Kogiso, Sawano, Itto, and Sugimoto, "Identification Procedure for McKibben Pneumatic Artificial Muscle Systems." presented in IROS 2013

1. IEEE/RSJ International Conference on
Intelligent Robots and Systems
October 7-12, 2012
Vilamoura, Algarve, Portugal
Identification Procedure for
McKibben Pneumatic Artificial
Muscle Systems
K. Kogiso, K. Sawano, T. Itto, and K. Sugimoto
Nara Institute of Science and Technology (NAIST), Japan
Oct. 10, 2012 @ WedBT5, 9:30 to 9:45 am, Regular Session, Gemini 2, Tivoli Marina Vilamoura
13年1月30日水曜日

2. Outline
Introduction
Modeling of PAM system
Identification Procedure
Experimental Validation
Extension
Conclusion
Active Link, Co.
2
13年1月30日水曜日

3. Introduction
McKibben Pneumatic Artificial Muscle (PAM)
rubber tube mesh
Advantage for application Disadvantage for modeling & control
high power/weight ratio complex & nonlinear system (hydrodynamics)
flexibility empirical approximation or linearization
3
13年1月30日水曜日

4. Introduction
Motivation
Mathematical modeling of PAM is a challenging issue. L0 L l
nonlinearities such as hysteresis, hydrodynamics, friction,... solenoid
PDC
valve
valve
difficulty to explain validity of approximation or linearization.
dependence on what kind of a valve you use. air
compressure M
M
PAM system = PAM + proportional directional control (PDC) valve 0.3
Mathematical modeling of PAM system w/ constant weight. 0.2
[Itto, Kogiso: Hybrid modeling of McKibben pneumatic artificial muscle systems, IEEE ICIT&SSST, 2011]
ε
formulates model structure based on dynamics, but 0.1
M = 3 [kg]
requires complete try and errors for identifying parameters. M = 6 [kg]
M = 9 [kg]
0
hysteresis loop
100 200 300 400 500 600 700
pressure [kPa]
4
13年1月30日水曜日

5. Introduction
Motivation
Mathematical modeling of PAM is a challenging issue. L0 L l
nonlinearities such as hysteresis, hydrodynamics, friction,... solenoid
PDC
valve
valve
difficulty to explain validity of approximation or linearization.
dependence on what kind of a valve you use. air
compressure M
M
PAM system = PAM + proportional directional control (PDC) valve 0.3
Mathematical modeling of PAM system w/ constant weight. 0.2
[Itto, Kogiso: Hybrid modeling of McKibben pneumatic artificial muscle systems, IEEE ICIT&SSST, 2011]
ε
formulates model structure based on dynamics, but 0.1
M = 3 [kg]
requires complete try and errors for identifying parameters. M = 6 [kg]
M = 9 [kg]
0
hysteresis loop
Outcomes 100 200 300
pressure
400
[kPa]
500 600 700
a parameter identification procedure supported by analysis of the mathematical model,
which contributes to reduce the cost for try and errors.
an identified model validated by comparison with several experimental data,
which well simulates behaviors of a practical PAM system.
an extension to a model expressing the PAM system over a specified weight range,
which is realized by interpolation of some dominant parameters in terms of a weight. 4
13年1月30日水曜日

6. Modeling
Dynamics of PAM system
[Itto, Kogiso, IEEE ICIT&SSST 11]
switched system with 64 nonlinear subsystems
x(t) = f (x(t), u(t)) if x(t) 2 S
˙
y(t) = h(x(t))
T
x := [✏ ✏ P ]T y := [✏ F ]
˙
S := {x 2 <3 | (x)  0} 2 {1, 2, · · · , 64}
5
13年1月30日水曜日

