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# CON2 - Chapter 2 Mathematical Modelling Electrical Systems...

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Chapter 2 Mathematical Modelling Electrical Systems R L C Voltage (e), Current (i)

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i R = e / R i C = C de / dt i L = (1/L) e dt
Mechanical Systems M K B velocity ( v) force ( f )

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f K = K v dt f M = M dv/dt f B = Bv
Force - current analogy C M 1/L K 1/R B

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Integro - differential equations by analogy approach free body diagram approach
Electro mechanical example + - R L M K B + f i e S v

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current i force f velocity v back emf e b voltage source e S
e s

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e s i f
e s i v f

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e s i v f e b
(e s - e b ) - + e s e b i f v

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Equations (e s -e b ) = Ri + L di/dt K 1 i = f f = M dv/dt+Bv+K vdt e b = K 2 v
input output x y y = K x K

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In our example i f K 1 v e b K 2
How do we represent (e s - e b ) = Ri + L di / dt + - e s e b e Let e s - e b = e Then Error Detector

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Take Laplace Transform of e = Ri + L di/dt ? Now e i
E (s) = R I (s) + Ls I (s) = (R + Ls) I(s) assuming ?

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Assuming Initial Conditions to be Ignored/zero
? E I I(s)/E(s) = 1 /( R + Ls) = G 1 (s) (say)

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Similarly F ? v
Mechanical Components M + f B K

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Free body diagram approach: M x f Kx x B
f Kx x B x M x B - Kx - f x M = + + =

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Analogy approach B M K f g x
? = 1 / ( Ms + B + K /s ) = G 2 (say) F ? v

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e s + e G 1 i f v K 1 G 2 K 2 - e b
Block Diagram Reduction X 2 / X 1 = G 1 X 3 / X 2 = G 2 Therefore X 3 / X 1 = G 1 . G 2 G 1 x 2 G 2 x 1 x 3

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e s + e G 1 G 2 K 1 v K 2 - e b
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CON2 - Chapter 2 Mathematical Modelling Electrical Systems...

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