Biological Cybernetics

Biol. Cybern. 64, 87-94 (1990)

9 Springcr-Verlag 1990

Optimal task performance of antagonistic muscles* M. N. Oi~uzt6reli 1 and R. B. Stein 2 Department of Mathematics and 2 Division of Neuroscience, University of Alberta, Edmonton, Alberta, Canada Received April 6, 1990/Accepted in revised form July 4, 1990

Abstract. Movements against a variety of loads are

relatively invariant in form. These movements are controlled in general by antagonistic groups of muscles. In this paper optimal control strategies are computed for coupling antagonistic muscles so as to minimize deviations from a desired trajectory. Simulations are presented for linear and nonlinear "decision functions" linking control of the two muscles for a variety of movements in a way that may be compared with experimental observations.

I Introduction

Because of the simple leverage of many joints, muscles tend to be arrayed in groups which oppose each other's actions in that they pull the limb segment in opposite directions at the joint. The control of such antagonistic muscles or groups of muscles so as to optimize a movement is still not clear. In rapidly moving an inertial load, a rather stereotyped triphasic pattern of activity has been described (HaUett et al. 1975) in which the prime mover (agonist) and the muscle opposing its action (antagonist) alternate their activity in time (agonist, antagonist, agonist). Hogan and his colleagues have argued that movements about a single joint (Hogan 1984) or multiple joints (Hogan and Flash 1987) are carried out in such a way so that a kinematic variable, jerk, which is the derivative of acceleration is minimized in a least mean squares sense. Although movements against a variety of loads tend to show a smooth acceleration to a single peak velocity and a similar smooth deceleration, a variety of mathematical expressions fit the time course as well as predictions from the minimum jerk hypothesis (Stein et al. 1988). * This work was partly supported by the Natural Sciences and Engineering Research Council of Canada Grant NSERC OGP-4345 and by the Medical Research Council of Canada Grant MRC PG-47 through the University of Alberta

Nonetheless, under a variety of inertial, viscous and elastic loads, the movement showed approximately the same form, suggesting that the central nervous system adjusted its control strategy to produce a similar pattern of movement. Under isometric conditions (when muscles attempt to contract against an immovable object) either muscle of an antagonist pair can be activated singly at a variety of levels up to its maximal value or the two muscles can be co-contracted at submaximal levels (Smith 1981; DeLuca and Mambrito 1987). Although muscles have complex, non-linear properties which are still not completely understood (e.g., Bobet et al. 1990); Weiss et al. (1988) showed that up to the maximum levels of activation of antagonistic muscles the mechanical properties could he well approximated by a linear system with a constant inertia and viscosity and stiffness which depended on the level of activity in the muscle groups. In this paper we will extend previous work on single muscles (Stein et al. 1986) or antagonistic groups of muscles (O~uzt6reli and Stein 1982, 1983). We consider the optimal performance of tasks by antagonistic muscles in the sense that they follow with minimum error a prescribed trajectory. To keep the mathematics tractable we will assume that the two antagonistic muscles can he approximated by linear elements and that the control strategies for the antagonistic muscles are linked by simple linear or nonlinear functions. II Mathematical description of the basic model

The basic model for a pair of antagonistic muscles is shown in Fig. 1 (cf. O~uztrreli and Stein 1982). Some of the notation is indicated in the figure and further details are listed below: t, time (s); j, index for the muscles, j = 1, 2; aj(t), active state of t h e j t h muscle at time t; f(t), external force at time t; u(t), uj(t), deviations from the equilibrium positions at time t; Bj, viscosity of the active element of the j t h muscle; D, viscosity or external damping of the load; Ke, stiffness of the external elastic elements in the load; K;j, stiffness

88 and a22 = - D / M ,

a21 = - ( K e + Kil + K,2)/M,

a24 = K J M ,

a2a = K , / M , a3~ = K , , / B ~ ,

-.

[

+ K~O/BI ,

(II.5)

'," I

ut(t ~ -

a33 = - ( K ,

....

i ......

L

I

t

9.

a., = K,~IB2,

/

b3, =

t

~ Kil

M2SC'E• J

. . . . . . . . .

1/Bl,

a.. = -(K,2 + Kp2)IS2, b42 =

1/B2,

the system (II.3) can be written in the form -~1 = X 2 ,

Yc2 = a21x I + a22x2 + a23x3 + a24x4 + ~b,

(II.6) ~ Ki2

Y2 It)

"~3 ~ a31xl "[- a33x3 ~ b31Vl , 3r4 = a41x I + R44X2 - - b42v 2 ,

for t > O, subject to the initial condition ~(0) = 0 _

(Xk(O) = 0,

......................................