7. Modeling
Dynamics of PAM system
[Itto, Kogiso, IEEE ICIT&SSST 11]
switched system with 64 nonlinear subsystems
x(t) = f (x(t), u(t)) if x(t) 2 S
˙
y(t) = h(x(t))
T
x := [✏ ✏ P ]T y := [✏ F ]
˙
S := {x 2 <3 | (x)  0} 2 {1, 2, · · · , 64}
dynamic equation (w/ friction [Kikuue, IEEE TRO 06] )
8
> F (P, ✏, t) M g Ff (t)
>
<
M L¨(t) =
✏ K(L0 L(1 ✏(t)))3 , if ✏(t)  L LL0 ,
>
>
:
F (P, ✏, t) M g F (t), f otherwise,
8
> cv L✏(t) + cc sgn(✏(t)),
˙ ˙ if ✏(t) 6= 0,
˙
>
>
< c , if ✏(t) = 0 and Fo (t) > cc ,
˙
c
Ff (t) =
> Fo (t), if ✏(t) = 0 and Fo (t) 2 [ cc , cc ],
> ˙
>
:
cc , if ✏(t) = 0 and Fo (t) < cc ,
˙
5
13年1月30日水曜日

8. Modeling
Dynamics of PAM system PAM volume [Kagawa, CEP 97], [Minh, Mechatronics 10]
[Itto, Kogiso, IEEE ICIT&SSST 11] V (t) = D1 ✏(t)2 + D2 ✏(t) + D3
switched system with 64 nonlinear subsystems
x(t) = f (x(t), u(t)) if x(t) 2 S
˙
y(t) = h(x(t))
T
x := [✏ ✏ P ]T y := [✏ F ]
˙
S := {x 2 <3 | (x)  0} 2 {1, 2, · · · , 64}
dynamic equation (w/ friction [Kikuue, IEEE TRO 06] )
8
> F (P, ✏, t) M g Ff (t)
>
<
M L¨(t) =
✏ K(L0 L(1 ✏(t)))3 , if ✏(t)  L LL0 ,
>
>
:
F (P, ✏, t) M g F (t), f otherwise,
8
> cv L✏(t) + cc sgn(✏(t)),
˙ ˙ if ✏(t) 6= 0,
˙
>
>
< c , if ✏(t) = 0 and Fo (t) > cc ,
˙
c
Ff (t) =
> Fo (t), if ✏(t) = 0 and Fo (t) 2 [ cc , cc ],
> ˙
>
:
cc , if ✏(t) = 0 and Fo (t) < cc ,
˙
5
13年1月30日水曜日

9. Modeling
Dynamics of PAM system PAM volume [Kagawa, CEP 97], [Minh, Mechatronics 10]
[Itto, Kogiso, IEEE ICIT&SSST 11] V (t) = D1 ✏(t)2 + D2 ✏(t) + D3
switched system with 64 nonlinear subsystems
contraction force [Tondu, IEEE CSM 00], [Kang, ICRA 09]
x(t) = f (x(t), u(t)) if x(t) 2 S
˙  n ⇣ ⌘ o2
Cq2 Pg (t)
F (P, ✏, t) = APg (t) at a C q1 1 + e ✏(t) as
y(t) = h(x(t))
T
x := [✏ ✏ P ]T y := [✏ F ]
˙
S := {x 2 <3 | (x)  0} 2 {1, 2, · · · , 64}
dynamic equation (w/ friction [Kikuue, IEEE TRO 06] )
8
> F (P, ✏, t) M g Ff (t)
>
<
M L¨(t) =
✏ K(L0 L(1 ✏(t)))3 , if ✏(t)  L LL0 ,
>
>
:
F (P, ✏, t) M g F (t), f otherwise,
8
> cv L✏(t) + cc sgn(✏(t)),
˙ ˙ if ✏(t) 6= 0,
˙
>
>
< c , if ✏(t) = 0 and Fo (t) > cc ,
˙
c
Ff (t) =
> Fo (t), if ✏(t) = 0 and Fo (t) 2 [ cc , cc ],
> ˙
>
:
cc , if ✏(t) = 0 and Fo (t) < cc ,
˙
5
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10. Modeling
Dynamics of PAM system PAM volume [Kagawa, CEP 97], [Minh, Mechatronics 10]
[Itto, Kogiso, IEEE ICIT&SSST 11] V (t) = D1 ✏(t)2 + D2 ✏(t) + D3
switched system with 64 nonlinear subsystems
contraction force [Tondu, IEEE CSM 00], [Kang, ICRA 09]
x(t) = f (x(t), u(t)) if x(t) 2 S
˙  n ⇣ ⌘ o2
Cq2 Pg (t)
F (P, ✏, t) = APg (t) at a C q1 1 + e ✏(t) as
y(t) = h(x(t))
T pressure change in a PAM [Richer, JDSMC 00]
x := [✏ ✏ P ]T y := [✏ F ]
˙ ˙
˙ RT V (t)
S := {x 2 <3 | (x)  0} 2 {1, 2, · · · , 64} P (t) = k1 m(t)
˙ k2 P (t)
V (t) V (t)
dynamic equation (w/ friction [Kikuue, IEEE TRO 06] )
8
> F (P, ✏, t) M g Ff (t)
>
<
M L¨(t) =
✏ K(L0 L(1 ✏(t)))3 , if ✏(t)  L LL0 ,
>
>
:
F (P, ✏, t) M g F (t), f otherwise,
8
> cv L✏(t) + cc sgn(✏(t)),
˙ ˙ if ✏(t) 6= 0,
˙
>
>
< c , if ✏(t) = 0 and Fo (t) > cc ,
˙
c
Ff (t) =
> Fo (t), if ✏(t) = 0 and Fo (t) 2 [ cc , cc ],
> ˙
>
:
cc , if ✏(t) = 0 and Fo (t) < cc ,
˙
5
13年1月30日水曜日

