pzmap(H) pole and zeros switched?

% Characteristic Polynomial a(s) = a0*s^n + a1*s^(n-1) + ... + an-1*s + an

as = [a0 a1 a2 a3]; d = 1;

a0 = 1; a1 = 1; a2 = 1; a3 = 1;

r = roots(as)

A = tf(as,d)

[p,z] = pzmap(A)

pzmap(A)

please check the coherency, thanks

Answer 1

[Jeevan Joishi]

For a given equation say A(s)/B(s), the pole are those occurrences where the equation evaluates to Infinity, and the zeros are basically the roots of the equation.

In your case, the equation is: a(s) = s^3 + s^2 + s + 1.

Only occurrences where the equation a(s) tends to Infinity is when 's' is itself Infinite and hence MATLAB does not give you a specific pole. Zeros happen when a(s) evaluates to 0 which is the case with -1,+i,-i.

MATLAB's output are as expected. Leave a comment with your expected output.