Transcript Advanced Control Systems (ACS) - Dr. Imtiaz Hussain

Advanced Control Systems (ACS)
Lecture-8
S-plane Design
Dr. Imtiaz Hussain
Assistant Professor
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
1
Lecture Outline
2
Lead Compensation
• Generally Lead compensators are represented by
following transfer function
𝐺𝑐 𝑠 =
𝑇𝑠+1
𝐾𝑐 𝛼
𝛼𝑇𝑠+1
,
(0 < 𝛼 < 1)
• or
𝐺𝑐 𝑠 = 𝐾𝑐
1
𝑠+
𝑇
1
𝑠+
𝛼𝑇
,
(0 < 𝛼 < 1)
3
𝐺𝑐 𝑠 =
𝑠+1
3
𝑠+10
(𝛼 = 0.1)
,
Pole-Zero Map
Bode Diagram
1
Magnitude (dB)
0
0.5
0
-0.5
-1
-10
-5
-10
-15
-20
60
Phase (deg)
Imaginary Axis
Lead Compensation
-8
-6
-4
Real Axis
-2
0
30
0
-2
10
-1
10
0
10
1
10
2
3
10
10
Frequency (rad/sec)
4
Example-1
• Consider the position control system shown in following
figure.
• It is desired to design an Electronic lead compensator Gc(s)
so that the dominant closed poles have the damping ratio
0.5 and undamped natural frequency 3 rad/sec.
5
Step-1 (Example-1)
• Determine the characteristics of given system using root loci.
C ( s)
10
 2
R( s) s  s  10
• The damping ratio of the closed-loop
poles is 0.158.
• The undamped natural frequency of
the closed-loop poles is 3.1623
rad/sec.
• Because the damping ratio is small,
this system will have a large
overshoot in the step response and
is not desirable.
6
Step-2 (Example-1)
• From the performance specifications, determine the
desired location for the dominant closed-loop poles.
• Desired performance Specifications are:
 It is desired to have damping ratio 0.5 and undamped natural
frequency 3 rad/sec.
n2
C ( s)
9
 2
 2
2
R( s) s  2n s  n s  3s  9
s  1.5  j 2.5981
7
Step-3 (Exampl-1)
• To calculate the angle of deficiency apply Angle Condition at desired
closed loop pole.
Desired Closed Loop Pole
s  1.5  j 2.5981
 d  180  120  100.8
-2
 d  40.89
-1
100.8o
-2
120o
-1
8
Step-5 (Exampl-1)
Solution-1
• Solution-1
– If we choose the zero of the
lead compensator at s = -1 so
that it will cancel the plant pole
at s =-1, then the compensator
pole must be located at s =-3.
40.89
9
Step-5 (Example-1)
Solution-1
• The pole and zero of compensator are determined as
𝐺𝑐 𝑠 = 𝐾𝑐
1
𝑇
1
𝑠+𝛼𝑇
𝑠+
=
• The Value of 𝛼
determined as
𝑠+1
𝐾𝑐
𝑠+3
can be
40.89
yields
1
=1
𝑇=1
𝑇
yields
1
=3
𝛼 = 0.333
𝛼𝑇
10
Step-6 (Example-1)
Solution-1
• The Value of Kc can be
determined using magnitude
condition.