11. Modeling
Dynamics of PAM system PAM volume [Kagawa, CEP 97], [Minh, Mechatronics 10]
[Itto, Kogiso, IEEE ICIT&SSST 11] V (t) = D1 ✏(t)2 + D2 ✏(t) + D3
switched system with 64 nonlinear subsystems
contraction force [Tondu, IEEE CSM 00], [Kang, ICRA 09]
x(t) = f (x(t), u(t)) if x(t) 2 S
˙  n ⇣ ⌘ o2
Cq2 Pg (t)
F (P, ✏, t) = APg (t) at a C q1 1 + e ✏(t) as
y(t) = h(x(t))
T pressure change in a PAM [Richer, JDSMC 00]
x := [✏ ✏ P ]T y := [✏ F ]
˙ ˙
˙ RT V (t)
S := {x 2 <3 | (x)  0} 2 {1, 2, · · · , 64} P (t) = k1 m(t)
˙ k2 P (t)
V (t) V (t)
net mass flow rate of PDC valve
dynamic equation (w/ friction [Kikuue, IEEE TRO 06] ) m(t) = ↵(t)mi (t) (1 ↵(t))mo (t)
˙ ˙ ˙
8 8 r
⇣ ⌘k 1
> F (P, ✏, t) M g Ff (t)
> >
> A Pp k
> 0 tank R k+1 2
k+1
< >
>
>
>
T
⇣ ⌘ kk 1
>
M L¨(t) =
✏ K(L0 L(1 ✏(t)))3 , if ✏(t)  L LL0 , >
>
< if P (t)  2
k+1 Ptank ,
>
> mi (t) =
˙ r
: >
>
>
q ⇣ ⌘k
1 ⇣ ⌘ kk 1
F (P, ✏, t) M g F (t), f otherwise, > A0 Pp
>
>
>
tank
T
2k P (t)
R(k 1) Ptank 1 P (t)
Ptank
>
> ⇣ ⌘ kk 1
>
8 : if P (t) > 2
Ptank ,
> cv L✏(t) + cc sgn(✏(t)),
˙ ˙ if ✏(t) 6= 0,
˙ 8 r
k+1
>
> > ⇣ ⌘k 1
k+1
< c , if ✏(t) = 0 and Fo (t) > cc ,
˙
> A P (t) k
> 0p
>
>
2
R k+1
c >
>
T
⇣ ⌘ kk 1
Ff (t) = >
>
> 2
> Fo (t), if ✏(t) = 0 and Fo (t) 2 [ cc , cc ],
> ˙ < if P (t) k+1 Pout ,
>
: mo (t) =
˙
> q ⇣ ⌘1
r
⇣ ⌘ kk 1
>
>
cc , if ✏(t) = 0 and Fo (t) < cc ,
˙ > A0 P (t)
>
> p 2k Pout k
R(k 1) P (t) 1 Pout
P (t)
>
>
T
⇣ ⌘ kk 1
>
>
: if P (t) 2
< Pout .
k+1
5
13年1月30日水曜日

