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controlppt(29.8 -3.9)

# controlppt(29.8 -3.9) - Chapter 3 Comparison of closed loop...

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Comparison of closed loop & open loop Control systems Chapter 3

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Ref Source i f R f , L f G e g R a v t i a Load feed back resistance high enough to avoid loading
Equations : V f (field voltage) = (V r - V b ) K A V b = K V t , V t is the terminal voltage of the generator I f = V f / ( R f + sL f ) E g = K g I f V t = E g - I a R a

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V r - V b K A V f 1 R f + sL f K g K E g = V t I f + Case 1 No load ( I a = 0) So V t = E g
K A K g V t (s) / V r (s) = ---------- R f + sL f Case 1 (a): open loop case 1 (b) : closed loop V t (s) / V r (S) = K A K g ----------------------- R f + sL f + K K A K g

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Substitute numerical values K A = 20 V / V K g = 50 V / A R f = 100 ohms L f = 200 Henries K = 0 . 1
Final value Therom f (t) F (s) f (t) / t = Limit sF(s) s 0 (t) (t)

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With a sudden change in v r from 0 to 25 V, v r (t) = 25 u(t) V r (s) = 25 / s
open loop V t (s) = (v r (s) )(10 / ( 1+2 s )) = (25 / s ) ( 10 / (1 + 2 s)) closed loop v t (s) = ( 25 / s)( 5 / ( 1+s )) V t ( ) = 125 V V t ( ) = 250 V

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Open loop Gain(G) = 10 Closed loop Gain[G/(1+ GH)] = 5
To make the output same as 250 V in both cases, for open loop v r = 25 V for closed loop v r = 50 V

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V r (s) = 25 / s V t (s) = (25 / s)(10 /( 1+2s)) = 250 ((1 / s) - 1 /( s+0.5)) v t (t) = 250 ( 1- e - 0. 5 t ) = 250 ( 1 - e - t /2 ) Open loop Step response
V r (s) = 50 / s V t (s) = (50 / s )(5 /( 1 + s)) = 250( (1 / s )- 1 /( s + 1)) v t (t) = 250 ( 1 - e -t ) = 250 ( 1 - e -t / 1) Closed loop Step response

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0 50 100 150 200 250 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 x1 x2

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