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ME_446_-_Lec_05-Laplace_Transformation

# ME_446_-_Lec_05-Laplace_Transformation - Laplace...

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Unformatted text preview: Laplace Transformation 446 - 5 Prof. Neil A. Duffie University of Wisconsin-Madison Wisconsin- 5 Neil A. Duffie, 1996 All rights reserved. 1 Laplace Transformation L[f(t)] = 0 f(t)e -st dt s is a new, complex variable Transform: Differential equations Algebraic equations Functions of time (step, impulse, sine, etc.) 5 2 DC Motor-Amplifier System Motor- (t),(t) v(t) e(t),i(t) Motor torque T(t) (t) 5 (t) 3 Mechanical System Variables v(t) (t),(t) e(t),i(t) (t) = rotational position (t) = rotational velocity T(t) = motor torque 5 T(t) (t) (t) 4 Mechanical System Parameters v(t) (t),(t) e(t),i(t) J = motor inertia Kt = torque constant 5 T(t) (t) (t) 5 Electrical System Variables v(t) (t),(t) e(t),i(t) i(t) = motor current e(t) = amplifier output voltage v(t) = amplifier input voltage 5 T(t) (t) (t) 6 Electrical System Parameters v(t) (t),(t) e(t),i(t) R = motor resistance L = motor inductance Ke = back emf constant 5 T(t) (t) (t) Kv = tachometer gain 7 Motor-Amplifier Equations Motor- 5 8 Transforms of Equations 5 9 Transformed Model - Velocity 5 10 Transformed Model - Position 5 11 Unit Step Function f(t) 1 u(t) t 0 L[u(t)] = 0 u(t)e -st dt L[u(t)] = 5 0 e -st dt 12 Unit Step Function f(t) 1 u(t) t 0 -st L[u(t)] = e -s 0 -s -s0 L[u(t)] = e-s - e -s 5 1 L[u(t)] = s 13 Unit Impulse Function (t) t 0 1 t "strength" (Area) = 1 1 1 t (t) (t) t (t) t 5 t 0 t 0 t t 14 0 t 1 t f(t) Unit Impulse Function (t) Area = 1 t 0 t L[(t)] = 0 (t)e -st dt t L[(t)] = lim 5 t0 0 1 e -st dt t 15 1 t f(t) Unit Impulse Function (t) strength = 1 t 0 t t -st L[(t)] = lim e t0 -st 0 L[(t)] = lim e 5 - e -s0 = lim 1 - e -st t0 t0 -st st Use L'Hopital's rule 16 -st 1 t f(t) Unit Impulse Function (t) strength = 1 0 t t Differentiate numerator and denominator with respect to t L[(t)] = lim ses t0 5 -st = lim e -st t0 17 L[(t)] = 1 1 f(t) Exponential Function f(t) = e -t t 0 -t L[e ] = 0 e -t e -st dt e - (1 0 L[e ] = 5 -t + s )t dt 18 1 f(t) Exponential Function f(t) = e -t t 0 5 - (1 + s )t L[e -t ] = e - (1 + s ) - (1 + s ) - (1 + s )0 - e L[e -t ] = e - (1 + s ) - (1 + s ) L[e -t ] = 1 1 = +s s + 1 0 19 Nonlinear Tank System Inlet flow h(t) qi(t) (t) Tank Valve Tank area = A q0(t) Outlet flow 5 Nonlinear behavior of valve flow: q0 (t) = k(t) h(t) 20 Nonlinear Tank Model qs (t) = qi (t) - qo (t) qs (t) = A dh(t) dt qo (t) = k(t) h(t) A A 5 dh(t) = qi (t) - k(t) h(t) dt dh(t) + k(t) h(t) = qi (t) dt 21 Linearized Model of Valve Flow qo = k h qi = qo q 2 h= i k qo (t) qo + qo ((t) - ) + .h qo h (h(t) - h ) .h qo (t) qo + k h ((t) - ) - 5 k (h(t) - h ) 2 h 22 Linearized Tank Model qo (t) qo + k h ((t) - ) + k (h(t) - h ) 2 h qo (t) k h + k h(t) - k h + qo (t) k h(t) + A 5 k k h(t) - h 2 h 2 h k 1 h(t) - k h 2 h 2 dh(t) k 1 qi (t) - k h(t) - h(t) + k h 2 h dt 2 23 Transformed Tank System Model dh(t) k 1 A + h(t) qi (t) - k h(t) + k h 2 2 h dt A 1 dh(t) k + h(t) qi (t) - k h(t) + k h u(t) 2 h 2 dt A sH(s) - h(0 + ) + ( ) k H(s) 2 h 1 1 Q i (s) - k h(s) + k h 2 s How would tank respond to a change in qi(t)? To a change in (t)? Need solution! 5 24 ...
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