12. Analysis
Dominant parameters
switched system with 64 nonlinear subsystems
x(t) = f (x(t), u(t)) if x(t) 2 S
˙
y(t) = h(x(t))
x := [✏ ✏ P ]T y := [✏ F ]T
˙
6
13年1月30日水曜日

13. Analysis
Dominant parameters M : mass of weight [kg]
D0 : natural diameter of PAM [m]
switched system with 64 nonlinear subsystems
L0 : natural length of PAM [m]
x(t) = f (x(t), u(t)) if x(t) 2 S
˙
D1 D2 D3 : coefficients for PAM volume [m^3]
y(t) = h(x(t)) Ptank : source absolute pressure [Pa]
x := [✏ ✏ P ]T y := [✏ F ]T
˙ Pout : atmospheric pressure [Pa]
k : specific heat ratio for air [-]
R : ideal gas constant [J/kg K]
T : absolute temperature [K]
K : coefficient of elasticity [N/m^3]
✓ : initial angle btw braided thread
& cylinder long axis [deg]
Cq1 : correction coefficient [-]
Cq2 : correction coefficient [1/Pa]
cc : Coulomb friction [N]
A0 : orifice area of PDC valve [m^2]
k1 k2 : polytropic indexes [-]
cv : viscous friction coefficient [Ns/m]
6
13年1月30日水曜日

14. Analysis
Dominant parameters M : mass of weight [kg]
D0 : natural diameter of PAM [m]
switched system with 64 nonlinear subsystems
L0 : natural length of PAM [m]
x(t) = f (x(t), u(t)) if x(t) 2 S
˙
D1 D2 D3 : coefficients for PAM volume [m^3]
y(t) = h(x(t)) Ptank : source absolute pressure [Pa]
x := [✏ ✏ P ]T y := [✏ F ]T
˙ Pout : atmospheric pressure [Pa]
k : specific heat ratio for air [-]
R : ideal gas constant [J/kg K]
Analysis result:
T : absolute temperature [K]
For the PAM system model, : coefficient of elasticity [N/m^3]
K
its steady-state behavior is characterized by : initial angle btw braided thread
✓
& cylinder long axis [deg]
parameters: K ✓ Cq1 Cq2 cc : correction coefficient [-]
Cq1
and its transient behavior is characterized by Cq2 : correction coefficient [1/Pa]
cc : Coulomb friction [N]
parameters: A0 k1 k2 cv
A0 : orifice area of PDC valve [m^2]
k1 k2 : polytropic indexes [-]
Hint: as t ! 1 , then params left or not.
cv : viscous friction coefficient [Ns/m]
6
13年1月30日水曜日

15. Analysis
Dominant parameters M : mass of weight [kg]
D0 : natural diameter of PAM [m]
switched system with 64 nonlinear subsystems
L0 : natural length of PAM [m]
x(t) = f (x(t), u(t)) if x(t) 2 S
˙
D1 D2 D3 : coefficients for PAM volume [m^3]
y(t) = h(x(t)) Ptank : source absolute pressure [Pa]
x := [✏ ✏ P ]T y := [✏ F ]T
˙ Pout : atmospheric pressure [Pa]
k : specific heat ratio for air [-]
R : ideal gas constant [J/kg K]
Analysis result:
T : absolute temperature [K]
For the PAM system model, : coefficient of elasticity [N/m^3]
K
its steady-state behavior is characterized by : initial angle btw braided thread
✓
& cylinder long axis [deg]
parameters: K ✓ Cq1 Cq2 cc : correction coefficient [-]
Cq1
and its transient behavior is characterized by Cq2 : correction coefficient [1/Pa]
cc : Coulomb friction [N]
parameters: A0 k1 k2 cv
A0 : orifice area of PDC valve [m^2]
k1 k2 : polytropic indexes [-]
Hint: as t ! 1 , then params left or not.
cv : viscous friction coefficient [Ns/m]
6
13年1月30日水曜日

