本文仅供学习使用
本文参考:
B站:CLEAR_LAB
笔者带更新-运动学
课程主讲教师:
Prof. Wei Zhang
课程链接 :
https://www.wzhanglab.site/teaching/mee-5114-advanced-control-for-robotics/
机器人——运动能力、计算能力、感知决策能力 的机电系统
Steady-state error :
e
s
s
=
lim
?
t
→
∞
θ
e
(
t
)
e_{\mathrm{ss}}=\underset{t\rightarrow \infty}{\lim}\theta _{\mathrm{e}}\left( t \right)
ess?=t→∞lim?θe?(t)
Precent overshoot : P.O.
Rise time / Peak time :
Settling time : T s T_{\mathrm{s}} Ts?
详细推导见 : (待补充)
详细推导见 : (待补充)
Dynamic equation of fully-acuated robot (with external force) :
{
τ
=
M
(
q
)
q
¨
+
c
(
q
,
q
˙
)
q
˙
+
g
(
q
)
+
J
T
(
q
)
F
e
x
t
y
=
h
(
q
)
\begin{cases} \tau =M\left( q \right) \ddot{q}+c\left( q,\dot{q} \right) \dot{q}+g\left( q \right) +J^{\mathrm{T}}\left( q \right) \mathcal{F} _{\mathrm{ext}}\\ y=h\left( q \right)\\ \end{cases}
{τ=M(q)q¨?+c(q,q˙?)q˙?+g(q)+JT(q)Fext?y=h(q)?
q
∈
R
n
q\in \mathbb{R} ^n
q∈Rn : joint positions (generalized coordinate)
τ
∈
R
n
\tau \in \mathbb{R} ^n
τ∈Rn : joint torque (generalized input)
y
y
y : output (variable to be controlled) —— can be any func of
q
q
q , e.g.
y
=
q
,
y
=
[
T
(
q
)
]
∈
S
E
(
3
)
y=q,y=\left[ T\left( q \right) \right] \in SE\left( 3 \right)
y=q,y=[T(q)]∈SE(3)
Joint-space vs. Task-space control
Joint-space :
y
(
t
)
=
q
(
t
)
y\left( t \right) =q\left( t \right)
y(t)=q(t) , i.e. , want
q
(
t
)
q\left( t \right)
q(t) to track a given
q
d
(
t
)
q_{\mathrm{d}}\left( t \right)
qd?(t) joint reference
Task-space :
y
(
t
)
=
[
T
(
q
(
t
)
)
]
∈
S
E
(
3
)
y\left( t \right) =\left[ T\left( q\left( t \right) \right) \right] \in SE\left( 3 \right)
y(t)=[T(q(t))]∈SE(3) denotes end-effector pose/configuration, we want
y
(
t
)
y\left( t \right)
y(t) to track
y
d
(
t
)
y_{\mathrm{d}}\left( t \right)
yd?(t)
Actuation models:
Velocity source :
u
=
q
˙
u=\dot{q}
u=q˙? —— directly control velocity
Acceleration sources :
u
=
q
¨
u=\ddot{q}
u=q¨? —— directly control acceleration
Torque sources :
u
=
τ
u=\tau
u=τ —— directly control torque
Acutation model make sense if for ant given
u
u
u , the joint velocity
q
˙
\dot{q}
q˙? can immediatly reach
u
u
u
Motion Control Problem
Design u u u to set y y y track desired reference y d y_{\mathrm{d}} yd?
