Automatic control systems
II.
Basic knowledge of compensation
Brief overview
Compensation
While the variety of compensation schemes is great the classical threeterm controller design has been found to be particularly simple and
effective. This compensation scheme has historically implemented
using analogue mechanical, pneumatically and electrical controllers.
Today the compensation is implemented in a software because the
controllers are microprocessor based. An extra steps is required to
convert the analogue signal to digital information and the digital
information to analogue signal. Nowadays in the industrial area the
most popular A/D and D/A converter contain 12 bits or 11 bits.
The sample time
If the sampling time is much more less than the sum of time constants of the
process than the theory and design methodology is the same in digital and
analogue technologies.
If one wants the digital PIDT1 compensator seems to be such as analogue must
choose appropriate sample time period.
Grey model
(Tj are the time constants of the model.)
Black model energizing step signal
(Ts is the settling time or time constant
of the approximate model.)
Relationship between Ts and Tj :
T
j
50
T 0
T


150
Ts
T
T 0  s
250
750
Ts  5 *  Tj
j
Structure of classical three-term controller
1
sTD
G c (s)  K c {1 

}
1
sTi 1 
sTD
AD
AD 
TD
4
T
h(t )
t
1
e
r

yM
m  n!
1
sTI
KC
w
sTD
1 0.1sTD
u
h(t )
t
h(t )
t

1
de(t )
dw ( t )
u(t )  K C{e( t )   e(t )dt  TD
 0.1TD
}
TI 0
dt
dt
AD  10
Structure of classical three-term controller in
sampled controller
T0 n
e(n)  e(n  1)  0.1y(n)
u(n  1)  K c {e(n)   e(i)  TD
}
TI 0
T0  0.1TD
1
e
r

T0 n
 e(i)
Ti i 0
0.1w (n  1)
TD
T0  0.1TD
yM
Kc
w
u
Real combination of the three kinds
Proportional P
u(t )  KC e(t )
Gc ( s)  KC
u (n  1)  KC e(n)
Proportional-Integral PI
1
Gc ( s )  K C {1 
}
sTI

1
u (t )  KC {e(t )   e(t )dt}
TI 0
T0 n
u(n  1)  KC {u(n)   u(i)}
TI 0
Proportional-Derivative PD
Gc ( s)  K C {1 
s0.9TD
1  sTD
}  K D{
}
1  0.1sTD
1  0.1sTD
u ( t )  K C {e( t )  TD
u (n  1)  K C {u (n )  TD
u (n )  u (n  1)  0.1w (n )
}
T0  0.1TD
de( t )
dw ( t )
 0.1TD
}
dt
dt
Integral I
1
Gc ( s ) 
sTI

