Sideways velocity control 1D

To investigate the consequences of sideways velocity control a configuration is chosen which in fact is a one dimensional  reduction of the complete model. The gain for rotation control KrB has been chosen zero and the gain Kr for distance control has been chosen zero too (KrB=0 and Kr=0). The female is supposed to move with constant velocity VF along a straight line. The male body axis is chosen perpendicular to the female path. Therefore in this case the male is moving parallel to the path of the female, as given in the figure below.


In the figure the female (indicated by green dots at time k*T) is moving with constant velocity VF to the right. The male direction is –π/2. When the control system is stable, after  some time Kt*θefilt(k)=VF and therefore Kt* θe(k)=VF for k large.

The algorithm for this configuration with KrB=0 and Kr=0 is reduced from the 2D to the 1D case.  xM , vM and xL are real numbers now.

Define startconditions for k=1

for k=2 to Nend

xM(k)=xM(k-1)+ vM(k-1)*T


θefilt(k)=( exp(-T/τf))* θefilt(k-1)+(1- exp(-T/τf))* θe(k).

vM(k)=(exp(-T/τ))*vM(k-1)+(1- exp(-T/τ))* Kt*θefilt(k)


The stability of this control algorithm will now be investigated.

To be able to apply the linear theory, the expression θe(k)=asin((xF(k)-xM(k))/|r|)

is linearized using the approximation asin(α)=α  for small α. Using this approximation θe(k)=(xF(k)-xM(k))/|r|. The continuous block diagram is given below.


This third order system will eventually oscillate with increasing Kt. Harmonic oscillation with frequency ω rad/s and constant amplitude will occur when

atan(ω.τf)+atan(ω.τ)+π/2= π  or  atan(ω.τf)+atan(ω.τ)= π/2  (1) and


Solution of (1) is  ω2 = 1/(τ.τf).

Substitution in the second relation delivers   Kt=|r|*(1/τ+1/τf).

Creative Commons License
This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.