1 Controller Optimization according to the Modulus

Optimum

GK(s) F0(s) xw

-

The goal of applying a control loop usually is to get the control value x equal to thereference value w.

x(t) ≡ w(t)

X(s) ≡ W (s)

⇒ FW (s) =X(s)

W (s)≡ 1

for all s = jω ω ← R.

The goal of applying modulus optimum ist to achieve a transfer function of the closed(!)control loop which is constant on value 1 between zero frequency and a limit frequency ashigh as possible.

In technical control systems transfer functions usually are ”minimal phase” transferfunctions, i.e. all poles and zeros have a negative real component.

In these systems it is sufficient to know the amplification characteristic - in case ofa ”minimal phase” transfer function the phase characteristic is then defined as well.

|FW (s)| ≡ 1 results in FW (s) ≡ 1

1

Counter-Example:

all-pass network filter: G(s) =1− Ts

1 + Ts

⇒ |G(jωs)| = 1

∠G(jωs) = −2 arctan(Tω)

Im{s}

Re{s}

polezero

1T

1T

-

When dealing with electrical drives or inverters for control issues, the assumption of thissystem being a ”minimal phase” system usually is ok.

2

2 2nd Order Controlled System

2.1 Modulus Optimum

PI-control of a 1st order controlled system fulfills Nyquist criterion without any problems.

-1 V

FS

FS

FSF

R

.

FR

FS

X*

-

X

FK

2nd order controlled system:

T2

T1

FS(s) =1

(T2s + 1)(T1s + 1)

convention: T1 > T2 > T3 . . .

GK

FS

X*

-

XF

W

XX*

FK

FK = FR · FS

FR(s) is chosen to get:|FW (jω)| = 1

FW (jω) =FS(jω)FR(jω)

1 + FS(jω) · FR(jω)

3

Example:Controlled System FS(s) has one dominant time constant and several minor time constants

FS(s) =Ks

(1 + T1s)∏

(1 + Tvs)

≈ Ks

(1 + T1s)(1 + TEs)with TE =

∑

ν

Tv

Chosen: PI controller

FR(s) = KR

1 + T1s

T1s

with the requirement of T1 > TE!

Counter-example: Electrical and mechanical time constants of an electrical drive might beclose to each other ⇒ PI control does not work any more.

⇒ FK(s) = FR(s) · F0(s)

=KR ·KS

sT1(1 + TEs)

FW (s) =FK(s)

1 + FK(s)=

KS ·KR

KS ·KR + T1s + TET1s2

In the case of FW (s) ≡ 1 x would be equal to x∗!For transfer functions with ”minimal phase”

FW (s) = 1 results from |FW (jω)| = 1

FW (jω) =KS ·KR

KS ·KR + jωT1 + (jω)2Te · T1

|FW (jω)|2 =K2

S ·K2R

(KS ·KR − ω2TET1)2 + T 21 ω2

=K2

S ·K2R

K2S ·K2

R + ω4T 2ET 2

1 + ω2(T 21 − 2TET1KSKR)

4

for small ω : ω4T 2ET 2

1 ≈ 0

⇒ |FW (jω)|2 = 1 for

ω2(T 21 − 2TET1KSKR) = 0

T1 = 2TEKSKR

KR = T1

2TEKS

In fact the complete closed control loop not a proportional amplifier for all frequencies,but a first order low pass filter in the superposed control loop.

Summary:

Choose controller FR(s):

• to compensate the dominant time constant of the controlled system.

Requirement: A dominant time constant does exist.

• to make the transfer function of the closed control loop constant in a wide frequencyrange:

|FW (jω)| = 1

FW (jω) =F0(jω)GK(jω)

1 + F0(jω)GK(jω)

is valid in any case.