- Depending on our assumption on u / y u/y u/y
output y y y —— 6大基本问题
y ? q ∈ R n y\leftrightarrow q\in \mathbb{R} ^n y?q∈Rn - joint variable : Joint space motion control (Velocity-resolved Joint-space control ; Acceleration-resolved Joint-space control ; Torque-resolved Joint-space control ; )
y ? [ T ( q ) ] ∈ S E ( 3 ) y\leftrightarrow \left[ T\left( q \right) \right] \in SE\left( 3 \right) y?[T(q)]∈SE(3) or y = f ( q ) y=f\left( q \right) y=f(q) - task space variable - e.g. origin of end-effector frame : Task space motion control (Velocity-resolved Task-space ; Acceleration-resolved Task-space ; Torque-resolved Task-space ; )
Linear control / feedback lineariazation
Each joints’ velocity q ˙ i \dot{q}_{\mathrm{i}} q˙?i? can be directly controlled
Good approximation for hydraulic actuators
Common approxiamtion of the outer-loop control for the Inner / outer loop control setup
Joint-space ‘dynamics’ : single integrator q ˙ = u \dot{q}=u q˙?=u
Joint-space tracking becomes standard linear tracking control problem : u = q ˙ d + K 0 q ¨ ? q ~ ˙ + K 0 q ¨ = 0 u=\dot{q}_{\mathrm{d}}+K_0\ddot{q}\Rightarrow \dot{\tilde{q}}+K_0\ddot{q}=0 u=q˙?d?+K0?q¨??q~?˙?+K0?q¨?=0 , where q ~ = q d ? q \tilde{q}=q_{\mathrm{d}}-q q~?=qd??q is the joint position error. —— stable if e i g ( ? K 0 ) ∈ O L H P eig\left( -K_0 \right) \in OLHP eig(?K0?)∈OLHP
The error dynamic is stable if ? K 0 -K_0 ?K0? is Hurwitz
For task space control , y = [ T ( q ) ] y=\left[ T\left( q \right) \right] y=[T(q)] needs to track y d y_{\mathrm{d}} yd? , y y y can be ant function of q q q, in particular , it can represents position and/or the end-effector frame
Taking derivatives of
y
y
y , and letting
u
=
q
˙
u=\dot{q}
u=q˙? , we have :
y
˙
=
J
a
(
q
)
u
\dot{y}=J_{\mathrm{a}}\left( q \right) u
y˙?=Ja?(q)u
Note that
q
q
q is function of
y
y
y through inverse kinematics (
q
=
I
K
(
y
)
q=IK\left( y \right)
q=IK(y))
So the above dynamics can be written in terms of
y
y
y and
u
u
u only. The detailed form can be quite complex in general
y
˙
=
J
a
(
I
K
(
y
)
)
u
\dot{y}=J_{\mathrm{a}}\left( IK\left( y \right) \right) u
y˙?=Ja?(IK(y))u
We can design outer-loop controller as if we can directly control
y
˙
\dot{y}
y˙?
y
˙
=
v
y
=
y
˙
d
+
K
(
y
d
?
y
)
?
p
l
u
g
??
i
n
??
y
˙
=
v
y
??
y
~
˙
=
?
K
y
~
\dot{y}=v_{\mathrm{y}}=\dot{y}_{\mathrm{d}}+K\left( y_{\mathrm{d}}-y \right) \overset{plug\,\,in\,\,\dot{y}=v_{\mathrm{y}}\,\,}{\Longrightarrow}\dot{\tilde{y}}=-K\tilde{y}
y˙?=vy?=y˙?d?+K(yd??y)?pluginy˙?=vy??y~?˙?=?Ky~?
We can select
K
K
K such that
?
K
-K
?K is Hurtwiz , object of inner loop : determine
u
=
q
˙
u=\dot{q}
u=q˙? such that
y
˙
≈
v
y
\dot{y}\approx v_{\mathrm{y}}
y˙?≈vy?
System(2) is nonlinear system , a commeon way is to break it into inner-outer loop , where the outer loop directly control velocity of y y y, and the inner loop tries to find u u u to generate desired task space velocity
Outer loop : y ˙ = v y \dot{y}=v_{\mathrm{y}} y˙?=vy? , where control v y = y ˙ d + K 0 y ~ v_{\mathrm{y}}=\dot{y}_{\mathrm{d}}+K_0\tilde{y} vy?=y˙?d?+K0?y~? , resulting in task-space closed-loop error dynamics: y ~ ˙ + K 0 y ~ = 0 \dot{\tilde{y}}+K_0\tilde{y}=0 y~?˙?+K0?y~?=0
Above task space tracking relies on a fictitious control v y v_{\mathrm{y}} vy? , i.e. , it assumes y ˙ \dot{y} y˙? can be arbitrarily controlled by selecting appropriate u = q ˙ u=\dot{q} u=q˙? , which is true if J a J_{\mathrm{a}} Ja? is full-row rank
Inner loop : Given
v
y
v_{\mathrm{y}}
vy? from the outer loop, find the joint velocity control by solving
{
min
?
u
∥
v
y
?
J
a
(
q
)
u
∥
2
+
r
e
g
u
l
a
r
i
z
a
t
i
o
n
??
t
e
r
m
s
u
b
j
.
t
o
??