(𝑠 + 1) 10
𝐾𝑐
𝑠 + 3 𝑠(𝑠 + 1)
10
𝐾𝑐
𝑠(𝑠 + 3)
=1
𝑠=−1.5+𝑗2.5981
=1
40.89
𝑠=−1.5+𝑗2.5981
𝑠(𝑠 + 3)
𝐾𝑐 =
10
= 0.9
𝑠=−1.5+𝑗2.5981
𝑠+1
𝐺𝑐 𝑠 = 0.9
𝑠+3
11
Final Design Check
Root Locus
Solution-1
Root Locus
5
5
0.158 3.16
3
Imaginary Axis
0.5
0
0
0.5
3
0.158 3.16
-5
-4
-3
-2
-1
Real Axis
10
𝐺(𝑠) =
𝑠(𝑠 + 1)
0
1
-5
-4
-3
-2
-1
0
1
Real Axis
9
𝐺𝑐 𝑠 𝐺(𝑠) =
𝑠(𝑠 + 3)
12
Step-5 (Exampl-1)
Solution-2
• Solution-2
-2
40.89
-1
90o
49.2o
-3
-2
-1
13
Solution-2
Step-5 (Exampl-1)
• Solution-2
-2
40.89
-1
90o
49.2o
-3
-2
-1
𝑠 + 1.5
𝐺𝑐 𝑠 = 1.03
𝑠 + 3.6
14
Exampl-2
• Design a lead compensator for following system.
4
s ( s  2)
• The damping ratio of closed loop poles is 0.5 and natural
undamped frequency 2 rad/sec. It is desired to modify
the closed loop poles so that natural undamped
frequency becomes 4 rad/sec without changing the
damping ratio.
15
Example-3
• Consider the model of space vehicle control system
depicted in following figure.
• Design lead compensator such that the damping ratio
and natural undamped frequency of dominant closed
loop poles are 0.5 and 2 rad/sec.
16
Lag Compensation
• Lag compensation is used to improve the steady state error
of the system.
• Generally Lag compensators are represented by following
transfer function
𝐺𝑐 𝑠 =
𝑇𝑠+1
𝐾𝑐 𝛽
𝛽𝑇𝑠+1
• Or
𝐺𝑐 𝑠 = 𝐾𝑐
1
𝑇
1
𝑠+
𝛽𝑇
𝑠+
,
,
(β > 1)
(β > 1)
• Where 𝐾 𝑐is gain of lag compensator.
17
Lag Compensation
𝐺𝑐 𝑠 =
𝑠+10
3
𝑠+1
(𝛽 = 10)
,
Bode Diagram
Pole-Zero Map
30
Magnitude (dB)
1
0.5
20
15
10
5
0
Phase (deg)
0
-0.5
-1
-10
25
-8
-6
-4
Real Axis
-2
0
-30
-60
-2
10
0
10
Frequency (rad/sec)
2
10
18
Lag Compensation
• Consider a lag compensator Gc(s), where
1
𝐺𝑐 𝑠 = 𝐾𝑐
𝑠+𝑇
1
𝑠+𝛽𝑇
,
(β > 1)
• If we place the zero and pole of the lag compensator very close to
each other, then at s=s1 (where s1 is one of the dominant closed
loop poles) then the magnitudes
1
𝑠1 +
𝑇
and
1
𝑠1 +
𝛽𝑇
are almost
equal, or
1
𝑠+
𝑇 ≅𝐾
𝐺𝑐 (𝑠1 ) = 𝐾𝑐
𝑐
1
𝑠+
𝛽𝑇
19
Lag Compensation
• To make the angle contribution of the lag portion of the
compensator small, we require
1
𝑠+
𝑇
−5° < 𝑎𝑛𝑔𝑙𝑒
1
𝑠+
𝛽𝑇
< 0°
• This implies that if gain 𝐾𝑐 of the lag compensator is set equal to 1,
the alteration in the transient-response characteristics will be very
small, despite the fact that the overall gain of the open-loop
transfer function is increased by a factor of 𝛽, where 𝛽>1.
20
Lag Compensation
• If the open loop transfer function of the uncompensated system is
G(s), then the static velocity error constant Kv of the uncompensated
system is
𝐾𝑣 = lim 𝑠𝐺(𝑠)
𝑠→0
• Then for the compensated system with the open-loop transfer
function Gc(s)G(s) the static velocity error constant 𝐾𝑣 becomes
𝐾𝑣 = lim 𝑠𝐺𝑐 𝑠 𝐺(𝑠) = 𝐾𝑣 lim𝐺𝑐 𝑠
𝑠→0
𝑠→0
1
𝑠+
𝑇 =𝐾 𝐾𝛽
𝐾𝑣 = 𝐾𝑣 lim𝐾𝑐
𝑣 𝑐
1
𝑠→0
𝑠+
𝛽𝑇
21
Design Procedure
• The procedure for designing lag compensators by the rootlocus method may be stated as follows.
