Laplace transforms:DC Motor circuit: Difference between revisions

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<math>v_m(t) = k \omega(t)</math>
<math>v_m(t) = k \omega(t)</math>


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


<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>
<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>
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<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>
<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s)</math>

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

<math>i(0) = \lim_{s\rightarrow \infty} s\int_0^\infty i(t) e^{-st} u(t) dt = \frac{\infty}{e^{\infty}} = 0</math>

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

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s)</math>

Revision as of 11:52, 21 October 2009

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.

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

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

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

Take the Laplace transform.

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

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