# Transfer function of a speed of a dc motor

Discussion in 'The Projects Forum' started by tobman3, Mar 2, 2011.

1. ### tobman3 Thread Starter New Member

Mar 2, 2011
5
0
Can someone please help me with the transfer function of speed of a dc motor. Its goint to be; w/v w= omega while v=voltage. Thanks

2. ### DickCappels Moderator

Aug 21, 2008
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If it is a permanent magnet DC motor (the only kind I have tested), the RPM's (proportional to your omega) will be linearly proportional to applied voltage IF THERE IS NO FRICTION. Friction will tend to make the curve tend toward being more flat at higher voltages.

So, first order, it is a straight line.

3. ### DrNick Active Member

Dec 13, 2006
110
2
The voltage to speed transfer function of a PM field DC motor is:

omega(s)/Ea(s) = 1/Ke * 1/(s^2 + R/L *s + Ke*Kt/(J*L) )

Where,

Ke is the back emf constant

Kv is the torque constant (this is equal to Ke if done in SI units.)

J is the polar moment of inertia of the motor/load.

L is the armature inductance

R is the armature resistance

In steady state, yes the speed is directly proportional to applied armature voltage; this is not the case for dynamic situations as both the back emf and armature currents cross couple to form a more complex relationship as described above.

...

**NOTE** This ignores viscous damping and mechanical loss, which can typically be neglected for small machines (as I assume you are dealing with)

4. ### tobman3 Thread Starter New Member

Mar 2, 2011
5
0
Hi, i did a little bit of research and i got a transfer function which is;

omega(s)/Voltage = k/[(R+Ls)(Js+B)+k^2]

Do u know the difference between the two

Thanks

5. ### DrNick Active Member

Dec 13, 2006
110
2
B the viscous damping term due to the shaft coupling is included in the euqation you ahve given, however after doing a little litmus test, I do not believe the equation you have given is correct. As stated before, the equation i gave neglects damping caused by windup in the coupler. If you remove the B you are left with:

K/ ( (Ls +R) sJ +K^2)

so, if we multiply through we get

K/ ( J*L*s^2 + J*R*s +K^2 )

Then if J*L is pulled out on the denominator we get

K/J*L * 1/( s^2+R/L*s + K^2/(J*L))

After doing this algebra, it appears that your transfer function is incorrect. The DC gain you should get (regardless of characteristic polynomial, ac gain) you should get 1/K...this is not the case. There may have been an algebraic error in the transfer function you propose...Another way to look at this is that K is in units of V/rad/s. Since the AC gain is technically dimensionless in this form having a gain of K does not make sense, SOO, yoou are then saying

omega/v [rad/s / Volt ] = K [ Volt/ rad/s * AC GAIN [dimensionless]