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Unit-3Time Response of feedback control systemsQ1) The open loop transfers function of a system with unity feedback gain G (S) = 20 / S2 + 5S + 4. Determine the ξ, Mp, tr, tp.A1) Finding closed loop transfer function,C (S) / R (S) = G (S) / 1 + G (S) + H (S)As it is unity feedback so, H(S) = 1C(S)/R(S) = G(S)/1 + G(S)= 20/S2 + 5S + 4/1 + 20/S2 + 5S + 4C(S)/R(S) = 20/S2 + 5S + 24Standard equation for second order system,S2 + 2ξWnS + Wn2 = 0We have,S2 + 5S + 24 = 0Wn2 = 24Wn = 4.89 rad/sec2ξWn = 5(a). ξ = 5/2 x 4.89 = 0.511(b). Mp% = e-∏ξ / √1 –ξ2 x 100= e-∏ x 0.511 / √1 – (0.511)2 x 100Mp% = 15.4%(c). tr = ∏ - φ / Wdφ = tan-1√1 – ξ2 / ξφ= tan-1√1 – (0.511)2 / (0.511)φ = 1.03 rad.tr = ∏ - 1.03/WdWd = Wn√1 – ξ2= 4.89 √1 – (0.511)2 Wd = 4.20 rad/sectr = ∏ - 1.03/4.20tr = 502.34 msec(d). tp = ∏/4.20 = 747.9 msec Q2) A second order system has Wn = 5 rad/sec and is ξ = 0.7 subjected to unit step input. Find (i) closed loop transfer function. (ii) Peak time (iii) Rise time (iv) Settling time (v) Peak overshoot.A2) The closed loop transfer function isC(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2= (5)2 / S2 + 2 x 0.7 x S + (5)2C(S)/R(S) = 25 / S2 + 7s + 25(ii). tp = ∏ / WdWd = Wn√1 - ξ2= 5√1 – (0.7)2= 3.571 sec(iii). tr = ∏ - φ/Wdφ= tan-1√1 – ξ2 / ξ = 0.795 radtr = ∏ - 0.795 / 3.571tr = 0.657 sec(iv). For 2% settling timets = 4 / ξWn = 4 / 0.7 x 5ts = 1.143 sec(v). Mp = e-∏ξ / √1 –ξ2 x 100Mp = 4.59% Q3) The open loop transfer function of a unity feedback control system is given by G(S) = K/S (1 + ST). Calculate the value by which k should be multiplied so that damping ratio is increased from 0.2 to 0.4?A3)

C(S)/R(S) = G(S) / 1 + G(S)H(S) H(S) = 1

C(S)/R(S) = K/S(1 + ST) / 1 + K/S(1 + ST)

C(S)/R(S) = K/S(1 + ST) + K

C(S)/R(S) = K/T / S2 + S/T + K/T

For second order system,

S2 + 2ξWnS + Wn2

2ξWn = 1/T

ξ = 1/2WnT

Wn2 = K/T

Wn =√K/T

ξ = 1 / 2√K/T T

ξ = 1 / 2 √KT

forξ1 = 0.2, for ξ2 = 0.4

ξ1 = 1 / 2 √K1T

ξ2 = 1 / 2 √K2T

ξ1/ ξ2 = √K2/K1

K2/K1 = (0.2/0.4)2

K2/K1 = 1 / 4

K1 = 4K2

Q4) Consider the transfer function C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2 Find ξ, Wn so that the system responds to a step input with 5% overshoot and settling time of 4 sec?A4)Mp = 5% = 0.05Mp = e-∏ξ / √1 –ξ20.05 = e-∏ξ / √1 –ξ2Cn 0.05 = - ∏ξ / √1 –ξ2-2.99 = - ∏ξ / √1 –ξ28.97(1 – ξ2) = ξ2∏20.91 – 0.91 ξ2 = ξ20.91 = 1.91 ξ2ξ2 = 0.69(ii). ts = 4/ ξWn4 = 4/ ξWnWn = 1/ ξ = 1/ 0.69Wn = 1.45 rad/sec Q5) For the CLTF G(s) =

. Find Kp, Kv and Ka?A5) Kp =

G(s)

T(s) =

= 1

Kv = SG(s)

= S = 0

Ka = s2 G(s)

= s2

= 0

Q6) For a unity feedback system G(s) =

an input t3u(t) is applied. Find the steady state error?A6) ess=

r(t) = t3u(t)R(s) = 6/s4H(s) =1

ess=

= 5/3

Q7) For the OLTF with unity feedback is G(s)=

. Determine the damping ratio, maximum overshoot, rise time?A7) The CLTF will be T(s) =

T(s) =

For second order system,

S2 + 2ξWnS + Wn2

wn = = 5.1

2ξWn = 5

i) Damping Ratio = 0.49

ii) Maximum overshoot Mp = e-∏ξ / √1 –ξ2 x 100

= e-∏x0.49 / √1 – (0.49)2

= 17.1%

iii) Rise Time tr = ∏ - φ/Wd

Wd = Wn√1 - ξ2

= 5.1 √1 – (0.49)2

= 4.45 sec

φ= tan-1√1 – ξ2 / ξ = 1.059 rad

tr = ∏ - φ/Wd

= ∏ - 1.059/4.45

=468.53msec

Q8) Define and derive the time domain specifications for a second order system?A8)1) Rise Time (tp): - The time taken by the output to reach the already status value for the first time is known as Rise time.C(t) = 1-e-

wnt/

2 sin (wdt+ø)Sin (wd +ø) = 0Wdt +ø = n

tr =n

-ø/wdfor first time so, n=1.

2) Peak Time (tp) The peak value attained by the output is called peak time. The time required by the output to reach this value is lp.d(cct) /dt = 0 (maxima)d(t)/dt = peak valuetp = n