Discrete-time weight updates in neural-adaptive control

Abstract

Typical neural-adaptive control approaches update neural-network weights as though they were adaptive parameters in a continuous-time adaptive control. However, requiring fast digital rates usually restricts the size of the neural network. In this paper we analyze a delta-rule update for the weights, applied at a relatively slow digital rate. We show that digital weight update causes the neural network to estimate a discrete-time model of the system, assuming that state feedback is still applied in continuous time. A Lyapunov analysis shows uniformly ultimately bounded signals. Furthermore, slowing the update frequency and using the extra computational time to increase the size/accuracy of the neural network results in better performance. Experimental results achieving link tracking of a two-link flexible-joint robot verify the improved performance.

Appendix: discrete-time transformation

We are interested in the discrete-time transformation of a continuous nonlinear system given by

$$ \dot{{\bf z}} = {\bf A}(\varvec{\theta}) {\bf z} + {\bf H}(\varvec{\theta},\dot{\varvec{\theta}},\varvec{\theta}_d,\dot{\varvec{\theta}}_d,\ddot{\varvec{\theta}}_d) + {\bf B}(\varvec{\theta}) {\bf u}_k $$

(82)

We want to integrate the expression from time t = t ₁ to t = t ₂, where t ₂ − t ₁ = T. Let us denote terms like ${\bf A}({\varvec{\theta}}(t))$ simply as A(t) for ease of presentation. Now in anticipation of handling the time-varying matrix A(t), let us rewrite the equation as

$$ \dot{{\bf z}} = {\bf A}(t_1) {\bf z} + [{\bf A}(t)-{\bf A}(t_1)] {\bf z} + {\bf H}(t) + {\bf B}(t) {\bf u}_k $$

(83)

Premultiply by $\exp(-{\bf A}(t_1) t),$ and let us define $\Updelta {\bf A} = {\bf A}(t)-{\bf A}(t_1);$ then

$$ e^{-{\bf A}(t_1) t} \dot{{\bf z}} - e^{-{\bf A}(t_1) t} {\bf A}(t_1) {\bf z} =e^{-{\bf A}(t_1) t}[ \Updelta {\bf A} {\bf z} + {\bf H}(t) + {\bf B}(t) {\bf u}_k ] $$

(84)

$$ \frac{d e^{-{\bf A}(t_1) t} {\bf z}}{dt} = e^{-{\bf A}(t_1) t}[ \Updelta {\bf A} {\bf z} + {\bf H}(t) + {\bf B}(t) {\bf u}_k ] $$

(85)

Let us integrate the left-hand side of (85).

$$ \int\limits_{t_1}^{t_2} \frac{d e^{-{\bf A}(t_1) \tau} {\bf z}}{d\tau} d\tau = e^{-{\bf A}(t_1) t} {\bf z}(t) \Bigg\vert_{t_1}^{t_2} \\ = e^{-{\bf A}(t_1) t_2} {\bf z}(t_2) - e^{-{\bf A}(t_1) t_1} {\bf z}(t_1) $$

By integrating and multiplying both sides by $\exp({\bf A}(t_1) t_2)$ we have

$$ {\bf z}(t_2) - e^{{\bf A}(t_1)(t_2- t_1)} {\bf z}(t_1) = \int\limits_{t_1}^{t_2} e^{{\bf A}(t_1)(t_2- \tau)} [ \Updelta {\bf A}(\tau) {\bf z} + {\bf H}(\tau) + {\bf B}(\tau) {\bf u}_k ] d \tau $$

(86)

Choose t ₁ = kT, t ₂ = (k + 1)T, and denote z _k  = z(kT), A _k  = A(kT), and B _k  = B(kT)

$$ {\bf z}_{k+1} = e^{T {\bf A}_k} {\bf z}_k + \int\limits_{kT}^{(k+1)T} e^{{\bf A}_k((k+1)T- \tau)}[ \Updelta {\bf A}(\tau) {\bf z} + {\bf H}(\tau)+ {\bf B}(\tau)] d \tau $$

(87)

By defining $\Updelta {\bf B}(t) = {\bf B}(t) - {\bf B}_{k},$ then we can write

$$ {\bf z}_{k+1} = {\bf a}_k {\bf z}_k + {\bf b}_k {\bf u}_k + \int\limits_{kT}^{(k+1)T} e^{{\bf A}_k((k+1)T- \tau)} [ \Updelta {\bf A}(\tau) {\bf z} + {\bf H}(\tau) + \Updelta {\bf B}(\tau) {\bf u}_k] d \tau $$

(88)

where (Fig. 8)

$$ {\bf a}_k= e^{T {\bf A}_k} $$

(89)

$$ {\bf b}_k = \int\limits_{kT}^{(k+1)T} e^{{\bf A}_k((k+1)T- \tau)}d \tau {\bf B}_k $$

(90)