Characteristics of transfer functions
2. Characteristics of transfer functions
2.1 Modeling of Dynamic systems
Modeling of dynamic systems is concerned with cause and effect relationship
between inputs (X) and outputs (Y). The simplest model has only one input and
output (SISO = Single input, Single Output). The input and the output are
variables of real process with physical dimension.
In the actual application the input and output are assigned to a limited
area (Xmin, Xmax and Ymin, Ymax). The following expressions show how you can
involve the model equations dimensionless:
X
(
t
)
d
i
m
X
d
i
m
[
]
[
]
m
i
n
X
(
t
)
=
×
1
0
0
<2.1.a.>
Xd
i
m
Xd
i
m
[]m
[]
m
a
x
i
n
Y
(
t
)
d
i
m
Y
d
i
m
[]
[]
m
i
n
Y
(
t
)
=
×
1
0
0
<2.1.b.>
Y
d
i
m
Y
d
i
m
[]
[]
m
a
x
m
i
n
Using dimensionless variables the model equations will involve time as
a variable, primarily in the form of derivatives. There are a lot of methods to
involve these differential equations if the process can be modeled as a LTI
(Linear Time Invariant) system.
Note: The system is linear if the superposition is available. The superposition means if you energized the system an arbitrary input signal X1 and
the response output signal is Y1 after energized the system an other arbitrary
input signal X2 and the response output signal is Y2 than energized the system
X1+X2 signal the response will be Y1+Y2.
The Laplace transform of a function of time, f(t), is written as the
¥
t
()
s=òf(
te
)-s
d
tand the initial conditions are zero:
integral F
0
df()
t
=sFs
()
<2.2.>
dt
The Laplace transform converts the system from time domain to sdomain. The s-domain implicit contains the frequency domain. The Laplace
transform is a linear transformation consequently
z
(
t
)
=×
a
x
(
t
)
+
axt
×
()than Z
(
s
)
=×
a
X
(
s
)
+×
a
X
(
s
)
.
11
2
2
1
1
2
2
Characteristics of transfer functions
Supposing that the process operates in steady-state at the working point
one (X1, Y1) when the input signal start to change (X1+x(t)). The output
response is (Y1+y(t)). You see on the Fig. 2.1, the initial conditions of x(t) and
y(t) signals are zero, and if the x(t) and y(t) signals are sufficiently small than
the process can be modeled as an LTI system. Final the process will operate in
steady-state at the working point two (X2, Y2).
Y
y(t)
WP2
WP1
t
X
x(t)
t
Figure 2.1. Steady-State characteristics and dynamic response
Hereinafter we will examine only LTI system. The dynamic behavior of
LTI system can be written linear time invariant differential equitation.
n
n
1
d
y
(
t
)
d
y
(
t
)
d
y
(
t
)
a
+
a
+
L
+
a
+
a
y
(
t
)
=
n
n
1
1
0
d
t
d
t
d
t
<2.3.a.>
m
m
1
d
x
(
t
)
d
x
(
t
)
d
x
(
t
)
=
b
+
b
+
L
+
b
+
b
x
(
t
)
m
m
1
1
0
d
t
d
t
d
t
Note: The system is proper if n ³ m and strictly proper if n > m .
The Laplace transform is:
n
n
1
a
s
Y
(
s
)
+
a
s
Y
(
s
)
+
L
+
a
s
Y
(
s
)
+
a
Y
(
s
)
=
n
n
1
1
0
<2.3.b.>
m
m
1
=
b
s
X
(
s
)
+
b
s
X
(
s
)
+
L
+
b
s
X
(
s
)
+
b
X
(
s
)
m
m
1
1
0
Arranged this equation:
m m
1
b
sb
+
s
+
L
+
b
s
+
b
Y
(
s
)
m
m
1
10
=
G
(
s
)
=
n
n
1
X
(
s
)
a
s
+
a
s+
L
+
a
s
+
a
n
n
1
10
<2.4.a.>
Characteristics of transfer functions
The G(s) is called transfer function.
The other form of the transfer function is the so called zero-pole form.
