# How to check if the poles of transfer function are stable?

$$H(z)=\frac{az}{[z-(b+ja)][z-(b-ja)]}$$

The two poles are looking like this: $z_{\inf, 1}=b+ja, z_{\inf, 2}=b-ja$

I know they must be inside the unit circle: $|z_{\inf,i}|<1$

So I replace the term: $|b+ja|<1$ by the first pole, but how I must continue?

• your poles are in complex plane, so you have to find their magnitude and the magnitude must be smaller than one for stability. Sep 7 '17 at 9:00
• mind telling us what "$z_\text{inf}$" means? Sep 8 '17 at 1:42
• @robertbristow-johnson I think your edit might mask part of the problem: perhaps OP misunderstood the inequetion and his/her mistake prevented him/her from moving forward with the equation... Sep 8 '17 at 1:44
• yeah, maybe. i dunno. Sep 8 '17 at 1:44

It's not $|z_{\inf,i}<1|$ it's $|z_{\inf,i}|<1$ which means that, as @Mohammad Mohammadi said in the comments, your pole on a complex plane must be inside the unit circle.

Analytically, you get the magnitude with $$r = |z| = \sqrt{a^2+b^2}$$ If $r<1$, your pole is stable.

Note that $z_1 = b + ja$ and $z_2 = b-ja$ have the same magnitude, which means that if one is stable, the other is stable as well (in the complex plane, they are symmetrical to the horizontal axis).

# Illustration # Matlab code for the illustration

%%
clear
close all
clc;

% STABLE

a = 0.73;
b = 0.24;

P1 = a + b*1i; % Pole 1
P2 = a - b*1i; % Pole 2

r = sqrt(a^2+b^2); % Magnitude

% UNSTABLE

a2 = 1.23;
b2 = 0.84;

r2 = sqrt(a2^2+b2^2);

P12 = a2 + b2*1i; % Pole 1
P22 = a2 - b2*1i; % Pole 2

zplane([0 0; 0 0], [P1 P12; P2 P22]); % Plot the poles on a complex plane
legend('', '', ...
['z = ' num2str(a) ' \pm ' num2str(b) 'i (r = ' num2str(r) ')'],...
['z = ' num2str(a2) ' \pm ' num2str(b2) 'i (r = ' num2str(r2) ')'], ...
'Unit Circle (r=1)')