16. Analysis
Dominant parameters M : mass of weight [kg]
D0 : natural diameter of PAM [m]
switched system with 64 nonlinear subsystems
L0 : natural length of PAM [m]
x(t) = f (x(t), u(t)) if x(t) 2 S
˙
D1 D2 D3 : coefficients for PAM volume [m^3]
y(t) = h(x(t)) Ptank : source absolute pressure [Pa]
x := [✏ ✏ P ]T y := [✏ F ]T
˙ Pout : atmospheric pressure [Pa]
k : specific heat ratio for air [-]
R : ideal gas constant [J/kg K]
Analysis result:
T : absolute temperature [K]
For the PAM system model, : coefficient of elasticity [N/m^3]
K
its steady-state behavior is characterized by : initial angle btw braided thread
✓
& cylinder long axis [deg]
parameters: K ✓ Cq1 Cq2 cc : correction coefficient [-]
Cq1
and its transient behavior is characterized by Cq2 : correction coefficient [1/Pa]
cc : Coulomb friction [N]
parameters: A0 k1 k2 cv
A0 : orifice area of PDC valve [m^2]
k1 k2 : polytropic indexes [-]
Hint: as t ! 1 , then params left or not.
cv : viscous friction coefficient [Ns/m]
Note, no couplings btwn the two param groups.
6
13年1月30日水曜日

17. Achievement
Identification procedure
-5
x 10
7
PAM volume:
PAM volume
6
measurable or known in advance V (t) =
V [m ]
3
5
D1 ✏(t)2 + 4
M D0 L0 D1 Ptank Pout k T R D2 ✏(t) + D3 3
2
contraction ratio
0 0.1 0.2 0.3
ε
steady-state behavior transient behavior
Determines the parameters value: Determines the parameters value:
K ✓ Cq1 Cq2 cc A0 k1 k2 cv
until model maker satisfies. until model maker satisfies.
7
13年1月30日水曜日

18. Achievement
Identification procedure
-5
x 10
7
PAM volume:
PAM volume
6
measurable or known in advance V (t) =
V [m ]
3
5
D1 ✏(t)2 + 4
M D0 L0 D1 Ptank Pout k T R D2 ✏(t) + D3 3
2
contraction ratio
0 0.1 0.2 0.3
ε
steady-state behavior transient behavior
Determines the parameters value: Determines the parameters value:
K ✓ Cq1 Cq2 cc A0 k1 k2 cv
until model maker satisfies. until model maker satisfies.
When satisfied?
Since determination of values is subjective, observe a trend by parameter variation.
7
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19. Info. for observing
Trend by parameter variation
0.3
0.3 0.3
0.2
0.2 0.2
cq  increases cc increases
ε
ε
ε
cq  increases 0.1
0.1 0.1
cc = 0
cq  = 0.8 cq  = - 1/0.01 x10-6 cc = 4.875/P (oo) x105
cq  = 0.99 cq  = - 1/0.083 x10-6 0 cc = 9.75 / P (oo) x105
0 0
cq  = 1.2 cq  = - 1/0.2 x10-6
100 200 300 400 500 600 700 100 200 300 400 500 600 700 100 200 300 400 500 600 700
pressure [kPa] pressure [kPa] pressure [kPa]
param in contraction force param in contraction force Coulomb friction coefficient
0.26 0.26 0.26
ε
ε
ε
0.24 0.24 0.24
k  increases k  increases
c v increases A increases
0.22 0.22 0.22
0 10 20 0 10 20 0 10 20
pressure [kPa]
pressure [kPa]
550
pressure [kPa]
550 550
-6
k = 1.0, k = 1.0 c v = 10 A = 0.058 x10
-6
k = 1.0, k = 1.4 A increases A = 0.099 x10
500 k = 1.4, k = 1.0 500 c v = 500 500 -6
c v = 1000 A = 0.176 x10
k = 1.4, k = 1.4
450 k  increases k  increases 450 450
0 10 20 0 10 20 0 10 20
time [s] time [s] time [s]
polytropic indexes viscous friction coefficient max orifice area 8
13年1月30日水曜日

20. Experimental Validation
PAM system setup for model validation
proportional
directional
control valve
How to validate:
Input a step signal to the PDC valve, and
check steady-state and transient responses of simulation and experiment.
9
13年1月30日水曜日