Arranged the 2.4.a:
bsz
(+
)
L
(
sz
+
)
m
1
m
G
(
s
)
=
<2.4.b.>
a
(
sp
+
)
L
(
sp
+
)
n
1
n
bm
where zj are the zeros and pi are the poles, and the gain is signed Kzp = a .
n
The Bode form of the transfer function is generated from the zero-pole
form. Arranged the 2.4.b if all roots are real numbers:
1
1
(
1
+
s)
L
(
1
+
s )
bzz
×
×
L
z
z
z
1
m
G
(
s
)
=m12 m
<2.4.c.>
1
1
ap
×
×
p
L
p
n
1
2
n
(
1
+
s)
L
(
1
+
s )
p
p
1
n
1
1
where - z = t j and - p = Ti are the time constants of the transfer function.
j
i
If there is a complex conjugate pair of roots than first they are
multiplied and after arranged it, for example:
1 1
1
12
1
1
(
1)
+
s(
1
+
s
)
=
1
+
s
(+
)
+
s×
<2.4.d>
p
p
p
p
1 p
2
1
2 p
1
2
where
1 1
× =T is the time constants of the transfer function and
p1 p1
1 1
+ =2
××
D
Twhere D is named damping ratio. If D < 1 than there is a
p
1 p
2
complex conjugate pair of roots, else the roots are real numbers.
If you define this transfer function in MATLAB you should give a
concrete value the polynomial coefficients of numerator and denominator. For
example:
2
s
+
4
.
1
G
(
s
)
= 2
1
<2.5.>
3
.
5
s+
1
.
7
s
+
2
.
3
The syntax of the tf command the next:
tf([num],[den])
where num is the polynomial coefficient in descending order in degree of
numerator and den is the polynomial coefficient in descending order in degree
of denominator.
Characteristics of transfer functions
The square brackets aren’t required if the polynomial’s order is zero,
otherworld it is a constant.
Type the expression 2.5 into a MATLAB’s M-file:
G1=tf([2 4.1],[3.5 1.7 2.3])
Use the Save File and Run item of Debug menu! On the Command window
the following answer appears.
Transfer function
2s+4.1
--------------3.5s^2+1.7s+2.3
You define the zero-pole form of a transfer function in MATLAB.
Of course you must give a concrete value the roots! The syntax of the zpk
command the next:
zpk([zeros of num],[poles of den],gain)
If you have the G1 variable in transfer function form and you want to
use the pole-zero form of this transfer function it is easy to convert it. If the G1
variable is being it is enough to type the next:
G1=zpk(G1)
Use the Save File and Run item of Debug menu!
Zero/pole/gain
0.57143(s+2.05)
-------------------(s^2+0.7857s+0.6571)
Of course if the G1 variable is being in zero-pole form you can convert
it using the G1=tf(G1) command.
Note: Unfortunately there isn’t command to define a transfer function
in Bode form.
The syntax of the all commands whose operand has transfer function
type (impulse, step, bode, nyquist, and so on) are the following:
command(name of transfer function). For example step(G1)
which figures the unit step response of the G1(s) transfer function. It is possible
to see more transfer function together typing step(G1,G2).
Characteristics of transfer functions
2.2 Block diagram representation
The basic block diagram element, called block, is a box with one input and one
output (Fig. 2.2.a.). The direction of flow from input to output is indicated by
arrows. This box contains the transfer functions, G(s), expressing the
dynamical relationship between its input and output.
Note: Sometimes we use the block to represent the steady-state
relationship or the non-linear behavior.
When a signal goes two destinations you can use the takeoff point as
shown in Fig. 2.2.b. The summing junction is used to represent the addition or
subtraction of signals as shown in Fig. 2.2.c. These three graphical objects are
enough to represent more complex linear system. The arrows show the
direction of the signal flow.
a.
b.
c.
y(s)
y(s)
x(s)
x(s)
y(s)
x(s)
x(s) ± y(s)
G(s)
y(s)
±
Figure 2.2. Graphical elements of block diagram
The great advantages of the block diagram representation are that they
clearly show the cause and the effect relationship of system components and
how the elements of the system are interconnected.
The advantages of the s-domain is to give simply method to deduce the
transfer function of a complicated time-domain linear system. The s-domain
functions are known as block’s transfer functions.
