MU Electrical Engineering (Semester 7)
Control System - 2
December 2015
Total marks: --
Total time: --
INSTRUCTIONS
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary


1 (a) Compare lag and lead compensator along with electrical equivalent circuit and pole-zero plot in s-plane.
5 M
1 (b) Explain in brief different forms of industrial PID controllers.
5 M
1 (c) Write a short note on modelling of the sampler in digital control system.
5 M
1 (d) Explain start/stop interlocking circuit in PLC programming with the help of suitable example.
5 M
1 (e) Explain the scan cycle of execution in PLC.
5 M
1 (f) Consider the system defined by x˙=Ax+Bu  y=Cxwhere A=[101120003]  B=[001]  C=[110] Transform the system equation into (i) controller canonical form and (ii) observer canonical form.
5 M

2 (a) Give a unit feedback system with G(s)=1S(S+1)(0.5s+1) Use frequency response method to design lead compensator to yield (i) Steady state error to a unit ramp input is less than 0.05 (ii) Phase margin more than 45 degree.
10 M
2 (b) Design a state variable feedback controller to yield a 20.8% overshoot and a settling time of 4 second for a plant. x˙=Ax+Bu  y=Cx+DuWhere, A=[010001156]  B=[001]  C=[1010]  D=[0]
10 M

3 (a) Write a short note on Proportional Band (PB) in case of PID controller.
5 M
3 (b) Explain the stability of Digital control system.
5 M
3 (c) Consider a linear system described by the differential equation y¨+2y˙+y=u˙+u with x1=y, x2=y˙u
Determine whether this system is observable or not?
10 M

4 (a) Design a integral controller for the plant x˙=[0179]x+[01]uy[4  1]x to yield a step response with 10% overshoot, a peak time of 2 seconds and zerd steady state error.
10 M
4 (b) Given a Z.O.H. in cascade with G1=(S)=s+2s+1 Find the sampled data transfer function, G(z) if a sampling time T is 0.5 second.
10 M

5 (a) Find T(z)=C(z)/R(z) for the given block diagram if T=0.3 sec.

10 M
5 (b) Draw and explain the PLC ladder diagram for manufacturing of 5 mH and 10mH inductor coils. When 5 mH inductor is produced, the machine makes 400 revolutions to wind the coil. If the inductor is produced 10 mH, the machine makes 800 revolutions before stopping.
10 M

6 (a) Explain relay logic instruction in PLC.
5 M
6 (b) Write a short note on PLC processor unit.
10 M
6 (c) Explain AC input module of PLC.
5 M



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