Control system with equation C = A*x + B*dx/dt

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 2K views
zsero
Messages
12
Reaction score
1

Homework Statement


This question came up in a computer science / robotics exam and I still don't know the solution for it. I figured out that it's classical mechanics related, so I thought this might be the best place to ask it.

Suppose a control system is described by the equation C = A*x + B*dx/dt, where B is proportional to the mass of the robot. The behaviour of the system can be characterised by the steady state (e.g. the asymptotic velocity of the robot) and the half-life time of the decrease of the distance to the steady state. Explain how the steady state and the half-life change if the mass of the robot is doubled.

Homework Equations


C = A*x + B*dx/dt

The Attempt at a Solution


I've figured out that the steady state doesn't change, as it happens when dx/dt is 0, thus B is not affecting the solution.

And this is how far I understand it. Can you explain to me, what kind of movement is this, what is the real-life meaning of the steady state and half-life for this movement and that how to calculate the change in the half-life?
 
Physics news on Phys.org
tiny-tim said:
hi zsero! :smile:

write it Bdx/dt = C - Ax, then solve it by "separating the variables" :wink:

OK, I arrive at the following equation:

-B/A*ln(x) = c*t + k1

My problem here is that I don't understand the meaning of the equations and that what is asked by half-life and steady state.

How can I get the half-time from this equation?
 
Thanks for the help! Actually, I'm not a Physics student, so I don't really know, but I'd guess it's the exponential delay. So if B doubles then the constant becomes half, thus the half-life becomes double. Is this correct?