Sos, Fopz, Sopz

SOS – General Form

• General second order system

Standard form

SRSaunders - WSU - ChE 441 141

𝑎2

𝑑2𝑦(𝑡)

𝑑𝑡2+ 𝑎1

𝑑𝑦(𝑡)

𝑑𝑡+ 𝑎𝑜𝑦(𝑡) = 𝑏𝑢(𝑡)

𝜏2𝑑2𝑦(𝑡)

𝑑𝑡2+ 2𝜏𝜁

𝑑𝑦(𝑡)

𝑑𝑡+ 𝑦(𝑡) = 𝐾𝑢(𝑡)

𝜏 =𝑎2

𝑎𝑜, 2𝜏𝜁 =

𝑎1

𝑎𝑜, 𝐾 =

𝑏

𝑎𝑜

𝜏2𝑠2𝑌 𝑠 + 2𝜏𝜁𝑠𝑌 𝑠 + 𝑌 𝑠 = 𝐾𝑈(𝑠)

𝑌 𝑠 =𝐾

𝜏2𝑠2 + 2𝜏𝜁𝑠 + 1𝑈(𝑠)

SOS Parameters

• Three key parameters

Gain K

Natural period τ

Damping coefficient ζ (Zeta)

• Existence and nature of oscillation are

characterized by ζ and τ

• Poles?


𝑝1,2 =−2𝜏𝜁 ± 4𝜏2𝜁2 − 4𝜏2

2𝜏2=

−𝜁 ± 𝜁2 − 1

𝜏

Lets look at 3 test cases and their

response to a unit step…

K 10 10 10

τ2 40 42.25 13

2τζ 25 13 3


0 10 20 30 40 50 60 70 80 90 1000

5

10

15

time

y(t)

K=10,t2=40, 2tz=25

K=10,t2=42.25, 2tz=13

K=10,t2=40, 2tz=3

Which is :

Underdamped?

Overdamped?

Critically Damped?

SOS Dynamics

Damping Coefficient ζ

• Damping coefficient characterizes qualitative

process response

• Case 1: 0 < ζ < 1

Poles are real or imaginary?

Displays oscillation

Has OVERSHOOT

• progression past the final value,

• followed by a return to the

• steady state

Underdamped


0 20 40 60 80 1000

5

10

15

time

y(t)

K=10,t2=40, 2tz=3

SOS Dynamics


• Case 2: ζ = 1

Poles are real or

imaginary?

Fastest approach to

final value w/o

overshoot

Critically damped

• Case 3: ζ > 1

Poles are real or

imaginary?

Slower response than

case 2

Overdamped system


0 20 40 60 80 1000

2

4

6

8

10

time

y(t)

Overdamped

Critically Damped

SOS Dynamics


• Case 4: ζ = 0

• Pole are real or imaginary?

• Oscillatory response

• with no damping

• Frequency of oscillation :

• 1/τ => period = τ

• Case 5: ζ < 0

• UNSTABLE


0 10 20 30 40 50 60 70 80 90 1000

5

10

15

20

25

30

time

y(t)

Oscillatory No Damping

Unstable

FOS vs. SOS

• Step response comparison

SOS (w/ no zeros) always more

sluggish than FOS

SOS has an “S” shape response

TOS is even more sluggish

System FOS SOS

Final Value AK AK

Initial Value 0 0

Initial

SLOPE

Finite, non-

zero0


Underdamped SOS

• Rise time

Time to the first crossing of the final steady state

value


tr=12

𝑡𝑟 =𝜏

𝛽𝜋 − 𝜙

𝛽 = 1 − 𝜁2

𝜙 = tan−1𝛽

𝜁A=1, K=10, τ2=40, 2τζ=3

Underdamped SOS

• Period

Time between successive oscillation peaks


𝑇 =2𝜋𝜏

𝛽𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛

𝜔 =𝛽

𝜏𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛

A=1, K=10, τ2=40, 2τζ=3

Underdamped SOS

• Decay Ratio

A measure of the rate of oscillation decay

• Overshoot


a1=4.64

a2=1.0𝐷𝑅 =𝑎2

𝑎1= 𝑒

−2𝜋𝜁𝛽

A=1, K=10, τ2=40, 2τζ=3𝑂𝑆 = 𝐴𝑒

−𝜋𝜁

1−𝜁2

Underdamped SOS

• Settling Time – time at which the output

enters (and remains within) a percentage of

the final value

Often 90%, 95% or 99% settling time


t90%=49

t95%=69

t99%=123

A=1, K=10, τ2=40, 2τζ=3

FOS - Lead-lag Systems

• System with a “proper” transfer function

Gain K

Zero -1/𝜏𝑎

Pole -1/τ

Lead-to-lag ratio ρ = 𝜏𝑎𝜏

• Lead arises from the zero, lag from the pole.


