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.

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

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 Equation *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.

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

Transforming yields the following.

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

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*.

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

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

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

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

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

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.

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

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.

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

Answers

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

$\omega (\infty )=$