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.