1
u (t )   e(t )dt
TI 0
Proportional-Integral-Derivative PID
TI  4TD
T0 n
u(n  1)   e(i)
TI 0
Tuning controllers
• Which type of PID is the best solution?
• Which parameters are optimal ?
The type of PID depend on transfer function of the plant (process model)
The optimal parameters depend on which is the required performance
specification of the feedback control.
(Looking for these parameters in time domain the step response of process
field and in frequency domain the Bode or the Nichols diagram are used)
If the simple feedback loop with compensation in cascade is unsatisfactory for
the required performance specification than must modify the structure of
control loop or exchange the classical PID scheme of control algorithm.
(The most popular are the cascade and the feedforward control.)
Compensation techniques
Depends on the identification of the process field
transfer function
• Step response of the process field:
Black-box model and recommendations for the controller type
and value of the parameters.
• Frequency response of the process field:
Black-box model, and fitting of P, PI, PDT1, or PIDT1
controller‘s frequency response to the process field transfer
function.
• Pole replacement:
Grey/box model. Poles of the process field transfer function is
known.
Modelled the process from reaction curve
by dead-time first order transfer function HPT1
ym 1
Gp ( s) 
e sTu
u 1  sTg
ym , u
ym
u
KP
Tg
Tu
0
3.3
I
t
PID
Tu
55
7.8
PI
Tg
P or PI
Suggested parameters for HPT1
minimum settling time and without overshoot
system performance with following trajectory of reference signal
Recommended by Chien-Hrones-Reswick
KC
TI
TD
1 Tg
K p Tu
P
0.3
PI
0.36
PID
0.6
1 Tg
K p Tu
1.2Tg
1 Tg
K p Tu
Tg
0.5Tu
This parameters based on the absolute integral criterion of performance.
There are any other parameter commendation
The identified HPT1 model from the
measured step response of process field
The files are in the TDcompensationLecture2 library.
The model seems to be self-adjusting nature. (Without integral
effect) First we define the final value. (The measured value saved
as FinalValueStepresponse_SA.fig)
Then using denser sampling the time constants can be edit.
(stepresponse_SA.fig)
Parameters: Kp=0.72, Tg =10.7sec., Tu =0.9sec.
The suggested compensation is PI, because the ratio of Tg/Tu
is almost 12. (Recommendation of Piwinger)
(The result is saved as FeedbackStepResponse_SA.fig)
Modelled the process from reaction curve
by “n” order transfer function PTn
1
G p ( s)  K p
(1  sT )n
ym , u
Ym  ym
70%
U  u
30%
10%
t10
t30
t
t70
Determination of system parameters
First the order “n” of the system is determined by the time percent co-efficient are best fitted
The time percent co-efficient n=
1
2
3
4
5
6
t10
t30
0.30
0.48
0.58
0.63
0.87
0.70
t10
t70
0.09
0.22
0.31
0.37
0.42
0.45
0.36
1.10
1.91
2.76
3.63
5.52
1.20
2.44
3.62
4.76
5.89
7.01
Dominant time T 
t30
T1
t70
T2
T1  T2
2
Suggested parameters for PTn
minimum settling time and without overshoot
system performance with following trajectory of reference signal
Ti
P
n=1
Kc
20
Kp
PI
n=1
1 n2
3

Kp n
Kp
2n
1
T T
n3
2
PI
n=2,3
n2 1
n Kp
2n
T
n2
PID
n=4,5
1 n
Kp n2
2n
T
n 1
I
n=6
2nT
Td
T
5
The identified nT1 model from the
measured step response of process field
The files are in the TDcompensationLecture2 library.
The model seems to be self-adjusting nature. (Without integral
effect) First we define the final value. (The measured value saved
as FinalValueStepresponse_SA.fig)
Then using denser sampling to define the time values which
belong to special response values. (nTstepresponse_SA.fig)
Parameters: t110%=1.94sec., t230% =4.04sec., t370% =10.1sec.
The counted order n =2. dominant time constant T =14.5sec.,
suggested compensation is PI,. (n less than 3.)
(The result is saved as nTfeedbackStepresponse_SA.fig)
Modelled the process from reaction curve
by dead-time integral first order transfer function HIT1
1
1
G p ( s) 
sTI sTg  1
ym , u
u
Tg
TI
t
Suggested parameter for IT1 from Friedlich
minimum settling time and without overshoot
system performance with following trajectory of reference signal
If the measured value has got noise, then the suggested compensation is P,
other case PDT1. The PIDT1 is only used if it is necessary to use second
type feedback control.
The identified IT1 model from the
measured step response of process field
The files are in the TDcompensationLecture2 library.
The model seems to be integrating nature. (With integral effect)
First we define the time constants. (stepresponse_I.fig)
Parameters: Tg=6.5sec., TI =2.7sec.
Assuming that the measured value noise-free, the suggested
compensation is PDT1, the counted compensation
parameters: KC =0.21, TD =6.5sec., T =0.7sec.
(The result is saved as FeedbackStepresponse_I.fig)
Analysis of Bode diagram
(The files in the ExampleFDcompansatinLecture2 library)
The Bode plot is popular, because many properties of
process field are well read.
Recorded the measured values or the identified model the
following compensation technique are also used.
• It must be concluded if the process field has or hasn't
got integral effect. It possible from the phase plot.
• It is necessary to determine how many time constants
belongs to the process field and how close to each other
these time constants. It possible from the Bode plot.
Without integral effect
The integral effect is possible to determine from the phase
plot of process field transfer function. (without integral effect if
sufficiently low frequencies the phase shift is nearly zero.)
In this case the most commonly used structure is the PI, if
the process field model has got more than three poles
relatively close to each other (inside two decade), then offered
to use the PIDT structure.
The quality properties of the closed loop depends on the
phase margin of the G0(s) open loop transfer function.
(If the phase margin “pm” higher or equal 90 degree the closed loop
system without overshoot, if pm between 70 and 90 degree and there are
three or more time constant and none of them are dominant, then the
closed loop system can have got a small overshoot, but otherwise not.)
Transfer function
PI compensation
Europeanofstructure

