Transcript Design system with Bode Hany Ferdinando Dept. of Electrical Eng. Petra Christian University General Overview  Bode vs Root Locus design  Information from open-loop freq.

Design system with Bode
Hany Ferdinando
Dept. of Electrical Eng.
Petra Christian University
General Overview
 Bode vs Root Locus design
 Information from open-loop freq. response
 Lead and lag compensator
Design System with Bode - Hany Ferdinando
2
Bode vs Root Locus
 Root Locus method gives direct
information on the transient response of
the closed-loop system
 Bode gives indirect information
 In control system, the transient response
is important. For Bode, it is represented
indirectly as phase and gain margin,
resonant peak magnitude, gain crossover,
static error constant
Design System with Bode - Hany Ferdinando
3
Open-loop Freq. Response
 Low-freq. region indicates the steady-state
behavior of the closed-loop system
 Medium-freq. region indicates the relative
stability
 High-freq. region indicates the complexity
of the system
Design System with Bode - Hany Ferdinando
4
Lead and Lag Compensator
 Lead Compensator:
 It yields an improvement in transient response
and small change in steady-state accuracy
 It may attenuate high-freq. noise effect
 Lag Compensator:
 It yields an improvement in steady-state
accuracy
 It suppresses the effects of high-freq. noise
signal
Design System with Bode - Hany Ferdinando
5
Lead Compensator
1
s
Ts  1
T
Gc ( s)  K c
 Kc
1
Ts  1
s
T
(0 <  < 1)
The minimal value of a is limited by the
construction of lead compensator. Usually, it
is taken to be about 0.05
Design System with Bode - Hany Ferdinando
6
Lead Compensator
1.
Ts  1
Define Kc = K then Gc ( s)  K
then
Ts  1
Gc ( s)G ( s)  K
Ts  1
Ts  1
Ts  1
G( s) 
KG ( s) 
G1 ( s)
Ts  1
Ts  1
Ts  1
Determine gain K to satisfy the
requirement on the given static error
constant
2. With gain K, draw the Bode diagram and
evaluate the phase margin
Design System with Bode - Hany Ferdinando
7
Lead Compensator
3. Determine the phase-lead angle to be
added to the system (fm), add additional
5-12o to it
1
4. Use sin f m  1   to determine the
attenuation factor . Find wc in G1(s) as
|G1(s)| = -20 log(1/√) and wc is 1/(√T)
5. Find zero (1/T) and pole (1/T)
Design System with Bode - Hany Ferdinando
8
Lead Compensator
6. Calculate Kc = K/
7. Check the gain margin to be sure it is
satisfactory
Design System with Bode - Hany Ferdinando
9
Lead Compensator - example
4
G( s) 
s( s  2)
 It is desired that the Kv is 20/s
 Phase margin 50o
 Gain margin at least 10 dB
Design System with Bode - Hany Ferdinando
10
Lead Compensator - example
Ts  1
Ts  1
Ts  1
Gc ( s )G ( s)  K
G(s) 
KG ( s) 
G1 ( s )
Ts  1
Ts  1
Ts  1
Ts  1
Ts  1
lim sGc ( s )G ( s )  lim s
G1 ( s )  lim s
KG ( s )
s 0
s 0 Ts  1
s 0 Ts  1
Ts  1
4K
lim
s
 2 K  20
s 0 Ts  1 s ( s  2)
K = 10
Design System with Bode - Hany Ferdinando
11
Lead Compensator - example
Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 18 deg (at 6.17 rad/sec)
Magnitude (dB)
50
0
Phase (deg)
-50
-90
-135
-180
-1
10
0
1
10
10
2
10
Frequency (rad/sec)
Design System with Bode - Hany Ferdinando
12
Lead Compensator - example
 Gain margin is infinity, the system requires
gain margin at least 10 dB.
 Phase margin is 18o, the system requires
50o, therefore there is additional 32o for
phase margin
 It is necessary to add 32o with 5-12o as
explained before…it is chosen 5o  37o
Design System with Bode - Hany Ferdinando
13
Lead Compensator - example
 Sin 37 = 0.602, then  = 0.25
 |G1(s)| = -20 log (1/√)
o
40
1
 20 log
jw c ( jw c  2)
0.25
40
 6.02dB  0.5
jw c ( jw c  2)
wc = 8.83 rad/s  new crossover freq.
