How can I get the transfer function G(s) = (X2(s)-W(s))/W(s)

According to the site below, how can I get the transfer function G(s) = (X2(s)-W(s))/W(s)

Control Tutorials for Matlab and Simulink

%%Parameters
m1 = 2500;   % (kg)
m2 = 320;    % (kg)
k1 = 80000;  % (N/m)
k2 = 500000; % (N/m)
b1 = 350;    % (N*s/m)
b2 = 15020;  % (N*s/m)

%%Transfer Function (Open Loop)
% Displacement Of Sprung Mass G(s) = X1(s)/W(s)
num1 = [(0) (0) (b1*b2) (b1*k2+b2*k1) (k1*k2)];
den1 = [(m1*m2) (m1*b1+m1*b2+m2*b1) (m1*k1+m1*k2+m2*k1+b1*b2) (b1*k2+k1*b2) (k1*k2)];
G1 = tf(num1,den1);

% Suspension Deflection G(s) = (X1(s)-X2(s))/W(s)
num2 = [(0) (-m1*b2) (-m1*k2) (0) (0)]; 
den2 = [(m1*m2) (m1*b1+m1*b2+m2*b1) (m1*k1+m1*k2+m2*k1+b1*b2) (b1*k2+k1*b2) (k1*k2)];
G2 = tf(num2,den2);

% Tire Deflection G(s) = (X2(s)-W(s))/W(s)
% num3 = [];
% den3 = [];
% G3 = tf(num3,den3);

[Azzi Abdelmalek]

I can't open your link. Can you clarify your question?

Answer 1

[Star Strider]

It seems to me that you just copy it from the website and paste it to a tab in your Editor, just like I copied it and pasted it here:

M1 = 2500;
M2 = 320;
K1 = 80000;
K2 = 500000;
b1 = 350;
b2 = 15020;

s = tf('s');
G1 = ((M1+M2)*s^2+b2*s+K2)/((M1*s^2+b1*s+K1)*(M2*s^2+(b1+b2)*s+(K1+K2))-(b1*s+K1)*(b1*s+K1));
G2 = (-M1*b2*s^3-M1*K2*s^2)/((M1*s^2+b1*s+K1)*(M2*s^2+(b1+b2)*s+(K1+K2))-(b1*s+K1)*(b1*s+K1));

If you want to know how it was derived, that is described in detail on the page.

[Bob]

Thanks for your answer.

This site gives only the description on how to get the transfer function for suspension deflection:

G1(s) = (X1(s)-X2(s))/U(s) % Input: U(s) Output: X1(s)-X2(s)

G2(s) = (X1(s)-X2(s))/W(s) % Input: W(s) Output: X1(s)-X2(s)

But I can't understand how to get the Tire Deflection:

G(s) = (X2(s)-W(s))/W(s)

[Star Strider]

I don’t see where the tire dynamics are specifically modeled anywhere, so you would have to do that yourself.

The wheel-tire systme seems to be the spring-dashpot system characterised by K2 and b2, both connected to M2. The input would be W and the output would be X2. That is likely as close as you can get to modeling them independently.

You would likely have to model and solve the entire system first, then use those data to model the wheel-tire system specifically.

[Bob]

Thank you for your answer.

[Star Strider]

My pleasure.

When I looked at it again, the wheel-tire system looks like a simple spring-mass-damper that could be modeled by a second-order linear ordinary differential equation. (You could solve that on paper easily with Laplace transforms, or with the Symbolic Math Toolbox, or code it as an anonymous function to use ode45.)

I’m not a mechanical engineer, but spring-mass-damper systems seem to be modeled by the same dynamics whether the spring is in series or in parallel with the dashpot. (In an electrical circuit, series systems would be modeled differently than parallel systems, because of Kirchoff’s laws.)

Answer 2

[Thomas John]

A=[(1100*s+2200) -(1100*s+2200) 0;

-(1100*s+2200) (1100*s^2+3300*s+4400) -(1100*s+2200);

0 -(1100*s+2200) (1100*s^2+3300*s+4400)];

A2=[(1100*s+2200) -(1100*s+2200) 0;

U 0 0;

0 -(1100*s+2200) (1100*s^2+3300*s+4400)];

i need to find det(A2)/det(A) but i end up with error

[Star Strider]

Post this as a new Question. It is not an Answer to the existing post.

I will delete it and this Comment in a few hours.

[Walter Roberson]

If you had done

s = tf('s')

then you cannot proceed because tf does not permit unresolved variables such as U.

So use the symbolic toolbox

syms s U

and proceed. You will probably want to simplify()

The result will be a typical transfer function but one that has a gain of U/1100. You cannot bring that back as a tf for Control Systems purposes as the toolbox does not permit unresolved parameters.

The toolbox does permit tunable parameters, which always have a current value but the value is easily changed, including possibly automatically by tuning procedures.