ut 1 t 0 t 0 L ut 1s c Ramp Signal The ramp signal increase linearly with time

Ut 1 t 0 t 0 l ut 1s c ramp signal the ramp signal

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u(t) = 1 t ≥ 0 t < 0 L [u(t)] = 1/s ( c) Ramp Signal:-
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The ramp signal increase linearly with time from initial value of zero at t= 0 as shown in fig.13 below Diagram:- r(t) = At t ≥ 0 =0 t<0 A is the slope of the line. The Laplace transform of ramp signal is L[r(t)] = R(s) = A/s 2 (d) Parabolic Signal:- The instantaneous value of a parabolic signal varies as square of the time from an initial value of zero t=0. The signal representation in fig, 14 below. Diagram:- r(t) = At 2 t≥0
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=0 t<0 Then Laplace Transform is given as R(s) = L[At 2 ] = 2A/S 3 ** if no error, then E(t) = 0 , R(t) = c(t) Output is tracking the input. Steady state Errors signal (ess) :- ( t ) e ss = t e(t) Using final values theorem e(t) s->0 = e ss = lt s->0 S E (s) e ss = lt s->0 S[R(s)/(1+G/(s))] So, for different input, e ss will be different. Inputs still be 1) R(s) = unit step 2) R(s) = parabolic 3) R (s) = Ramp 1) R(s) = Unit step R(t) = u(t) R(s) = 1/s e ss = lt s->0 s[ R(s)/(1+G(s))] =lt s->0 s[(1/s)/(1+G(s))] e ss =lt s->0 1/(1+G(s)) =1/(1+lt s->0 G(s) ) K p =lt s->0 G(s) (position error coefficient) e ss = 1/1+k p 2)R(s) = Ramp
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R(t) = t R(s) = 1/s 2 e ss = lt s->0 s[ R(s)/(1+G(s))] = lt s->0 s[(1/s 2) /(1+G(s))] = lts ->0 1/(s(1+G(S)) = lt s->0 1/(s+sG(s)) e ss = lt s->0 1/ sG (s) K v = lt s->0 sG(s) velocity error coefficient e ss = 1/K v (3) Parabolic R(t) = t 2 R(s) = 1/ s 3 e ss = lt s->0 s[ R(s)/(1+G(s))] = lt s->0 s(1/s 3 )/(1+G(s)) = lt s->0 1/(S 2 +S 2 G(s)) e ss = lt s->0 1/(s 2 G(s)) K a =lt s->0 s 2 G(s) Acceleration error coefficient e ss =1/k a I/p error coefficient (1)Unit step e ss = 1/(1+k p ) k p =lt s->0 G(S) (2)Ramp. e ss = 1/k v k v =lt s->0 sG(s) (3)Parabolic e ss = 1/k a k a =lt s->0 s 2 G(s)
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Now, finding errors in type 0,1,2 system Type 0:- G(s) = 1/((s+a) (s+b)) (a) K p :- K p = lt s->0 G(s) = lt s->0 1/((s+a)(s+b)) K p =1/ab e ss = 1/(1+k p ) = 1/(1+(1/ab)) 0 Output not tracking the input (b) K v :- K v = lt s->0 sG(s) =lt s->0 s /((s+a) (s+b)) =lt s->0 s/[s2+(a+b)s+ab] K v = 0 e ss = 1+k v = Output not tracking the input (3) K a :- k a = lt s->0 s 2 G(s) = lt s->0 s 2 /((s+a) (s+b)) k a = 0 e ss = 1/k a = Output not tracking the input
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(2)Type –I: - G(s) = 1/(s(s+a) (s+b)) (a) K p :- K p = lt s->0 sG(s) = lt s->0 1/(s(s 2 +as+ab)) K p = 1/0 = e ss = 1/(1+k p ) = 0 Output is tracking the input (b) K v :- K v = lt s->0 s G(s) = lt s->0 1/((s+a) (s+b)) Kv = 1/ab e ss = 1/k v = ab 0 Output is not tracking the input (C) K a :- K a = lt s->0 s 2 G(s) =lt s->0 s 2 /(s(s+a)(s+b)) =lt s->0 s/((s+a)(s+b)) K a = 0 e ss = 1/k a = ∞ Output is not tracking the input. 3) Type 2:- G(s) = 1+/(s 2 (s+a) (s+b))
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a) K p :- K p = lt s->0 G(s) =lt s->0 1/(s 2 (s+a) (s+b)) K p = e ss = 1/1+k p = 0 Output tracking the input. b) K v :- K v = lt s->0 sG(s) = lt s->0 1/(s(s+a) (1+b)) k v = ∞ e ss =1/∞ = 0 Output tracking the input (c) k a :- K a = lt s->0 s 2 G(s) =lt s->0 1/((s+a) (s+b)) k a = 1/ab e ss = 1/(1/ab) ≠0 Output not tracking the input e ss Type 0 Type 1 Type 2 Unit step ≠0 0 0 Ramp ≠0 0 Parabolic ≠0 2.2 Time response analysis of First Order Systems Now calculating errors of when order is varied 1)Order 1 : OLTF G(s) = 1/TS
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