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Control ENG HW_Part_26

# Control ENG HW_Part_26 - R1 Ra RR A H 1 s R1 Ra τ1 = 1 a 1...

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Unformatted text preview: R1 Ra RR A H 1 ( s ) R1 + Ra ; τ1 = 1 a 1 = [τ 1s + 1] R1 + Ra Q( s) and from (2 ) we get R2 Ra R2 R + R R H1 ( s) a 1 τ =RA = 1 [ τ 1s + 1](τ 2 s + 1) 2 2 2 Q( s) putting the numerical values of parameters 2 H1 ( s) 3 = Q( s) 4 s + 1 3 2 H 2 (s) 3 = Q(s) 4 s + 1(s + 1) 3 8.1 A step transfer change of magnitude 4 is introduced into a system having the Y ( s) 10 =2 X ( s ) s + 1.6s + 4 Determine (a) % overshoot (b)Rise time (c)Max value of Y(t) (d)Ultimate value of Y(t) (e) Period of Oscillation. Given X ( s ) = 4 s Y ( s) = The transfer function is 40 s ( s + 1.6s + 4) 2 2.5 Y ( s) 10 × 0.25 = = 2 0.25s + 0.4 s + 1) X ( s ) 0.2( s 2 ) + ( 1.6 ) s + 1) 4 τ 2 = 0.25 ; τ = 0.5 and 2ξ τ = 0.4 ξ= 0.4 = 0.4 (< 1 = system is underdamped ) 2(0.5) we find ultimate value of Y(t) 40 s 40 = = 10 S →0 s ( s + 1.6 + 4) 4 Lt Y (t ) = Lt sY ( s ) = Lt t →∞ S →0 2 thus B= 10 now, from laplace transform tables ξt − 1 Y (t ) = 101 − e τ sin(α + φ ) 1−ξ 2 where α = 1−ξ 2 τ , φ = tan − 1−ξ 2 ξ ...
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Control ENG HW_Part_26 - R1 Ra RR A H 1 s R1 Ra τ1 = 1 a 1...

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