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

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?

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    $\begingroup$ your poles are in complex plane, so you have to find their magnitude and the magnitude must be smaller than one for stability. $\endgroup$ – Mohammad M Sep 7 '17 at 9:00
  • $\begingroup$ mind telling us what "$z_\text{inf}$" means? $\endgroup$ – robert bristow-johnson Sep 8 '17 at 1:42
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    $\begingroup$ @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... $\endgroup$ – Florent Sep 8 '17 at 1:44
  • $\begingroup$ yeah, maybe. i dunno. $\endgroup$ – robert bristow-johnson Sep 8 '17 at 1:44
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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

Illustration of pole stability

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)')
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