contr systems ppt07e (Time domain analysis, 2nd order...

Class 07“Time domain analysis”

parte II - 2nd order systems

cbsas)s(G

2 ++α=

outputinput

Second order systems

of the type

S

Time domain analysis – 2nd order systems ______________________________________________________________________________________________________________________________________________________________________________________

S cbsas2 ++

α

cbsas)s(G

2 ++α=

Second order systems

of the type


outputinput

a

cs

a

bs

a

)s(R

)s(Y

2 ++

α

=

Second order systems:

Koωn2

ωn2

that is:

2ζωn

cbsas2 ++

α

cbsas)s(R

)s(Y2 ++

α=


outputinput

the transfer function can be rewritten as:

Ko = gain of the system

ζ = damping coefficient

ωn = natural frequency

2

nn

2

2

no

s2s

K

)s(R

)s(Y

ω+ζω+ω=

outputinput

2o n

2 2n n

K

s 2 s

ω+ ζω + ω


Besides these 3 above parameters, we also have

outputinput

2o n

2 2n n

K

s 2 s

ω+ ζω + ω

ωd = damping frequency

1012

nd ≤ζ<ζ−⋅ω=ω


Ko = gain of the system

ζ = damping coefficient

ωn = natural frequency

Example 1:

1s12s4

3

)s(R

)s(Y2 ++

=

Ko = 3 ζ = 1 ωn = 1

Example 2:

1s2s

3

)s(R

)s(Y2 ++

=

Ko = 3 ζ = 3 ωn = 0,5

poles are

real and

distinct

poles are

real and

repeated

poles:

s = –2,914

s = –0,086

poles:

s = –1

(duple)

ωd = 0


2s2s2

3

)s(R

)s(Y2 ++

=

Example 3:

Ko = 3 ζ = 0 ωn = ωd = 1

Example 4:

Ko = 1,5 ζ = 0,5 ωn = 1

1s

3

)s(R

)s(Y2 +

=

complex

conjugate

poles

complex

conjugate

poles

poles:

s = –0,5 ± 0,866j

poles:

s = ± j

(pure imaginary)

ωd = 0,866


2

nn

2s2s)s(p ω+ζω+=

Characteristic equation:

∆ = 4 ζ2 ω2n – 4 ω2

n =

= 4 ω2n (ζ2 – 1)

ζ > 1 → poles are real and distinct

ζ = 1 → poles are real and repeated

0 < ζ < 1 → poles are complex conjugate

∆ > 0 → (ζ2 – 1) > 0 → ζ2 > 1 → ζ > 1

∆ = 0 → (ζ2 – 1) = 0 → ζ2 = 1 → ζ = 1

∆ < 0 → (ζ2 – 1) < 0 → ζ2 < 1 → ζ < 1


What is the output?

(step response)

unit step input

outputinput

2o n

2 2n n

K

s 2 s

ω+ ζω + ω


2o n

2 2n n

KY(s) R(s)

s 2 s

ω= ⋅+ ζω + ω

and since r(t) = unit step:

2o n

2 2n n

K 1Y(s)

ss 2 s

ω= ⋅+ ζω + ω

2o n

2 2n n

K

s 2 s

ω+ ζω + ω

outputinput


[ ])s(Y)t(y 1−= Lthe unit step response depends on the value of ζ:

a) 0 < ζ < 1 (under damping)

b) ζ = 1 (critical damping)

c) ζ > 1 (over damping)

2o n

2 2n n

K

s 2 s

ω+ ζω + ω

outputinput


0t,tsen1

tcos1K)t(y d2

d

tn

o >

ω⋅

ζ−ζ+ω−= ζω−

e

So, in the case 0 < ζ < 1 (under damping) the unit step response

is:

[ ])s(Y)t(y 1−= L

2

nd 1 ζ−⋅ω=ωwhere

( damping frequency )

