pzmap(H) pole and zeros switched?

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Ale Crb
Ale Crb 2016-4-1
回答： Jeevan Joishi 2016-4-5
% 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

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回答（1 个）

Jeevan Joishi
Jeevan Joishi 2016-4-5
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.

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