21. Experimental Validation
Comparison: steady state behavior in e vs P
0.30 0.30
0.25
M = 4.0 [kg] 0.25
M = 5.0 [kg]
0.20 0.20
0.15 0.15
ε
ε
0.10 0.10
0.05 0.05
experiment experiment
model fixed at M=4 model fixed at M=5
0 0
model parametried by M model parametried by M
0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700
P [kPa] P [kPa]
experiment 0.30 experiment
0.30
model fixed at M=7 model fixed at M=8
0.25 model parametried by M 0.25 model parametried by M
0.20 0.20
0.15 0.15
ε
ε
0.10 0.10
0.05 0.05
0
M = 7.0 [kg] 0
M = 8.0 [kg]
0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700
P [kPa] P [kPa]
10
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22. Experimental Validation
Comparison: transient behavior in e vs t & P vs t
M = 4.0 [kg]
11
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23. Extension to M-‐‑‒parameterized Model
Interpolation over [1, 9] in weight
1.15 18
1.1 16
1.05 14
1
12
Cq1 [-]
10
cc [N]
0.95
8
0.9
6
0.85
4
0.8 Cq1 (M ) = 0.1573 log M + 0.7974 2 cc (M ) = 1.7353M + 0.1422
0.75 0
1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
M [kg] M [kg]
x 10 7
8 40
7 38
0.7915M 0.2296
K(M ) = 109600 exp 36
✓(M ) = 39.984M
6
34
5
K [N/m]
32
θ [rad]
4
30
3
28
2 26
1 24
0 22
1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
M [kg] 12
M [kg]
13年1月30日水曜日

24. Extension to M-‐‑‒parameterized Model
Interpolation over [1, 9] in weight
1.15 18
1.1 16
1.05 14
1
12
Cq1 [-]
10
cc [N]
0.95
8
0.9
6
0.85
4
0.8 Cq1 (M ) = 0.1573 log M + 0.7974 2 cc (M ) = 1.7353M + 0.1422
0.75 0
1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
M [kg] M [kg]
x 10 7
8 40
7 38
0.7915M 0.2296
K(M ) = 109600 exp 36
✓(M ) = 39.984M
6
34
5
K [N/m]
32
θ [rad]
4
30
3
28
2 26
1 24
0 22
1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
M [kg] 12
M [kg]
13年1月30日水曜日

25. Extension to M-‐‑‒parameterized Model
Comparison: steady state behavior in e vs P
0.30 0.30 0.30
0.25
M = 4.0 [kg] 0.25
M = 4.5 [kg] 0.25
M = 5.0 [kg]
0.20 0.20 0.20
0.15 0.15 0.15
ε
ε
ε
0.10 0.10 0.10
0.05 0.05 0.05
experiment experiment
model fixed at M=4 experiment model fixed at M=5
0 0 model parametried by M 0
model parametried by M model parametried by M
0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700
P [kPa] P [kPa] P [kPa]
experiment 0.30 experiment 0.30 experiment
0.30
model fixed at M=7 model parametried by M model fixed at M=8
0.25 model parametried by M 0.25 0.25 model parametried by M
0.20 0.20 0.20
0.15 0.15 0.15
ε
ε
ε
0.10 0.10 0.10
0.05 0.05 0.05
0
M = 7.0 [kg] 0
M = 7.5 [kg] 0
M = 8.0 [kg]
0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700
P [kPa] P [kPa] P [kPa]
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26. Extension to M-‐‑‒parameterized Model
Comparison: transient behavior in e vs t & P vs t
M = 4.5 [kg] M = 7.5 [kg]
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27. Conclusion
Summary
a mathematical model of a PAM system (PAM + PDC valve) with a constant weight,
which involves 11 measurable parameters and 9 need-to-be-identified parameters.
a parameter identification procedure supported by analysis of the mathematical model,
which contributes to reduce the cost for try and errors in finding the 9 parameter values.
an identified model validated by comparison with several experimental data,
which well simulates behaviors of a practical PAM system.
a mathematical model expressing the PAM system over a specified weight range,
which is also identifiable by using the proposed procedure plus interpolation.
Future works
an automatic identification procedure that appropriately determines parameter values
based on experimental sample data.
an antagonistic layout of PAMs as an actuator to realize position/force controls appropriate
for applications of power-assist systems or rehabilitation/training exoskeleton systems.
a model reduction technique in case of practical use of many PAMs.
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28. Thank you for your kind attention.
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13年1月30日水曜日