When is it needed?
• Proportional control does not meet
specifications (both transient response
and steady-state error).
• Lead meets transient response
specifications.
• Need a lag section to reduce the steadystate error.
Lag-lead Design
M. Sami Fadali
Professor of Electrical Engineering
University of Nevada, Reno
1
General form
2
Design Options
1.    1
• Lag-lead compensator
2.   = 1
3
i. Select a suitable value for .
ii. Design procedure (skip).
4
1.    1
Example
• Design a lead compensator to improve the
transient response .
• Improve the steady-state response by
adding a lag compensator.
• Because the lag compensator will cause a
deterioration in the transient response, it is
advisable to use a somewhat conservative
lead design.
Design a lag-lead compensator for the
system to obtain closed-loop dominant
poles with a damping ratio of 0.5, an
undamped natural frequency
of 4 rad/s,
and reduce the steady-state error by a
factor of 5.
5
Solution
: Lead Design
6
Solution
From the lead presentation, we have
: Lag Design
Type 1: The lead section increases
so we need
by
and the lead compensator
7
Check the time response and, if necessary,
adjust the design to meet the specs.
8
2.i
Step Response
• Choose a value for  =1/  sufficiently small to
provide the desired specifications.
System: Closed Loop r to y
I/O: r to y Step Response
Peak amplitude: 1.18
Overshoot (%): 17.6
At time (seconds): 0.719
1.4
1.2
•
A m p litu d e
1
System: Closed Loop r to y
I/O: r to y
Settling time (seconds): 1.22
0.8
• Design a lead compensator with the selected .
• For a conservative design, use  = 0.1 and lead
compensator angle
0.6
0.4
0.2
0
provides the necessary improvement in
steady-state error.
• For pole-zero cancellation (cancel pole at  )
0
0.2
0.4
0.6
0.8
1
Time (seconds)
1.2
1.4
1.6
1.8
2
9
2.i   = 1 (Page 2)
10
Example
• Add a lag section with  = 1/.
• Check the error constant and the pole
locations for the design.
• Check the time response and modify the
design if necessary to meet the design
specifications.
• Compensator for pole-zero cancellation
Design a lag-lead compensator for the system
to obtain closed-loop dominant poles with a
damping ratio of 0.5, an undamped natural
of 4 rad/s, and reduce the
frequency
steady-state error by a factor of 3. Use the
passive circuit subject to the constraint
11
12
Solution
Solution (Page 2) Easier Design
• Calculate the desired closed-loop pole location
>> g=zpk([],[0,-2],4);scl=-2+j*2*sqrt(3)
scl =
-2.0000 + 3.4641i
• The angle of the loop gain at the desired closedloop pole location is
>> phi=pi-angle( evalfr(g,scl))
phi =
0.5236
Can use to calculate pole and zero locations.
• Try  = 0.2,  = 5. Meets design
specifications.
• Cancel pole with zero: zero at 2
• Compensator pole at 2/  = 10
• MATLAB root locus plot: Gain =24.9 for
• Lag Compensator:
13
MATLAB Commands
Root Locus: Lead Compensation
>> alpha=0.2; % alpha=1/beta
Root Locus
10
9
Imaginary Axis (seconds-1)
>> a=-real(scl)/10;
System: untitled1
Gain: 24.9
Pole: -5 + 8.64i
Damping: 0.501
Overshoot (%): 16.2
Frequency (rad/s): 9.98
8
7
6
14
>> gc=zpk(-2,-2/alpha,1)*zpk(-a,-alpha*a,24.6);
Zero/pole/gain:
5
4
3
2
1
0
-1
-12
15
-10
-8
-6
-4
Real Axis (seconds-1)
-2
0
2
24.6 (s+2) (s+0.5)
---------------(s+10) (s+0.1)
16
Compensated System
2.ii   = 1: Design Procedure
Adjust gain to improve the transient response:
Reduce gain to 12 (compromise between
settling and overshoot)
System: Closed Loop r to y
I/O: r to y
Step Response
Peak amplitude: 1.07
Overshoot (%): 7.09
At time (sec): 0.647
Amplitude
1.5
• Calculate the error constant for the system
with proportional control and calculate the
needed gain to meet the steady-state
error requirements.
• Obtain the ratio of the pole distance to
the zero distance for the lead section using
System: Closed Loop r to y
I/O: r to y
Settling Time (sec): 2.95
1
0.5
0
0
0.5
1
1.5
2
2.5
3
Time (sec)
Pole-Zero Map
Imaginary Axis
5
0
17
-5
-5
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
18
0
Real Axis
2.ii   = 1 Proc. (Page 2)
2.ii   = 1 Proc. (Page 3)
• Calculate the angle contribution of the lead
section
.
• Obtain the pole and zero locations for the
lead section using
1
cos( )  r 1

tan( z )
sin( )
z   n 
d cos   r 1 
sin  
r  cos( )
1

tan( p )
sin( )
p   n 
d r  cos 
sin  
19
• Add a lag section with
where and are the pole and zero of the
lead section.
• Check the error constant and the pole
locations for the design.
• Check the time response and modify the
design if necessary to meet the design
specifications.
20