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1 Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Ph: CLASS TEST 08-9 ELECTRONICS ENGINEERING Subject : Control System Date of test : 30/07/08 Answer Key. (c) 7. (b) 3. (b) 9. (a) 5. (b). (d) 8. (c) 4. (b) 0. (b) 6. (c) 3. (b) 9. (a) 5. (d). (b) 7. (c) 4. (c) 0. (b) 6. (b). (b) 8. (b) 5. (c). (d) 7. (a) 3. (a) 9. (b) 6. (d). (d) 8. (b) 4. (b) 30. (b)

2 CT-08 EC Control System 7 Detailed Explanations. (c) t r π φ ω d where φ tan ξ ξ and ω d ωn ξ. (d) characteristic equation (si A) s 0 (si A) ( s )( s ) 0 s comparing with second order characteristic equation ω n ξω n 3 s 3s ξ.5.06 thus the system is over damped. 3. (b) Number of infinite semicircle thus number of pole at origin 4. (c) For the circuit shown G(s) α s R RC R R R s R R RC R R 3Ω Ω 5.50 R Ω 5. (c) Slope at highest frequency ( number of poles number of zeros) 0 db/decade 40 db/decade

3 8 Electronics Engineering 6. (d) Cs () Rs () or C(s) s 5 s s 5 Taing inverse Laplace transform. we get, c(t) 0 t e 5 or τ sec 7. (b) Comparing with standard transfer function of a compensator Here, τ 0.04 ατ 0. or α φ α sin 30 α 8. (c) characteristic equation G(s)H(s) 0 s 4 3s 3 (4s ) 0s 9 0 using Routh s tabular form S 4 S 3 S S S Number of sign change Hence, two poles lies on the RHS. 9. (a) The intersection point centroid Σp Σz ( p z ) magnitude (b) Characteristic equation is given by s 6s 6 0 s ξω n s ω n 0

4 CT-08 EC Control System 9 ω n 4 rad/s ξ 3 ξ 0.75 underdamped system For underdamped system T sec ξωn 3 settling time of the system for % tolerance band 4T 4.33 sec 3. (d) The poles of the three systems will have same real part and two poles for each system indicate a second order system. The envelope of the second order system with unit step input is governed by e ξω n t. Thus the second order system will have same envelope for the following systems. ct () 3 t. (d) Since the phase margin is the phase required for the system to be unstable, or the phase required to mae the total phase of the system equal to 80 at gain crossover frequency. 3. (b) G(s)H(s) s ( st ) Phase crossover frequency for GH(jω) 90 tan ω pc T 80 jω ( jωt) is given by tan ω pc T 90 ω pc Since the phase crossover frequency is not finite thus gain margin of the system does not exist. We can also see this from the nyquist and root locus plot which are shown in the figure below. 4. (b) A system is always stable if its root locus lies on the left side of (jω) axis.

5 0 Electronics Engineering 5. (d) Characteristic matrix I G(s)H(s) 0 s s s 0 s s 0 s s s s ( s 3) ( s ) T(s) Gs () I Gs () Hs () G(s) [I G(s)H(s)] Here, T(s) s 3 s s s s s s s s 5s ss ( ) 3s 9s 4 ( s ) ( s ) s 5s ( ss ) 3s 6. (b) From the characteristic equation the location of open loop poles and zeros are P j P j P 3 3 θ P Z thus angle of departure 80 φ θ d φ θz θp φ tan () 90 tan (/) φ 45 tan (/) θ d tan (/) 08.4 θ z P P 3 θ P jω j j σ 7. (a) comparing it with the standard equation T(s) T(s) ωn ξω n ω Cs () Rs () s s n 0 s ( 0 ) s 0 D Rs () 0 Cs () ss ( ( 0 )) D

6 CT-08 EC Control System ω n 0 0 D ξω n ξ 0 D D (b) Put s (p ) then the system becomes: (p ) 3 3( ) (p ) (7 5) (p ) p 3 3p p( ) 4 0 The Rooth s array is 3 p ( ) p 3 4 p p 4 For stability: 3 > 0 and > 0 > > 0 > 0.53 and <.5 > (a) G(s) ( s)0 ss ( ) 0 0s T(s) s s( 0 ) 0 ω n rad/sec ξω n 0 ξωn 0 ξπ ξ now, M p % 9.49% e ξ 0.6 thus 0.8 v lim sg( s) 5 s 0 e ss 0. v 0. (b) By calculating the characteristic equation, we get G(s) H(s) 0 K s ( ) 3 ( 4) s 0

