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
1. Basic Linear Control Design
1.1 Error Response
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→∞limθe(t)
Precent overshoot : P.O.
Rise time / Peak time :
Settling time : T s T_{\mathrm{s}} Ts
1.2 Standard Second-Order Systems
详细推导见 : (待补充)
1.3 Second-Order Response Characteristics
详细推导见 : (待补充)
1.4 State-Space Controller Design
- Eigenvalue assignment : Find control gain K K K such that e i g ( A − B K ) = e i g d e s i r e d eig\left( A-BK \right) =eig_{\mathrm{desired}} eig(A−BK)=eigdesired
- Solvability : We can always find such K K K if ( A , B ) \left( A,B \right) (A,B) is controllable ( r a n k ( m c ) = n rank\left( m_{\mathrm{c}} \right) =n rank(mc)=n)
- How to choose desired eigs? —— refer to 2nd-order system
specification (P.O. T s T_{\mathrm{s}} Ts T p T_{\mathrm{p}} Tp) ⇒ a r t \overset{art}{\Rightarrow} ⇒art dominant poles + other poles ⇒ \Rightarrow ⇒ e i g d e s i r e d eig_{\mathrm{desired}} eigdesired ⇒ s c i e n c e \overset{science}{\Rightarrow} ⇒science K K K
2. Motion Control Problems
2.1 Robotic Motion Control Problem
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)Fexty=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)
- Motion Control Problems : Let
y
y
y track given reference
y
d
y_{\mathrm{d}}
yd
often times q d q_{\mathrm{d}} qd is given by planner represented by polynomials , so that q ˙ d , q ¨ d \dot{q}_{\mathrm{d}},\ddot{q}_{\mathrm{d}} q˙d,q¨d can be easily obtained
2.2 Variations in Robot Motion Control
-
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
3. Motion Control with Velocity/Acceleration as Input
3.1 Velocity-Resolved Control
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
3.2.1 Velocity-Resolved Joint Space Control
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+K0q¨⇒q~˙+K0q¨=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
3.2.2 Velocity-Resolved Task Space Control
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
- Let v y v_{\mathrm{y}} vy be virtual control y ˙ = v y \dot{y}=v_{\mathrm{y}} y˙=vy design v y v_{\mathrm{y}} vy to track y d y_{\mathrm{d}} yd (same as above)
- Find actual control u u u such that J a ( I K ( y ) ) u ≈ v y J_{\mathrm{a}}\left( IK\left( y \right) \right) u\approx v_{\mathrm{y}} Ja(IK(y))u≈vy
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˙=vyy~˙=−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+K0y~ , resulting in task-space closed-loop error dynamics: y ~ ˙ + K 0 y ~ = 0 \dot{\tilde{y}}+K_0\tilde{y}=0 y~˙+K0y~=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˙maxqmin⩽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
3.2 Acceleration-Resolved Control
3.2.1 Acceleration-Resolved Control in Joint Space
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+K1q~˙+K0q~=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−K0E−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
3.2.2 Acceleration-Resolved Control in Task Space
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+K1y~˙+K0y~⇒y~¨+K1y~˙+K0y~=0
—— PD control , stable if
[
0
E
−
K
0
−
K
1
]
\left[ \begin{matrix} 0& E\\ -K_0& -K_1\\ \end{matrix} \right]
[0−K0E−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¨maxVel: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
4. Motion Control with Torque as Input and Task Space Inverse Dynamics
4.1 Recall Properties of Robot Dynamics
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)=∑JiT[Ii]6×6Ji∈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=∑k21(∂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
4.2 Computed Torque Control
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¨+K1y~˙+K0y~⇒q~¨+K1q~˙+K0q~=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+K1q~˙+K0q~)+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+K1y~˙+K0y~
Inner loop : sekect torque control
u
=
τ
u=\tau
u=τ by
{
min
u
∥
a
y
−
J
˙
a
(
q
)
q
˙
−
J
a
(
q
)
M
−
1
(
u
−
C
q
˙
−
g
)
∥
2
s
u
b
j
.
t
o
:
C
o
n
s
t
r
a
i
n
t
s
\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}
⎩
⎨
⎧minu
ay−J˙a(q)q˙−Ja(q)M−1(u−Cq˙−g)
2subj.to:Constraints
If
J
a
J_{\mathrm{a}}
Jais invertible and we don’t impose additional torque constraints, analytical control law can be easily obtained ——
u
=
(
J
a
(
q
)
M
−
1
)
−
1
(
a
y
−
J
˙
a
(
q
)
q
˙
.
.
.
)
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˙...)
4.3 Inverse Dynamics Control
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:
{
min
u
∥
a
y
−
J
˙
a
(
q
)
q
˙
−
J
a
M
−
1
(
u
−
C
q
˙
−
g
)
∥
2
s
u
b
j
.
t
o
:
u
−
⩽
u
⩽
u
+
\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˙−JaM−1(u−Cq˙−g)
2subj.to:u−⩽u⩽u+
optimization variable
u
∈
R
n
u\in \mathbb{R} ^n
u∈Rn文章来源:https://www.toymoban.com/news/detail-773233.html
This is equivalent to the following more popular form:
{
min
u
,
q
¨
∥
a
y
−
J
˙
a
q
˙
−
J
a
q
¨
∥
2
s
u
b
j
.
t
o
:
M
q
¨
+
C
q
˙
+
g
=
u
u
−
⩽
u
∈
R
n
⩽
u
+
\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}
⎩
⎨
⎧u,q¨min
ay−J˙aq˙−Jaq¨
2subj.to:Mq¨+Cq˙+g=uu−⩽u∈Rn⩽u+
optimization variable
u
,
q
¨
∈
R
n
u,\ddot{q}\in \mathbb{R} ^n
u,q¨∈Rn文章来源地址https://www.toymoban.com/news/detail-773233.html
到了这里,关于[足式机器人]Part4 南科大高等机器人控制课 CH12 Robotic Motion Control的文章就介绍完了。如果您还想了解更多内容,请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章,希望大家以后多多支持TOY模板网!