Design System with Bode - Hany Ferdinando
14
Lead Compensator - example
 wc = 1/(√T), then 1/T = wc√  4.415 
zero = 4.415
 Pole = 1/(T) = wc/√ = 17.66
 Kc = K/ = 10/0.25 = 40
1
s
4
T
Gc ( s )G ( s )  K c
1 s ( s  2)
s
T
s  4.415 4
Gc ( s )G ( s )  40
s  17.66 s ( s  2)
Design System with Bode - Hany Ferdinando
15
Lead Compensator - example
Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 49.6 deg (at 8.83 rad/sec)
Magnitude (dB)
50
0
-50
Phase (deg)
-100
-90
-135
-180
-1
10
0
10
1
10
2
10
3
10
Frequency (rad/sec)
Design System with Bode - Hany Ferdinando
16
Lead Compensator - example
Clear variable
clear;
Gain from Kv
K = 10;
Original system
Plot margin
Get Gm & Pm
num = 4; den = [1 2 0];
margin(K*num,den)
[Gm,Pm] = margin(K*num,den)
New Pm in deg
new_Pm = 50 - Pm + 5;
New Pm in rad
new_Pm_rad = new_Pm*pi/180;
alpha
From calculation
Get zero & pole
Compensator gain
Plot final result
alpha = (1 - sin(new_Pm_rad))/(1 + sin(new_Pm_rad))
Wc = 8.83;
zero = Wc*sqrt(alpha); pole = Wc/sqrt(alpha);
Kc = K/alpha;
figure; margin(conv(4*Kc,[1 zero]),conv([1 2 0],[1 pole]))
Design System with Bode - Hany Ferdinando
17
Lag Compensator
1
s
Ts  1
T
Gc ( s)  K c 
 Kc
1
Ts  1
s
T
( > 1)
With b > 1, its pole is closer to the origin than
its zero is
Design System with Bode - Hany Ferdinando
18
Lag Compensator
1. Assume
1
s
Ts  1
T
Gc ( s)  K c 
 Kc
1
Ts  1
s
T
Kc = K
Ts  1
Ts  1
Ts  1
Gc ( s)G( s)  K
G(s) 
KG ( s) 
G1 ( s)
Ts  1
Ts  1
Ts  1
Calculate gain K for required static error
constant or you can draw it in Bode
diagram
Design System with Bode - Hany Ferdinando
19
Lag Compensator
2. if the phase margin of KG(s) does not
satisfy the specification, calculate fm! It is
fm = 180 – required_phase_margin
(don’t forget to add 5-12 to the required
phase margin). Find wc for the new fm!
3. Choose w = 1/T (zero of the
compensator) 1 octave to 1 decade
below wc.
Design System with Bode - Hany Ferdinando
20
Lag Compensator
4. At wc, determine the attenuation to bring
the magnitude curve down to 0 dB. This
attenuation is equal to -20 log (). From
this point, we can calculate the pole,
1/(T)
5. Calculate the gain Kc as K/
Design System with Bode - Hany Ferdinando
21
Lag Compensator - example
4
G( s) 
s( s  1)(0.5s  1)
 It is desired that the Kv is 5/s
 Phase margin 40o
 Gain margin at least 10 dB
Design System with Bode - Hany Ferdinando
22
Lag Compensator - example
Gc ( s)G( s)  K c 
Ts  1
1
Ts  1
1
K
Ts  1 s( s  1)(0.5s  1)
Ts  1 s( s  1)(0.5s  1)
K = Kc
K v  lim sGc ( s)G( s)  sK
s 0
Ts  1
1
Ts  1 s( s  1)(0.5s  1)
With Kv = 5, K = 5
Design System with Bode - Hany Ferdinando
23
Lag Compensation - example
Bode Diagram
Gm = -4.44 dB (at 1.41 rad/sec) , Pm = -13 deg (at 1.8 rad/sec)
100
Magnitude (dB)
50
0
-50
-100
-150
-90
Phase (deg)
-135
-180
-225
-270
-2
10
-1
10
0
10
1
10
2
10
Frequency (rad/sec)
Design System with Bode - Hany Ferdinando
24
Lag Compensation - example
 The required phase margin is 40 ,
o




therefore, fm = -180o + 40o + 10o = -130o
Angle of G1(s) is -130o, wc = 0.49 rad/s
w = 1/T = 0.2wc = 0.098 rad/s
At wc, the attenuation to bring down the
magnitude curve to 0 dB is -18.9878 dB
(rounded to -19 dB)
From -20log() = -19,  is 8.9125
Design System with Bode - Hany Ferdinando
25
Lag Compensation - example
 Pole of the compensator is 1/(T), with 1/T
= 0.098, the pole is 0.011
 Kc = K/b, and Kc = 0.561
Design System with Bode - Hany Ferdinando
26
Lag Compensation - example
Gain
K = 5;
num = 1;
den = conv([1 1 0],[0.5 1]);
Plot margin
Get gain and phase margin
Find new phase margin
margin(K*num,den)
[Gm,Pm] = margin(K*num,den);
new_Pm = (-180 + 40 + 10)*pi/180; %rad
wc = 0.49;
att = -(20*log10(5/abs(i*wc*(i*wc+1)*(i*0.5*wc+1))))
beta = 10^(att/-20)
zero = 0.2*wc
pole = zero/beta
Kc = K/beta
Design System with Bode - Hany Ferdinando
27