Is my ODE solution correct for car suspension system?

[exidez]

Homework Statement

http://img27.imageshack.us/img27/2193/matlapassig2fig.jpg
http://img132.imageshack.us/img132/7126/matlapassig2.jpg

Homework 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

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 doesn't look anything like the one given in the question...Have i done something wrong?

 

[tanujkush]

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

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?

 

[exidez]

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?

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

 

[tanujkush]

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

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.

 

[exidez]

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.

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?

 

[tanujkush]

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?

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.

 

[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...

 

[tanujkush]

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...

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