# Car Suspension, coupled ODE

1. Oct 1, 2009

### exidez

1. The problem statement, all variables and given/known data
http://img27.imageshack.us/img27/2193/matlapassig2fig.jpg [Broken]
http://img132.imageshack.us/img132/7126/matlapassig2.jpg [Broken]

2. Relevant equations
My impulse response is entirely different which make me believe that i have messed up my ODE. I have taken the laplace tranform to get it into the from of y to f

3. The attempt at a solution
$$m_{s}\frac{d^{2}y}{dt^{2}}-c(\frac{dy}{dt} - \frac{dy1}{dt}) + k(y - y1)=0$$
$$m_{u}\frac{d^{2}y1}{dt^{2}}+c(\frac{dy1}{dt} - \frac{dy}{dt}) + k(y1 - y) +k_{t}(y1-f(t))=0$$

Laplace Transform:

$$Y(S)(m_{s}s^{2}-sc+k)+Y1(S)(sc-k)=0$$
$$Y1(S)(m_{u}s^{2}+sc+k+k_{t})+Y(S)(-sc-k)-F(S)k_{t}=0$$

Rearranging First Eqn:

$$Y(S)\frac{m_{s}s^{2}-sc+k}{sc-k}=-Y1(S)$$

Substitution:

$$-Y(S)\frac{m_{s}s^{2}-sc+k}{sc-k}(m_{u}s^{2}+sc+k+k_{t})+Y(S)(-sc-k)=F(S)k_{t}$$

$$-Y(S)(s^{4}(m_{u}m_{s})+s^{3}cm_{s}+s^{2}m_{s}k+s^{2}m_{s}k_{t}-s^{3}cm_{u}-s^{2}c^{2}-sck-sck_{t}+s^{2}m_{u}k+csk+k^{2}+kk_{t}+Y(S)(-c^{2}s^{2}+k^{2}=F(S)(sck_{t}-kk_{t}$$

$$Y(S)(-s^{4}(m_{u}m_{s})-s^{3}(cm_{s}-cm_{u})-s^{2}(m_{s}k+m_{s}k_{t}+m_{u}k+s(ck_{t})-kk_{t})=F(S)(sck_{t}-kk_{t})$$

So to get the transform from y to f we want this right?
$$\frac{sck_{t}-kk_{t}}{-s^{4}(m_{u}m_{s})-s^{3}(cm_{s}-cm_{u})-s^{2}(m_{s}k+m_{s}k_{t}+m_{u}k+s(ck_{t})-kk_{t}}$$

When substituting the values in from the question my impulse response done through matlab doesnt look anything like the one given in the question...

Have i done something wrong?

Last edited by a moderator: May 4, 2017
2. Oct 1, 2009

### tanujkush

Hey, at first look, I found these equations to be fishy: is the f(t) a force being applied on the sprung mass m_s? If so, then shouldn't the expression f(t) appear in the equation of motion of m_s as opposed to m_u as you have formulated?

3. Oct 1, 2009

### exidez

I did think of that, but being a car suspension system, it will be traveling on a road which will have the forces acting on m_u. So i put f(t) as the reaction force due to the road...

much like this

Mind you, i am not strong in mechanics, so there could be other flaws in it too

4. Oct 1, 2009

### tanujkush

Even so, the way you have used f(t) in the expressions, it appears to be the road displacement rather than a road reaction force. If that is the case then your formulation seems correct. If however, f(t) is a force on the sprung mass (like you showed in your first illustration), then f(t) should just appear as a force in the first equation.

5. Oct 1, 2009

### exidez

So, with that being said

$$m_{s}\frac{d^{2}y}{dt^{2}}-c(\frac{dy}{dt} - \frac{dy1}{dt}) + k(y - y1)=f(t)$$
$$m_{u}\frac{d^{2}y1}{dt^{2}}+c(\frac{dy1}{dt} - \frac{dy}{dt}) + k(y1 - y) +k_{t}(y1)=0$$

I will see how this spans out in the impulse response latter but just as a check, do those new ODE's seem correct?

6. Oct 1, 2009

### tanujkush

Ok so assuming downward direction as positive and also assuming that the spring damper push the sprung mass upwards(i.e. the spring gets compressed), here are the ODEs:
$$F = ma$$
$$f(t)-k(y - y1)-c(\frac{dy}{dt} - \frac{dy1}{dt}) = m_{s}\frac{d^{2}y}{dt^{2}}$$
and
$$c(\frac{dy}{dt} - \frac{dy1}{dt}) + k(y - y1) -k_{t}(y1)=m_{u}\frac{d^{2}y1}{dt^{2}}$$

So your equations are correct except for the sign on the damper force in the first equation.

7. Oct 1, 2009

### exidez

doing what i did before i gte the transfer function to be

$$\frac{s^{2}m_{u}+sc+k+k_{t}}{s^{4}(m_{u}m_{s})+s^{3}(cm_{s}+cm_{u})+s^{2}(m_{s}k+m_{s}k_{t})+s(ck_{t})+kk_{t}}$$

plugging in the values i get an unstable system and the impulse response is exponentially increasing

for k = 30 N/mm (and also kt) i put in the value of 30*10^-3 right ?
I have tried it both with 30 and 30*10^-3 and it is still the same response...

and i thought this was going to be a simple question...

8. Oct 6, 2009

### tanujkush

k (and kt) are in N/mm so their values in SI units would be 30N/mm or 30x10^3 N/m