2o n

2 2n n

K

s 2 s

ω+ ζω + ω

outputinput


ω⋅


tsen1

tcos1K)t(y d2

d

tn

o e

unit step response :

2o n

2 2n n

K

s 2 s

ω+ ζω + ω

outputinput


unit step response:

ω⋅


tsen1

tcos1K)t(y d2

d

tn

o e


( )[ ] 0t,t11K)t(y n

tn

o >ω+⋅−= ζω−e

In the case ζ = 1 (critical damping) the unit step response is:

2o n

2 2n n

K

s 2 s

ω+ ζω + ω

[ ])s(Y)t(y 1−= L

outputinput


2o n

2 2n n

K

s 2 s

ω+ ζω + ω


outputinput

( )[ ]t11K)t(y n

tn

o ω+−= ζω−e



( )[ ]t11K)t(y n

tn

o ω+−= ζω−e


0t,pp12

1K)t(y2

tp

1

tp

2

n21

o >

−⋅

−ζω+= ee

In the case ζ > 1 (over damping) the unit step response is:

[ ])s(Y)t(y 1−= L

( )11p 2

n

2

nn2,1 −ζ±ζω−=−ζωζω−= m

whereSystem has

real poles

2o n

2 2n n

K

s 2 s

ω+ ζω + ω

outputinput


unit step response:

−⋅

−ζω+=

2

tp

1

tp

2

n

pp121K)t(y

21

o

ee

2o n

2 2n n

K

s 2 s

ω+ ζω + ω

outputinput


unit step response:

−⋅

−ζω+=

2

tp

1

tp

2

n

pp121K)t(y

21

o

ee


ζ = 1

ζ = 2

unit step

response


ζ = 4

ζ = 12

unit step

response


The under damping case


In the case 0 < ζ < 1,

the unit step response

can have many different forms, depending on the

values of ζ (damping coefficient),

ωn

(natural frequency) e Ko (gain)

0t,tsen1

tcos1K)t(y d2

d

tn

o >

ω⋅


e

Observe that ωd

depends on ζ and ωn

2

nd 1 ζ−⋅ω=ω (damping frequency)


ζ = 0,132

ωn = 0,57

ζ = 0,1

ωn = 2

unit step

response


ζ = 0,25

ωn = 1

ζ = 0,5

ωn = 0,2

unit step

response


ζ = 0,65

ωn = 2

ζ = 0,72

ωn = 0,8

unit step

response


ζ = 0,8

ωn = 1,4

ζ = 0,85

ωn = 0,7

unit step

response


ζ = 0

ωn = 0,2

unit step

response


Now let us concentrate in the

case 0 < ζ < 1 and calculate

some parameters.


calculate some parame-

ters/variables for y(t)

the unit step response

of the 2nd order system.


steady state output

yss

yss = steady state response

( steady state output )

)s(Rs2s

K)s(Y

2

nn

2

2

no ⋅ω+ζω+

ω=

o

2

nn

2

2

no

0s

ostss

K

s

1

s2s

sKlim

)s(Yslim)t(ylimy

=

=⋅ω+ζω+

ω=

=⋅==

→

→∞→

yss = Ko

s

1)s(R =


oss Ky =


rising time

tr

tr rising timetime it takes the step response y(t)

to reach the final value yss = Ko for

the first time.


oo Ktsen1

tcos1K)t(y rd2

rd

t

rrn =

ω⋅


e

tr = rising time

0tsen1

tcos rd2

rdrtn =

ω⋅

ζ−ζ+ωζω−

e

1

ζζ−−=ω

2

rd

1)t(tg

n

d

ζωω−=

Is the instant of time that y(t) reaches the final value Ko

for the first time.


tr = rising time

Depends on the values of ζ ( damping coefficient ),

and of ωn ( natural frequency )

d

2

d

n

d

r

1arctgarctg

tω

ζζ−−

=ω

ζωω−

=

tr = arctg(– ωd /ζω n) / ωd


d

n

d

r

arctg

tω

ζωω−

=


peak time

tp

tp peak time is the instant of time in

which the step response y(t) reaches

the first peak.