:
C
o
n
s
t
r
a
i
n
t
s
??
o
n
??
u
??
,
e
.
g
.
{
q
˙
min
?
?
u
?
q
˙
max
?
q
min
?
?
q
+
u
Δ
t
?
q
max
?
\begin{cases} \min _{\mathrm{u}}\left\| v_{\mathrm{y}}-J_{\mathrm{a}}\left( q \right) u \right\| ^2+regularization\,\,term\\ subj.to\,\,: Constraints\,\,on\,\,u\,\,, e.g.\begin{cases} \dot{q}_{\min}\leqslant u\leqslant \dot{q}_{\max}\\ q_{\min}\leqslant q+u\varDelta t\leqslant q_{\max}\\ \end{cases}\\ \end{cases}
?
?
??minu?∥vy??Ja?(q)u∥2+regularizationtermsubj.to:Constraintsonu,e.g.{q˙?min??u?q˙?max?qmin??q+uΔt?qmax???
Inner-loop is essentially a differential IK controller
One can also use the pseudo-inverse control
u
=
J
a
?
v
y
u={J_{\mathrm{a}}}^{\dagger}v_{\mathrm{y}}
u=Ja??vy?
Joint acceleration cna be directly controlled , resulting in double-integrator dynamics q ¨ = u \ddot{q}=u q¨?=u . Given q d q_{\mathrm{d}} qd? reference , we want q → q d q\rightarrow q_{\mathrm{d}} q→qd? (double integartor)
Joint-space tracking becomes standard linear tracking control problem for double-integrator system:
u
=
q
¨
d
+
K
1
q
~
˙
+
K
0
q
~
=
0
,
q
~
∈
R
n
u=\ddot{q}_{\mathrm{d}}+K_1\dot{\tilde{q}}+K_0\tilde{q}=0,\tilde{q}\in \mathbb{R} ^n
u=q¨?d?+K1?q~?˙?+K0?q~?=0,q~?∈Rn
—— PD control , closed-loop system , where
q
~
=
q
d
?
q
\tilde{q}=q_{\mathrm{d}}-q
q~?=qd??q is the joint position error.
Stablility condition : Let
x
=
[
q
~
q
~
˙
]
∈
R
2
n
x=\left[ \begin{array}{c} \tilde{q}\\ \dot{\tilde{q}}\\ \end{array} \right] \in \mathbb{R} ^{2n}
x=[q~?q~?˙??]∈R2n ,
[
0
E
?
K
0
?
K
1
]
[
q
~
q
~
˙
]
,
x
˙
=
A
x
\left[ \begin{matrix} 0& E\\ -K_0& -K_1\\ \end{matrix} \right] \left[ \begin{array}{c} \tilde{q}\\ \dot{\tilde{q}}\\ \end{array} \right] ,\dot{x}=Ax
[0?K0??E?K1??][q~?q~?˙??],x˙=Ax
closed-loop system is stable . if
e
i
g
(
A
)
∈
O
L
H
P
eig\left( A \right) \in OLHP
eig(A)∈OLHP or
A
A
A is Hurwitz
For task space control ,
y
=
[
T
(
q
)
]
∈
S
E
(
3
)
y=\left[ T\left( q \right) \right] \in SE\left( 3 \right)
y=[T(q)]∈SE(3) needs to track
y
d
y_{\mathrm{d}}
yd?
Note : For
y
=
f
(
q
)
y=f\left( q \right)
y=f(q)
y
˙
=
J
a
(
q
)
q
˙
\dot{y}=J_{\mathrm{a}}\left( q \right) \dot{q}
y˙?=Ja?(q)q˙? and
y
¨
=
J
˙
a
(
q
)
q
˙
+
J
a
(
q
)
q
¨
?
y
¨
=
J
˙
a
(
q
)
q
˙
+
J
a
(
q
)
u
?
\ddot{y}=\dot{J}_{\mathrm{a}}\left( q \right) \dot{q}+J_{\mathrm{a}}\left( q \right) \ddot{q}\Rightarrow \ddot{y}=\dot{J}_{\mathrm{a}}\left( q \right) \dot{q}+J_{\mathrm{a}}\left( q \right) u\Leftarrow
y¨?=J˙a?(q)q˙?+Ja?(q)q¨??y¨?=J˙a?(q)q˙?+Ja?(q)u? nonlinear dynamics
Following the same inner-outer loop strategy deiscussed before . Introduce virtual control , a y a_{\mathrm{y}} ay? such that y ¨ = a y \ddot{y}=a_{\mathrm{y}} y¨?=ay? , we can design controller for a y a_{\mathrm{y}} ay? to let y → y d y\rightarrow y_{\mathrm{d}} y→yd?