sTI  1
1 
G C (s)  G PI (s)  K C 1 
  KC
sTI
 sTI 
There are two variables.
First step we chose the KC = 1, and TI = 1 rad/sec. values!
s 1
G PI (s) 
s
In case of PIDT must be: TI > 4TD and TD > 5T conditions.
PI compensation
Tenfold value of ωI the amplitude gain nearly 0. and the phase shift -5,7°
The PI compensation process
• Have to plotted the process field G PF ( j) Bode plot.
• On the phase plot must be looking for the future gaincrossover frequency which is corresponding the following
phase shift: ps = pm + 5,7 - 180.
(The suggested phase margin if the process field order higher than 3
and the break point are relatively close to each other is minimum 70
degree.)
• A tenth of this frequency is ωI, and TI is the reciprocal of
ωI.
sTI  1
G E ( j) (grey model).
• Plots the Bode diagram of g0 
sTI
• On this g0 Bode plot have to be looking for the frequency
which is corresponding the chosen phase margin, next
have to read the gain at this frequency. The KC controller
Bode plot of GPF(jω)
It can be seen the rounding is permitted, but must be documented!
(Saved as DiagBodePF_SA)
Black model plot GPF(jω) and GPI(jω)
Based on the above figure 10wI = 0.5 rad/sec., and so wI = 0.05 rad/sec.
Creating the reciprocal value: TI = 20 sec. The g0 is figured with KC = 1:
The gain of the controller converted from dB: KC = 5.8
Step response of the closed-loop
There aren’t overshoot, and steady-state error
With integral effect PDT1 compensation
European structure
sTD 
s(TD  T)  1

G C (s)  G PDT (s)  K C 1 
  KC
sT  1
 sT  1 
Case PDT1 compensation must be TD > 5T.
There are three variables. The first step you choose KC = 1, TD = 0.9 rad/sec,
and T = 0.1 rad/sec.
s 1
G PDT (s) 
0.1s  1
Bode plot of PDT1
The φmax phase shift depends on the AD differential gain. Present example
AD = 9, and so φmax = 54.9°.
The principle of compensation of PDT
First we chose a considers appropriate phase margin!
On the phase plot of process field must look for the chosen phase
margin and the corresponding frequency will be the future ωC gaincrossover frequency. At the chosen phase margin the phase shift is
ps = pm – 54.9° - 180°.
(Put this frequency equate with ωmax, which belongs to the maximum positive
phase shift of GPDT(jω) and assuming AD =9, then ωD =ωmax/3 ωT =3*ωmax.)
The controller gain KC value to be chosen so, that at the future ωC
gain-crossover frequency will be unit the K0 loop-gain.
(The reciprocal value of the amplitude gain at the future gain-crossover
frequency on the amplitude plot of the g0 (the g0 is the G0(jω) open-loop
transfer function with KC = 1 value) will be the actual KC.)
The GPF(jω) Bode diagram
The (max )  D  T and assuming that AD = 9, so
ωT= 0.756 rad/sec and ωD= 0.084 rad/sec.
Determination of TD and T with AD═9
Based on above figure
wD = 0.0.084 rad/sec. and wT = 0.756 rad/sec. and so
TD = 11.9 sec., and T = 1.3 sec.
The open-loop transfer function with KC = 1 g0 is :
G PDT (s) 
12.4s  1
1.3s  1
On the Bode plot of g0 should look for the kC gain which
corresponding the phase margin = 65°. (Grey model)
The next figure (black model we put each other the GC(jω) and GPF(jω)
bode plots.)
Bode plot of GPF(jω) and GPDT(jω)
The gain 10.3-6.35═3.95dB, KC = -3.95dB which is converted KC = 0.63
Step response of feedback system
Possible, but not necessary additional tuning.