Analytical solution for PI controllers:

FK = V · Tis + 1

Tis· 1

(T2s + 1)(T1s + 1); Ti = T1

FK =1

TiKs(T2s + 1); TiK =

Ti

V

FW =FK

1 + FK

=ZN

1 + ZN

=Z

Z + N

FW =1

TiKT2s2 + TiKs + 1≡ 1

(s

ω0

)2

+ 2D sω0

+ 1

5

ω0 =1√

TiKT2

2D

ω0= TiK

D =1

2

TiK√TiKT2

=1

2

√

TiK

T2

!=

1

2

√2

TiK = 2T2 =Ti

V=

T1

V

⇓

V = T1

2T2Ti = T1

↪→ identical to T2 ≡ TE

V ≡ KR ·KS

2.2 P controller

GK

FS

X*

-

XF

W

XX*

FK

demand:|Fδ(jω)| ≈ 1

for ω = 0 . . . ωW , ωW →∞ (as far as possible)

|F|

1w

w

6

standardized form:

Fδ =1

(s

ω0

)2

+ 2D sω0

+ 1

FR = V

FK = FRFS =V

(T2s + 1)(T1s + 1)=

N

D=

Numerator

Denominator

FW =FK

1 + FK

=ND

1 + ND

=N

D + N

=V

T1T2s2 + (T1 + T2)s + 1 + V

Eigenvalues: Roots of the characteristic equation:

T1T2s2 + (T1 + T2)s + 1 + V = 0

s1,2 =−(T1 + T2)±

√

(T1 + T2)2 − 4T1T2(1 + V )

2T1T2

Discussion of Root Locus:influence of the controller: root locus with V as parameter.

1T

1

-

1T

2

-

crit. D s

45

7

characteristic points of the root locus:

• V = 0: s1 = − 1T1

s2 = − 1T2

•√

(T1 + T2)2 − 4 · T1T2(1 + V ) = 0: s1 = s2 = −12

(1T1

+ 1T2

)

• else: conjugated complex poles

conclusions: (P controller)

• no stability problems

• damping decreases with increasing V

• the more ”distance” there is between T1 and T2, the bigger must be parameter V togain acceptable dynamics of the closed control loop(critical damping)

1T1

-1T2

-

sK ss

1

s2

W

by control

an optimized controller provides good position of the eigenvalues of the closed contol loop.

8

controller optimization :There is only a single parameter: V

T1T2s2 + (T1 + T2)s + 1 + V = 0

s1,2 =−(T1 + T2)±

√

(T1 + T2)2 − 4T1T2(1 + V )

2T1T2

• 1st solution1

s1,2 =−(T1 + T2)±

√

(T1 + T2)2 − 4T1T2(1 + V )

2T1T2

Re(s1)!= Im(s1)→ 45◦ only numerator of s1,2 is to be treated:

⇒ (T1 + T2)2 = −(T1 + T2)

2 + 4T1T2(1 + V )

1 + V =2(T1 + T2)

2

4T1T2

V =T 2

1 + T 22

2T1T2

• 2nd solution2 standardized form ⇒ one term is 1 (see above)

FW =V

V +1T1T2

V +1s2 + T1+T2

V +1s + 1

=VW

(s

ω0

)2

+ 2D sω0

+ 1

VW =V

V + 1

ω0 =

√

V + 1

T1T2

2D

ω0=

T1 + T2

v + 1⇓

D =1

2

T1 + T2√

T1T2(V + 1)

increasing V means:

– ω0 →∞– D → 0

– VW → 1

1usual (aperiodic damping))2modulus optimum

9

modulus optimum:

2D2 = 1

⇒ D =1

2

√2

w00,5

1

1

22

( )qD

1

in the case of critical damping:

1. ITAE(integral of time adapted error) }

are fulfilled2. modulus optimum

pre-setting:

D = Dcrit. =1

2

√2 =

1

2

T1 + T2√

T1T2(V + 1)

(T1 + T2)2

2T1T2

= V + 1

V =(T1 + T2)

2

2T1T2− 1 =

T 21 + T 2

2

2T1T2

for example:

T1 = 4T2

V =16 + 1

8=

17

8≈ 2

VW =V

V + 1=

2

3

10

0,67

1

t

eV

( )¥ » ¬

+

FH

IK33%

1

1

dynamic behaviour: ok but not extremely goodstationary deviation too big

e

-

e 6= 0 is control signal ⇒ e cannot be 0!

