Ohhh

Smith & Corripio, 3rd edition

%TO := %

%CO := %

Problem 6-1. Second-order loop with proportional controller.

D(s) R(s) + E(s)
-

G2(s) G1(s)
+ +

Gc(s)

M(s)

C(s)

G1 ( s) = Problem parameters: K := 0.10

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)
%TO %CO τ 1 := 1min

K

Gc( s) = Kc τ 2 := 0.8min

(a) Closed loop transfer function and characteristic equation of the loop.
Kc⋅ =C( s) R( s)

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1) (τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1)
2

K

1 + Kc⋅

K

=

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) + Kc⋅ K

Kc⋅ K

Characteristic equation: Closed-loop transfer function:

τ 1 ⋅ τ 2 ⋅ s + τ 1 + τ 2 s + 1 + Kc⋅ K = 0 C( s) R( s)
2

(

)

=

0.1Kc 0.8s + 1.8s + 1 + 0.1Kc
2

Characteristic equation:

0.8s + 1.8s + 1 + 0.1Kc = 0

(b) Values ofthe controller gain for which the response is over-damped, critically damped, and under-damped
Roots of the characteristic equation: −1.8 + 1.8 − 4 ⋅ 0.8⋅ 1 + 0.1Kc 2 ⋅ 0.8
2

r1 =

(

)

=

−1.8 1.6

+

2  1.8  − 1 + 0.1Kc  0.8  1.6 

The response is critically damped when the term in the radical is zero:

Critically damped: Over-damped (real roots):

 1   1.8  0.8 Kccd := − 1 0.1   1.6  
2

2  1.8  − 1 + 0.1Kc = 0  0.8  1.6 

%CO Kccd = 0.125 %TO

%CO Kc < 0.125 %TO

%CO Under-damped: Kc > 0.125 %TO

The loop cannot be unstable for positive gain because, • for real roots the radical cannot be greater than the negative term, so both roots are negative • for complex conjugate roots the real part is always negative, -1.8/1.6, or -(τ1 +τ2)/2τ1 τ2 This is true for all positive values of the time constants and the product K. cK.

(c) Equivalent time constants for different values of the gain:
%CO Kc := 0.1 %TO τ e1 = −1 r1 −1 r2 (over-damped, two equivalent time constans) τ e1 := 2⋅ τ 1⋅ τ 2 τ e1 = 0.935 min

(τ 1 + τ 2) − ( τ 1 + τ 2) (τ 1 + τ 2) + ( τ 1 + τ 2)

2

− 4 ⋅ τ 1 ⋅ τ 2 ⋅ 1 + Kc⋅ K

( (

) )

τ e2 =

τ e2:=

2⋅ τ 1⋅ τ 2
2

τ e2 = 0.847 min

− 4 ⋅ τ 1 ⋅ τ 2 ⋅ 1 + Kc⋅ K

%CO Kc := 0.125 %TO τ e1 = −1 r1 −1 r2

(critically damped, two equal real time constants) τ e1 := 2⋅ τ 1⋅ τ 2 τ e1 = 0.889 min

(τ 1 + τ 2) − ( τ 1 + τ 2) (τ 1 + τ 2) + ( τ 1 + τ 2)
τ 1⋅ τ 2 1 + Kc⋅ K ζ :=

2

− 4 ⋅ τ 1 ⋅ τ 2 ⋅ 1 + Kc⋅ K

( (

) )

τ e2 = %CO Kc := 0.2 %TO

τ e2 :=

2⋅ τ 1⋅ τ 2
2

τe2 = 0.889 min

− 4 ⋅ τ 1 ⋅ τ 2 ⋅ 1 + Kc⋅ K

(under-damped, time constant and damping ratio) τ s + 2ζ ⋅ τ ⋅ s + 1 = τ 1⋅ τ 2 1 + Kc⋅ K
2 2

s +

2

τ1 + τ2 1 + Kc⋅ K

s+1

Match coefficients:

τ :=

τ1 + τ2

2 ⋅ τ ⋅ 1 + Kc⋅ K

(

)

τ = 0.886 min ζ = 0.996

(d) Steady-state offset for a unit step change in set point.
Final value theorem: lim
t→∞

Y( t) =

lim s⋅Y( s)
s→0

R( s) =

1 s

(Table 2-1.1)

%CO Kc := 0.1 %TO

lim s⋅
s→0

Kc⋅ K τ 1 ⋅ τ 2 ⋅ s + τ 1 + τ 2 ⋅ s + 1 + Kc⋅ K offset := ( 1 − 0.0099)%TO
2

1

(

)

s

→ 9.9009900990099009901 ⋅ 10

-3

offset = 0.99 %TO

Kc⋅ K %CO 1 -2 Kc := 0.125 lim s⋅ → 1.2345679012345679012 ⋅ 10 2 %TO s → 0 s τ 1 ⋅ τ 2 ⋅ s + τ 1 + τ 2 ⋅ s + 1 + Kc⋅ K

( (

) )

%CO Kc := 0.2 %TOlim s⋅
s→0

offset := ( 1 − 0.01235 )%TO Kc⋅ K τ 1 ⋅ τ 2 ⋅ s + τ 1 + τ 2 ⋅ s + 1 + Kc⋅ K offset := ( 1 − 0.01961 )%TO
2

offset = 0.988 %TO 1 s → 1.9607843137254901961 ⋅ 10
-2

offset = 0.98 %TO

These are very large offsets because the loop gains are so small.
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Smith & Corripio, 3rd edition
Problem 6-2. Inverse-response second-order system with proportional...