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). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, J_m = 0.01 kg*m², and B = e^-4N*m*s.

Solution

Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

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 Equation *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.

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

Transforming yields the following.

$T(s)=J_{m}s(\mathrm{\Omega }(s)-\omega (0))+B\mathrm{\Omega }(s)$

Recall T(t) = k i(t), and so T(s) = k I(s). Again the initial value theorem will yield ω(0) = 0. This gives us Equation *2*.

$kI(s)=(J_{m}s+B)\mathrm{\Omega }(s)*2*$

Solve Equation *2* for I(s) and substitute that into Equation *1*.

$V_s(s)=(R+Ls)\frac{(J_{m}s+B)}{k}\mathrm{\Omega }(s)+k\mathrm{\Omega }(s)$

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

$\mathrm{\Omega }(s)=\frac{\frac{k}{J_{m}L}}{s^2+(\frac{B}{J_m}+\frac{R}{L})s+\frac{RB+k^2}{J_{m}L}}V(s)*3*$

Now in order to apply the final value theorem we let V_s(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

${\omega }_{ss}=\underset{s\to 0}{\lim }s\mathrm{\Omega }(s)=\frac{k}{RB+k^2}K=\omega (\mathrm{\infty })$

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

$I(s)=\frac{(J_{m}s+B)}{k}\left[\frac{\frac{k}{J_{m}L}}{s(s^2+(\frac{B}{J_m}+\frac{R}{L})s+\frac{RB+k^2}{J_{m}L})}\right]K=\frac{\frac{s}{L}+\frac{B}{J_{m}L}}{s(s^2+(\frac{B}{J_m}+\frac{R}{L})s+\frac{RB+k^2}{J_{m}L})}K$

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

$i_{ss}=\underset{s\to 0}{\lim }sI(s)=\frac{B}{RB+k^2}K=i(\mathrm{\infty })$

Answers

Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

$\omega (\mathrm{\infty })=14.91radians/second$

$i(\mathrm{\infty })=5.46Amperes$

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Checked by: Andrew Hellie