Blocks in cascade
Resulting transfer function is given by multiplying the two transfer function of
the blocks.
z(s)
x(s)
G1(s)
y(s)
G2(s)
y
(
s
)
=
G
(
s
)
=
G
(
s
)
×
G
(
s
)
1
2
x
(
s
)
Figure 2.3. Block in cascade
Characteristics of transfer functions
Blocks in parallel
Resulting transfer function is given by adding the two transfer function of the
blocks.
x(s)
y(s)
G1(s)
±
y
(
s
)
=
G
(
s
)
=
G
(
s
)
±
G
(
s
)
1
2
x
(
s
)
G2(s)
Figure 2.4. Block in parallel
Blocks in feedback
Numerator of the resulting transfer function equals to the block in the forward
branch, and the denominator is given one and plus-minus the blocks in the loop
in cascade.
x(s)
y(s)
G1(s)
G
(
s
)
y
(
s
)
1
=
G
(
s
)
=
x
(
s
)
1
±
G
(
s
)
×
G
(
s
)
1
2
G2(s)
Figure 2.5. Block in parallel
In generally if the feedback is negative the denominator of resulting
transfer function contains the plus sign.
2.3 Basic transfer functions
The nature of a transfer function can be proportional, integral, or differential.
The speed of response of the system depends on its time constants. Some
transfer functions have special nature, when the response of the block repeats
the input signal with delayed it in time. It is called delay time, transport time,
or dead time.
The most complicated transfer function of a linear system can be built
only six basic transfer functions connected them cascade, parallel, and
feedback. These are the proportional tag signed by P, the integral tag signed by
Characteristics of transfer functions
I, the differential tag signed by D, the dead time tag signed by H, accordingly a
first order tag signed PT1 and the second order tag signed PT2.
We will define the transfer functions and differential equations, and plot
and analyze the unit step responses and bode diagrams of these basic tags.
P proportional tag
The nature of the proportional tag is that the output repeats the shape of the
input signal, only gains it. The sign of the gain is K.
()=Kxt
×()
The differential equitation: yt
GP(s)= K
The transfer function
If you want to define this transfer function in MATLAB you must give
a concrete value of the gain K. The name of the proportional tag transfer
function is GP ( s ) . For example let be K = 2.4 and the name is GP. Type it an
M-file:
K=2.4;
GP=tf(K,1)
Use the Save File and Run item of Debug menu! Two variables (K, GP) come
into being in the Workspace and the result of the command lines is highlight on
the Command window:
Transfer function
2.4
Figure 2.6. Unit step response of proportional tag
Characteristics of transfer functions
If the GP transfer function type variable is available in the Workspace
you can create the figure of impulse or step response in time-domain of the
proportional tag, typing the impulse or step command.
step(GP)
The response signal is on the Fig. 2.6. With the Data Cursor you can see
the value of the gain.
The Bode or the Nyquist diagram of a transfer function can be drawn by
typing the bode or nyquist command.
bode(GP)
Figure 2.7. Bode diagram of proportional tag
You can read the gain’s value on the Fig. 2.7 in decibel. Typing the
7.6
10 20
expression on the Command window you will receive 2.988 as the
gain’s value for result which is 2.4 for the engineers.
Note: MATLAB automatically offers an area of time or frequency for
the figures. You can change it. For example:
t=0:0.05:2.5;
step(GP,t)
Characteristics of transfer functions
I integral tag
The nature of the integral tag is that the output increases or decreases while the
input signal isn’t zero. The KI integral coefficient or TI integral time constant
describes the speed of the changes in the output.
yt
()
()
t =K
()
td
t or T
=xt
()
òx
The differential equitation: y
I×
I×
dt
K
1
I
Gs
I()= =
The transfer function
s s
T
I
When you define this transfer function in MATLAB you must give a
concrete value of the integral coefficient KI or integral time constant TI, and a
name of the integral tag’s transfer function GI ( s ) . For example be the value of
[ec.]and the name is GI! Type it an M-file:
the integral time constant TI=2.4s
TI=2.4;
GI=tf(1,[TI 0])
Use the Save File and Run item of Debug menu! Two variables (TI, GI) come
into being in the Workspace and the result of the command lines is highlight on
the Command window:
Transfer function
1
---2.4s
Type the step(GI) command to see the unit step response (Fig. 2.8).
On the Fig. 2.8 we show how you can count the value of TI from arbitrary two
1
point of line. Counting of KI is similar, because of next expression: KI = T .
