A Mechanical System
J Inertia
v
mAngular velocity τ External torque
b
1Rot. friction coefficient K
1Torsional stiffness
R Wheel radius
M Translational mass K
2Translational stiffness b
2Tran. friction coefficient x Position of the final point
R
M J
b
1b
2K
2x
K
1w t
R
M J
b
1b
2K
2x
K
1w t
• The system can be graphically represented as follows:
1
a
3
c
4 5
2
b d e
T J
w b
1T
1T
4v
5T = R
K
1F
1M
v
mb
2F
bK
2F
2v
d• This scheme has been obtained using the following source code:
**, Gr, Si, Sn, Si, As, Si, POG, Si, EQN, Si, SLX, Si rT, 1, a, Fn, T=10, An, -90, Sh,-0.3
rJ, 1, a, Kn, J, En, w, Sh, 0.3 - -, 1, 2
- -, a, b
rB, 2, b, Kn, b 1, Fn, T 1, Sh,-0.2 CB,[2;3],[b;c], Kn,E2=R*E1
mK, 3, 4, Kn, K 1, Fn, F 1 - -, c, d
mM, 4, d, Kn, M, En, v m
mB, 4, d, Kn, b 2, Fn, F b, Sh, 0.5
mK, 4, 5, Kn, K 2, Fn, F 2, Ln, 1.5, Tr, 0.3 - -, d, e, Ln, 1.5
mV, 5, e, En, -v d, Fn, F d, Pin,1
• The corresponding POG block scheme is:
w
1T
1 J
w
b
1T
1R
R
K
1F
11 M
v
mb
2F
b1 s 1
s 1
s 1
s
K
2F
2F
dv
d• The POG state space equations L ˙ x = Ax + Bu and y = Cx + Du are:
J 0 0 0
0
K11
0 0
0 0 M 0 0 0 0
K12
| {z }
L
˙x =
−b
1−R 0 0
R 0 −1 0
0 1 −b
2−1
0 0 1 0
| {z }
A
w F
1v
mF
2
| {z }
x
+
1 0 0 0 0 0 0 −1
| {z }
B
u
"
w
1F
d#
| {z }
y
=
"
1 0 0 0 0 0 0 1
#
| {z }
C
x +
"
0 0 0 0
#
| {z }
D
"
T v
d#
| {z }
u
• The new parameters:
-- Matlab commands --- (POG_Mechanical_system_J_K_T_M_K_Par_SLX_m.m) ---
%%%%%%%%%%%% SYSTEM PARAMETERS %%%%%%%%%%%%%%%%%%%%%%%%%%
J=1*kg*(30*cm)^2; % 2. Inertia. Internal parameter.
b_1=1*Nm/(10*rpm); % 3. Angular friction. Internal parameter.
R=40*cm; % 5. Parameter. Transformer/Gyrator.
K_1=5*Newton/(1*mm); % 6. Stiffness. Internal parameter.
M=10*kg; % 7. Mass. Internal parameter.
b_2=50*Newton/(2*meters/sec); % 8. Friction. Internal parameter.
K_2=20*Newton/(1*cm); % 9. Stiffness. Internal parameter.
%%%%%%%%%%%% INPUT VALUES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
T=10*Nm; % 1. Torque. Input value.
v_d=0; % 10. Velocity. Input value.
---
• The use of the units helps a correct choice of the parameters.
• The Simulink file “POG Mechanical system J K T M K Par SLX.slx” :
• With the new parameter one obtains the following simulation results:
0 1 2 3 4 5
0 10 20 30 40 50
F_1 [Newton]
Teeth Force
0 1 2 3 4 5
−10
−5 0 5 10 15
w_1 [rpm]
Angular velocity
0 1 2 3 4 5
−0.15
−0.1
−0.05 0 0.05 0.1 0.15 0.2
w_m [meters/sec]
Linear velocity
Time [s]
0 1 2 3 4 5
0 0.5 1 1.5 2 2.5
x [cm]
Mass Position
Time [s]
• The Backlash between the teeth of the gear can be considered substituting the stiffness K
1with a lookup table:
F
1(x)
x x
b−x
bK
1K
1• The Coulomb friction has the following non linear shape:
F
1(x)
x N
c−N
cb
b
• Backlash and Coulomb friction can be modeled using Simulink blocks:
• Using the following parameters:
-- Matlab commands --- (POG_Mechanical_system_J_K_T_M_K_NL_SLX_m.m) ---
%%%%%%%%%%%% SYSTEM PARAMETERS %%%%%%%%%%%%%%%%%%%%%%%%%%
J=1*kg*(30*cm)^2; % 2. Inertia. Internal parameter.
b_1=1*Nm/(10*rpm); % 3. Angular friction. Internal parameter.
R=40*cm; % 5. Parameter. Transformer/Gyrator.
K_1=5*Newton/(1*mm); % 6. Stiffness. Internal parameter.
x_b=1*mm; % Backlash amplitude
X_in =[ -2*x_b -x_b x_b 2*x_b]; % Lookup table input F_out=[-K_1*x_b -x_b x_b K_1*x_b]; % Lookup table output M=10*kg; % 7. Mass. Internal parameter.
Nc=2*Newton; % Amplitude of the Coulomb friction b_2=50*Newton/(2*meters/sec); % 8. Friction. Internal parameter.
K_2=20*Newton/(1*cm); % 9. Stiffness. Internal parameter.
%%%%%%%%%%%% INPUT VALUES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
T=2*Nm; % 1. Torque. Input value.
w_T=2*pi*1; % Input frequency Ph_T=pi/2; % Initial phase
v_d=0; % 10. Velocity. Input value.
---
one obtains the following simulation results:
0 1 2 3 4 5
−10
−5 0 5 10 15
F_1 [Newton]
Teeth Force
0 1 2 3 4 5
−3
−2
−1 0 1 2 3 4
w_1 [rpm]
Angular velocity
0 1 2 3 4 5
−0.03
−0.02
−0.01 0 0.01 0.02 0.03
w_m [meters/sec]
Linear velocity
Time [s]
0 1 2 3 4 5
−0.4
−0.3
−0.2
−0.1 0 0.1 0.2 0.3
x [cm]
Mass Position
Time [s]
