How Do You Apply Laplace Transforms to Vehicle Suspension Analysis?

AI Thread Summary
The discussion focuses on applying Laplace transforms to analyze vehicle suspension dynamics, specifically how the body displacement 'y' responds to various inputs like step, impulse, and ramp. The user derives the differential equation governing the system and formulates the transfer function, Y(S)/X(S) = (cs + k) / (ms^2 + cs + k). There is clarification on notation, emphasizing the correct use of lowercase 's' in the Laplace domain. The conversation also touches on the physical implications of changing wheel displacement with step functions and the necessity of finding inverse Laplace transforms for time-domain analysis. Overall, the user seeks confirmation on their calculations and understanding of the process.
AnkleBreaker
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Homework Statement


question.png

Diagram for a vehicle suspension is given. Displacement of wheel is given by 'x' and and displacement of body is 'y'.
Spring constant, k = (7*10^4) Nm
Damping coefficient, c = (3*10^3) N/m/s
mass,m = 250kg
a) Make a Laplace Transform of system and utilize it to predict 'y' in response to various inputs (step, impulse, ramp)

Homework Equations


Sprint force, Fs = kx
Damping force, Fd = cv
F = ma

The Attempt at a Solution


In this question,
y(t) = output
x(t) = input
y(t)/x(t) = transfer function

∑F = ma (but at equillibrium, a = 0)

Therefore ∑F = 0

(ignoring force by damper for now and resolving forces)
question2.png

Fs = mg
Fs = k[ (original spring length, L0) - (new spring length, X0) ] = mg
Fs = k[ L0 - ( X0 + y(t) - x(t) ) ]
After simplifying...
Fs = k( L0 - X0 ) - k( y(t) -x(t) )
∑F = -mg + k( L0 - X0 ) - k( y(t) -x(t) )
Because mg = k( L0 - X0 ) [proven above]
∑F = - k( y(t) -x(t) ) = ma
m(d^2y/dt^2) + ky(t) = kx(t)

Accounting for the damping force
Fd = -c[ (dy/dt) - (dx/dt) ]

Therefore
∑F = - k( y(t) -x(t) ) - c[ (dy/dt) - (dx/dt) ] = m(d^2y/dt^2)
m(d^2y/dt^2) + c(dy/dt) + ky(t) = kx(t) + c(dx/dt) <---- Differential equation

Applying Laplace transformation...
Y(s)ms^2 + Y(S)cs + Y(S)k = X(S)k + X(s)cs
Y(S)[ms^2 + cs + k] = X(S)[k + cs]

Therefore,
Y(S)/X(S) = (cs + k) / (ms^2 + cs + k)

I have completed the question up to that point. It would be appreciated if someone could check my working and let me know if my answers are correct or there is something wrong.

I'm also having trouble completing this part of the question:
"it to predict 'y' in response to various inputs (step, impulse, ramp)"

Am I correct in assuming, this is how you do it?
Y(S) = [ (cs + k) / (ms^2 + cs + k) ] * X(S)
(cross multiply X(S))

And then you substitute values for X(S).
Ex:
unit step, X(S) = 1/s
ramp, X(S) = 1/s^2

Is that how its done?

Any and all advice will be much appreciated. Thank you
 
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AnkleBreaker said:
Ex:
unit step, X(S) = 1/s
You should write: X(s) = 1/s, because what is (upper case) S?

Another thing is: Is it physically possible to change the displacement of the wheel, X(s), by a stepfunction? What will the velocity, s*X(s), or the forces be at that instance ?
A ramp or a sine wave is ok.

I have not checked your work, but in principle it seems to be correct.
 
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Instead of calculating in the time domain, why don't you sketch the system directly in the Laplace domain ( see attached ).
 

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Hesch said:
You should write: X(s) = 1/s, because what is (upper case) S?

Another thing is: Is it physically possible to change the displacement of the wheel, X(s), by a stepfunction? What will the velocity, s*X(s), or the forces be at that instance ?
A ramp or a sine wave is ok.

I have not checked your work, but in principle it seems to be correct.

Thank you so much for your reply

"You should write: X(s) = 1/s, because what is (upper case) S?"

I was under the impression that once laplace transform is applied, (lower case) t is converted to (upper case) S
y(t) ---> Y(S)
x(t)----> X(S)

"What will the velocity, s*X(s), or the forces be at that instance ?"

The question doesn't explicitly say anything about velocity or forces. They just want to see how the displacement of the mass 'y' varies with each input of x
Each input meaning unit step, ramp step and impulse

Again, thank you for your reply.
 
Hesch said:
Instead of calculating in the time domain, why don't you sketch the system directly in the Laplace domain ( see attached ).
Thank you very much for your reply.

I looked at the attached pdf. I calculated in the time domain, because in the second part of the question, we have to create a simulink model, for which I thought I would need the differential equation.

Your attached document has a final Laplace model similar to the equation I got. Do you think my working is correct?

Thank you for your reply
 
AnkleBreaker said:
I was under the impression that once laplace transform is applied, (lower case) t is converted to (upper case) S
You may choose 'S' or 's': X(S) = 1/S or X(s) = 1/s. Both makes sense.

But writing X(S) = 1/s doesn't, because the righthand side is not a function of S, but a function of s, which is the complex variable in the Laplace domain.
In the time domain, you wouldn't write: F(T) = sin(t) ?

AnkleBreaker said:
The question doesn't explicitly say anything about velocity or forces. They just want to see how the displacement of the mass 'y' varies with each input of x
Each input meaning unit step, ramp step and impulse
I'm not a mathematician, but I think there are some rules, regarding in which cases the Laplace transform is defined. So I just stick to reasonable sense and in which cases e.g. the amount of force will be finite. Then I feel 'safe'.
AnkleBreaker said:
Your attached document has a final Laplace model similar to the equation I got. Do you think my working is correct?
Yes.
 
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Hesch said:
You may choose 'S' or 's': X(S) = 1/S or X(s) = 1/s. Both makes sense.

But writing X(S) = 1/s doesn't, because the righthand side is not a function of S, but a function of s, which is the complex variable in the Laplace domain.
In the time domain, you wouldn't write: F(T) = sin(t) ?I'm not a mathematician, but I think there are some rules, regarding in which cases the Laplace transform is defined. So I just stick to reasonable sense and in which cases e.g. the amount of force will be finite. Then I feel 'safe'.

Yes.

"But writing X(S) = 1/s doesn't, because the righthand side is not a function of S, but a function of s, which is the complex variable in the Laplace domain.
In the time domain, you wouldn't write: F(T) = sin(t) ?"

I understand now. Thank you.

I'm partway through finding the responses. Just to clarify, all I have to do is substitute X(s) by 1/s (unit step), 1 (impulse), 1/s^2, and then simplify and then find the inverse laplace (to convert it back into a time domain). Am I correct?

Thank you for your response.
 

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