I
Type the bode(GI) command to see the Bode diagram (Fig. 2.9)! On
the Fig. 2.9 you can see the dialog box shown by right mouse click, the items
of Characteristic and Minimum Stability Margins are chosen. The gain and
phase crossover frequencies are highlighted after using these items. (The Fig
2.9 only has gain crossover frequency)
The phase margin belongs to the gain crossover frequency and indicated
on the Bode phase diagram. After left mouse click on the point a dialog box
appears with detailed information. The value of the gain crossover frequency in
rad./sec. is equal the value KI in 1/sec.
Characteristics of transfer functions
Figure 2.8. Unit step response of integral tag
Figure 2.9. Bode diagram of integral tag
Characteristics of transfer functions
D differential tag
The nature of the differential tag is that the output is zero while the input signal
is a constant. The KD differential coefficient or TD differential time constant
describes the gain of the speed of the changes in the input.
d
x
(
t
)
d
x
(
t
)
(
t
)
=
K
× =×
T
The differential equitation: y
D
D
d
t
d
t
G
()
s=
Ks
×
=
s
T
The transfer function
D
D
D
Figure 2.10. Bode diagram of differential tag
When you define this transfer function in MATLAB you must give a
concrete value of the differential coefficient KD or differential time constant
TD, and a name of the differential tag’s transfer function GD ( s ) . For example
D=0.4s
[ec.]and the name is GD!
be the value of the integral time constant T
Type it an M-file:
TD=0.4;
GD=tf([TD 0],1)
Use the Save File and Run item of Debug menu! Two variables (TD, GD)
come into being in the Workspace and the result of the command lines is
highlighted on the Command window:
Characteristics of transfer functions
Transfer function
0.4s
If we type the step(GI) command the MATLAB sends an error
messages on The Command window. It isn’t able to plot the theoretic Direct
delta δ(t) signal.
Type the bode(GD) command to see the Bode diagram! The
reciprocal value of the gain crossover frequency in rad./sec. equals the
differential time constant value in sec. shown it on the Figure 2.10.
H dead time tag
The nature of the dead time tag is that the output repeats the shape of the input
signal, only delays it. The sign of delay time is τ.
(
t)=
1
(
t-t)
×x
(
t t)
The differential equitation: y
- st
G
The transfer function
H(s)=e
When you define this transfer function in MATLAB you must give a
concrete value of the delay time constant τ and a name of the dead time tag’s
transfer function GH ( s ) . For example be the value of Tau= 0.1 and the name
is GH. The transfer function of dead time is transcendent and so it needs a
special tf command. Type the following an M-file:
Tau=0.1;
GH=tf(1,1,’inputdelay’,0.1)
Use the Save File and Run item of Debug menu! Two variables (Tau, GH)
come into being in the Workspace and the result of the command lines is
highlight on the Command window:
Transfer function
Exp(-0.1*s)*(1)
Note: Instead of 1,1 part of this tf command you can type arbitrary
transfer function.
Type the step(GH) command to see (Fig. 2.11) the unit step
response. On the Fig. 2.11 we show how you can read the value of Tau.
Note: We plot the Fig. 2.11 define the t=0:0.001:0.8 time variable and
type the tf(GH,t) command.
Type the bode(GI) command to see the Bode diagram (Fig. 2.12)!
On the Fig. 2.12 we show how you can read the value of Tau.
On the Fig. 2.12 you can see the very strong phase shift caused by the
dead time.
Characteristics of transfer functions
Figure 2.11. Unit step response of dead time tag
Figure 2.12. Unit step response of dead time tag
Characteristics of transfer functions
PT1 first order tag
The nature of the first order tag is that the output increases with decreasing
speed until the output reached the final value. The T time constant describe the
required settling time.
d
y
(
t)
+y
(
t)=x
(
t)
The differential equitation: T
d
t
1
G
()
s=
P
T
1
The transfer function
sT+1
When you define this transfer function in MATLAB you must give a
concrete value of the time constant T, and a name of the first order tag’s
transfer function GPT1(s) . For example be the value of the time constant
T= 4[sec.]and the name is GT1! Type it an M-file:
T=4;
GT1=tf(1,[T 1])
Use the Save File and Run item of Debug menu! Two variables (T, GT1)
come into being in the Workspace and the result of the command lines is
highlight on the Command window:
Transfer function
1
---4s+1
Type the step(GT1) command! On the Fig. 2.13 we show where you
can read the value of T and one percent settling time Ts1%.
