CLTF CsRs 11TS CE 1Ts 0 S 1T Cs Rs GS1Gs Cs Rs1Ts So Calculating value of cs ct

# Cltf csrs 11ts ce 1ts 0 s 1t cs rs gs1gs cs rs1ts so

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CLTF C(s)/R(s) = 1/(1+TS) CE : 1+Ts = 0 S= -1/T C(s) = R(s) [G(S)/(1+G(s))] C(s) = R(s)/(1+Ts) So, Calculating value of c(s) , c(t) for different inputs a) Unit s tep :- R(t) = u(t) R(s) = 1/s C(s) = 1/(s(1+sT)) 1/(s(1+sT)) = A/s + B/(1+Ts) 1 = A(1+sT)+ Bs AT +B = 0 A = 1 B= -T C(s) = 1/s + (-T)/(1+TS) = 1+ (-T)/(1-(-T)s) = 1-1/(s+1/T) C(t) = [ 1-e -t/T ]  b) Ramp:- R(t) = t R(s) = 1/s 2 C(s) = R(s) [G(s)/(1+G(s))] C(s) = 1/s 2 *1/(1+TS) 1/(s 2 (1+TS)) = A/(s) + B/s 2 + c/(1+TS) 1 = A(s+s 2 T) + B(1+Ts) +Cs 2 TA +C = 0 A+TB =0 B =1 A = -T C = T 2 C(s) = -T/s + 1/s 2 +T 2 /(1+sT) =(-T) u(t) + tu(t) + T e -t/T u(t) = [ -T + t +Te -t/T ] u(t) C(s) = t- T+ Te -t/T e ss = lt t->∞ R(t) – c(t) = lt t->∞ [ t- t+ T-Te –t/T ] =lt t->∞ T[1-e -t/T ] e ss = T The less the value of T the less is the error. Inteference:- From both the Cases input unit step, ramp Unit step Ramp r(t)=1 c(t) = 1-e -t/T. r(t) = t e ss = e -t/T (r/t)-(t) c(t) = t- T+te -t/T e ss = e -tlT e ss = T- Te -t/T for t->∞:- for t->∞:- e ss =0 e ss = T In both the cases (values of T) must be as small as possible ( so, that e -t/T ) must be as small as possible which gives us G(s) = 1/Ts CLTF= 1/(1+TS) 1 st order system Poles must be situated as far as possible from origin i.e deeper and deeper into the left half of s-plane. Thus, we get less error. 2.3 Time response analysis of second order systems 2)ORDER – 2 OLTF G(s) = k /(s (1+Ts)) CLTF c(s)/R(s) = k / (k + s(1+Ts)) = k / [s 2 T+ s +k] Comparing above equation with standard 2 nd order equation CLTF = ω n 2 /s 2 +2 εω n s+ ω n 2. Standard 2 nd order equation CE : s 2 +2 εω n s+ ω n 2 =0 S = [-2 εω n ± (2 εω n ) 2 -4 ω n 2 ]/2 = [2 εω n ±2 ω n ε ¿ 2 -1)]/2 s =-ξω n ±ω n ξ 2 -1 s = -ξω n ±ω n ξ 2 -1 s= -ξω n ±ω n -(1-ξ 2 ) Standard eqn:- T(s) = ω n 2 /s 2 +2ξω n s+ω n 2 Our eqn T(s) = (K/T)/s 2 +1/T s+ K/T ω n 2 = k/T 2ξω n = 1/T k/T = 1/T ξ k/T = 1/2T ξ= 1/2T* T/k ξ= 1 2kT Graphically showing the position of loops s 1 ,s 2 for offered of As the characteristic Equation(location of pols ) is dependent only on r (ω n constant for a given system) S 1 = -ξω n + j ω n 1-ξ 2 S 2 = -ξω n - jω n 1-ξ 2 CASE 1: ( ξ =0) S 1 =jω n , S 2, =-jω n Diagram:- Undamped CASE 2 : (0< ξ <1) S 1 = -ω n + jω n 0.75 n /2 +jω n (0.26) S 2 = -ω n /2 – jω n (0.86) Diagram:- CASE:3 ( ξ =1) S 1 = S 2 = -ξω n = -ω n Diagram :- CASE4:- ( ξ >1) S 1 = -2ω n +jω n -3 = -2ω n – jω n (1.73) S 1 = -3.73ω n S 2 = -2ω n –jω n -3 =-2ω n + jω n (1.73) = -0.27ω n Diagram:- Overdamped All practical systems are 2 order so, if R(e) = R(t) R(s) = 1 :. CLTF = 2 nd order st eqn and hence already the system is possible. CLTF = c(s)/ R(s) = wn 2 /s 2 +2ξwns+wn 2 Now calculating c(s) , c(t) for different values of input 1) Impulse I/p R(t) = r(t) R(s) = 1 C(s) = R(s) wn 2 /s 2 +2ξswns+wn 2 C(s) = wn 2 /s 2 +2ξwns+wn 2 Under this i/p (R(t) = l(t) the output varies with different values of r. so, CONDITION 1 ( Q s =0) C(s) s 2 +wn 2 =0 S= jwn C(t) = w n sinwnt Diagram:- Sinwnt Diagram:- As there in no damping i.e oscillations at t= 0 are some at t – infinity so, called UNDAMPED CONDITION 2: 0<r<1 c(s) = R(s) wn 2 /s 2 +2ξwns+wn 2 R(t) = s(t) R(s) =1 C(s) = wn 2 /s 2 +2ξwns+wn 2 CE S 2 +2ξwns+wn 2 =0 S 2 , S 1 = -ξwn ±jwn 1-ξ 2  #### You've reached the end of your free preview.

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• Two '10
• DRWEW
• LTI system theory, Impulse response, Transient response

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