𝐺 𝑠 =𝐾(𝜏𝑎𝑠 + 1)

𝜏𝑠 + 1

FOS – Lead-Lag Systems

• Partial Fraction Expansion

• General Form


(𝜏𝑎𝑠 + 1)

𝜏𝑠 + 1= 𝐴 +

𝐵

𝜏𝑠 + 1

𝜏𝑎𝑠 + 1 = 𝐴 𝜏𝑠 + 1 + 𝐵

𝑠1: 𝜏𝑎𝑠 = 𝐴𝜏𝑠

𝐴 =𝜏𝑎

𝜏= 𝜌

𝑠0: 1 = 𝐴 + 𝐵𝐵 = 1 − 𝐴 = 1 − 𝜌

𝐺 𝑠 =𝐾(𝜏𝑎𝑠 + 1)

𝜏𝑠 + 1= 𝐾𝜌 +

𝐾 1 − 𝜌

𝜏𝑠 + 1

Lead-Lag Step Response

• Observations

For very small t (t->0, i.e., use initial value

theorem) y(0)=KρA

• Discontinuous jump in the output signal

• Effect of the zero

For very large t (t->∞, i.e., use final value

theorem) y(∞)=KA

• Effect of lag term

Behavior of y(t) is a big function of ρ


𝑌 𝑠 =𝐾 𝜏𝑎𝑠 + 1

𝜏𝑠 + 1𝑈 𝑠 =

𝐾𝜌𝐴

𝑠+

𝐾 1 − 𝜌 𝐴

𝑠(𝜏𝑠 + 1)

Lead-Lag Systems – Effect of ρ

• Case 1: 0 < 𝜏𝑎 < τ (0 < ρ < 1)

Discontinuous Jump @ moment of

step

𝜏𝑎-> 0, becomes more of a “pure”

FOS and lag dominates


A=1, K=5, 𝜏𝑎=0.5, τ=1y(t

)

0


• Case 2: 𝜏𝑎 = τ (ρ = 1)

Pole-zero cancellation

Pure gain system G(s)=K

Discontinuous jump @

moment of step


A=1, K=5, 𝜏𝑎=0.5, τ=0.5

y(t

)

0


• Case 3: 𝜏𝑎 > τ (ρ > 1)

Overshoot

Lead dominates

Discontinuous jump @

moment of step


A=1, K=5, 𝜏𝑎=1, τ=0.5

y(t

)

0


• Case 4: 𝜏𝑎 < 0 < τ (ρ < 0)

Inverse response

Initial move away from the SS Value


A=1, K=5, 𝜏𝑎=-1, τ=0.5

y(t

)

t0

0

FOS In Parallel

𝐺1 𝑠 =𝐾1

𝜏1𝑠 + 1

𝐺2 𝑠 =𝐾2

𝜏2𝑠 + 1

𝑌 𝑠 = 𝑌1 𝑠 + 𝑌2 𝑠

= 𝐺1 𝑠 𝑈 𝑠 + 𝐺2 𝑠 𝑈 𝑠

= 𝑈 𝑠 𝐺1 𝑠 + 𝐺2 𝑠

= 𝑈 𝑠 𝐺(𝑠)

𝐺 𝑠 = 𝐺1 𝑠 + 𝐺2 𝑠 =𝐾1

𝜏1𝑠 + 1+

𝐾2

𝜏2𝑠 + 1

SOS with

Zeroes!


=𝐾1(𝜏2𝑠 + 1) + 𝐾2(𝜏1𝑠 + 1)

(𝜏1𝑠 + 1)(𝜏2𝑠 + 1)

=(𝐾1 + 𝐾2)

(𝐾1𝜏2 + 𝐾2𝜏1)𝐾1 + 𝐾2

𝑠 + 1

(𝜏1𝑠 + 1)(𝜏2𝑠 + 1)

=(𝐾1𝜏2 + 𝐾2𝜏1)𝑠 +(𝐾1 +𝐾2)

(𝜏1𝑠 + 1)(𝜏2𝑠 + 1)=

𝐾1𝜏2𝑠 + 𝐾1 + 𝐾2𝜏1𝑠 + 𝐾2

(𝜏1𝑠 + 1)(𝜏2𝑠 + 1)

