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.

$\mathrm{\Omega }(s)={ℒ}[\omega (t)]/v_s(s)$

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.

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

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

$i(0)=\underset{s\to \mathrm{\infty }}{\lim }s\underset{0}{\overset{\mathrm{\infty }}{\int }}i(t)e^{-st}u(t)dt=\frac{\mathrm{\infty }}{e^{\mathrm{\infty }}}=0$

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

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

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.

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