Laplace transforms:DC Motor circuit: Difference between revisions

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The torque is proportional to the armature current.
The torque is proportional to the armature current.


<math>T(t) = k i(t)</math>
<math>T(t) = k i(t) \, </math>


Similarly, relating mechanical (''T(t)&omega;(t)'') and electrical (''v<sub>m</sub>(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''v<sub>m</sub>(t)'') and the angular velocity (''&omega;(t)'').
Similarly, relating mechanical (''T(t)&omega;(t)'') and electrical (''v<sub>m</sub>(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''v<sub>m</sub>(t)'') and the angular velocity (''&omega;(t)'').


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


<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s)</math>
<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).
At this point we can use the initial value theorem to find i(0).
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Substituting i(0) into the transformed differential equation gives us Equation *1*.
Substituting i(0) into the transformed differential equation gives us Equation *1*.


<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1*</math>
<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>


Repeat the process with the analogous mechanical differential equation. Here J<sub>m</sub> is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.
Repeat the process with the analogous mechanical differential equation. Here J<sub>m</sub> is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.
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Transforming yields the following.
Transforming yields the following.


<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s)</math>
<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>


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


<math>k I(s) = (J_ms + B) \Omega(s) *2*</math>
<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>


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


<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>
<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>


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


<math>i_{ss} = \lim_{s \to 0} sI(s) = \fracP\{B}{RB + k^2} K</math>
<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

Revision as of 06:51, 22 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

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.

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 Equation *1*.

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

Transforming yields the following.

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*.

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

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

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

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.

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