Outer-loop dynamics :
y
¨
=
a
y
\ddot{y}=a_{\mathrm{y}}
y¨?=ay? , with
a
y
a_{\mathrm{y}}
ay? being the outer-loop control input
a
y
=
y
¨
d
+
K
1
y
~
˙
+
K
0
y
~
?
y
~
¨
+
K
1
y
~
˙
+
K
0
y
~
=
0
a_{\mathrm{y}}=\ddot{y}_{\mathrm{d}}+K_1\dot{\tilde{y}}+K_0\tilde{y}\Rightarrow \ddot{\tilde{y}}+K_1\dot{\tilde{y}}+K_0\tilde{y}=0
ay?=y¨?d?+K1?y~?˙?+K0?y~??y~?¨?+K1?y~?˙?+K0?y~?=0
—— PD control , stable if
[
0
E
?
K
0
?
K
1
]
\left[ \begin{matrix} 0& E\\ -K_0& -K_1\\ \end{matrix} \right]
[0?K0??E?K1??] Hurwitz
Inner-loop : given
a
y
a_{\mathrm{y}}
ay? from outer loop , find the “best” joint acceleration:
{
min
?
u
∥
a
y
?
J
˙
a
(
q
)
q
˙
?
J
a
(
q
)
u
∥
2
+
r
e
g
u
l
a
r
i
z
a
t
i
o
n
??
t
e
r
m
s
u
b
j
.
t
o
??
:
C
o
n
s
t
r
a
i
n
t
s
??
o
n
??
u
??
\begin{cases} \min _{\mathrm{u}}\left\| a_{\mathrm{y}}-\dot{J}_{\mathrm{a}}\left( q \right) \dot{q}-J_{\mathrm{a}}\left( q \right) u \right\| ^2+regularization\,\,term\\ subj.to\,\,: Constraints\,\,on\,\,u\,\,\\ \end{cases}
?
?
??minu?
?ay??J˙a?(q)q˙??Ja?(q)u
?2+regularizationtermsubj.to:Constraintsonu?
——
u
u
u : optimization variable ,
J
˙
a
(
q
)
,
q
˙
,
q
\dot{J}_{\mathrm{a}}\left( q \right) ,\dot{q},q
J˙a?(q),q˙?,q - known
{
A
c
c
??
:
q
¨
min
?
?
u
?
q
¨
max
?
V
e
l
??
:
q
˙
min
?
?
q
+
u
Δ
t
?
q
˙
max
?
\begin{cases} Acc\,\,: \ddot{q}_{\min}\leqslant u\leqslant \ddot{q}_{\max}\\ Vel\,\,: \dot{q}_{\min}\leqslant q+u\varDelta t\leqslant \dot{q}_{\max}\\ \end{cases}
{Acc:q¨?min??u?q¨?max?Vel:q˙?min??q+uΔt?q˙?max??
Mathematically , the above problem is the same as the Differential IK problem
At any given time , q ˙ , q \dot{q},q q˙?,q can be measured , and then y , y ˙ y,\dot{y} y,y˙? can be computed, which allows us to compute outer loop control a y a_{\mathrm{y}} ay? and inner loop control u u u
For fully actuated robot :
τ
=
M
(
q
)
q
¨
+
C
(
q
,
q
˙
)
q
˙
+
g
(
q
)
\tau =M\left( q \right) \ddot{q}+C\left( q,\dot{q} \right) \dot{q}+g\left( q \right)
τ=M(q)q¨?+C(q,q˙?)q˙?+g(q)
M
(
q
)
=
∑
J
i
T
[
I
i
]
6
×
6
J
i
∈
R
n
×
n
M\left( q \right) =\sum{{J_{\mathrm{i}}}^{\mathrm{T}}\left[ \mathcal{I} _{\mathrm{i}} \right] _{6\times 6}J_{\mathrm{i}}}\in \mathbb{R} ^{n\times n}
M(q)=∑Ji?T[Ii?]6×6?Ji?∈Rn×n
There are many valid difinitions of
C
(
q
,
q
˙
)
C\left( q,\dot{q} \right)
C(q,q˙?) , typical choice for
C
C
C include:
C
i
j
=
∑
k
1
2
(
?
