I have closed loop - discrete model. This is how it looks in matlab:

enter image description here

I need to compute analitically output $y(t)$ (same as in the Matlab's scope) having $Y(z)$.



Input is sine wave so Z transform for sine is:

$U(z) = \frac{z\sin(\omega T)}{z^2-2zcos(\omega T)+1}$


$G(z) = \frac{G_o(z)}{1+G_o(z)} = \frac{kz^3}{(a+k)z^3+bz^2+cz+d} = \frac{kz^3}{(a+k)(z-z_1)(z-z_2)(z-z_3)}$


$\omega$, $k$, $a$, $b$, $c$, $d$ - are given parameters.


$Y(z) = \frac{z\sin(\omega T)kz^3}{(z^2-2zcos(\omega T)+1)(a+k)(z-z_2)(z-z_3)(z-z_4)}$

I saw on Wolfram Alpha that roots of $z^2-2zcos(\omega T)+1$


$z_0 = cos(\omega T)-isin(\omega T)$

$z_1 = cos(\omega T)+isin(\omega T)$

and rest of roots ($z_2$, $z_3$, $z_4$) I compute using roots function in matlab from $G(z)$.

$z_2 = 0.9889+0.022i$

$z_3 = 0.9889-0.022i$

$z_4 = 0.8784$

The residue method is what I know but other would be good too.

For example:

$res_{z=z_0}=\lim_{z\to z_0}(z-z_0)Y(z)z^{n-1}=\lim_{z\to z_0}\frac{z\sin(\omega T)kz^3z^{n-1}}{(a+k)(z-z_1)(z-z_2)(z-z_3)(z-z_4)}=lim_{z\to z_0}\frac{\sin(\omega T)kz^{n+3}}{(a+k)(z-cos(\omega T)-isin(\omega T))(z-z_2)(z-z_3)(z-z_4)}=\frac{\sin(\omega T)k[cos(\omega T)-isin(\omega T)]^{n+3}}{(a+k)(-2isin(\omega T))(cos(\omega T)-isin(\omega T)-z_2)(cos(\omega T)-isin(\omega T)-z_3)(cos(\omega T)-isin(\omega T)-z_4)}$

$y(t)$ as you know is sum of residues for all poles.

In order to plot graph from my function $y(t)$ I have to get rid of imaginary parts of equation. How can I do this?

Yeah it's boring because it's a homework.

  • 1
    $\begingroup$ you can use a partial fraction decomposition on Y(z) and then match the fractions against a table of common sequences. This is the way we used to do it in uni. Further, there seems to be an error in the residuum calculation. The denominator in the second line doesn't have a matching term for ${i}~sin(\omega T)$. $\endgroup$ Commented Jan 1, 2017 at 15:16
  • $\begingroup$ Thanks. There was bracket missing there after $isin(\omega T)$ or something else? $\endgroup$
    – mdkdy
    Commented Jan 1, 2017 at 15:37
  • $\begingroup$ When you replace $z^2-2z \cos(\omega T) + 1$ with $(z-z_1)(z-z_0)$ you don't seem to add $(z-z_0)$ to the denominator. Only $(z-z_1)$. Where did the other term go? $\endgroup$ Commented Jan 1, 2017 at 15:44
  • $\begingroup$ Yes because after $lim_{z \to z_0}$ there is ${(z-z_0)}$ so it can be reduced. That seem to be how equation of $res$ look like. $\endgroup$
    – mdkdy
    Commented Jan 1, 2017 at 15:50


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