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I'm doing an analysis of a PMDC motor and I don't understand something about the transfer function.

I have the following equation: \$(K_t\frac{1}{((s* L_A) + R_A)})(V_I(s)-k_b\omega(s))= (s^2J_m+sb_m)\theta(s) \$ which solved for \$ \frac{\theta(s)}{V_I(s)}\$ gives the result:

\$\frac{K_t}{s((s* L_A) + R_A)(sJ_m+b_m)+K_tK_b)}\$

I wanted to ask, where did \$\omega(s)\$ go? I'm sure it should be out of the result, but I can't figure why.

Thank you.

EDIT: I forgot to mention this. The equation is taken from the transformations of the electrical and mechanical components of a PMDC motor.

\$I_A=(\frac{1}{((s* L_A) + R_A)})(V_I(s)-k_b\omega(s))\$ as the equation of the electrical part of the motor, with \$V_I\$ as the voltage input, \$k_b\omega(s)\$ as the EMF. \$L_A and R_A\$ are the electrical resistance and inductive impedance.

Then, \$K_t I_A=(s^2J_m+sb_m)\theta(s)\$ is the equation for the net torque of PMDC motors which combined with the one above gives the first equation.

I've been trying the think about it and it might be related to \$\omega(s)= \frac{\partial \theta(s)}{\partial t}\$ but I'm not sure.

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  • \$\begingroup\$ It might help if you explained all the parts of the original equation or left a link. \$\endgroup\$
    – Andy aka
    Commented Sep 19, 2014 at 12:23
  • \$\begingroup\$ I've added some explanation. I hope it is correct. \$\endgroup\$
    – Naeriel
    Commented Sep 19, 2014 at 12:50
  • \$\begingroup\$ Rate of change of angle is frequency \$\endgroup\$
    – Andy aka
    Commented Sep 19, 2014 at 13:03
  • \$\begingroup\$ Is there a difference between \$K_t\$ (numerator) and \$k_t\$ (denominator)? Or are the both supposed to be \$K_t\$? \$\endgroup\$
    – Eric
    Commented Sep 19, 2014 at 13:03
  • \$\begingroup\$ @Brad it's supposed to be the same. My fault. \$\endgroup\$
    – Naeriel
    Commented Sep 19, 2014 at 13:11

1 Answer 1

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I think the key to your problem is the equation \$\omega(s) = \frac{\partial\theta(s)}{\partial{t}}\$. You need to re-write that in the Laplace domain: \$\omega(s) = s\theta(s)\$. Once you substitute that in for \$\omega(s)\$, the math works out.

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  • \$\begingroup\$ I though so after giving it a closer look. Thank you for the answer. \$\endgroup\$
    – Naeriel
    Commented Sep 19, 2014 at 13:18

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