P control of a low pass system results in

1. guaranteed stability

2. decreasing damping factor (D) with increasing controller gain (V )

3. the more ”distance” between T1 and T2 the higher is critical gain Vcrit

11

2.3 Differential Controllers

basic idea: control signal derived from de(t)dt

, 2 types: PD, PID

FR

FS

-

1. PD controller

FR

V VT

T

d

d¢

w

FR = V · Tds + 1

T ′

ds + 1

FK = V · Tds + 1

T ′

ds + 1· 1

(T2s + 1)(T1s + 1)

FK =V

(T ′

ds + 1)(T1s + 1)

12

1T

1

-1T

2

-

pole shift to the left:

(a) more dynamic behaviour

(b) more control power

Td = T2, T ′

d ' 0, 1Td

last example:

T1 = 4T2

T1 = 40T ′

d

V =T 2

1 + T ′2d

2T1T ′

d

=402 + 1

2 · 40≈ 20 ; Vδ =

V

V + 1=

20

21= 0, 95

e(∞) = 5%

dynamic behaviour

ω0 =

√

V + 1

T1T2

ω(PD)0

ω(P )0

=

√

20 + 1

(2 + 1) · 0, 1 =

√

210

3=√

70 = 8, 4 T2 = 10T ′

d

why not factor 10? (T2 → T ′

d = 0, 1T2)phase shift by T1 did not have any influence so far but now it has

13

2. PID control

P

I

D

= PD controller with an I channel in parallel (more accurate, but worse in dynamics( like P → PI))

FR = V · Tis + 1

Tis· Tds + 1

T ′

ds + 1

low gain at ω ≈ ωd, high gain elsewhere

t

VT

T

d

d¢ w(t)

FR

VT

T

d

d¢

» w d

no phase shift

w

fast leading

14

(a) pole shift to the left: more accuracy

(b) pole shift to the right: more dynamics

1T

1

-1T

2

-1T'

d

-

FK =1

TiKs(T ′

ds + 1)

controller with D channel:

(a) fast response is only possible when there is enough control ”power” (small signalbehaviour - large signal behaviour)

(b) increasing harmonics (inverters, digital sensors, e.g position sensors)

alternative solution (if possible): make use of (i.e sense) a derivation of the con-trolled value from the controlled system itself

e.g.

L

( )=

»

»C iC

iC = C · du

dt

15

-

x*

-

x

Tdx

dt

T

same FK(s) as with PID control, but thereis no differentiation process in the controller.

2.4 Symmetrical optimum

In the case of T1 > 7 · T2 a pole shift does not change the situation significantly.

1T

1

-1T

2

-

s

PI controller replaces pole at −1/T1 by pole in the origin.

1T

2

-

s

This cannot work, if one of the poles already is in the origin (or close by)!

16

1

Ti

The most extreme situation is:

1

T1s + 1→ 1

Ts

-

V,Ti T2 Tm

G (s)K

F (s)0

FK = V · Tis + 1

Tis· 1

(T2s + 1)Tms=

V

TiTms2· Tis + 1

T2s + 1= F1 · F2

FK

-1,0

T = T impossiblei 2

not stable

17

control loop with double integrator

Requirements for stability and sufficient damping (Nyquist criterion)

|FK(jωd)| = 1 arg(FK(jωd)) = −π + Ψd

arg(FK) = arg(F1) + arg(F2)

= −π + arg(F2) = −π + Ψd

Ψd = arg(F2(jωd))

FK

-1,0

d

d

If T2 6= Ti, the characteristic of F2 is:

F2 s

wd

w

1

j

T2

Ti

18

Demand: Ti → shoud be as small as possible espacially as possible at ϕ = Ψd otherwisethe response on disturbancesis slow.→ ϕ = arg(F2) should be max.

F2(s) =Tis + 1

T2s + 1

F2(jω) =1 + jωTi

1 + jωT2

ϕ = arg(F2) = arctan ωTi − arctan ωT2

1T

1

-

1T

i

-

w

+-

for any ω the resuling angle is positiv!

dϕ

dω|ω=ωd

=Ti

1 + ω2dT

2i

+−T2

1 + ω2dT

22

!= 0 (Maximum!)

Ti(1 + ω2dT

22 )− T2(1 + ω2

dT2i ) = 0

Ti − T1 − (Ti − T1)ω2dTiT2 = 0

ω2dTiT2 = 1

ωd =1

√

TiT2

19

needed: Ti

mathematical solution:

Ψd = arg(F2)

Ψd = arctan ωdTi − arctan ωdT2

Ψd = arctan

√

Ti

T2− arctan

√

T2

Ti

optimized PI controller: calculation of Ti based on Ψd; then calculation of VHow to calculate Ti from Ψd