• We will assume that the uncompensated system meets the
transient-response specifications by simple gain adjustment.
• If this is not the case then we need to design a lag-lead
compensator which we will discuss in next few classes.
22
Example-4
• Consider the system shown in following figure.
• The damping ratio of the dominant closed-loop poles is
0.491. The undamped natural frequency of the dominant
closed-loop poles is 0.673 rad/sec. The static velocity error
constant is 0.53 sec–1.
• It is desired to increase the static velocity error constant Kv to
about 5 sec–1 without appreciably changing the location of
the dominant closed-loop poles.
23
Example-4 (Step-1)
• The dominant closed-loop poles of given system are
s = -0.3307 ± j0.5864
24
Example-4 (Step-3)
• The static velocity error constant of the plant (𝐾𝑣 ) is
1.06
𝐾𝑣 = lim𝑠𝐺(𝑠) = lim 𝑠
𝑠→0
𝑠→0
𝑠 𝑠+1 𝑠+2
= 0.53𝑠 −1
• The desired static velocity error constant ( 𝐾𝑣 ) of the
compensated system is 5𝑠 −1 .
𝐾𝑣 = lim 𝑠𝐺𝑐 𝑠 𝐺(𝑠) = 𝐾𝑣 lim𝐺𝑐 𝑠
𝑠→0
𝑠→0
1
𝑠+
𝑇 =𝐾 𝐾𝛽
𝐾𝑣 = 𝐾𝑣 lim𝐾𝑐
𝑣 𝑐
1
𝑠→0
𝑠+
𝛽𝑇
25
Example-4 (Step-3)
1
𝑇 =𝐾 𝐾𝛽
𝐾𝑣 = 𝐾𝑣 lim𝐾𝑐
𝑣 𝑐
1
𝑠→0
𝑠+
𝛽𝑇
𝑠+
𝐾𝑣 = 𝐾𝑣 𝐾𝑐 𝛽
5 = 0.53𝛽
𝛽 = 10
26
Example-4 (Step-4)
• Place the pole and zero of the lag compensator
𝐺𝑐 𝑠 =𝐾𝑐
• Since 𝛽 = 10, therefore
𝐺𝑐 𝑠 =𝐾𝑐
1
𝑠+
𝑇
1
𝑠+
𝛽𝑇
1
𝑇
0.1
𝑠+
𝑇
𝑠+
27
Example-4 (Step-4)
Solution-1
• Place the zero and pole of the lag compensator at s=–0.05
and s=–0.005, respectively.
• The transfer function of the lag compensator becomes
𝐺𝑐 𝑠
𝑠+0.05
=𝐾𝑐
𝑠+0.005
• Open loop transfer function is given as
𝐺𝑐 𝑠
𝑠+0.05
1.06
𝐺(𝑠)=𝐾𝑐
𝑠+0.005 𝑠(𝑠+1)(𝑠+2)
𝐺𝑐 𝑠
𝐾(𝑠+0.05)
𝐺(𝑠)=
𝑠(𝑠+0.005)(𝑠+1)(𝑠+2)
𝑤ℎ𝑒𝑟𝑒 𝐾 = 1.06𝐾𝑐
28
Example-4 (Step-5)
Solution-1
• Root locus of uncompensated and compensated systems.
• New Closed Loop poles
are
𝑠 = −0.31 ± 𝑗0.55
29
Example-4 (Step-6)
Solution-1
• The open-loop gain K is determined from the magnitude
condition.
