Transfer Function of a DC Motor.
Consider the model presented in Figure 10:
Figure 10. Model for a DC Machine [4]
Let’s determine the Transfer Function of the DC Motor from Figure 10. Since the current-carrying armature is rotating in a magnetic field, its voltage is proportional to its speed. That is the back electromotive force as it was established in equation 7:
Equations 12
Taking the Laplace Transform we get:
Equations 13
The torque developed by the motor is proportional to the armature current, as it was said in Equations 9:
Equations 14
Transforming every impedances of Figure 10 into their Laplace Transform equivalent , we find the voltage equation for the loop around the armature circuit:
Equations 15
Now, we substitute Equations 13 y 14 en 15:
Equations 16
We need Tm in terms of 
in order to find 
. That can be get using the equivalent model for mechanical loading on a motor as shown in Figure 11:
Figure 11. Typical equivalent mechanical loading for a DC Machine [4]
Where Jm and Dm are mechanical constant which can be derived from a typical configuration such as:
Figure 12. A DC Motor driving a rotational mechanical load [4]
Considering Figure 12, Jm and Dm are:
Equations 17
Now, from Figure 11 we can find the relationship between Tm and :
Equations 18
Substituting Equations 18 in 16 we get:
Equations 19
In the most cases La is too small compared with Ra, so Equations 19 can be simplified and rearrange as:
Equations 20
Now from Equations 20 we obtain the Transfer Function for a DC Motor as follow:
Equations 21
The electrical constants of the motor Kt y Kb can be found with the following relations:
Equations 22
Where Tstall, Ea y Wno-load, use to be derive from a Graphic Speed Vs Torque such as:
Figure 13. Torque-speed curves with an armature voltage Ea as a parameter [4]
As an example, consider the case of Figure 14:
Figure 14. Torque-speed curves and system example [4]
Hence:
And using the gear ratio N1/N2=1/10:
[4] Control Systems Engineering, Norman Nise
Written by: Larry Francis Obando – TSU
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