ω⋅

ζ−ωζ+ωω−−

+

ω⋅

ζ−ζ+ω⋅ζω==′

ζω−

ζω−

rd2

drdd

t

rd2

rd

t

n

tcos1

tsen

tsen1

tcosKdt

dyy

n

n

o

e

e

peak time yss = Ko ( gain )

ζ ( damping coefficient ),

ωn ( natural frequency )

ωn


)tcostsen

tsen1

tcosKdt

dy

dndd

d2

n

2

dn

tn

o

ω⋅ζω−ωω+

+ω⋅

ζ−ωζ+ωζω⋅= ζω−

e

ζ−ζ−ω+

ζ−ωζ⋅ω⋅⋅= ζω−

2

2

n

2

n

2

d

t

1

)1(

1tsenK n

o e

01

tsenK2

nd

tn

o =ζ−

ω⋅ω⋅⋅= ζω−e


peak time

0dt

dy =

d

ptωπ=

0tsen d =ω

L,3,2,,0td πππ=ω

tp = π / ωd


peak time

d

ptωπ=


overshoot

Mp

Mp overshoot

it is the percentage above of

the final value yss that is

reached by the first peak.


The overshoot Mp can be expressed as a

value between 0 and 1.


Or, instead, it can be expressed as a

value between 0% and 100%.


0 ≤ overshoot Mp ≤ 1

or

0% ≤ overshoot Mp ≤ 100%


o

o

ζ1

t

o

pK

K)(senζ

)(cos1K

M2

pnζ −

π−π−

=−

ω−e

overshoot

ss

ssmaxp

y

yyM

−=o

op

pK

K)t(yM

−=ou

( )

o

o

/

oop

K

KKKM

dnζ −+=

ωπω−e

– 1 0

yss = Ko ( gain )

ζ ( damping coefficient ),

ωn ( natural frequency )


o

2

op

K

KM

ζ1

πζ

−

−

= e

2

p

ζ1

πζ

M−

−

= e

Mp depends only on ζ ( damping coefficient )


overshoot

%100M2

p

ζ1

πζ

×= −

−

e

overshoot

Mp depends only on ζ ( damping coefficient )

2

p

ζ1

πζ

M−

−

= eor


2

p

ζ1

πζ

M

−

−

= e


settling time

ts

ts settling time

is the time required for the step

response y(t) to reach and remain

within a given error band around

the final value yss.


This error band can

be of 5% above and

5% below of the

final value yss.


or, instead, from

2% above and 2%

below the final

value yss.


The settling time is obtained from the equations of ye(t), the

curves that encompass y(t).

[ ] 0t,1K)t(yt

en

o >±= ζω−e


n

s

3%)5(t

ζω=

n

s

4%)2(t

ζω=

That is, the settling time ts is obtained by calculating

ye(ts) ≈ 1,05 Ko

for the case of ts with 5% tolerance, and

ye(ts) ≈ 1,02 Ko

for the case of ts with 2% tolerance,

obtaining the following values:


n

s

3%)5(t

ζω=

n

s

4%)2(t

ζω=

ts(5%) = 3 / ζωn

ts(2%) = 4 / ζωn

hence:

and

Análise no domínio do tempo - Sistemas de 2ª ordem______________________________________________________________________________________________________________________________________________________________________________________

Note that the settling time

tr is inversely proportional

to ζωn, which is the

distance of the real part of

the poles to the origin.

settling time

n

s

3%)5(t

ζω=


n

s

4%)2(t

ζω=


However, if ζ < 0 then:

Note that we have seen cases in which

the system is unstable

0 < ζ < 1

ζ = 1

ζ > 1

that is:

ζ > 0


ζ < 0 → unstable system (an example)


ζ < 0 → unstable system (another example)


Thank you!

Felippe de Souza

[email protected]

Departamento de Engenharia Eletromecânica

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