G2

G1

U(s) Y(s)+

+

SOS with Zeroes

• 2-pole, 1 zero system:

• Output step response:


𝐺 𝑠 =𝐾 𝜏𝑎𝑠 + 1

𝜏1𝑠 + 1 𝜏2𝑠 + 1

𝑌 𝑠 =𝐾 𝜏𝑎𝑠 + 1

𝜏1𝑠 + 1 𝜏2𝑠 + 1

𝐴

𝑆= 𝐾𝐴

𝐵

𝑠+

𝐶

𝜏1𝑠 + 1+

𝐷

𝜏2𝑠 + 1

𝐵 = 1

𝐶 = −𝜏1 𝜏1 − 𝜏𝑎

𝜏1 − 𝜏2

𝐷 =𝜏2 𝜏2 − 𝜏𝑎

𝜏1 − 𝜏2

𝑦 𝑡 = 𝐴𝐾 1 −𝜏1 − 𝜏𝑎

𝜏1 − 𝜏2𝑒

−𝑡𝜏1 +

𝜏1 − 𝜏𝑎

𝜏1 − 𝜏2𝑒

−𝑡𝜏2

SOS with Zeroes – Step Response

A=1, K=10, 𝜏1=5, 𝜏2=10


>

SOS – Case Evaluations

• Let τ1

< τ2, τ

a> 0

Unless otherwise noted

• Case 1: τa

> τ2

Overshoot

• Case 2: τa

= τ1

or τa

= τ2

Pole-zero cancellation

Yields a FOS


𝐺 𝑠 =𝐾 𝜏𝑎𝑠 + 1

𝜏1𝑠 + 1 𝜏2𝑠 + 1=

𝐾

𝜏𝑖𝑠 + 1

>

SOS – Case Evaluations

• Case 3: 0 < τa

< τ2

Resembles a FOS until τa

<< τ1

• Case 4: τa

< 0

Always displays inverse response


Case Summary – FOS τa

Values Key Observations

0 < τa < τJump at t=0 toward

y(∞)

τa = τ

Pure gain system

(pole-zero

cancellation)

τa > τ Overshoot

τa < 0 < τ Inverse response


• Discontinuous jump @ t=0 for all cases

• Let τ1

< τ2, τ

a> 0

• Unless otherwise noted

• No discontinuous jump @ t=0

Case Summary – SOS τa


0 < τa < τ2 Similar to FOS

τa = τ1 or τ2

FOS

(pole-zero cancellation)

τa > τ2 Overshoot

τa < 0 Inverse response


Case Summary – SOS ζ


ζ < 0 Unstable

ζ = 0Underdamped

oscillates forever

0 < ζ < 1Overshoot and

underdamped

ζ = 1 Critically damped

ζ > 1Overdamped –

sluggish


Inverse Respone

• When a process output initially moves in a

direction opposite to its steady state value

followed by a return to steady state

Effect of (at least) two opposing processes are

different timescales

Occurs when τa

< 0 in single-zero systems

y(t) crosses the zero axis (in deviation variables)

in response to a step input

• Where does this show up?


A Two Timescale Exercise

• Given the following block diagram and

transfer functions, calculate G(s)


𝐺1 𝑠 =5

10𝑠 + 1𝐺2 𝑠 =

−1

1𝑠 + 1

𝐺 𝑠 = 𝐺1 𝑠 + 𝐺2 𝑠

=5

10𝑠 + 1−

1

𝑠 + 1

=5 𝑠 + 1 − (10𝑠 + 1)

(10𝑠 + 1)(𝑠 + 1)

=4 −

54

𝑠 + 1

10𝑠2 + 11𝑠 + 1

G2

G1

U(s) Y(s)+

+

FOS in Parallel

• Two FOS in Parallel

• Let:

|K1|> |K

2|

K1

and K2

be opposite signs

τ1

> τ2

(G2

is faster than G1)

• Consequences

Fast process => Initial response

Slow Process => final response (due to higher

gain)


𝐺(𝑠) =(𝐾1 + 𝐾2)

(𝐾1𝜏2 + 𝐾2𝜏1)𝑠𝐾1 + 𝐾2

𝑠 + 1

(𝜏1𝑠 + 1)(𝜏2𝑠 + 1)

Real Inverse Response

• Drum Boiler

Used for steam generation


Heat

Source

Steam

Cold Water

Documents

Sos, Fopz, Sopz