1. LAG LEAD COMPENSATOR

2. 7.6 Root Locus Design Example. Consider rigid satellite control system
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
motor
compensator
R(s)
A(s) + K –
motor
compensator K
R(s) – + KV
R(s) j –j
The open loop-function is now:

3. 7.6 Root Locus Design And the RL is
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 j –j
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.

4. PHASE LEAD DESIGN
The TF of 1st order phase lead compensator is given by
The angle criteria is
(all zero angles) – (all pole angles) = r
consider the RL of the uncompensated is j –j
2 poles
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
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 θ1 θz θp
The product of KKc can be considered as single gain parameter p0 z0 –j
2 poles

5. Analytical Phase Lead Design
We can choose K=1 leaving 3 unknown. ao can be chosen arbitrarily
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
The object of the design is to choose a0, a1, and b1 such that given s1
That is we design the compensator places the pole at s1. Here we have 4 unknowns with only 2 equations (magnitude and angle).

6. Controller Design Using Computer The following material can be downloaded from
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([ ])

7. Controller Design Using Computer
The program will produce the root locus chart as follows

8. 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

9. 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