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POG dynamic model of the considered mechanical transmission: F

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Capitolo 0. INTRODUCTION 3.1

Mechanical transmission: state space model

• Mechanical transmission:

• POG dynamic model of the considered mechanical transmission:

F - 

? 1 s

? 1 M1

?

x˙1

 -

 

b

6

6 - -

 -

6 1 s 6

K

6

Fe

- 

- 

? 1 s

? 1 M2j

?

x˙2

 -

 

b2

6

6

- -- 1 R -

ω2

 1 

R  -

Kj

s

6

6

τ

- 

- 

? 1 s

? 1 J1

ω?1

 -

 

d1

6

6 - -

τd

• State vector (output power variables):

x = 

1 Fe2 τ ω1

T

• The force F is the system control input. The torque τd is an external disturbance input.

• The output variable is the angular velocity ω1 of the last rotational element.

Zanasi Roberto - System Theory. A.A. 2015/2016

(2)

Capitolo 3. DYNAMIC MODELING 3.2

• The equivalent mass M2j can be expressed as follows:

M2j = M2 + J2

R2

• State space equations:







M11 1 K

e M2j2

1 Kj τ˙ J1 ω˙1







| {z } L ˙x

=







−b1−1 0 0 0 1 0 −1 0 0 0 1−b2R1 0 0 0 1

R 0 −1

0 0 0 1−d1







| {z }

A





 x˙1

Fe2

τ ω1







| {z } x

+







1 0 0 0 0 0 0 0 0 −1







| {z }

B

F τd



|{z}u

y = 

0 0 0 0 1 

| {z }

C

x that is

( L ˙x = A x + B u y = C x

( x = L˙ −1A x + L−1B u y = C x

dove L =







M1 0 0 0 0 0 1

K 0 0 0

0 0 M2j 0 0 0 0 0 1

Kj 0 0 0 0 0 J1







• The symmetric part of matrix A is a function of the dissipative parameters b1, b1, R and d1. The skew-symmetric part of matrix A is a function only of the “connection” coefficients.

• Possible values for the parameters to be used for simulations in Ma- tlab/Simulink environment:

M1 = 0.6*Kg; % First mass

b1 = 2*N/(40*m/sec); % Linear friction coefficient of the first mass K = 100*N/(1*cm); % Stiffness of the first spring

M2 = 1*Kg; % Second mass

b2 = 1*N/(50*m/sec); % Linear friction coefficient of the first mass R = 10*cm; % Radius of the wheel

J2 = 150*gr*(12*cm)^2; % Inerzia of element J2

Kj = 100*N/(0.1*rad); % Stiffness of the torsional spring J1 = 190*gr*(10*cm)^2; % Inerzia of element J1

d1 = 10*N*m/(100*rad/sec); % Linear friction coefficient of element J1 M2j = M2+J2/(R^2); % Equivalent translational mass

Zanasi Roberto - System Theory. A.A. 2015/2016

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