General question on making state space equations for a DC motor. In the link below, why are both the electrical and dynamics equations 3 and 4 used to make a state space eqn.8, instead of combining them together into a single? You could combine them by setting one of the equations to theta_dot, and inserting into the theta_dot the other equation.


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    \$\begingroup\$ Ahh, those are coupled differential equations. You can't just treat them like algebraic equaitons where you set theta dot equal to y and solve. (I know next to nothing about driving DC motors.) \$\endgroup\$ Commented Aug 23, 2014 at 0:28
  • \$\begingroup\$ Why can't you? In general coupled equations are left separate because it's difficult to set equal to one variable [eg one equation has sin(theta_dot_dot), another one has tan(theta_dot) ] so they're left separate. But if you can directly say "theta_dot=..." and then plug that into the second equation, it seems like it should work. \$\endgroup\$
    – J B
    Commented Aug 24, 2014 at 3:33
  • \$\begingroup\$ Hi @PB, OK I guess you can do that. I'm not sure it helps you though. I think the rotational speed is one of the parameters you are interested in, so you don't want to eliminate it. Is this for a class? (Maybe a TA or professor can help more than I can.) \$\endgroup\$ Commented Aug 24, 2014 at 14:09

2 Answers 2


You have a two state system, so (generically) you cannot replace the coupled two state system (Equation (8) in your link) by a single first order differential equation.

However, since the system is completely controllable, you can transform the system into controller canonical form and write the system as a single second order differential equation. In this case we can read off the equation from the transfer function.

You can see from Equation (7) that we have the system \$ ((J s+b)(Ls + R) + K^2) \hat{\omega} = K \hat{V}\$, where \$\hat{\omega}, \hat{V}\$ are the Laplace transforms of the rotational speed and input voltage respectively. This gives the following differential equation \$ JL \ddot{\omega}(t) + (JK+bL)\dot{\omega}(t) + (K^2+bK) \omega(t) = K v(t) \$.

It is not clear that this is more convenient that the state space form, but perhaps gives more insight into the rotational dynamics.


You could combine them by setting one of the equations to theta_dot, and inserting into the theta_dot the other equation

In differential equations you solve for \$\theta(t)\$, not for \$\dot{\theta}(t)\$, and as their values are interconnected the solution is time dependent. For linear differential equations, like the ones in the link, the solution is an exponential function.

In general you could, however, do as you say, but you would get mathematically stuck with a coupled differential equation that's quite difficult to solve. Start by fixing \$\dot{\theta}(t)\$:

$$\dot{\theta}(t) = \frac{K i(t) - J\ddot{\theta}(t)}{b} $$

and then substitute in the other one, getting:

$$ L \frac{di(t)}{dt} + Ri(t) = V - K\frac{K i - J\ddot{\theta}(t)}{b}$$

You need \$\ddot{\theta}(t)\$ to find \$i(t)\$. You can't proceed.

The most common approach to solve the interconnected system for its states' dynamics is to stack the multiple linear differential equations in a single state space (eqs. 8-9 in the link) and solve it in its matrix form.

Your intuition can however be used to find the steady-state (final values, after dynamics) of the system. This is done by setting all derivative to 0 and solving for the states' final values.

  • \$\begingroup\$ Your comment makes more sense, thanks for writing it out. Regarding your comment "You need θ¨(t) to find i(t). You can't proceed". That second equation is exactly what i was thinking of (Ldi...=V-K(.../b) ; it would have otherwise had theta_dot instead of double-dot. You'd need theta_dot then as well to find i(t), which doesn't seem simpler? \$\endgroup\$
    – J B
    Commented Sep 6, 2014 at 7:16
  • \$\begingroup\$ I see what you mean, but mind that the matrix equation, in that form, solves both equations simultaneously. They are interconnected. It's not that we look for theta_dot rather than theta_dotdot first, but write the whole problem in a different way and solve it altogether. Do you understand my point, now? \$\endgroup\$
    – raggot
    Commented Sep 9, 2014 at 10:15

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