# Calculating discrete-time transfer function

I have a continuous function $$G_p(s) = {1\over s^2}$$ which I am trying to combine with a zero order hold (with a sampling time of 1 second) to produce a discrete function. I started by combining the plant with the hold to obtain $${(1-e^{-sT})\over s^3}$$ I know $${(1-e^{-sT})}$$ is equal to

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

$${1/s^3}$$ is equal to $${T{^2}z^{-1}(1+z^{-1})\over (1-z^{-1})^3}$$ where T = 1. Next, I multiplied these together to obtain, $${z^{-1}(z-1)(1+z^{-1})\over z(1-z^{-1})^3}$$ which cancels down to $${z^{-1}+z^{-2}\over z^3-2z^{2}+z}$$

When I convert the continuous function to a discrete function in matlab, I obtain $${0.5z+0.5\over z^{2}-2z+1}$$

Why am I getting a different result than matlab? Is there something wrong with my method?

• That is the correct pulse TF. How do the results differ?
– Chu
May 19, 2017 at 7:41

My Z table doesn't have 1/s3. It has 2/s3. The Z transform for 2/s3 is;

$${T{^2}z(z+1)\over (z-1)^3}$$

To make this work with 1/s3, you have to take multiply this transform by 1/2. Using this and your equation;

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

gives the Matlab result.

Here is a link to a second table that confirms the transform I have; Laplace to Z transform table with 1/s^3

• The OP has divided by 2, and has the correct pulse TF.
– Chu
May 19, 2017 at 22:21
• Chu, I've edited my answer with a link to a online table that confirms 1/S^3 is z(z+1)/(z-1)^3. This table has 1/s^3 instead of the one in a my book that has 2/s^3. It's on the 4th row of the second page. Are you saying the OP's z-transform is equivalent to the ones in the tables? I'm missing something because I can't get them to be equal. May 19, 2017 at 23:48
• The s and z transforms of $t^2$ are $\frac{2}{s^3}$ and $\frac{Tz(z+1)}{(z-1)^3}$
– Chu
May 20, 2017 at 6:57