M
i
j
?
q
k
+
?
M
i
k
?
q
j
?
?
M
j
k
?
q
i
)
C_{\mathrm{ij}}=\sum_k^{}{\frac{1}{2}\left( \frac{\partial M_{\mathrm{ij}}}{\partial q_{\mathrm{k}}}+\frac{\partial M_{\mathrm{ik}}}{\partial q_{\mathrm{j}}}-\frac{\partial M_{\mathrm{jk}}}{\partial q_{\mathrm{i}}} \right)}
Cij?=∑k?21?(?qk??Mij??+?qj??Mik????qi??Mjk??)
For the above defined
C
C
C , we have
M
˙
?
2
C
\dot{M}-2C
M˙?2C is skew symmetric
For all valid
C
C
C, we have
q
˙
T
[
M
˙
?
2
C
]
q
˙
=
0
\dot{q}^{\mathrm{T}}\left[ \dot{M}-2C \right] \dot{q}=0
q˙?T[M˙?2C]q˙?=0
These properties play improtant role in designing motion controller
For fully-actuated robot, we have
M
(
q
)
?
0
M\left( q \right) \succ 0
M(q)?0 and
q
¨
\ddot{q}
q¨? can be arbitrarily specified through torque control
u
=
τ
u=\tau
u=τ
q
¨
=
M
?
1
(
q
)
[
u
?
C
(
q
,
q
˙
)
q
˙
?
g
(
q
)
]
\ddot{q}=M^{-1}\left( q \right) \left[ u-C\left( q,\dot{q} \right) \dot{q}-g\left( q \right) \right]
q¨?=M?1(q)[u?C(q,q˙?)q˙??g(q)]
we know how to design controller if
u
=
q
¨
u=\ddot{q}
u=q¨?
Thus , for fully-acuated robot, torque controlled case can be reduced to the acceleration-resolved case
Outer loop:
q
¨
=
a
q
\ddot{q}=a_{\mathrm{q}}
q¨?=aq? with joint acceleration as control input
a
q
=
q
¨
+
K
1
y
~
˙
+
K
0
y
~
?
q
~
¨
+
K
1
q
~
˙
+
K
0
q
~
=
0
a_{\mathrm{q}}=\ddot{q}+K_1\dot{\tilde{y}}+K_0\tilde{y}\Rightarrow \ddot{\tilde{q}}+K_1\dot{\tilde{q}}+K_0\tilde{q}=0
aq?=q¨?+K1?y~?˙?+K0?y~??q~?¨?+K1?q~?˙?+K0?q~?=0
Inner loop : since
M
(
q
)
M\left( q \right)
M(q) is square and nonsingular , inner loop control
u
u
u can be found analytically:
u
=
M
(
q
)
(
q
¨
d
+
K
1
q
~
˙
+
K
0
q
~
)
+
C
(
q
,
q
˙
)
q
˙
+
g
(
q
)
u=M\left( q \right) \left( \ddot{q}_{\mathrm{d}}+K_1\dot{\tilde{q}}+K_0\tilde{q} \right) +C\left( q,\dot{q} \right) \dot{q}+g\left( q \right)
u=M(q)(q¨?d?+K1?q~?˙?+K0?q~?)+C(q,q˙?)q˙?+g(q)
The control law is a function of
q
,
q
˙
q,\dot{q}
q,q˙? and the reference
q
d
q_{\mathrm{d}}
qd?. It is called computed-torque control.
The control law also relies on system model
M
,
C
,
g
M,C,g
M,C,g if these model information are not accurate, the control will not perform well.
y
=
f
(
q
)
,
y
¨
=
J
˙
a
(
q
)
q
˙
+
J
a
(
q
)
M
?
1
(
u
?
C
?
g
)
y=f\left( q \right) ,\ddot{y}=\dot{J}_{\mathrm{a}}\left( q \right) \dot{q}+J_{\mathrm{a}}\left( q \right) M^{-1}\left( u-C-g \right)
y=f(q),y¨?=J˙a?(q)q˙?+Ja?(q)M?1(u?C?g)
Idea easily extends to task space :
y
˙
=
J
a
(
q
)
q
˙
\dot{y}=J_{\mathrm{a}}\left( q \right) \dot{q}
y˙?=Ja?(q)q˙? and
y
¨
=
J
˙
a
(
q
)
q
˙
+
J
a
(
q
)
q
¨
\ddot{y}=\dot{J}_{\mathrm{a}}\left( q \right) \dot{q}+J_{\mathrm{a}}\left( q \right) \ddot{q}
y¨?=J˙a?(q)q˙?+Ja?(q)q¨? ——
τ
=
u
=
τ
,
q
¨
=
M
?
