I've been tasked with showing that the pulse transfer function G(z) of the following plant \begin{equation}G_p(s) = {12\over (s+1)(s+4)} \end{equation} being sampled and held by a zero order hold with a sampling period T = 0.2s is equal to \begin{equation}G(z) = {0.1745z^{-1}+0.1249z^{-2}\over1-1.268z^{-1}+0.3678z^{-2} } \end{equation}

I started by combining the plant with the hold to obtain \begin{equation} {12(1-e^{-sT})\over s(s+4)(s+1)} \end{equation}

I know \begin{equation} {(1-e^{-sT})} \end{equation} is equal to

\begin{equation} {(z-1)/z} \end{equation}

and I split the remaining equation into partial fractions, found the z transform of each, and recombined them to obtain,

\begin{equation}{0.173z^{-1}+0.126z^{-2}\over1-2.27z^{-1}+1.64z^{-2}-0.368z^{-3} } \end{equation} combined with the z transform of the hold, this equals

\begin{equation}G(z) = {0.173-0.047z^{-1}-0.126z^{-2}\over z-2.27+1.64z^{-1}-0.368z^{-2} } \end{equation}

This is not equal to the transfer function I am supposed to obtain- Does anyone see what I am doing wrong?


2 Answers 2


You have used the correct expression for the zero-order hold, and the transfer function is indeed: \begin{equation} H(s) = \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)} \end{equation} However, it's unclear how you proceeded from here. It looks like you've simply replaced s by z (z is actually equal to esT) and then split the function into partial fractions.

Alternatively, you can first find h(t) and then find H(z):

\begin{eqnarray} H(s) &=& \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)} \\ &=& (1-e^{-sT}) \left \{ \dfrac{3}{s} - \dfrac{4}{s+1} + \dfrac{1}{s+4} \right \} \end{eqnarray} Taking the Inverse Laplace Transform: \begin{eqnarray} h(t) &=& f(t)u(t) - f(t-T)u(t-T) \end{eqnarray} where \begin{equation} f(t) = 3 - 4e^{-t} + e^{-4t} \end{equation} H(z) can now be found from h(t). You can refer to this link for a table of formulae: http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html

  • \$\begingroup\$ After splitting the function up into partial fractions and obtaining \begin{eqnarray} H(s) &=& \ (1-e^{-sT}) \left \{ \dfrac{3}{s} - \dfrac{4}{s+1} + \dfrac{1}{s+4} \right \} \end{eqnarray} , I found the z-transform of each part of the equation using a table of z transforms, so \begin{equation} {\dfrac{3}{s} = \dfrac{3}{1-z^{-1}} } \end{equation} \begin{equation} {\dfrac{4}{s+1} = \dfrac{4}{1-e^{-T}z^{-1}} } \end{equation} \begin{equation} {\dfrac{1}{s+4} = \dfrac{1}{1-e^{-4T}z^{-1}} } \end{equation} \begin{equation} {(1-e^{-sT}) = \dfrac{z-1}{z}} \end{equation} \$\endgroup\$
    – Ca01an
    Commented May 14, 2017 at 20:02
  • \$\begingroup\$ I then combined them to find H(z) \$\endgroup\$
    – Ca01an
    Commented May 14, 2017 at 20:03

There's a common factor of \$\small (z-1)\$ in the numerator and denominator of your final expression. Cancel these and you're left with:

$$\small G(z)=0.17\frac{z(z+0.73)}{z^2-1.29z+0.38}$$

compare with your required target expression: \begin{equation}\small G(z) = {0.1745z^{-1}+0.1249z^{-2}\over1-1.268z^{-1}+0.3678z^{-2} } =0.1745\frac{z(z+0.7158)}{z^2-1.268z+0.3678}\end{equation}

Note: I've worked with positive powers of z (easier to type into root solver!) and rounded all calculations to 2 d.p.


To illustrate, consider the simple process TF: \$\small G_p(s)=\large\frac{1}{s+1}\$, and sample/hold: \$\small G_H(s)=\large\frac{1-e^{-sT}}{s}\$.

The combined TF is: $$\small G(s)=\frac{(1-e^{-sT})}{s}\times \frac{1}{(s+1)}\small=(1-e^{-sT})\frac{1}{s(s+1)}=(1-e^{-sT})\left( \frac{1}{s}-\frac{1}{s+1} \right)$$

Taking z-transforms: $$\small G(z)=\frac{z-1}{z}\left( \frac{z}{z-1}-\frac{z}{z-a} \right)$$ where \$\small a=e^{-T}\$

If the bracket is evaluated first, we have:

$$\small G(z)= \frac{(z-1)}{z}\times\frac{z(1-a)}{(z^2-(1+a)z+a)}=\frac{(1-a)(z-1)}{(z^2-(1+a)z+a)} $$ ...and the \$\small (z-1)\$ factor in the denominator is not apparent.

However, multiplying through by \$\frac{(z-1)}{z}\$ first, gives the simplified form: $$\small G(z)=1-\frac{z-1}{z-a}=\frac{1-a}{z-a} $$

  • \$\begingroup\$ Do you have any idea how the common factor of (z-1) got in there? Is it just a problem with my math? \$\endgroup\$
    – Ca01an
    Commented May 14, 2017 at 20:05
  • \$\begingroup\$ The ZOH z-TF is \$\frac{z-1}{z}\$, and the z-transform of \$\frac{1}{s}\$ is \$\frac{z}{z-1}\$, so the 's' in the denominator of Gp(s) puts a (z-1) factor in the denominator of G(z) \$\endgroup\$
    – Chu
    Commented May 14, 2017 at 20:21
  • \$\begingroup\$ ...I'll add a note to my answer to illustrate what's happening. \$\endgroup\$
    – Chu
    Commented May 14, 2017 at 20:42
  • \$\begingroup\$ Thank you for illustrating it, this makes things clearer. \$\endgroup\$
    – Ca01an
    Commented May 15, 2017 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.