𝐾(𝑠 + 0.05)
𝑠(𝑠 + 0.005)(𝑠 + 1)(𝑠 + 2)
=1
𝑠=−0.31+𝑗0.55
𝐾 = 1.0235
• Then the compensator gain 𝐾𝑐 is determined as
𝐾 = 1.06𝐾𝑐
𝐾
𝐾𝑐 =
= 0.9656
1.06
30
Example-1 (Step-6)
Solution-1
• Then the compensator transfer function is given as
𝑠 + 0.05
𝐺𝑐 𝑠 = 0.9656
𝑠 + 0.005
31
Example-4 (Final Design Check)
Solution-1
• The compensated system has following open loop transfer
function.
𝐺𝑐 𝑠
1.0235(𝑠+0.05)
𝐺(𝑠)=
𝑠(𝑠+0.005)(𝑠+1)(𝑠+2)
• Static velocity error constant is calculated as
𝐾𝑣 = lim 𝑠𝐺𝑐 𝑠 𝐺(𝑠)
𝑠→0
1.0235 𝑠 + 0.05
𝐾𝑣 = lim 𝑠[
]
𝑠→0 𝑠 𝑠 + 0.005 𝑠 + 1 𝑠 + 2
1.0235 0.05
𝐾𝑣 =
= 5.12𝑠 −1
0.005 1 2
32
Example-4 (Step-4)
Solution-2
• Place the zero and pole of the lag compensator at s=–0.01
and s=–0.001, respectively.
• The transfer function of the lag compensator becomes
𝐺𝑐 𝑠
𝑠+0.01
=𝐾𝑐
𝑠+0.001
• Open loop transfer function is given as
𝐺𝑐 𝑠
𝑠+0.01
1.06
𝐺(𝑠)=𝐾𝑐
𝑠+0.001 𝑠(𝑠+1)(𝑠+2)
𝐺𝑐 𝑠
𝐾(𝑠+0.01)
𝐺(𝑠)=
𝑠(𝑠+0.001)(𝑠+1)(𝑠+2)
𝑤ℎ𝑒𝑟𝑒 𝐾 = 1.06𝐾𝑐
33
Example-4 (Step-5)
Solution-2
• Root locus of uncompensated and compensated systems.
Root Locus
6
• New Closed Loop poles
are
Compensated System
4
Imaginary Axis
𝑠 = −0.33 ± 𝑗0.55
Actual System
2
0
-2
-4
-6
-8
-6
-4
-2
Real Axis
0
2
4
34
Example-5
• Design a lag compensator for following unity
feedback system such that the static velocity error
constant is 50 sec-1 without appreciably changing the
closed loop poles, which are at 𝑠 = −2 ± 𝑗 6.
35
Lag-Lead Compensation
• Lag-Lead compensators are represented by following transfer
function
𝐺𝑐 𝑠 = 𝐾𝑐
1
𝑇1
𝛾
𝑠+
𝑇1
𝑠+
1
𝑇2
1
𝑠+
𝛽𝑇2
𝑠+
, (γ > 1 𝑎𝑛𝑑 β > 1)
• Where Kc belongs to lead portion of the compensator.
Lag-Lead Compensation
𝐺𝑐 𝑠 =
𝑠+1
𝐾𝑐
𝑠+2
𝑠+0.4
𝑠+0.1
Example-6 (Case-1)
• Consider the control system shown in following figure
• The damping ratio is 0.125, the undamped natural frequency is 2
rad/sec, and the static velocity error constant is 8 sec–1.
• It is desired to make the damping ratio of the dominant closed-loop
poles equal to 0.5 and to increase the undamped natural frequency
to 5 rad/sec and the static velocity error constant to 80 sec–1.
• Design an appropriate compensator to meet all the performance
specifications.
Example-6 (Case-1)
• From the performance specifications, the dominant closed-loop
poles must be at
𝑠 = −2.50 ± 𝑗4.33
• Since
4

 235
s ( s  0.5) s  2.50 j 4.33
• Therefore the phase-lead portion of the lag–lead compensator
must contribute 55° so that the root locus passes through the
desired location of the dominant closed-loop poles.