1
[
u
?
C
?
g
]
\tau =u=\tau ,\ddot{q}=M^{-1}\left[ u-C-g \right]
τ=u=τ,q¨?=M?1[u?C?g]
Outer loop : y ¨ = a y \ddot{y}=a_{\mathrm{y}} y¨?=ay? and a y = y ¨ d + K 1 y ~ ˙ + K 0 y ~ a_{\mathrm{y}}=\ddot{y}_{\mathrm{d}}+K_1\dot{\tilde{y}}+K_0\tilde{y} ay?=y¨?d?+K1?y~?˙?+K0?y~?
Inner loop : sekect torque control
u
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u=τ by
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\begin{cases} \min _{\mathrm{u}}\left\| a_{\mathrm{y}}-\dot{J}_{\mathrm{a}}\left( q \right) \dot{q}-J_{\mathrm{a}}\left( q \right) M^{-1}\left( u-C\dot{q}-g \right) \right\| ^2\\ subj.to\,\,: Constraints\,\,\\ \end{cases}
?
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??minu?
?ay??J˙a?(q)q˙??Ja?(q)M?1(u?Cq˙??g)
?2subj.to:Constraints?
If
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J_{\mathrm{a}}
Ja?is invertible and we don’t impose additional torque constraints, analytical control law can be easily obtained ——
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u=\left( J_{\mathrm{a}}\left( q \right) M^{-1} \right) ^{-1}\left( a_{\mathrm{y}}-\dot{J}_{\mathrm{a}}\left( q \right) \dot{q}... \right)
u=(Ja?(q)M?1)?1(ay??J˙a?(q)q˙?...)
The computed-torque controller above is also canned inverse dynamics control
Forward dynamics : given τ \tau τ to compute q ¨ \ddot{q} q¨? —— from torque to motion
Inverse dynamics : given desired acceleration a q a_{\mathrm{q}} aq?, we inverted it to find the required control by u = M a q + C q ˙ + g u=Ma_{\mathrm{q}}+C\dot{q}+g u=Maq?+Cq˙?+g
Task space case can be viewed as inverting the task space dynamics —— Given a y a_{\mathrm{y}} ay? ( y y y task space) , find τ \tau τ such that y ¨ = a y \ddot{y}=a_{\mathrm{y}} y¨?=ay?
With recent advances in optimization , it is often preferred to do ID with quedratic program
For example, above equation can be viewed as task-space ID. We can incorporate torque contraints explicitly as follows:
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\begin{cases} \min _{\mathrm{u}}\left\| a_{\mathrm{y}}-\dot{J}_{\mathrm{a}}\left( q \right) \dot{q}-J_{\mathrm{a}}M^{-1}\left( u-C\dot{q}-g \right) \right\| ^2\\ subj.to\,\,: u_-\leqslant u\,\,\leqslant u_+\,\,\\ \end{cases}
?
?
??minu?
?ay??J˙a?(q)q˙??Ja?M?1(u?Cq˙??g)
?2subj.to:u???u?u+??
optimization variable
u
∈
R
n
u\in \mathbb{R} ^n
u∈Rn
This is equivalent to the following more popular form:
{
min
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\begin{cases} \underset{u,\ddot{q}}{\min}\left\| a_{\mathrm{y}}-\dot{J}_{\mathrm{a}}\dot{q}-J_{\mathrm{a}}\ddot{q} \right\| ^2\\ subj.to\,\,: \begin{array}{c} M\ddot{q}+C\dot{q}+g=u\\ u_-\leqslant u\in \mathbb{R} ^n\,\,\leqslant u_+\,\,\\ \end{array}\\ \end{cases}
?
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??u,q¨?min?
?ay??J˙a?q˙??Ja?q¨?
?2subj.to:Mq¨?+Cq˙?+g=uu???u∈Rn?u+???
optimization variable
u
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q
¨
∈
R
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u,\ddot{q}\in \mathbb{R} ^n
u,q¨?∈Rn