Fig. I. Basic model for a pair o f antagonistic muscles. Each muscle consists o f an active state element producing a force a(t), as well as a dashpot with viscosity B and parallel and internal series springs with stiffnes Kr and K~ respectively. The muscles work against a generalized load consisting o f a mass M, a dashpot with viscosity D and external spring with stiffness K~. The u(t) measure distances from steady-state positions

o f elastic elements internal to the j t h muscle; Kpj, stiffness o f elastic elements in parallel with the active state element o f the j t h muscle; M, mass. We assume that external and internal forces are applied only for positive values of t: (t < 0),

(II.1)

and that the system is in the resting position for t ~< O: u(t) = u,(t) = u2(t )

=

0,

fi(t) ~---ul(t) = gl2(t) : / l ( t ) ~.0

(t ~I 0. Putting

a l = Vl,

=f/M,

a2 -~"/.)2,

Id2=X4,

V = (vt, v2),

X ~ - ( X I , X 2 , X3, X4), (II.4)

My,

j = 1, 2},

(II.8)

will be designated as the control region, where M t and M2 are two positive constants. A control v = v(t) will be called admissible if v(t) E [1 and is piecewise continuous for 0 ~< t ~< T. The set of all admissible controls will be denoted by ~/r: Y" = {v(t) [ v = v(t) is admissible}.

(II.9)

When a pair of antagonistic muscles are engaged to perform a preassigned task, the active states aj = vj ( j = 1, 2) are coordinated in such a way that the resulting movement o f the system becomes as close as possible to the trajectory of the task. This coordination is achieved by a special decision process in the brain which correlates the strength o f one o f the active states when the strength o f the other active state has been determined. This decision process will be represented by an equation of the form v2 = ~(Vl)

= (K~ + K,, + K,=)u(O + D~(t) + Mii(t)

U-.~-XI, 1.l..-~--X2, U I = X 3 ,

(II.7)

For each given v = v(t) - (Vl (t), v2(t)), pieccwise continuous for 0 ~ t ~ T, the dynamical system (II.6) has a unique solution x = x(t) -= (xl(t), x2(t), x3(t), x4(t), ), 0 ~< t ~ T, which satisfies the initial condition (II.7). As is customary, i = ~(t) and u = u(t) will be called the state and control variables, respectively. Considering the limitations in the actions o f the active states, a j - vj ( j = 1, 2), the set ~'~ -m--{(Vl, V2) [ 0 ~ Vj ~

a, (t) = a2(t) = f ( t ) = 0

k = 1, 2, 3, 4 ) .

2__ j

.......................

.......................

~.-~_~-.- _ ~

uus_c~

(or

Vl = ~(v2)).

(II.10)

Note that these decision functions are single-valued. The objective of the present paper is the construction o f an admissible control under which the antagonistic muscles with a decision function o f the form (II.10) will perform a preassigned task in the best way in some sense. This optimal problem will be formulated and solved in the next sections.

89

We can easily show that

III Optimization problem

Let the pair of antagonistic muscles with a decision function of the form (II.10), considered in the previous section, be assigned to a task of movement by the central nervous system or by an external mechanism. Let the task be described by a pattern function p = p(t), continuous for 0 < t ~< T, and p(0) = 0. The task is to make x~ = xt(t) as close as possible to the function p = p(t) for 0 ~< t ~< T, and such that

x1,~+ 3 ~ C~x,,. + C2x2,~ + C3x3,n + C , x , , . + Cs, n

x, ( T) = p( T) .

(74 = h2(3 + ha22 + ha44)a24,

(III.1)

Recall that x l ( 0 ) = p ( 0 ) . We measure the closeness of Xl(t), under an admissible control v = v ( t ) = ( v l ( t ) , v2(t)) (v2 = D(v0), to p(t) by the performance index

(IV.6)

"~ C6/)1, n - - C 7 u 2 , n , where

C1 = 1 + 3h2a21 + h3(a21a22 + a23a31 + a24a41), C 2

=

h(3 + 3ha22 + h2a22 + h2a2l),

C 3 = h2(3 + ha22 -~- ha33)a23 , (W.7)

C5, ~ = h2[~b~+ i + $n(2 + ha22)], C6 = h3a23b31, C7 = h3a24b42.

T

~(v) = ~ [x,(t) - p ( t ) l 2 dt (v ~ : ) . (III.2) 0 Then, our optimization problem can be formulated as follows: Find an admissible control v* = v*(t) with v* = ~ ( v * ) in such a way that the functional ~(v) will assume its minimal value:

re_in~(v) = ~(v*)

(v* r Y').

(III.3)

v e "g"

Putting Yn = C l x l , . +

C2X2,

n "Jr- C 3 x 3 ,

n ,dr C 4 x 4 ,

n

(IV.8)

"~- C 5 , n - - P n + 3 "

Equation (IV.6) can be written in the form Xl, n+ 3 -- Pn+ 3 ~ Y~ + C6vi,.

--

CTU2,

(IV.9)

n "

We now write the performance index ~(v) as T

The solution of this optimization problem will be discussed in the next section using the techniques of the discrete dynamic programming (Bellman 1957; Dreyfus and Law 1977).