Example-6 (Case-1)
• The phase-lead portion of the lag–lead compensator becomes
𝐾𝑐
1
𝑠+
𝑇1
𝛾
𝑠+
𝑇1
𝑠+0.5
= 𝐾𝑐
𝑠+5.02
• Thus 𝑇1 = 2 and 𝛾 = 10.04.
• Next we determine the value of Kc from the magnitude
condition:
(𝑠 + 0.5)
4
𝐾𝑐
𝑠 + 5.02 𝑠(𝑠 + 0.5)
𝑠(𝑠 + 5.02)
𝐾𝑐 =
4
=1
𝑠=−2.5+𝑗4.33
= 5.26
𝑠=−2.5+𝑗4.33
Example-6 (Case-1)
• The phase-lag portion of the compensator can be designed as
follows.
• First the value of 𝛽 is determined to satisfy the
requirement on the static velocity error constant
𝐾𝑣 = lim 𝑠𝐺𝑐 𝑠 𝐺(𝑠)
𝑠→0
80 = lim 𝑠
𝑠→0
1
25.04 𝑠 +
𝑇2
𝑠 𝑠 + 5.02
1
𝑠+
𝛽𝑇2
80 = 4.988𝛽
𝛽 = 16.04
Example-6 (Case-1)
• Finally, we choose the value of 𝑇2 such that the following
two conditions are satisfied:
Example-6 (Case-1)
• Now the transfer function of the designed lag–lead
compensator is given by
𝐺𝑐 𝑠 = 6.26
𝑠+0.5
𝑠+5.02
𝑠+0.2
𝑠+0.0127
PID
• PID Stands for
– P  Proportional
– I  Integral
– D  Derivative
44
Four Modes of Controllers
• Each mode of control has specific advantages and
limitations.
• On-Off (Bang Bang) Control
• Proportional (P)
• Proportional plus Integral (PI)
• Proportional plus Derivative (PD)
• Proportional plus Integral plus Derivative (PID)
45
On-Off Control
• This is the simplest form of control.
Set point
Error
Output
46
Proportional Control (P)
• In proportional mode, there is a continuous linear relation
between value of the controlled variable and position of the
final control element.
𝑟(𝑡)
𝑒(𝑡)
𝑏(𝑡)
-
𝑐𝑝(𝑡) = 𝐾𝑝 𝑒(𝑡)
𝐾𝑝
𝑃𝑙𝑎𝑛𝑡
𝑐(𝑡)
𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙
𝐶𝑜𝑛𝑡𝑟𝑜𝑙
𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘
• Output of proportional controller is
𝑐𝑝(𝑡) = 𝐾𝑝 𝑒(𝑡)
• The transfer function can be written as
𝐶𝑝(𝑠)
= 𝐾𝑝
𝐸(𝑠)
47
Proportional Controllers (P)
• As the gain is increased the system responds faster to
changes in set-point but becomes progressively
underdamped and eventually unstable.
48
Proportional Plus Integral Controllers (PI)
• Integral control describes a controller in which the output
rate of change is dependent on the magnitude of the
input.
• Specifically, a smaller amplitude input causes a slower
rate of change of the output.
