Time Response Analysis (Part II)

Transcription

1 Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary axis (d) A pair of complex conjugate poles [GATE 988: Marks] Soln. For critically damped continuous time second order system roots of denominator are: s + ξω n s + ω n = 0 when ξ =, s + ω n s + ω n = 0 (s + ω n ) = 0 Root are s = ω n, ω n Double pole on real axis Option (b) 5. A second-order system has transfer function given by G(s) =. If S +8s+5 the system, initially at rest, is subjected to a unit step input at t = 0, the second peak in the response will occur at (a) π sec (c) π 3 sec (b) π 3 sec (d) π sec [GATE 99: Marks] Soln. G(s) = 5 S +8s+5 ω n = 5 ω n = 5 ξω n = 8

2 ξ = 8 0 = 0. 8 ω d = ω n ξ for nd peak n = 3 = 5 (0. 8) = 3 t p = nπ ω d = 3π 3 = π sec Option (a) 3. The poles of a continuous time oscillator are.. [GATE 994: Mark] Soln. The poles of a continuous time oscillator are imaginary 4. The response of an LCR circuit to a step input is (a) Over damped (b) Critically damped (c) Oscillatory If the transfer function has (i) poles on the negative real axis (ii) poles on the imaginary axis (iii) multiple poles on the positive real axis (iv) poles on the positive real axis (v) multiple poles on the negative real axis [GATE: 994: Marks] Soln. The response of an LCR circuit to a step input (a) over damped case ξ > () poles on the negative real axis C(s) R(s) = ω n s + ξ ω n s + ω n

3 Roots of denominator are s = ξ ω n ± ω n ξ (b) critically damped ξ = (5) C(s) R(s) = ω n s + ω n s + ω n = ω n (s+ω n ), s = ω n, ω n It has multiple poles on the negative real axis (c) oscillatory () poles on the imaginary axis ξ = 0 s = ± jω n Option a, b 5, c 5. Match the following codes with List-I with List-II: List I (a) Very low response at very high frequencies (b) Over shoot (c) Synchro-control transformer output List II (i) Low pass systems (iv) Phase-sensitive (ii) Velocity damping modulation (iii) Natural frequency (v) Damping ratio [GATE 944: Marks] Soln. (a) (b) very low response at very high frequencies low pass system over sheet Damping ratio (c) synchro control transformer output phase sensitive modulation Option: a, b 5, c 4

4 6. For a second order system, damping ratio (ξ), is 0 < ξ <, then the roots of the characteristic polynomial are (a) real but not equal (c) complex conjugates (b) real and equal (d) imaginary [GATE 995: Mark] Soln. ξ < underdamped system, roots of denominator s + ξ ω n s + ω n = 0 Are s = ξω n ± jω n ξ Roots are complex conjugate Option (c) 7. If L[f(t)] = (S+) then S +S+5 f(0+ ) and f( ) are given by (a) 0, respectively (c) 0, respectively (b). 0 respectively (d) /5, 0 respectively [Note: L stands for Laplace Transform of] [GATE 995: Mark] Soln. F(s) = (s+) s +s+5 f(0 + ) = lim s SF(s) = s +s s +s+5 + s = lim = +0 = s + s + 5 s +0 f( ) = lim s 0 SF(s) = s +s s +s+5 = 0 Option (b)

5 8. The final value theorem is used to find the (a) steady state value of the system output (b) initial value of the system output (c) transient behavior of the system output (d) none of these [GATE 995: Mark] Soln. The final value theorem is used to find the steady state value of the system. Option (a) 9. If F(s) = ω, then the value of lim f(t), S +ω t {where F(s)is the L[f(t)]} (a) cannot be determined (b) is zero Soln. F(s) = F(s) = ω s +ω ω s +ω (c) is unity (d) is infinite [GATE 998: Mark] has poles s ± jω (pure imaginary) it is oscillatory function hence final value lim t f(t) can not be determined 0. Consider a feedback control system with loop transfer function K( + 0.5s) G(s)H(s) = s( + s)( + s) The type of the closed loop system is (a) zero (c) two (b) one (d) three [GATE 998: Mark] Soln. G(s) H(s) = k(+0.5s) s(+s)(+s)

6 The number of poles at the origin of open loop transfer function given the type of the system. It is a type one system Option (b). The unit impulse response of a linear time invariant system is the unit step function u(t). For t > 0, the response of the system to an excitation e at u(t), a > 0 will be (a) ae at (b) (/a)( e at ) (c) a( e at ) (d) e at [GATE 998: Marks] Soln. H(s) = s System excitation r(t) = e at u(t) R(s) = s+a Response of the system C(s) = R(s) H(s) = s(s+a) C(s) = a [ s (s+a) ] C(t) = a [ e at ] Option (b). If the characteristic equation of a closed-loop system is s + s + = 0, then the system is (a) Overdamped (c) Under damped (b) Critically damped (d) undamped

7 Soln. Characteristic equation of closed loop system is s + s + = 0 Comparing with nd order equation s + ξω n + ω n = 0 ξω n = ξ = ω n ω n =, ω n =, ξ = which is less than It is case of under damped Option (c) 3. Consider a system with the transfer function (s) = s+6 ratio will be 0.5 when the value of K is (a) /6 (b) 3 (c) /6 (d) 6 Ks +s+6. Its damping [GATE 00: Mark] Soln. Transfer function G(s) = s+6 ks +s+6 Damping ratio ξ = 0. 5 Comparing with nd order equation s + ξω n s + ω n G(s) = s+6 k[s + s k +6 k ] ω n = 6 k, ω n = 6 k ξω n = k

