What the calc is this?

p=[1 -1 -2];

Roots(p)

i=0:20

yi=A1*z^i+A2*z^i;

stem(yi)

B=[1 1],

A=[1];

N=1:50;

Xn=cos(pi*n)

Yn=filter(b,a,xn);

Stem(n,yn)

[Arya]

hi, could you provide more details of what you are trying to do and what your query actually is ?

[Torsten]

This answer might help:

https://uk.mathworks.com/matlabcentral/answers/81270-is-matlab-solving-difference-equations

[DGM]

Google doesn't have this cached (maybe it got purged). Is there a way these things can be reverted from the database?

[Rena Berman]

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Answer 1

[Keerthi Reddy]

Hi Matheus, It is my understanding that you want to know what the above-mentioned calculation represents.

Here is the answer:

p = [1 -1 -2]; 
r = roots(p); 
disp(r); 

To calculate the roots of a polynomial 'p', you can use the “roots” function in MATLAB. The output will be the roots of the polynomial 'p'. You can go through this documentation to know more: Polynomial roots - MATLAB roots - MathWorks India

z = 0.9; % Choose a value for z 
A1 = 1; % Choose a value for A1 
A2 = 2; % Choose a value for A2 
i = 0:20; 
yi = A1*z.^i + A2*z.^i; 
stem(i, yi); 

The above code generates a sequence yi = A1*z^i + A2*z^i for i ranging from 0 to 20 and plots it using a stem plot. You can go through the documentation to know more: Plot discrete sequence data - MATLAB stem - MathWorks India

If you have a transfer function defined by numerator B and denominator A, and you want to filter a sequence xn using this transfer function, you can use the “filter” function. The following code does the same. B = [1 1];

A = [1]; 
n = 1:50; 
xn = cos(pi*n); 
yn = filter(B, A, xn); 
stem(n, yn); 

You can go through this documentation to know more: 1-D digital filter - MATLAB filter - MathWorks India.

Hope this helps .