Program for Impulsive Difference Equations

Write a MATLAB program to simulate the following impulsive difference equation
where , , , and , for
(a) Find values of y[n] (b) plot the solutions over the range, 1= n = 10.

2 件のコメント

Jon
Jon 2019 年 8 月 19 日
What have you tried to do so far? What problems have you encountered? Please include your code using the code button on the MATLAB answers toolbar.
Gokula Nanda
Gokula Nanda 2019 年 8 月 20 日
I am acquainted with the solution of difference equations. But no idea, how can i program the equations with impulse?

サインインしてコメントする。

 採用された回答

Jon
Jon 2019 年 8 月 20 日

1 投票

One way to approach this is to transform your difference equation in the single variable y to a system of two difference equations, which you can simply march forward in time.
So neglecting, for the moment the special condition when n = mk, you could define y1(n) = y(n) and y2(n) = y1(n-1), noting also that a(n), b(n) and c(n) are in fact not functions of n, but simple constants, and substituting for h(u) would give the system of equations
y1(n+1) = a*y1(n) - b*y2(n) - c*y2(n)^3
y2(n+1) = y1(n)
Assuming you have initial conditions for y you could then just march forward in a loop
a = -3/2
b = 1
c = 1/4
% assign initial conditions
y1(1)= y1ic % initial condition y for n=1
y2(1)= y2ic % initial condition y for n=0
% assign maximum iteration
nFinal = 11
% loop to march forward
preallocate
y1 = zeros(nFinal+1,1)
y2 = zeros(nFinal+1,1)
for n = 1:nFinal
y1(n+1) = a*y1(n) - b*y2(n) - c*y2(n)^3
y2(n+1) = y1(n)
end
% plot results (note y1 = y the variable you are interested in)
plot(y1) % with only one argument the x axis will just be the value of n
You will have to further modify your code to handle the special condition where n = m*k. I actually can't understand the details of what your are doing there from the description, for example I don't see where I_k is defined.
Hopefully this gives you a start on how to approach this.

4 件のコメント

Gokula Nanda
Gokula Nanda 2019 年 8 月 21 日
Here, I_k is any positive real contsant. After execution, there is a error message in the programming "undefined preallocated" and another error in "initial condition".
Jon
Jon 2019 年 8 月 21 日
編集済み: Jon 2019 年 8 月 21 日
I was just trying to give you an outline of an approach. The error "undefined preallocated" is due to an error I made typing in the example. There should be a comment % in front of the word preallocate, so change
% loop to march forward
preallocate
y1 = zeros(nFinal+1,1)
to
% loop to march forward
% preallocate
y1 = zeros(nFinal+1,1)
I don't know exactly what error you had regarding initial conditions, but likely it was that y1ic and y2ic are undefined. I didn't know what numerical values you wanted to assign as initial conditions, I was just putting those in as a place holder to show that you would need to assign initial conditions
You would need to add some lines something like (I just made up some numbers here to illustrate, you will have to assign values that make sense for your problem, maybe zero
a = -3/2
b = 1
c = 1/4
y1ic = 5.3
y2ic = 13.8
% assign initial conditions
y1(1)= y1ic % initial condition y for n=1
y2(1)= y2ic % initial condition y for n=0
As I said earlier, you will have to further modify this code to include treating your I_k. Hopefully you can see your way through doing that now that you have a start.
Gokula Nanda
Gokula Nanda 2019 年 8 月 21 日
Thank you very much.
Jon
Jon 2019 年 8 月 21 日
Your welcome. Glad to hear this worked. Good luck with your project

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

ヘルプ センター および File ExchangeCommunications Toolbox についてさらに検索

質問済み:

2019 年 8 月 19 日

コメント済み:

Jon
2019 年 8 月 21 日

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by