Classical Design in s-plane
• Compensator design in Time domain VS
Compensator design in Frequency
Domain.
• We will look at a simple second order
system to understand the transient
characteristics.
• We will plot the step responses in
MATLAB.
Time Responses for second order
system
Pole Locations
Second Order Underdamped
Responses
Transient Specifications
Step Responses of Second-order
underdamped system
Root Locus for a feedback system
Example 1
Example 1 (contd)
Matlab Code
• Num=poly([-3 -4])
• Den=poly([-1 -2])
• Ts1=tf(Num,Den)
• rlocus(Num,Den)
Example 2
• Find the exact point and gain where the
locus crosses the 0.45 damping ratio line.
• the exact point and gain where the locus
crosses the jw-axis.
• The range of K within which the system is
stable.
Root Locus for Example 2
Designing Compensators
Contd..
• A compensator or controller placed in the
forward path will modify the shape of the
loci if it contains additional poles and
zeros.
• We will consider two case studies to
understand compensator design in time
domain.
Case Study 1
• A control system has the following open-loop transfer
function.
• A PD compensator of the form
is introduced to achieve a performance specification of :
Overshoot : less than 5%
Settling Time (2%): less than 2 seconds
Original Controller
• The original controller may be considered to be a proportional controller of gain ‘K’.
• The root locus is shown in the next slide.
• For K=11.35, damping ratio=0.5 which has an overshoot of 16.3% (not acceptable).
• K=7.13, damping ratio=0.7 which has an overshoot of 4.6% (acceptable) but the settling time is not satisfied.
• We need a compensator.
PD Compensator Design
• The problem is where to place the zero on
the real axis. Possible locations include:
1) Between the poles s=0,-2 i.e. at s=-1
2) At s=-2 (pole zero cancellation)
3) Between the poles s=-2, -5 i.e. at s=-3
Option 1 (s=-1)
• The plant transfer function becomes:
• The root locus is shown in next slide:
• The pole at the origin and the zero at s=-1
dominate the response.
• The settling time is 3.9 seconds which is
outside the specifications.
Option 2 (s=-2)
• The cascaded compensator and plant transfer function is:
• The pole zero cancellation may be considered as a locus that starts at s=-2 and finishes at s=-2.
• The remaining loci breakaway at s=-2.49 and looks similar to a second order system.
• The compensator gain K1 that corresponds to zeta=0.7 is 12.8 and has an overshoot of 4.1% and settling time of 1.7 seconds. (witin specifications)
Option 3 (s=-3)
• The cascaded compensator and plant transfer function is:
• The real locus occurs between s=-5 and s=-3 and the root locus breaks away at s=-1.15.
• Since the root locus is further to the right than the previous option the transient response will be slower.
• The compensator gain that corresponds to zeta =0.7 is K1=5.3. The resulting overshoot is 5.3% and settling time is 3.1 seconds.
Conclusion
• Option 2 is the best
Matlab Code for Option 2
Case Study 2
• A ship roll stabilization system
Parameters
• Fin time constant= 1 seconds
• Ship roll natural frequency wn=1.414 rad/sec.
• Damping ratio = 0.248
• Steady sate gain :Ks=0.5
• Without stabilization the step response of the roll dynamics produces 45% overshoot and settling time of 10 seconds.
• The control system is required to provide a step response with an overshoot of less than 25% and a settling time of less than 2 seconds and zero steady state error.