7 Electronics Engineering (s ) 3 (s 4) K 0 s 4 7s 3 5s 3s (K 4) 0 s 4 5 (4 K) s s s K 3.4 (4 K) 0 0 s 0 4 K Range of K 4 < K < 0.4 a 4 b 0.4 b a 5.. (b) Transfer function C[sI A] B T (s) [ ] s 0 0 s 3 s 3 0 ( s )( s 3) 0 s [ ] ( s )( s 3) s [ ] T(s) ( s ) For unit step input U(s) s Y(s) s ( s ) e ss lim sy( s) s 0. (b) Before closing the switch s Cs () Rs () ζ and e ss s as a sr() s s s lim se( s) lim 0 0 GH ( s )

8 CT-08 EC Control System 3 e ss s 0 ( ) s s as lim s s as 3. (a) 4. (b) e ss a After closing the switch s Cs () Rs () s ( a ) s a ζ and e ss lim se( s) s 0 e ss a Therefore e ss and ζ both will increase. controllability matrix Q c [ B: AB ] for controllable Q c 0 αα αα α 0 α α α α α α 0 or α 0 observability matrix Q 0 C T : A T C T β β β β β for observability Q 0 0 or β (β β ) β β 0 β ββ ββ 0 or β 0 G(s) and H(s) ss ( 4) GH s 4s 0 ss ( ) SH T GH ss 4 ( ) at T H s j S ω ( ) 0 jω 4jω 0 ω 0.5 rad/sec 0 ω 4jω 0

9 4 Electronics Engineering 0 0 ( ) j j 4 T S H (b) The signal flow graph of the above system is U /s /s X X X X Y Ẋ U X Ẋ X X U The state model Chec for controllability: Y X X X U X 0 X X y [ 0] X Q c [B : AB] Q c Q c 0 and ρ(q c ) ρ(a) non-controllable Chec for observability Q 0 [C T : A T C T ] Observable Q 0 0 Q 0 0 and ρ(q 0 ) ρ(a) 6. (c) Brea away point can be calculated as dk 0 ds The characteristic equation for unity feedbac system is G(s) 0 4 K s s ( s ) 0

10 CT-08 EC Control System 5 or s ( s ) K s 4 or K dk ds 0 s ( s ) 4 s 4 (3 4 ) ( ) s s s s s 4 s 4 or 3 ss ( 8) s s ( s ) 0 or s 0, 4, 4 The brea point 4 Equal roots are at s 4 Hence K(s 4) (c) For second order system M PO 0.6 / e πξ ξ / e πξ ξ or ξ 0.5 Because of maximum overshoot occur at pea time t p or ωn ξ 5 3 π π π ω d ω ξ ω n (0.5) ω n 0 rad/sec for sinusoidal input, the frequency response is the steady state response ω r ωn ξ n rad/sec 8. (b) To unstablize the system, the critical point must be located between 0 to 0.5. Additional gain required 0.5 K K or 0 log K 6 db

11 6 Electronics Engineering 9. (b) Here, ω r 4 rad/sec ωn ξ ; ξ () m r.5...() ξ ξ from equation () 4ξ ( ξ ) 0.64 ξ ξ ξ ξ 0 ξ 0.8 and 0. ξ 0.894, for ξ ω n 5.6 rad/sec π θ π cos (0.447) τ r sec ωd (b) R ( GG G4) G3 C H 3 G H H R ( GG G4) G3 C G H H3 H H R ( GG G4) G3 C H H G HH 3 (G G G )G 4 3 C G (GG G 4)G3 R GH GH(H 3 GGH) 3 GGHH 3 4 If G G 3 5 G 9 G 4 H H 4 and H 3 6 C R ( 9 )5 ( 4)( )

Homework 7 - Solutions Homework 7 - Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the

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Introduction to Feedback Control Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System RULES This is a closed book, closed notes test. You are, however, allowed one piece of paper (front side only) for notes and definitions, but no sample problems. The top half is the same as from the first Contents Chapter Topic Page Chapter- Chapter- Chapter-3 Chapter-4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of