Laplace transforms:DC Motor circuit

Problem

Find the steady state current i(t) through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). The motor has input current i(t) and output angular velocity ω(t).

Solution

The torque is proportional to the armature current.

$T(t)=ki(t)$

Similarly, relating mechanical (T(t)ω(t)) and electrical (v_m(t)i(t)) power, the conservation of energy requires the same proportionality between the voltage across the motor (v_m(t)) and the angular velocity (ω(t)).

$v_{m}(t)=k\omega (t)$

We want to find the Laplace transfer function of the motor, and we define it as follows.

$\Omega (s)={\mathcal {L}}[\omega (t)]/v_{s}(s)$

Summing the voltages around the series circuit gives us our differential equation.

$v_{s}(t)=Ri(t)+L{\frac {di(t)}{dt}}+k\omega (t)$

Take the Laplace transform.

$V_{s}(s)=RI(s)+Ls(I(s)-i(0))+k\Omega (s)$

At this point we can use the initial value theorem to find i(0).

$i(0)=\lim _{s\rightarrow \infty }s\int _{0}^{\infty }i(t)e^{-st}u(t)dt={\frac {\infty }{e^{\infty }}}=0$

Substituting i(0) into the transformed differential equation gives us Eq *1*.

$V_{s}(s)=RI(s)+LsI(s)+k\Omega (s)*1*$

Repeat the process with the analogous mechanical differential equation. Here J_m is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction. Again the initial value theorem will yield ω(0) = 0.

$T(t)=J_{m}{\frac {d\omega (t)}{dt}}+B\omega (t)$

Transforming yields the following.

$T(s)=(J_{m}s+B)\Omega (s)$