Laplace transforms:DC Motor circuit: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
Line 14: Line 14:


<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>
<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>
Summing the voltages around the series circuit gives us our differential equation.
<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Revision as of 11:14, 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.

T(t)=ki(t)

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)).

vm(t)=kω(t)

We want to find the Laplace transfer function of the motor.

Ω(s)=[ω(t)]/vs(s)

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

vs(t)=Ri(t)+Ldi(t)dt+kω(t)