~(v) = ~ w(t) d t ,

(IV.lO)

0

where

OV.ll)

w(t) = [x I (t) -- p(t)] 2. Then, with to---O, 3N-- 1 tn+ I

IV Solution of the optimization problem

~(v) =

Let N be a sufficiently large integer. Put

h=T/3N,

tn=nh

IV.l)

(n=0,1,....,3N).

2

1,

3 N - - 3 tn

+ E

(IV.3)

,

and, according to (II.6), we have Xl,n

"~-

hxz, n

+

3

~ W(t) d t ,

n ~ 0

~ 1"~, (IV.2)

is an admissible control: v(t) ~ ~ . For large N O ~ (0n+ I -- On)/h

w(t) dt + I w(t) d t ,

tn

-- 1

f3N

Vn

w(t) d t ,

tl

(v, (0 = v~, n, V d 0 = V~,.),

tn ~< t ~< t.+

Xl, n+ 1 ~

I tn

=

We will denote the value of a function 0 = 0(t) at t = tn by On: 0n = O(tn). Let vn = (vl.., v2,,) be the control applied at t = t~. Then v(0 = vn

~ n~O

I

t3N

w(t) d t + 2

t 3 N -- 2

1 3~1 ~ ( v ) = g n - o rn(x., vn),

t2

rn(xn, v.)

ha33)x3.n + hb3~vl,~ ,

tn~ 3

w(t) dt

-

(n = 2 . . . . .

3 N - - 3), (IV.13)

tn

x4, n + ~ ~ ha41 xt, ~ + (I + ha44)x4, n - hb42v2, ~ ,

t 3 N -- 1

and, by (II.7), (III.l) and p(0)= 0, (k = 1, . . . . . 4), Po = 0,

tl

tl

(IV.4)

0

0 t4

r,(x,, v,) -- I w(t) + I w(t) d t ,

+ ha24x4, n + hdpn,

Xk, 0

(IV.12)

ro(Xo, Vo) - 2 ~ w(t) dt + ~ w(t) d t ,

,

(I +

,

t 3 N -- 1

0

ha31xL n +

"~

w(t) dt

where

x2,,,+ I ~ ha21xL n + (1 + ha22)x2, ~ + ha23x3, n

x3..+ 1 ~

I

I

r 3 N - - 2 ( X 3 N - - 2, V3N-- 2) "~-

w(0 d t ,

t3N --2 t3N

Xl,3N = P 3 N " (rV.5)

r3N_ l ( x 3 N - l, v3N- t) - 2

}" w(t) d t . t3N -- 1

90 We put (IV.14)

r3N(X3N, V3N) = 0 .

Then, applying the rectangular rule to the integrals involved, we find ro(Xo, Vo) ~ h[2(xi, o - - p o ) 2 + 3(Xl, 3 - - p 3 ) 2] (IV.15)

3h(yo + C 6 V l . 0 - C7v2, o) 2 ,

We must now consider the functional Eq. (IV.21) for the remaining cases n = 3N - 3, 3N - 4 . . . . . 1, 0. Since R. + ~(x. + 1) does not depend explicitly on %, the minimization operation in (IV.21) leads us to the minimization of r . ( x . , v.) with respect to v. = (vl.., v2..). The minimum is achieved when the partial derivatives of r . ( x . , v.) with respect to vL. and rE,. vanish simultaneously: d r . ( x . , v~)

rl(Xl,Vl)

~ h{(Xl, l--Pl)

,~ h{(x~.l

_p~)2 +

2+3(xl,4-p4)2]}

3(yl + C6Vl. i

OVl, ,,

(IV.16)

- - C71)2, 1 ) 2 } ,

a r . ( x . , v,,)

- 2C6(y. + C r v L ,, -- C7v2, ,,) = O, --

2C7(Yn + C6Vl,. - C702, n) -'--"O.

~V2, n

3h(xLn+3--Pn+3)2

r . ( x . , v.) ~

Since C6 and C7 are positive quantities, we find (IV.17)

3h(y,, + C 6 v l , . - C7v2,.)2 ,

y . + CrvI,~ - C71)2, n = 0 ,

or, after solving with respect to v2, n,

(n = 2 , . . . , 3 N - 3),

V2,. = C 7 1 [ y . + C6Vl,.] - A e . ( v I . . ) . r3N_2(X3N_2,

V3N_2),'~h(xI,3N_2--P3N_2)

r3N--I(X3N--1, V3N--1)~2h(Xl,3N-I--P3N--1)

2,

(IV.18)

2" (IV.19)

We now introduce 3N-- 1

R.(x.) =

min

~

{Vn ..... V3N -- I}

k=n

rk(xk,vk),

k =0, I..... 3 N -

(Vk ~ ,

(IV.20)

(n = 0 , 1. . . . .

3N-

(IV.21)

R3N(X3N) = 0 .

(IV.22)

For n = 3N -- 1 we have min

V3N -- 1 E

min

V3N -- 1 E

min

V3N _ 2 ~ ~'~

0 if Vk, n < O " if 0