Transcript Lag Compensation - Dr. Imtiaz Hussain

Modern Control Systems (MCS)
Lecture-16-17-18
Lag Compensation
Dr. Imtiaz Hussain
Assistant Professor
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture Outline
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+
,
,
• Where 𝐾 𝑐is gain of lag compensator.
(β > 1)
(β > 1)
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
Lag Compensation
• Consider the problem of finding a suitable compensation network
for the case where the system exhibits satisfactory transientresponse characteristics but unsatisfactory steady-state
characteristics.
• Compensation in this case essentially consists of increasing the
open loop gain without appreciably changing the transientresponse characteristics.
• This means that the root locus in the neighborhood of the
dominant closed-loop poles should not be changed appreciably, but
the open-loop gain should be increased as much as needed.
Lag Compensation
• To avoid an appreciable change in the root loci, the angle
contribution of the lag network should be limited to a small
amount, say less than 5°.
• To assure this, we place the pole and zero of the lag network
relatively close together and near the origin of the s plane.
• Then the closed-loop poles of the compensated system will be
shifted only slightly from their original locations. Hence, the
transient-response characteristics will be changed only slightly.
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 s1is one of the dominant closed
loop poles then the magnitudes 𝑠1 +
1
𝑇
and 𝑠1 +
equal, or
1
𝑠+
𝑇 ≅𝐾
𝐺𝑐 (𝑠1 ) = 𝐾𝑐
𝑐
1
𝑠+
𝛽𝑇
1
𝛽𝑇
are almost
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.
Lag Compensation
• If the pole and zero are placed very close to the origin, then the
value of 𝛽 can be made large.
• A large value of 𝛽 may be used, provided physical realization of the
lag compensator is possible.
• It is noted that the value of T must be large, but its exact value is
not critical.
• However, it should not be too large in order to avoid difficulties in
realizing the phase-lag compensator by physical components.
Lag Compensation
• An increase in the gain means an increase in the static error
constants.
• 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
Lag Compensation
• The main negative effect of the lag compensation is that
the compensator zero that will be generated near the
origin creates a closed-loop pole near the origin.
• This closed loop pole and compensator zero will generate a
long tail of small amplitude in the step response, thus
increasing the settling time.
Electronic Lag Compensator
• The configuration of the electronic lag compensator using
operational amplifiers is the same as that for the lead
compensator.
1
𝑠
+
𝐸𝑜 (𝑠) 𝑅4 𝐶1
𝑅1 𝐶1
=
𝐸𝑖 (𝑠) 𝑅3 𝐶2 𝑠 + 1
𝑅2 𝐶2
𝑇 = 𝑅1 𝐶1
𝛽𝑇 = 𝑅2 𝐶2
𝑅4 𝐶1
𝐾𝑐 =
𝑅3 𝐶2
𝑅2 𝐶2 > 𝑅1 𝐶1
Electronic Lag Compensator
• Pole-zero Configuration of Lag
Compensator
𝑅2 𝐶2 > 𝑅1 𝐶1
Electrical Lag Compensator
• Following figure shows lag compensator realized by
electrical network.
𝑅1
𝑅2
𝐶
𝐸2 (𝑠)
𝑅2 𝐶𝑠 + 1
=
𝐸1 (𝑠)
𝑅1 + 𝑅2 𝐶𝑠 + 1
Electrical Lag Compensator
𝐸2 (𝑠)
𝑅2 𝐶𝑠 + 1
=
𝐸1 (𝑠)
𝑅1 + 𝑅2 𝐶𝑠 + 1
𝑇 = 𝑅2 𝐶
𝑅1 + 𝑅2
𝛽=
>1
𝑅2
• Then the transfer function becomes
𝐸2 (𝑠)
𝑇𝑠 + 1
=
𝐸1 (𝑠)
𝛽𝑇𝑠 + 1
Electrical Lag Compensator
𝐸2 (𝑠)
𝑇𝑠 + 1
=
𝐸1 (𝑠)
𝛽𝑇𝑠 + 1
• If an RC circuit is used as a lag compensator, then it is
usually necessary to add an amplifier with an adjustable
gain 𝐾𝑐 𝛽 so that the transfer function of compensator is
𝐸2 (𝑠)
𝑇𝑠 + 1
= 𝐾𝑐 𝛽
𝐸1 (𝑠)
𝛽𝑇𝑠 + 1
1
𝑠+
𝐸2 (𝑠)
𝑇
= 𝐾𝑐
1
𝐸1 (𝑠)
𝑠+
𝛽𝑇
Mechanical Lag Compensator (Home Work)
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.
Design Procedure
• Step-1
– Draw the root-locus plot for the uncompensated system whose
open-loop transfer function is G(s).
– Based on the transient-response specifications, locate the
dominant closed-loop poles on the root locus.
Design Procedure
• Step-2
– Assume the transfer function of the lag compensator to be
given by following equation
𝐺𝑐 𝑠 =
𝑇𝑠+1
𝐾𝑐 𝛽
=𝐾𝑐
𝛽𝑇𝑠+1
1
𝑠+
𝑇
1
𝑠+
𝛽𝑇
– Then the open-loop transfer function of the compensated
system becomes Gc(s)G(s).
Design Procedure
• Step-3
– Evaluate the particular static error constant specified in the
problem.
– Determine the amount of increase in the static error constant
necessary to satisfy the specifications.
Design Procedure
• Step-4
– Determine the pole and zero of the lag compensator that
produce the necessary increase in the particular static
error constant without appreciably altering the original
root loci.
– The ratio of the value of gain required in the specifications
and the gain found in the uncompensated system is the
required ratio between the distance of the zero from the
origin and that of the pole from the origin.
Design Procedure
• Step-5
– Draw a new root-locus plot for the compensated system.
– Locate the desired dominant closed-loop poles on the root locus.
– (If the angle contribution of the lag network is very small—that
is, a few degrees—then the original and new root loci are almost
identical.
– Otherwise, there will be a slight discrepancy between them.
– Then locate, on the new root locus, the desired dominant closedloop poles based on the transient-response specifications.
Design Procedure
• Step-6
– Adjust gain of the compensator from the magnitude condition so
that the dominant closed-loop poles lie at the desired location.
– 𝐾𝑐 will be approximately 1.
Example-1
• 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.
Example-1 (Step-1)
• The dominant closed-loop poles of given system are
s = -0.3307 ± j0.5864
Example-1 (Step-2)
• According to given conditions we need to add following
compensator to fulfill the requirement.
𝐺𝑐 𝑠 =
𝑇𝑠+1
𝐾𝑐 𝛽
=𝐾𝑐
𝛽𝑇𝑠+1
1
𝑠+
𝑇
1
𝑠+
𝛽𝑇
Example-1 (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
Example-1 (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
Example-1 (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+
Example-1 (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
Example-1 (Step-5)
Solution-1
• Root locus of uncompensated and compensated systems.
• New Closed Loop poles
are
𝑠 = −0.31 ± 𝑗0.55
Example-1 (Step-5)
Solution-1
• Root locus of uncompensated and compensated systems.
Example-1 (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
Example-1 (Step-6)
• Then the compensator transfer function is given as
𝑠 + 0.05
𝐺𝑐 𝑠 = 0.9656
𝑠 + 0.005
Solution-1
Example-1 (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
Example-1 (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
Example-1 (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
Example-2
• 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.
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END OF LECTURE-16-17-18