Involving the differential equitation energized the PT1 tag by unit step
t
( )=0.6321
t = T the value is yT
( x(t) = 1(t) ) the result is yt
()=1- e T. At
T=
) 0
.9
8
1
7 and at t = 5×T the value
and at t = 4 ×T the value is y(4
T=
) 0
.9
9
3
3. You can find the T time constant and the two percent
is y(5
settling time Ts2% or the one percent settling time Ts1% on the curve of the step
response of first order tag as shown on the Fig. 2.13.
Type the bode(GT1) command! On the Fig. 2.14 you can read the T
time constant in second as the reciprocal value of the frequency in rad./sec. at
the -45º phase shift and also this frequency is called break-point frequency
where the amplitude value is -3 dB.
Characteristics of transfer functions
Figure 2.13. Unit step response of first order tag
Figure 2.14. Bode diagram of first order tag
Characteristics of transfer functions
PT2 second order tag
The nature of the second order tag is that the oscillatory behavior of the output
signal is possible. The T time constant the D damping ratio is required to
define a second order tag.
2
d
y
(
t
)
d
y
(
t
)
T2
+
2
×
D
×
T +=
y
(
t
)x
(
t
)
d
t
d
t
1
G
(
s
)
=2
P
T
2
The transfer function
T
s+
2
D
T
s
+
1
If you want to define this transfer function in MATLAB you must give
a concrete value of the time constant T and D damping ratio, and a name of the
first second tag’s transfer function GPT 2 (s) . At D ³ 1 condition the transfer
function of PT2 tag is decomposed two PT1 tags in cascade, but if D < 1 this
way is not possible. We will examine only the D < 1 condition with different
D values.
For example be the value of the time constant T= 2[sec.] and
The differential equitation:
D1 = 0.8 , D2 = 0.4 , D3 = 0.1, and the name is GT2! Type it an M-file:
T=2;
D1=0.8;
D2=0.4;
D3=0.1;
GT2D1=tf(1,[T^2 2*D1*T 1])
GT2D2=tf(1,[T^2 2*D2*T 1])
GT2D3=tf(1,[T^2 2*D3*T 1])
Use the Save File and Run item of Debug menu! The new variables (T, D1,
D2, D3, GT2D1, GT2D2, GT2D3) come into being in the Workspace and the
result of the command lines is highlight on the Command window:
Transfer function
1
-----------4s^2+3.2s+1
Transfer function
1
-----------4s^2+1.6s+1
Characteristics of transfer functions
Transfer function
1
-----------4s^2+0.4s+1
Type the step(GT2D1,GT2D2,GT2D3) command! Use the Save File and
Run item of Debug menu!
Figure 2.15. Unit step response of second order tag with different D values
You can see on Fig. 15 the smaller value of damping ratio causes higher
amplitude of the oscillatory behavior. We can’t determine the value of T time
constant and the D damping ratio.
Type the bode(GT2D1,GT2D2,GT2D3) command!
1
The wT =
is called natural frequency. The phase shift at the natural
T
frequency is -90º, accordingly you can count the T time constant finding the 90º phase shift on the Bode diagram of the second order tag. The Fig. 16 shows
it to you. You can also see on Fig. 16 the smaller value of damping ratio causes
higher peak value of the amplitude at the natural frequency and sharper
transition on the phase diagram.
Characteristics of transfer functions
Figure 2.16. Bode diagram of second order tag with different D values
2.4 Basic transfer functions in cascade
A lot of generally used transfer functions are composed from the basic transfer
functions in cascade. For example: HPT1 dead time first order, IT1 integral
first order, PT3 third order, and so on.
The HPT1 and IT1 are very popular to substitute a real measured curve
of a process.
Substitution with HPT1
The all transfer function without integral effect, other word have final value
can be substituted with HPT1, or HPT2, or PT3. The later two require
computer.
The transfer function of HPT1 is:
1 -s
u
G
(
s
)
=
eT
H
T
1
<2.6.>
s
T
+
1
g
where Tg is the approximated time constant and Tu is the approximated delayed
time.
If we have a reaction curve of the process energized it by a unit step
shape signal, the first fitting the line of maximum slope of the real curve.
Characteristics of transfer functions
Where this line crosses the beginning and the final value of the reaction curve
provides the Tu and the Tu + Tg approximated time values. See Fig. 2.17!
