How to Derive the Transfer Function of a Dual Mass-Spring System with Damping?

[kurt1288]

Lets see if anyone can help me with this.

I have to derive a transfer function for the following:

A small satellite with a moment of inertia J1 that has a instrument with a moment of inertia J2. The instrument is at the end of a small strut that has a stiffness constant of k and a damping coefficient of b. A torque is applied to the satellite.

I know that the satellite and instrument (and their respective moments of inertia) can be modeled as two masses with mass m1 and m2. The "strut" above is a spring in this case with the same k and b values. The torque described above can be modeled as a force applied to m1 here. x1 and x2 are the displacements of each mass.

I already have the final transfer function but I don't quite know how to derive it:

X(s)/F(s) =(bs+k)/(J1J2s^4+(J1+J2)bs^3+(J1+J2)ks^2 )

 

[LightHero]

Any help would be greatly appreciated, thanks!The transfer function is derived by writing down the governing equations of motion of the two masses m1 and m2:m1*x1''(t) + b*x1'(t) + k*x1(t) = F(t) (1)m2*x2''(t) + b*x2'(t) + k*x2(t) = -k*x1(t) (2)where x1(t) and x2(t) are the displacements of m1 and m2 respectively. Now, substituting x1(t) and x2(t) in the above equations and taking Laplace transforms, we get:(J1s^2 + bs + k)X1(s) = F(s) (3)(J2s^2 + bs + k)X2(s) = -kX1(s) (4)Solving for X1(s) and X2(s) from (3) and (4):X1(s) = F(s)/(J1s^2 + bs + k) (5)X2(s) = -k*F(s)/[(J1s^2 + bs + k)(J2s^2 + bs + k)] (6)Now, the total displacement is the sum of the displacement of the two masses:X(s) = X1(s) + X2(s)Substituting the values of X1(s) and X2(s) from (5) and (6), we get:X(s)/F(s) =(bs+k)/(J1J2s^4+(J1+J2)bs^3+(J1+J2)ks^2 )

 

[astrokreng]

To derive the transfer function, we can use Newton's second law of motion, which states that the sum of all forces acting on an object is equal to its mass times its acceleration. In this case, we have two masses, m1 and m2, and two displacements, x1 and x2. We can write two equations of motion:

For m1:
F - kx1 - bx1' = m1x1''

For m2:
-kx2 - bx2' = m2x2''

Where x1' and x2' represent the first derivative of x1 and x2 with respect to time, and x1'' and x2'' represent the second derivative. F is the external force applied to m1, which is equivalent to the torque applied to the satellite.

Next, we can use the relationship between force and displacement for a spring, F = -kx, and the relationship between force and velocity for a damper, F = -bx', to rewrite the equations as:

For m1:
F - kx1 - bx1' = m1x1''
F = -kx1 - bx1'

For m2:
-kx2 - bx2' = m2x2''
-kx2 = m2x2' + bx2'

Now, we can take the Laplace transform of both sides of the equations and solve for the ratio of the Laplace transforms of the displacements and the force:

For m1:
(s^2X1 - sx1(0) - x1'(0)) + (bsX1 - bx1(0)) + kX1 = F(s)
X1(s)/F(s) = (bs + k)/(s^2m1 + bs + k)

For m2:
(s^2X2 - sx2(0) - x2'(0)) + (bsX2 - bx2(0)) + kX2 = 0
X2(s)/F(s) = -k/(s^2m2 + bs)

Note that we set the initial conditions for displacement and velocity to be zero, as the system starts at rest.

Next, we can substitute the expressions for X1(s) and X2(s) into the equation for m1 to eliminate the intermediate variable F(s):

(s^2X1 - sx1(0) - x1'(0)) + (bsX1