearscale. This curve will approach a straight line for large

tif the system is second-order. Extend this line back

to t = 0, and note the value P1 where this line intersects

the Rpi scale. Now, 1 is the time at which the straight-line asymptote has the value 0.368P1.

opi

is

qR 1 100

Kq


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Now plot on the same graph a new curve which is the

difference between the straight-line asymptote and Rpi.If this new curve is not a straight line, the system is not

second-order. If it is a straight line, the time at which

this line has the value 0.368(P1-100) is numerically

equal to 2.

Frequency-response methods may also be used to find1 and 2.


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Step-

Response Test

for

OverdampedSecond-Order

Systems


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Frequency-

Response Test ofSecond-Order

Systems


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Dynamic System Exercise An underdamped 2nd-order system model has the following

transfer function:

Part 1: Using the properties and formulas for 2nd-order systems,

discuss the relationships between the step-response-

parameters rise time, settling time, and overshoot, and

the frequency-response-parameters bandwidth and peakamplitude as the model parameters vary. Use plots as

needed in your presentation.

2

n

2 2

n n

2

1,2 n n

1,2 d

KG(s) s 2 s

s i 1

s i


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Suggestion: Pick a base system. Generate 4 familiesof plots

d constant, vary constant, vary dn constant, vary

constant, vary n Show both time-response and frequency-responseplots. Include discussion.


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Part 2:

Investigate the effects on the time (step) response

and frequency response of adding a real pole or areal zero to the 2nd-order transfer function. The pole

and zero are added separately. In classical deign

using root-locus or frequency-response techniques,

real poles and zeros are added (lead, lag, lead-lag

controllers) to modify system dynamics, and so it is

important to have a good understanding of these

effects. Use plots as needed in your presentation.


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Suggestion: Pick a base second-order system.

Add a negative real pole (s + p) to the transferfunction and move the pole from the left towardsthe origin and describe its effect on the time-

response and frequency-response plots.

Add a negative real zero (s + z) to the transfer

function and move the zero from the left towardsthe origin and describe its effect on the time-

response and frequency-response plots.


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Part 3:

Now add a positive real zero to your base second-order system and evaluate the step response for thesystem. Explain your observations.

Physically, what might cause a transfer function to

have a right-half plane zero?


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Problem Solution

Base System

Effects of:

d = 1, = [0.5, 1, 5] Effects ofd:

= 1, d = [0.5, 1, 5]

Effects ofn:

= 0.707, n = [0.52, 2, 52] Effects of:

n = 2, = [0.866, 0.707, 0.5]

2

2G(s)

s 2s 2

d

1

1

2 22

dn

2 2 2 2 2

n n d

G(s)s 2 s s 2 s ( )

n2

0.707


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0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Amplitude

-40

-30

-20

-10

0

10

Magnitude(dB

)

100

101

-180

-135

-90

-45

0

Phase(deg)

Bode Diagram

Frequency (rad/sec)

= 0.5

Effects of varying

= 5

= 0.5

= 5

As increases:ts decreases

trdecreases

Mp decreasesBW increases

= 0.5

= 5

d = 1, = [0.5, 1, 5]


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-40

-30

-20

-10

0

10

Magnitude(dB)

100

101

-180

-135

-90

-45

0

Phase(deg)

Bode Diagram

Frequency (rad/sec)

0 1 2 3 4 5 6

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6Step Response

Time (sec)

Amplitude

d = 0.5

Effects of varying d

d = 5

d = 0.5

d

= 5

As d increases:ts is fixed

trdecreases

Mp increasesBW increases

d = 0.5

d = 5

= 1, d = [0.5, 1, 5]


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-70

-60

-50

-40

-30

-20-10

Magnitude(dB)

10-1

100

101

102

-180

-135

-90

-45

0

Phase(deg)

Bode Diagram

Frequency (rad/sec)

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Amplitude

n = 0.5

Effects of varying n

n = 5

n = 0.5

n = 5

As n increases:ts decreases

trdecreasesMp is fixed

BW increases

n = 0.5

n = 5

= 0.707, n = [0.52, 2, 52]


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-40

-30

-20

-10

0

10

Magnitude(dB)

100

101

-180

-135

-90

-45

0

Phase(deg)

Bode Diagram

Frequency (rad/sec)

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Amplitude

=0.5 Effects of varying

=0.866

= 0.5

=0.866

As increases:ts increases

trdecreasesMp increases

BW increases =0.5

=0.866

n = 2, = [0.866, 0.707, 0.5]


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Effect of an Additional LHP Pole

Base System

Additional Pole

22

G ss 2s 2

d

1

1

2 22

dn

2 2 2 2 2

n n d

G(s)

s 2 s s 2 s ( )

n2

0.707

2

3 2

2G(s)

ps 1 (s 2s 2)

1

ps (2p 1)s (2p 2)s 2

p [0,0.2, 1, 2]


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0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Amplitud

e

-100

-80

-60

-40

-20

Magnitude(dB)

10-1

100

101

102

-270

-180

-90

0

Phase(deg)

Bode Diagram

Frequency (rad/sec)

Effect of an Additional Pole

p [0,0.2, 1, 2]

increasing p

increasing p

increasing pAs p increases (pole gets

closer to the origin):

ts increases

trincreasesMp decreases to zero

BW decreases


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Effect of a LHP Zero

Base System

Add a Zero

22

G ss 2s 2

d

1

1

2 22

dn

2 2 2 2 2

n n d

G(s)

s 2 s s 2 s ( )

n2

0.707

2

2(zs 1)G(s)

(s 2s 2)

z [0,0.2, 1, 2]


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-60

-40

-20

0

Magnitude(dB)

10-1

100

101

-180

-135

-90

-45

0

45

Phase(deg)

Bode Diagram

Frequency (rad/sec)

0 1 2 3 4 5 6

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8Step Response

Time (sec)

Amplitud

e

Effect of a LHP Zero

z [0,0.2, 1, 2]

increasing z

increasing z

increasing zAs z increases (zero gets

closer to the origin):

ts increases

trdecreasesMp increases

BW increases


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Effect of a RHP Zero

Base System

Add a RHP Zero

22

G ss 2s 2

d

1

1

2 22

dn

2 2 2 2 2

n n d

G(s)s 2 s s 2 s ( )

n2

0.707

2

1 2 2 2

2 2 2 2

2G(s)

(s 2s 2)

2s 22 2s

G (s) (s 2s 2) (s 2s 2) (s 2s 2)

2 2s ( 2s 2)G (s)

(s 2s 2) (s 2s 2) (s 2s 2)

G(s) plus its derivative

G(s) minus its derivative


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0 1 2 3 4 5 6-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Amplitude G(s)

G1(s)

G2(s)


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0 1 2 3 4 5 6-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Amplitu

de 2

2

s 2s 2

2

2s

s 2s 2

2

2s 2

s 2s 2


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2 d O d D i S t R K C i 39

0 1 2 3 4 5 6-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2Step Response

Time (sec)

Amplitude

2 2s 2s 2

2

2s

s 2s 2

2

2s 2

s 2s 2

Documents

Second-Order Dynamic Systems KCC 2011