49
Proportional Plus Integral Control (PI)
𝐾𝑖 ∫
𝑒(𝑡)
𝑟(𝑡)
𝑏(𝑡)
𝐾𝑝
𝐾𝑖
𝑒(𝑡) 𝑑𝑡
𝐾𝑝 𝑒(𝑡)+
+
𝑐𝑝𝑖 𝑡
𝑃𝑙𝑎𝑛𝑡
𝑐(𝑡)
𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘
𝑐𝑝𝑖 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑖
𝑒 𝑡 𝑑𝑡
50
Proportional Plus Integral Control (PI)
𝑐𝑝𝑖 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑖
𝑒 𝑡 𝑑𝑡
• The transfer function can be written as
𝐶𝑝𝑖(𝑠)
1
= 𝐾𝑝 + 𝐾𝑖
𝐸(𝑠)
𝑠
51
Proportional Plus derivative Control (PD)
𝑑
𝐾𝑑
𝑑𝑡
𝑒(𝑡)
𝑟(𝑡)
𝑏(𝑡)
𝐾𝑝
𝑑𝑒(𝑡)
𝐾𝑑
𝑑𝑡
𝐾𝑝 𝑒(𝑡)+
+
𝑐𝑝𝑑 𝑡
𝑃𝑙𝑎𝑛𝑡
𝑐(𝑡)
𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘
𝑐𝑝𝑑
𝑑𝑒(𝑡)
𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑑 𝑑𝑡
52
Proportional Plus derivative Control (PD)
𝑐𝑝𝑑
𝑑𝑒(𝑡)
𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑑 𝑑𝑡
• The transfer function can be written as
𝐶𝑝𝑑(𝑠)
= 𝐾𝑝 + 𝐾𝑑 𝑠
𝐸(𝑠)
53
Proportional Plus derivative Control (PD)
• The stability and overshoot problems that arise when a
proportional controller is used at high gain can be mitigated by
adding a term proportional to the time-derivative of the error signal.
The value of the damping can be adjusted to achieve a critically
damped response.
54
Proportional Plus derivative Control (PD)
• The higher the error signal rate of change, the sooner the final
control element is positioned to the desired value.
• The added derivative action reduces initial overshoot of the
measured variable, and therefore aids in stabilizing the process
sooner.
• This control mode is called proportional plus derivative (PD) control
because the derivative section responds to the rate of change of the
error signal
55
Proportional Plus Integral Plus Derivative Control (PID)
𝑑
𝐾𝑑
𝑑𝑡
𝑒(𝑡)
𝑟(𝑡)
𝑏(𝑡)
𝐾𝑝
𝐾𝑑
𝑑𝑒(𝑡)
𝑑𝑡
𝐾𝑝 𝑒(𝑡) +
-
+
𝑐𝑝𝑖𝑑 𝑡
𝑃𝑙𝑎𝑛𝑡
𝑐(𝑡)
+
𝐾𝑖 ∫
𝐾𝑖
𝑒(𝑡) 𝑑𝑡
𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘
𝑐𝑝𝑖𝑑 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑖
𝑑𝑒(𝑡)
𝑒(𝑡) 𝑑𝑡 + 𝐾𝑑
𝑑𝑡
56
Proportional Plus Integral Plus Derivative Control (PID)
𝑐𝑝𝑖𝑑 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑖
𝑑𝑒(𝑡)
𝑒(𝑡) 𝑑𝑡 + 𝐾𝑑
𝑑𝑡
𝐶𝑝𝑖𝑑(𝑠)
1
= 𝐾𝑝 + 𝐾𝑖 +𝐾𝑑 𝑠
𝐸(𝑠)
𝑠
57
Proportional Plus Integral Plus Derivative Control (PID)
• Although PD control deals neatly with the overshoot and ringing
problems associated with proportional control it does not cure the
problem with the steady-state error. Fortunately it is possible to
eliminate this while using relatively low gain by adding an integral
term to the control function which becomes
58
The Characteristics of P, I, and D controllers
CL RESPONSE
RISE TIME
OVERSHOOT SETTLING TIME
S-S ERROR
Kp
Decrease
Increase
Small Change
Decrease
Ki
Decrease
Increase
Increase
Eliminate
Kd
Small
Change
Decrease
Decrease
Small
Change
59
Empirical Tuning (Zeigler-Nichol’s First Method)
• In the first method, we
obtain
experimentally
the response of the
plant to a unit-step
input.
• If the plant involves
neither integrator(s) nor
dominant
complexconjugate poles, then
such
a
unit-step
response curve may look
S-shaped
60
Zeigler-Nichol’s First Method
• This method applies if the response to a step input exhibits an
S-shaped curve.
• Such step-response curves may be generated experimentally
or from a dynamic simulation of the plant.
Table-1
61
Zeigler-Nichol’s Second Method
• In the second method, we first set 𝑇𝑖 = ∞ and 𝑇𝑑 = 0.
