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I have a double mass rotary model. The two masses are connected by an axis which together acts as a spring damper system of the fourth order. The system is actuated by a motor and the angle of rotation is measured by rotary encoders at each end of the mass. The input to the motor is milli-volts which then converts into corresponding torques. The output measured is angle of rotation of the mass. The actuation and the measurement is on the same side. The system looks similiar to this

enter image description here

The transfer function from output to input of the system is,

$Plant = \large {\frac{(4700 s^2 + 4393 s + 3.245 (10)^8)}{(s^4 + 7.574 s^3 + 1.202(10)^5 s^2)}}$

Is it possible to determine the mass of the system from the transfer function or bodeplot? How can I relate the voltage, torque, mass and angle of rotation?

idkfa
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Somanna Thapanda
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  • Do you have a free body diagram or derivation of the transfer function you can share? – willpower2727 Mar 28 '16 at 14:21
  • @willpower2727 I dont have the derivation. The transfer function was approximated through parametric identification of the system. I have no clue how to add an image in this dialog box. The system is more like the one in this link https://www.google.nl/search?docid=0uM_K1XrnG4lTM&biw=1920&bih=921&tbm=isch&tbs=simg:CAQSJgnhbwdF0U3V1hoSCxCwjKcIGgAMCxCOrv4IGgAMIdLjPytV65xu&sa=X&ved=0ahUKEwi6v6uy7uPLAhUGhQ8KHWsLC38Qwg4IGigA – Somanna Thapanda Mar 28 '16 at 17:04
  • Am I right in thinking your transfer function is output/input? In that case your mass spring damper system is 2nd order (the numerator), and the system mass should be 4700? – willpower2727 Mar 28 '16 at 17:35
  • Yes the transfer is from output to input. The denominator of the transfer function determines the order of the system right? The dynamics of the system has two masses so its a fourth order system. My question is how can I scale the system with voltage as input and angle of rotation as output – Somanna Thapanda Mar 28 '16 at 19:36
  • Yes, the denominator is typically the "system" and the numerator is the input. I understand there are two masses in your system, in that case you should have two transfer functions, one for each mass. Is the motor one of your masses? It would really help if you could share the differential equations governing your system. – willpower2727 Mar 28 '16 at 20:26
  • The differential equations are as follows. $$ $J_1 \ddot{x}_1$ &= tau + d$(\dot{\theta}_2 $- $\dot{\theta}_1)$ + k$(\theta_2-\theta_1)$ \ $J_2 \ddot{x}_2$ &= $d(\dot{x}_1$- $\dot{x}_2)$ + k$(x_1-x_2)$ $$

    J1 and J2 are moment of inertia. tau is the torque applied, theta is the angle of rotation

     – Somanna Thapanda Mar 29 '16 at 09:25
  • The differential equations are as follows. $$ $J_1 \ddot{\theta}_1$ &= tau + d$(\dot{\theta}_2 $- $\dot{\theta}_1)$ + k$(\theta_2-\theta_1)$ $J_2 \ddot{\theta}_2$ &= $d(\dot{theta}_1$- $\dot{theta}_2)$ + k$(theta_1-theta_2)$ $$

    J1 and J2 are moment of inertia. tau is the torque applied, theta is the angle of rotation.

     – Somanna Thapanda Mar 29 '16 at 09:31

1 Answers1

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I'm going to derive the transfer function symbolically from the differential equations you provided in the hopes that you might be able to extract the rotational moment of inertia (not the mass directly, you'd have to know the radius of the rotating masses) from the formula.

Here are the equations of motion as I interpreted them from your comment. J1 is the mass that the torque is applied to.

Equation of motion for mass 1

Equation of motion for mass 2

Taking the Laplace transform and writing in matrix form:

enter image description here

I will use Cramers' method to determine the transfer function for mass 1, the mass to which the torque is applied:

enter image description here

This transfer function Theta1/tT can be expressed as:

enter image description here

Comparing this to your experimentally determined transfer function indicates that the rotational moment of inertia for mass 2 is 4700, and d is 4393. Using algebra you can solve for J1:

enter image description here

Solving for J1 should give about 672.3891.

If you want to check my math, here is a simple Matlab code I used:

j1 = sym('J1');%moment of inertia mass 1
j2 = sym('J2');%moment of inertia mass 2
d = sym('d');%damping coefficient
k = sym('k');%spring constant
t = sym('T');%applied torque
s = sym('s');%laplace variable

%system matrix
A = [j1*s^2+d*s+k -(d*s+k);-(d*s+k) j2*s^2+d*s+k]

det(A)

Acramer = [t -(d*s+k);0 j2*s^2+d*s+k]

det(Acramer)
willpower2727
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  • I think the derivation is wrong. There are no extra terms in denominator. Below is the derivation, $$ m_1 \ddot{\theta}_1 = F + d(\dot{\theta}_2 - \dot{\theta}_1) + k(\theta_2-\theta_1) \ m_1 s^2 \theta(s) = F + ds(\theta_2-\theta_1)+k(\theta_2 - \theta_1) \ (m_1 s^2+ds+k)\theta_1 = F + (ds+k)\theta_2\ $$ similarly, $$ m_2 \ddot{\theta}_2 = d(\dot{\theta}_1 - \dot{\theta}_2) + k(\theta_1-\theta_2)\ m_2 s^2 \theta(s) = ds(\theta_1-\theta_2)+k(\theta_1 - \theta_2) \ (m_2 s^2+ds+k)\theta_2 = (ds+k)\theta_1\ $$ – Somanna Thapanda Mar 29 '16 at 16:23
  • Substituting $\theta2$ in the above equation and rearranging the terms we get,

    $$ (m_1 s^2+ds+k)\theta_1 = F + (ds+k)\frac{(ds+k)}{(m_2 s^2+ds+k)}\theta_1\ \frac{\theta_1}{F} = \frac{m_2 s^2 +ds + k}{m_1 m_2 s^4 +d(m_1 + m_2)s^3 + k(m_1+m_2)s^2} $$

    Oops! there is a mistake, replace m by j and F by $\tau$

     – Somanna Thapanda Mar 29 '16 at 16:37
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    By comparing the coefficients, $j_2$ = 4700 and $\frac{d(j_1+j_2)}{j_1 j_2} = 7.574$. we can get $j_1$ = 673. Is this the right way? – Somanna Thapanda Mar 29 '16 at 16:53
  • I did have a typo, thanks for pointing that out. You are right, I get 672.38 for mass 1. Remember this is moment of inertia, not mass, you'd need to know the shape of each mass (cylinder?) and the size to compute mass directly. – willpower2727 Mar 29 '16 at 17:05
  • I got until this point but it doesnt make sense. Both masses are of same size so the mass should almost be equal. This is not the case in here. – Somanna Thapanda Mar 29 '16 at 19:05
  • My guess is maybe there is noise or un-modeled elements in your system? Can you weigh and calculate the actual mass and moment of inertia? – willpower2727 Mar 29 '16 at 19:20
  • @SomannaThapanda or perhaps the other mass has a different density than the motor? – willpower2727 Mar 30 '16 at 18:47
  • We can approximate the equivalent mass of the system from the transfer function. The slope at the origin in a bodeplot represents the equivalent mass of the system. So by neglecting highest power of 's' we are left with $\frac{3.245(10)^8}{1.202(10)^5}$. Therefore the equivalent mass is given by $\frac{1}{\frac{1.202(10)^5}{3.245(10)^8}}$. – Somanna Thapanda Apr 12 '16 at 20:12