The ratio of the amplitude of reaction curve and input signal provides
the gain of the approximated model of process.
Figure 2.17. Approximated HPT1 model of a reaction curve
0
.8
3
=0
.8
3
As for the Fig. 2.17 the K=
, T
u=0.43sec., and
P
1
Tg=2.73sec.
and
so
the
approximated
transfer
function:
0
.
8
3 -0
4
3
s
Gs
()
=
e.
H
T
1
.
2
.
7
31
s
+
Substitution with IT1
The all transfer function witches have integral effect, other word haven’t final
value can be substituted with IT1, or HIT1, or IT2. The later two require
computer. The transfer function of IT1 is:
1 1
G
()
s=
I
T
1
<2.7.>
s
T
T
+
1
Is
g
where Tg is the approximated time constant and TI is the integral time constant.
Characteristics of transfer functions
If we have a reaction curve of the process energized it by a unit step
shape signal, the first is the extension of the linear section of the reaction curve.
Figure 2.18. Approximated IT1 model of a reaction curve
As for the Fig. 2.18 the TI = 2.5sec., and Tg = 4sec., and so the
1 1
()
s=
I
T
1
approximated transfer function: G
.
2
.54
ss
+
1
2.5 Transfer functions in parallel
The most wide-spread compensation is the PID. In the industrial area the P, I,
and D effect are used in parallel three channels, and differential first order DT1
is used instead of pure D tag.
Note: Often used solution is that P, I and D effects are generated in
feedback.
In the industrial area the ratio of the differential time TD constant and T
T
< D=A
<2
0The AD is called differential gain.
first order time constant is 5
D
T
There are differences between the American form and the European
form of the PIDT1 compensator are shown in Fig. 2.19.a and 2.19.b.
Characteristics of transfer functions
KC
KI +
s
K D ×s
sT + 1
1
+
+
Figure 2.19.a.
American form of PIDT1
1
sTI
+
+
+
KC
TD ×s
sT + 1
Figure 2.19.b.
European form of PIDT1
The effect of the two type of the PIDT1 compensator is same, if:
K
K
TI = C and TD = D
<2.8.>
KI
KC
Depending on the status of the switches of the channels there are P, I,
PI, PDT1, and PIDT1 compensator.
If the model of a process described by transfer function without integral
effect then the PI, PIDT1, maybe P compensator are practicable. The type of
the compensator depends on the ratio of the approximated time constant Tg and
the approximated delayed time Tu.
If the model of a process described by transfer function with integral
effect than the P, PDT1 maybe PIDT1 compensator are practicable.
Hereinafter we will use the European form of the PIDT1 compensator.
PI compensator
The most used compensator for plant without integral effect is the PI.
The transfer function
ì 1
ü 1
ï
ï
+
s
T
ï
I
G
(
s
)
=
K
1
+ï
=
K
í
ý
<2.8.>
P
I
C
C
ï
ï
T
s
T
I
I
ï s
ï
î
þ
Defining this transfer function in MATLAB requires a concrete value of
the integral time constant T=2.4 sec. and KC=2 gain.
Kc=2;
Gi=tf(1,[2.4 0])
Gpi=Kc*(1+Gi)
The unit step response of the PI compensator is shown on the Fig. 2.20.
Characteristics of transfer functions
Figure 2.20. Unit step response of PI compensator
Figure 2.21. Bode diagram of PI compensator
Characteristics of transfer functions
On the Fig. 2.20 you can read the KC = 2 at the beginning and the
K
é
1 ù
T
= C=2
.4
s
e
c
.It is clear
K
0
.8
3
2
I
,
and
so
from
formula
2.8
the
I=
ê
ú
e
c
.û
ës
K
I
the unit step response of a pure proportional P (Fig. 2.6) and integral I (Fig 2.8)
tag in parallel.
On the Fig. 2.21 you can read the KC = 6dB, and at the phase shift -45
6
0
.4
1
7
r
a
d
/s
e
c
.. The gain
degree the w
. The value
I=
2
0
K
=
1
0
=
1
.
9
9
5
;2
C
1
[1/sec.], and so TI =2.4sec.
rad/sec.]equals the value of
of w
I[
TI
PDT1 compensator
In the industrial area the most used compensator for plant with integral effect is
the P, but in the circuit technique The PDT1 compensator is often applied. The
DT1 is a pure differential D and first order T1 tag in cascade, and the PDT1 is
a proportional P and DT1 tag in parallel.