• Using the proportional control action only (as shown in
figure), increase Kp from 0 to a critical value Kcr at which
the output first exhibits sustained oscillations.
• If the output does not exhibit sustained oscillations for
whatever value Kp may take, then this method does not
apply.
62
Zeigler-Nichol’s Second Method
• Thus, the critical gain Kcr
and the corresponding
period Pcr are determined.
Table-2
63
Example-7
C (s)
K  sL

e
R ( s ) Ts  1
1
L
t
64
Example-7
Step Response
10
8
Amplitude
C( s )
10  2 s

e
R( s ) 3 s  1
6
4
2
0
0
5
10
15
Time (sec)
65
Example-8
• Consider the control system shown in following figure.
• Apply a Ziegler–Nichols tuning rule for the determination
of the values of parameters 𝐾𝑝 , 𝑇𝑖 and 𝑇𝑑 .
66
Example-8
• Transfer function of the plant is
1
𝐺 𝑠 =
𝑠(𝑠 + 1)(𝑠 + 5)
• Since plant has an integrator therefore Ziegler-Nichol’s
first method is not applicable.
• According to second method proportional gain is varied
till sustained oscillations are produced.
• That value of Kc is referred as Kcr.
67
Example-8
• Here, since the transfer function of the plant is known we can
find 𝐾𝑐𝑟 using
– Root Locus
– Routh-Herwitz Stability Criterion
• By setting 𝑇𝑖 = ∞ and 𝑇𝑑 = 0 closed loop transfer function is
obtained as follows.
𝐾𝑝
𝐾𝑝
𝐶(𝑠)
=
𝑅(𝑠) 𝑠 𝑠 + 1 𝑠 + 5 + 𝐾𝑝
68
Example-8
• The value of 𝐾𝑝 that makes the system marginally unstable so
that sustained oscillation occurs can be obtained as
𝑠 3 + 6𝑠 2 + 5𝑠 + 𝐾𝑝 = 0
• The Routh array is obtained as
• Examining the coefficients of first
column of the Routh array we find
that sustained oscillations will
occur if 𝐾𝑝 = 30.
• Thus the critical gain 𝐾𝑐𝑟 is
𝐾𝑐𝑟 = 30
69
Example-8
• With gain 𝐾𝑝 set equal to 30, the characteristic equation
becomes
𝑠 3 + 6𝑠 2 + 5𝑠 + 30 = 0
• To find the frequency of sustained oscillations, we substitute
𝑠 = 𝑗𝜔 into the characteristic equation.
(𝑗𝜔)3 +6(𝑗𝜔)2 +5𝑗𝜔 + 30 = 0
• Further simplification leads to
6(5 − 𝜔2 ) + 𝑗𝜔(5 − 𝜔2 ) = 0
6(5 − 𝜔2 ) = 0
𝜔 = 5 𝑟𝑎𝑑/𝑠𝑒𝑐
70
Example-8
𝜔 = 5 𝑟𝑎𝑑/𝑠𝑒𝑐
• Hence the period of sustained oscillations 𝑃𝑐𝑟 is
2𝜋
𝑃𝑐𝑟 =
𝜔
𝑃𝑐𝑟 =
2𝜋
5
= 2.8099 𝑠𝑒𝑐
• Referring to Table-2
𝐾𝑝 = 0.6𝐾𝑐𝑟 = 18
𝑇𝑖 = 0.5𝑃𝑐𝑟 = 1.405
𝑇𝑑 = 0.125𝑃𝑐𝑟 = 0.35124
71
Example-8
𝐾𝑝 = 18
𝑇𝑖 = 1.405
𝑇𝑑 = 0.35124
• Transfer function of PID controller is thus obtained as
1
𝐺𝑐 (𝑠) = 𝐾𝑝 (1 +
+𝑇𝑑 𝑠)
𝑇𝑖 𝑠
1
𝐺𝑐 (𝑠) = 18(1 +
+ 0.35124𝑠)
1.405𝑠
72
Example-8
73
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END OF LECTURE-8
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