8 ξ 6 k = k k = k or 6 k = k or k = 6 Option (c) 4. A casual system having the transfer function H(s) = s+ is excited with 0u(t). The time at which the output reaches 99% of its steady state value is (a).7 sec (c).3 sec (b).5 sec (d). sec [GATE 004: Marks] Soln. H(s) = s+ r(t) = 0μ(t) R(s) = 0 s C(s) = H(s) R(s) = 0 s(s+) C(s) = 5 s 5 s+ C(t) = 5[ e t ] Steady state value = 5 99% of the steady value reaches at

9 5[ e t ] = ( e t ) = e t = Or e t = 0. t = l n (0. ) t =. 3 sec Option (c) 5. The transfer function of a plant is (s) = 5 (s+5)(s +s+) approximation of T(s) using dominant pole concept is (a) (b) (s+5)(s+) 5 (s+5)(s+) Soln. Transfer function = 5 (s+5)(s +s+) (c) (d) 5 (s +s+). The second order (s +s+) [GATE 007: Marks] = 5 5( s 5 +)(s +s+) The poles nearer to imaginary axis dominates nature of time response and are called dominant poles. The factor that has to be eliminated should be in time constant form so, T(s) = Option (d) s +s+

10 6. Group I lists a set of four transfer functions. Group II gives a list of possible step responses y(t). Match the step responses with the corresponding transfer functions. Group I P = 5 s + 5 R = 36 s + s + 36 Q = 36 s + 0s + 36 S = 36 s + 7s + 49 y(t) y(t).. t t y(t) y(t) (a) P 3, Q, R 4, S (b) P 3, Q, R 4, S (c) P, Q, R 4, S 3 (d) P 3, Q 4, R, S t [GATE 008: Marks] t

11 Soln. The figure is a case of critically damped, Figure underdamped, Figure 3 undamped, figure 4 is overdamped The given transfer function P = Q = 5 s s + 0s + 36 R = S = 36 s + s s + 7s + 49 Comparing the given transfer function with ω n s + ξ ω n s+ω n, P = ω n = 5, ξ = 0 5 s +5 P is undamped, matched to figure 3 so P 3 Q = 36 s +0s+36, ω n = 6 ξ ω n = 0 ξ = 0 = Q is overdamped matched to fig. 4 Q 4 R = 36 s +s+36, ω n = 6 ξ ω n = ξ = 6 = R is critically damped matched to fig, R

12 S = 49 s +7s+49, ω n = 7 ξ ω n = 7 ξ = = 7 7 = 0. 5 S is underdamped matched to fig Option (d) P 3, Q 4, R, S 7. The unit step response of an under-damped second order system has steady state value of. Which one of the following transfer functions has these properties? (a).4 s +.59s+. (c).4 s.59s+. (b) 3.8 s +.9s+.9 (d) 3.8 s.9s+.9 [GATE 009: Marks] Soln. Steady state value = - The nd order T F = ω n s + ξ ω n s+ω n For underdamped system ξ < Option (c) and (d) are wrong because the system is unstable as ξ is negative In option (a) ω n =. =. 06 ξ ω n =. 59 ξ = =.

13 ξ > hence it is wrong In option (b) T. F = 3.8 s +.9s+.9 ω n =. 9 =. 38 ξ ω n =. 9 ξ =.9.38 = ξ < so system is underdamped Option (b) 8. The differential equation 00 d y dy 0 + y = x(t) describes a system dt dt with an input x(t) and an output y(t). The system, which is initially relaxed, is exited, by a unit step input. The output y(t) can be represented by the waveform. (a) (b) (c) (d) [GATE 0: Mark]

14 Soln. 00 d y dy dt 0 + y = x(t) dt Taking Laplace transform of both sides 00 s y(s) 0 s y (s) + y(s) = X(s) = s (00s 0s + )y(s) = s y(s) = Poles are at s = 0,, s(0s ) 0 0 Poles are on the right hand side of s plane so given system is unstable Option (a) represents unstable system. 9. The open-loop transfer function of a dc motor is given as ω(s) V a (a) = 0 + 0s when connected in feedback as shown below, the approximate value of K a that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is (a) (b) 5 (c) 0 (d) 00 [GATE 03: Marks] Soln. The open loop transfer function of a dc motor is given as ω(s) V a (s) = When connected in feedback is shown below. 0 +0s R(s) + - k a 0 + 0s ω(s)

15 Open loop transfer function = 0 k a +0 s Loop time constant τ 0l = 0 G(s) = k a s+ 0 g(t) = e t 0 Closed loop transfer function = H(s) = G(s) +G(s) H(s) = 0k a +0s = +(0k a ) +0s H(s) = 0k a +0s+0k a = h(t) = k a e ( 0k a +0s+0k a k a s+(k a + 0 ) ka+0. )t Time constant of closed loop system = (k a + 0 ) τ cl = τ ol 00 = = 0 00 = k a +0. k a +0. or k a = 0 0. = Option (c)

16 0. For the second order closed-loop system shown in the figure, the natural frequency (in rad/s) is U(s) s(s+4) Y(s) (a) 6 (b) 4 (c) (d) [GATE 04: Mark] Soln. Y(s) U(s) = 4 s(s+4) +4 s(s+4) = = 4 s(s+4)+4 4 s +4s+4 Comparing with nd order system transfer Function ω n s + ξ ω n s+ω n ω n = 4, or ω n = Option (c)