What the calc is this?
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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)
4 commentaires
DGM
le 29 Nov 2023
Google doesn't have this cached (maybe it got purged). Is there a way these things can be reverted from the database?
Réponses (1)
Keerthi Reddy
le 30 Juin 2023
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 .
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