The transfer functions of PDT1 compensator:
ì
ü
s
T
1
+
s
(
T
+
T
)
ï
ï
D
G
(
s
)
=+
K
1D
=
K
í
ý
<2.9.>
P
D
T
1
C
C
ïs
ï
T
+
1
s
T
+
1
ï
ï
î
þ
Defining this transfer function in MATLAB requires a concrete value of
the KC=2 gain and the differential time constant TD=0.9 sec. and the
differential gain AD=9, and so the time constant of the first order part of DT1 is
T=0.1 sec.
Kc=2;
Gdt=tf([0.9 0],[0.1 0])
Gpdt=Kc*(1+Gdt)
The Fig. 2.22. shows the unit step response of the PDT1 compensator.
.05»2at the final value, the peak
On the Fig 2.22 you can read the K
C=2
value Apeak = 20 of the amplitude at the beginning, and the time constant
T= 0.1sec. of the first order tag where the initial tangent line intersects the
final value.
From the Laplace final value theorem:
T
+
T
D
l
i
m
G
(
sK
)
=
=
A
<2.10.>
P
D
T
1
C
p
e
a
k
T
s
®
¥
From the formula 2.10:
Characteristics of transfer functions
AK
-C 2
0
2
p
e
a
k
T
=
T
=
0
.
1 =
0
.
9
s
e
c
.
D
K
2
C
<2.11.>
Figure 2.22. Unit step response of PDT1 compensator
The Fig. 2.23. shows the Bode diagram of the PDT1 compensator. On
the Fig 2.23 you can read the KC » 6dB at the beginning of the amplitude
3
.2
r
a
d
/s
e
c
. at the peak value of the
p
e
a
k=
curve and the frequency value w
positive phase shift. Convert the gain to real number KC = 2 .
w
×
w
We know that w
T= A
D×
D, and so
p
e
a
k= w
D
T and the w
wD=
wpeak
AD
×
w
and w
T= A
D
p
e
a
k
<2.12.>
Using the formula 2.12, assuming the differential gain AD = 9
w
.
2
1 1
p
e
a
k3
w
= ==
1
.
0
7/
r
a
d
s
e
c
T
= = =
0
.
9
3
s
e
c
.and
D
D
,
and
so
w
.
0
7
A
D1
D3
1
w
=
A
×=
w
×
3
.
2
=
9
.
6
r
a
d
/
s
e
c
.
=0
.1
s
e
c
.
, and so T=
T
D
p
e
a
k3
D
9
.6
Characteristics of transfer functions
Figure 2.23. Bode diagram of PDT1 compensator
2.6 Transfer functions in feedback
Sometimes you put the compensator in the feedback branch or create a
compensator with feedback of transfer function. For example:
-
G1 ( s ) 
2.5
s
G1( s)  1.5
Figure 2.24. Eliminate the integral effect of an actuator with proportional feedback
The transfer function:
2
.
5
G
(
s
)
2
.
5
G
(
s
)
= 1
= s =
2
.
5
1
+
G
(
s
)
G
(
s
)1
+
3
.
7
5
1
2
+×
1
.
5s
s
The G(s) is a first order tag.
<2.13.>
Characteristics of transfer functions
Note: In the industrial area it is used positioning an actuator or in the
two or three point controller.
2.7 Exercises
1.
3
.5
()=
1
Define the time constant and the gain of the Gs
s+1
.7
5transfer
function from the step response and from the Bode diagram! Save the figures
with comments!
2.
Define the transfer functions in cascade resulting from the following
figure, and indicate where to read the time constant and the integral time
constant on the figure of the unit step response and on the figure of Bode
diagram!
G21 ( s ) 
2.5
s
G22 ( s ) 
1
8s  1
3.
Define the transfer functions in parallel resulting from the following
figure, and indicate where to read the necessary values to define the gain, the
differential time constant and the first order time constant on the figure of the
unit step response and on the figure of Bode diagram!
G31 ( s)  2.5
G32 ( s ) 
2.7 s
0.3s  1
4.
Define the transfer functions in feedback resulting from the following
figure, and indicate where to read the necessary values!
G1 ( s)  5
G1 ( s ) 
1
2s  1