# Calculating the pulse transfer function

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

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

I know $${(1-e^{-sT})}$$ is equal to

$${(z-1)/z}$$

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

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

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

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

You have used the correct expression for the zero-order hold, and the transfer function is indeed: $$H(s) = \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)}$$ 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 $$f(t) = 3 - 4e^{-t} + e^{-4t}$$ 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

• 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 $${\dfrac{3}{s} = \dfrac{3}{1-z^{-1}} }$$ $${\dfrac{4}{s+1} = \dfrac{4}{1-e^{-T}z^{-1}} }$$ $${\dfrac{1}{s+4} = \dfrac{1}{1-e^{-4T}z^{-1}} }$$ $${(1-e^{-sT}) = \dfrac{z-1}{z}}$$ Commented May 14, 2017 at 20:02
• I then combined them to find H(z) 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: $$\small G(z) = {0.1745z^{-1}+0.1249z^{-2}\over1-1.268z^{-1}+0.3678z^{-2} } =0.1745\frac{z(z+0.7158)}{z^2-1.268z+0.3678}$$

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

$Addendum$

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: $$\small G(z)=1-\frac{z-1}{z-a}=\frac{1-a}{z-a}$$

• Do you have any idea how the common factor of (z-1) got in there? Is it just a problem with my math? Commented May 14, 2017 at 20:05
• 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)
– Chu
Commented May 14, 2017 at 20:21
• ...I'll add a note to my answer to illustrate what's happening.
– Chu
Commented May 14, 2017 at 20:42
• Thank you for illustrating it, this makes things clearer. Commented May 15, 2017 at 19:00