MODEL EXAMINATION ANSWER KEYEC6405- CONTROL SYSTEM
ENGINEERINGELECTRONICS AND COMMUNICATION ENGINEERINGPart-A1. The
negative feedback results in better stability in steady state and
rejects any disturbance signals2. Transfer function c(s)/R(s) = 3.
Time response C(t) = (2/3)-(2/3)e-3t4. 5. The magnitude of closed
loop transfer function with unit feed back can be shown in the form
of circle for every value of M. These circles are called M circles.
Closed loop phase plots are called as N circles.
6. P-Controller : It is device that produces a control signal
which is proportional to the input error signal. PI Controller : It
is device that produces a control signal consisting of two terms
one proportional to error signal and the other proportional to the
integral of error signal. PD controller : PD controller is a
proportional plus derivative controller which produces an output
signal consisting of two time -one proportional to error signal and
other proportional to the derivative of the signal.7. If the
Nyquist plot of the open loop transfer function G(s) corresponding
to the Nyquist control in the S-plane encircles the critical point
1+j0 in the counter clockwise direction as many times as the number
of right half S-plane poles of G(s), the closed loop system is
stable.8. System is unstable. One root lie on right half of s plane
, remaining two roots lie on left half of s plane
9. Advantages:High accuracy, minimum cost, The complex control
calculations can be performed very easily, Data processing with the
help of digital controllers is straight and fast,Disadvantages:The
sampling process tend to result in more errors, which may affect
the system performance. There are limitations on the signal
resolution due to the finite wordlength of the digital processors,
Designing the digital controllers is very much complex than
designin the analog controller for an equivalent level of
performance.10.
PART-B11.a)Case(i) R acting,D=0(8M)Step1: eliminating loop
1&2Step2:Combining all cascade blocksStep3: eliminating
feedback path.Ans TF c/r = G1G2G3/
1+G1+G3H1+G1G3H1+G1G2G3H2Case(ii) D acting R=0(8M)Step1:
Eliminating loop 1Step2:splitting summing pointStep3: combining
parallel blocksStep 4 : eliminating feedback pathAnsTF d/r=
G3(1+G1)/1+G1+G1G2G3H2+G3H1+G1G3H1
11.b)
12.a) wn=26 (2M) Zeta =0.49(2M)Maximum overshoot=17.1%(4M)Rise
time=468.53 msec(4M)Peak time=706.78msce(4M)
12.b)(i) type 2(4M) Order 4(ii)Error constants Kp=infinity(6M)
Kv = infinity Ka = k/6(iii) ess steady state error at parabolic
input=0 (6M)
13.a)Corner frequency wc1=5, wc2=10Magnitude plot (5M)Phase plot
(5M)Graph (4M)Result (2M)GM = infinityPM = 92 degree.13.b) Polar
plot(4M)Magnitude|G(jw)| = 1/ w(1+0.25w2) (1+16w2)Phase angle =
-90-tan-10.5w-tan-14wTable calculation(4M)Case(i)
k=0,227(4M)Case(ii)k=0.49(4M)
14.a)
14.b)(i) Constructing routh array (5M) Stability criterion (5M)
(ii) procedures for root locus(6M)Location of poles and zerosLocate
root locus on real axisAngle of asymptotes and centroidAngle of
departure and arrivalCrossing point on imaginary axis.
15.a) (i) TF= 1/s2+3s+6 (10) (ii) State space
representation(6M)
15.b) Finding state space representation(4M)Checking
controllability(kalmans test)(6M)System is totally state
controllableObservability check(6M)Sytem is totally observable.