Transcript Slide 1

LAG LEAD COMPENSATOR
7.6 Root Locus Design
Example.
Consider rigid satellite control system
A(s) +
compensator
motor
1
s2
K
–
R(s)
The RL is along the j axis, then the
system will always oscillate regardless the
value of K.
To obtain an acceptable design the
following system will be employed
compensator
1
KG ( s ) H ( s )  K 2
s
R(s) +

1
s
K
–
θ
motor
1
s
R(s)
j
+
KV
The open loop-function is now:
–j
 s  1/ Kv 
KG ( s) H ( s)  KK v 
2

s


7.6 Root Locus Design
And the RL is
j

1
Kv
–j
The rate feedback changes the gain to KKv
and add a zero at s = - 1/Kv . The system is
seen to be stable for all K and Kv positive.
Since there are 2 parameters K and Kv, the
poles theoretically can be placed
anywhere in the s plane. However it must
be remembered that the physical system
always has limitation.
The rate feedback is actually a PD
compensation with TF

1 
Gc ( s)  KK v s    ( K p  K d s)
 Kv 
A problem with PD compensator is that its gain increases with frequency. If
high frequency noise presents another compensator (phase lead) can be used to
limit the high frequency gain.
PHASE LEAD DESIGN
The TF of 1st order phase lead
compensator is given by
K c ( s  z0 )
Gc ( s) 
s  p0
Where |z0| < |p0|, and Kc, z0, and p0
are to be determined to satisfy the
design criteria. First before we
present the design procedure we
consider the CE of compensated
system
The angle criteria is
(all zero angles) – (all pole angles) = r
consider the RL of the uncompensated is
j
2 poles
–j
Adding the lead compensation the root
locus will shift the RL to the left. Suppose
S1 is on the RL then
θz – θp- 2θ1=-180
j
1  KGc ( s)G p ( s) H ( s) 
1
KK c ( s  z0 )
G p ( s) H ( s)
s  p0
The product of KKc can be
considered as single gain parameter
θz
θp
p0
z0
θ1
–j 2 poles
Analytical Phase Lead Design
For this procedure it is convenient
to express lead compensator as
ao can be chosen arbitrarily, a1 and
a2 is computed using (7.53) from the
reference book
a s  a0
Gc ( s )  1
b1s  1
The object of the design is to choose
a0, a1, and b1 such that given s1
KGc ( s)G p ( s) H ( s)
s s1
We can choose K=1 leaving 3
unknown. ao can be chosen arbitrarily
 1
That is we design the compensator
places the pole at s1. Here we have 4
unknowns with only 2 equations
(magnitude and angle).
a1 
b1 
sin   a0 G p ( s1 ) H ( s1 ) sin(  )
s1 G p ( s1 ) H ( s1 ) sin( )
sin(  )  a0 G p ( s1 ) H ( s1 ) sin( )
 s1 sin( )
Where
s1  s1 e j
G p ( s1 )G p ( s1 )  G p ( s1 )G p ( s1 ) e j
are known
Controller Design Using Computer
The following material can be downloaded from
http://www.engin.umich.edu/group/ctm/rlocus/rlocus.html
Using computer the design will be much easier. MATLAB®, give a
convenient way to design. Consider an open loop system which has a
transfer function of
Say our design criteria are 5% overshoot and 1 second rise time.
First we have to draw the root locus using MATLAB® program:
num=[1 7];
den=conv(conv([1 0],[1 5]),conv([1 15],[1 20]));
rlocus(num,den)
axis([-22 3 -15 15])
Controller Design Using Computer
The program will produce the root locus chart as follows
Choosing a value of K from the root locus
In our problem, we need an overshoot less than 5% (which
means a damping ratio Zeta of greater than 0.7) and a rise
time of 1 second (which means a natural frequency Wn
greater than 1.8). Enter in the Matlab command window:
zeta=0.7;
Wn=1.8;
sgrid(zeta, Wn)
This command will give chart as follows
To meet the requirement the pole must be located in the
shaded area.
Note: the blue shaded area is added to the original chart for
clarity