How to use previous answer in new calculation n times

Hi all

I am pretty new to Matlab and have been struggling with loops and arrays.

My issue is I am attempting to calculate yn where x in the original calculation becomes y1 in caclulation for y2 and so on.

ie:

x = [1;2] %orginal value

it then is multiplied by A= [0.3 -0.2; -0.6 -0.8]

and B=[-14;2] is added to it to become y1, then calculation is repeated and rather than using x, y1 used to calculate y2 and so on.

Any help would be greatly appreciated.

clear; clc; close all;

A=[0.3 -0.2; -0.6 0.8];

B=[-14;2];

x=[1;2];

y1=A*x+B

y2=A*y1+B

y3=A*y2+B

y4=A*y3+B

y5=A*y4+B

y6=A*y5+B

y7=A*y6+B

y8=A*y7+B

y9=A*y8+B

y10=A*y9+B

Answer 1

[Star Strider]

Perhaps something like this —

A=[0.3 -0.2; -0.6 0.8];
B=[-14;2];
x=[1;2];
y = x;
for k = 1:10
    y = A*y+B;
    ym(:,k) = y;                        % Y-Matrix: 'ym'
end

ym

ym = 2x10

-14.1000 -18.8300 -22.2210 -25.3835 -28.4553 -31.4548 -34.3857 -37.2497 -40.0484 -42.7832 3.0000 12.8600 23.5860 34.2014 44.5912 54.7462 64.6698 74.3673 83.8436 93.1039

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figure
plot((1:10), ym)
grid
legend('y(1)','y(2)', 'Location','best')

.

[Chris]

Thanks Star Strider - Works perfectly.

[Star Strider]

As always, my pleasure!

[David Goodmanson]

Hi Chris,

A key feature of the answer is that there are not nine different variables y1,y2, ... using all those variables is called dynamic variable naming and it is highly discouraged. Instead you have one matrix variable. Any of the previous variables, for example y4, can easily be accessed since it is ym(:,4). (The use of a colon here means every row in column 4).

[Chris]

thanks David

Answer 2

[Paul]

A=[0.3 -0.2; -0.6 0.8];
B=[-14;2];
x=[1;2];

sys = ss(A,B,eye(2),0,-1);
y = lsim(sys,ones(11,1),0:10,x);       % y0 = x
%y = step(sys,10) + initial(sys,x,10);  % y0 = x;  alternative method

figure
plot(0:10,y)