1

a

b

y

T2

Ti

a =1

2

(Ti

T2− 1

)

b = a + 1 =1

2

(Ti

T2+ 1

)

sin Ψ =a

b=

Ti

T2

− 1Ti

T2

+ 1

(Ti

T2

)

sin Ψ + sin Ψ =Ti

T2

− 1

(Ti

T2 − 1

)

(sin Ψ− 1) = −(1 + sin Ψ)

Ti

T2=

1 + sin Ψ

1− sin Ψ

Ti = T2

1 + sin Ψ

1 − sin Ψ

20

now: calculation of V

FK

-1

-j

wd

w

Yd

|FK(jωd)| = 1 ← at transition frequency

|FK(jωd)| =V

ω2dTiTm

·√

(ωdTi)2 + 1√

(ωdT2)2 + 1, ωd =

1√TiT2

(see page 19)

=V TiT2

TiTm

·

√√√√

Ti

T2

+ 1T2

Ti+ 1

=V T2

Tm

·

√√√√

Ti

T2

(1 + T2

Ti)

1 + T2

Ti

=V T2

Tm

·√

Ti

T2= V

√TiT2

Tm

= 1

V =Tm

√

TiT2

for each Ti(Ψd) there is a V

21

Re

Im

-1-2

-3-4

-1

-2

2

1,2

5

1

0,7

5

0,6

0,5

0,4

q=

w wd

The

resu

ltin

g o

pen

loop c

har

acte

rist

ic

22

The resulting open loop frequency charcteristic

1

(lg)

|F|

a a

+1

-1

-23

1-2

FK

wd

F2

j0

-pYd

2

wd

-p

2

1

Ti

1

T2

1

for symmetrical reasons:1

T2

= aωd = a2 1

Ti

thenTi = a2T2 ;

to calculate a:

FK(s) =V

TiTms2· Tis + 1

T2s + 1

=V

a2T2Tms2· a

2T2s + 1

T2s + 1, V =

Tm√TiT2

=Tm

aT2

=1

a3T 22 s2· a

2T2s + 1

T2s + 1

normalized frequency:

q =s

ωd

= s√

TiT2 = aT2

open loop:

FK(q) =aq + 1

aq2(

1aq + 1

)

23

closed loop:

Fg =FK

1 + FK

=N

N + D

Fg(q) =aq + 1

q3 + aq2 + aq1=

N(q)

D(q)︸ ︷︷ ︸

Eigenvalues: D(q) = 0;easy to realize: q = −1 is a solutionfurther solutions:

q2,3 = −a− 1

2± j

√

1−(

a− 1

2

)2

|q2,3| =

√(

a− 1

2

)2

+ 1−(

a− 1

2

)2

= 1

a -j

q3

a zero

a>3

q2

a a=1

j q

a=3

a>3

q=1pole 36

24

D2,3 >1

2

√2!

Fg(q) =aq + 1

q3 + aq2 + aq1=

N(q)

D(q)

Fg(q) =aq + 1

(q2 + (a− 1)q + 1)(q + 1)

!=

aq + 1

(q2 + 2Dq + 1)(q + 1)

2D = a− 1

a = 1 + 2D

Ti = a2T1

V =Tm

aT2

⇓aopt. = 2, 6

D, a optimized according to ITAE: Dopt. = 0, 7

Fg(q) =aq + 1

q3 + aq2 + aq + 1

1

t

w(t)» 30%

25

to compensate overshoot: filter in reference channel

in general:

FW =τ2q + 1

τ1q + 1

τi: normalized timeti optimized according to ITAE:

τ2 = −1 Dopt. = 0, 7

τ1 = a aopt. = 2, 6

-

1

1

a

26

a simple low pass filter would be too slow (acc. to Leonhardt)

0 2 4 6 8

wdt

0,5

1,0

1,3

w(t)

without reference filter

FFilter ( )qq

aq=

+

+

1

1

FFilter ( )qaq

=

+

1

1

symmetrical optimum

step responses

27

2.5 Limit between modulus optimum and symmetrical optimum

scheme

-

controller T1

T2

T3

Z1

Z2

xx*

PI controller

FR = V · Tis + 1

Tis

1T

1

-1T

2

-1T

3

-

s

1T1

-1T2

-

s

1T2

- 1T1

-

s

-

1

a Topt

2

2

a)

b)

28

a) modulus optimumT1 < a2

opt.T2

b) symmetrical optimumT1 > a2

opt.T2

aopt. = 2, 6

a2opt. = 6 . . . 7 = 6, 75

29