• Document: Figure. a. Block diagram repreentation o a ytem; b. block diagram repreentation o an interconnection o ubytem REVIEW OF THE LAPLACE TRANSFORM Table. Laplace tranorm table Table. Laplace tr...
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Figure 2.1 a. Block diagram representation of a system; b. block diagram representation of an interconnection of subsystems REVIEW OF THE LAPLACE TRANSFORM Table 2.1 Laplace transform table Table 2.2 Laplace transform theorems Figure 2.2 Block diagram of a transfer function MECHANICAL SYSTEMS TRANSFER FUNTIONS Table 2.4 Force-velocity, force- displacement, and impedance translational relationships for springs, viscous dampers, and mass Example: Find the transfer function X(s) / F(s) for the system given below Figure 2.15 a. Mass, spring, and damper system; b. block diagram Figure 2.16 a. Free-body diagram of mass, spring, and damper system; b. transformed free-body diagram We now write the differential equation of motion using Newton’s law d 2 x (t ) dx ( t ) M 2 + fv + Kx ( t ) = f ( t ) dt dt Taking the Laplace transform, assuming zero initial conditions Ms 2 X ( s ) + f v sX ( s ) + KX ( s ) = F ( s ) ( Ms 2 + f v s + K ) X ( s ) = F ( s ) Solving for transfer function yields X (s) 1 G (s) = = 2 F (s) Ms + fv + K [Sum of impedance]X(s) = [Sum of applied forces] Note that the number of equation of motion required is equal to number of linearly independent motion. Linear independence implies that a point of motion in a system can still move if all other points of motion are held still. Another name of the number of linearly independent motion is the number of degrees of freedom. Example : Find the transfer function X2(s)/F(s) Figure 2.17 a. Two-degrees-of- freedom translational mechanical system8; b. block diagram Figure 2.18 a. Forces on M1 due only to motion of M1 b. forces on M1 due only to motion of M2 c. all forces on M1 [M1s2 + (fv1+fv3)s + (K1+K2)] X1(s) – (fv3s+K2)X2(s) = F(s) -(fv3s +K2)X1(s) + [M2s2 + (fv2+fv3)s + (K2 + K3)]X2(s) = 0 Figure 2.19 a. Forces on M2 due only to motion of M2; b. forces on M2 due only to motion of M1; c. all forces on M2 The transfer function X2(s)/F(s) is [M1s2 + (fv1+fv3)s + (K1+K2)] – (fv3s+K2) ∆ = – (fv3s+K2) [M2s2 + (fv2+fv3)s + (K2+K3)] Note that Sum of impedance Sum of Sum of applied connected X (s) - impedance X2(s) = forces at 1 to the between X1 motion at X1 and X2 X1 Sum of impedance Sum of Sum of applied connected X (s) - impedance X1(s) = forces at 2 to the between X2 motion at X1 and X2 X2 Equations of Motion by Inspection Problem : Write, but not solve, the equation of motion for the mechanical system given below. Figure 2.20 Three-degrees-of-freedom translational mechanical system Using same logic, for M1 Sum of Sum of impedance Sum of Sum of connected X (s) - impedance X (s) - impedance X (s) = applied 1 2 3 to the between between forces at motion at X1 and X2 X1 and X3 X1 X1 and for M2 Sum of Sum of Sum of Sum of - impedance X (s)+ impedance X (s) - impedance X (s) = applied between 1 connected to 2 3 between forces at X and X 1 2 the motion at X2 X2 and X3 X2 Similarly, for M3 Sum of Sum of Sum of Sum of - impedance X (s)- impedance impedance applied 1 between X2(s) + X (s) = between connected to 3 X1 and X3 X2 and X3 the motion at forces at X3 X3 Anymore we can write the equations for M1, M2 and M3 Note that M1 has two springs, two viscous damper and mass associated with its motion. There is one spring between M1 and M2 and one viscous damper between M1 and M3. Equation for M1 [M1s 2 + ( f v1 + f v3 )s + (K1 + K 2 )]X 1 (s) − K 2 X 2 (s) − f v3 sX 3 (s) = 0 for M2 − K2 X1 (s) + [M 2 s 2 + ( f v2

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