How can I get system Matrix A,B and C for my state space system of Spring-Mass damper system of n blocks arranged in uniform way?

I am trying to generate equations of motion for Spring mass (damper) system consisting of N blocks. I want to write such a function that automatically creates equations of motion(that fills these matrices A,B,C) just by passing a value of number of masses in a function input arguement. This is what I want ideally. I know the pattern for equation of motion. My system matrix A is something like

[ -2k k 0 0... 0(n);

k -3k k 0.... 0(n);

0 k -3k k 0... 0(n);

0 0 k -3k k 0 .... 0(n);

.. 0(n);

.. 0(n);

0 0............. k -2k ]

where k is stiffness of the spring. This matrix has some kind of uniform pattern as you can see. How can I write a function to this?

[KALYAN ACHARJYA]

Sorry I didnot get the question, can you elaborate in context of maths way?

0(n)??

[Sandeep Parameshwara]

With n as my function input argument, what i need is n by n matrix as output. So 0(n) just tells that its last zero in n'th column. Sorry for the confusion. Its just a zero. I hope its clear now. So I am looking to generate n by n matrix with above pattern.

[Sandeep Parameshwara]

Here's a sample matrix to avoid any confusion with my question. Suppose I want to write a function to this pattern(similar to my question above) how do I write a general function?

Answer 1

[Dhananjay Kumar]

I think this should be very simple (Writing the generalized matrix in the image). You just need to know array indexing and for loop in MATLAB.

Take as input, a k vector, and a constant n.

Create a nxn zeros-only-matrix A with the help of zeros function.

For n>=3, loop through i= 2 to n-1, and update the matrix values according to the values in the image (you will need to write just 3 assgnment statements inside the for loop for the three values). For i=1 and i=n update the matrix values outside the loop (total 4 assignment statement).

For n<3, you can update the values in the matrix without loop.

Please first try yourself and then post your doubts here.

[Sandeep Parameshwara]

Hi Mr. Kumar, thanks for your reply. Yes, its simple and I should have tried harder before posting. I was able to code it something like this.

A=[];

A(1,:,:)=[-2*d d zeros(1,n-2) -2*k k zeros(1,n-2)];

for i=2:n-1

A(i,i-1,:)=d;

A(i,i,:)=-3*d;

A(i,i+1,:)=d;

A(i,n+i-1,:)=k;

A(i,n+i,:)=-3*k;

A(i,n+i+1,:)=k;

end

A(n,:,:)=[zeros(1,n-2) d -2*d zeros(1,n-2) k -2*k];

for i=n+1:2*n

A(i,i-n,:)=1;

end

In my case, I had operations involving concatenation of three dimensional array and hence was unclear. Anyway, thanks again.

[Kavya E]

What is d in the above code?

[Sandeep Parameshwara]

@Kavya, it's just some constant.