recurrent solution of difference equation

Hello,

We were tasked with solving the difference equation with recurrent solution

difference equation u(k) - 0.8u(k-1) - 0.16u(k-2) = e(k)

with initial conditions: e(k) = 0.5 sin2k for k ≥ 0

u(-1) = u(-2) = 0

I solve this on paper, beacuse we must first solve on paper then verify in MATLAB and I have problem to write a code for that. I tried to write some code, but my results are not the same from Matlab and on paper.

My code:

clc;

clear all;

close all;

n=15;

u=zeros(15,1);

u(1:2)=[0.017,0.0484]

for k=3:n

u(k)=0.5*sin(k*2)+0.8*u(k-1)+0.16*u(k-2)

end

So, can anyone help me with code?

Thank You in advance

Answer 1

[Sriram Tadavarty]

Hi Rasistlav,

You are in the correct way itself. Just use sind instead of sin function. Since, the values are treated as degrees, use sind function.

You can even replace the starting values with below code and use them directly. The complete modifications comprise to the code below

u(1) = 0.5*sind(2);
u(2) = 0.5*sind(2*2)+0.8*u(1);
for k=3:n
    u(k)=0.5*sind(k*2)+0.8*u(k-1)+0.16*u(k-2)
end

Hope this helps.

Regards,

Sriram

[John D'Errico]

A good catch here, in that e(k) uses sin as a function of degrees, not as radians which is the default for sin(x). But a quick check shows that this must be so to make e(1), and therefore u(1) to be the indicated value.

The advice to use the correct values for u(1) and u(2) is also important, because when you use only 3 significant digit approximations to the initial values, those small errors can accumulate, and in some cases, significantly magnified. As it turns out for this particular difference equation the coefficients are not such that they will tend to amplify small errors in the lower bits. So the problem must lie mainly in the forcing term, thus e(k).

[Rastislav Haska]

Hello Sriram,

it works now when I put d on sin, results are the same as on a paper, thank You very much.

Regard,

Rastislav