I am trying to solve a class exercise in which I am given the following, in Laplace domain:

$$G(s)=\dfrac{e^{-Ts}}{s+3}$$ $$H(s)=\dfrac{1}{s}$$

And I need to calculate $\dfrac{C(z)}{R(z)}$, which is defined as: $$\dfrac{C(z)}{R(z)}=\dfrac{G(z)}{1+H(z)G(z)}$$

In order to do so, I have to transform $G(s)$ and $H(s)$ into the $Z$ domain and operate to compute $\dfrac{C(z)}{R(z)}$.

I have tried to compute $G(z)$ and $H(z)$, and I think I have failed in calculating $G(z)$, because I am not getting the right result.

I obtained that $H(z)=\dfrac{1}{1-z^{-1}}$ (that I think is correct).

Therefore, my question is how to obtain $G(z)$, I know that $\boxed{e^{sT}=z}$, but I don't know how to relate it with it, because initially I obtained: $G(z)=\dfrac{z^{-1}}{1-e^{-3}z^{-1}}$ but apparently is not correct. What am I doing wrong?

Note: in the statement of the exercise I am told to consider the period of one second (T=1s).

Solution: $$\dfrac{C(z)}{R(z)}=\dfrac{(1-z^{-1})z^{-1}}{1-(1+e^{-3})z^{-1}+(\dfrac{2}{3}e^{-3}+\dfrac{1}{3})z^{-2}}$$

  • $\begingroup$ With that $H(z)$, it means you've used the substitution $s=1-z^{-1}$. Is that correct? A more usual transformation is $$s=\frac{2}{T} \frac{z-1}{z+1}$$. $\endgroup$ – Peter K. Apr 27 '16 at 14:17
  • $\begingroup$ @PeterK.: I used a Z-transform table, dii.unisi.it/~control/sdc/altro/TabellaTrasformataZ.pdf $\endgroup$ – Airman01 Apr 27 '16 at 14:21
  • $\begingroup$ OK! That should be OK. $\endgroup$ – Peter K. Apr 27 '16 at 14:28
  • $\begingroup$ Shouldn't it be $$G(z)=\dfrac{z^{\color{red}{-1}}}{1-e^{-3}z^{-1}}$$ ? You appear to have missed the minus sign in front of the $T$. $\endgroup$ – Peter K. Apr 27 '16 at 14:29
  • $\begingroup$ Yeah, there was a typo in $G(z)$ (I just fixed it), that was what I obtained. But it's wrong anyway. @PeterK. $\endgroup$ – Airman01 Apr 27 '16 at 14:32

We have $$G(z)=\dfrac{z^{-1}}{1-e^{-3}z^{-1}}$$ and $$H(z)=\dfrac{1}{1-z^{-1}}$$ and we want to calculate $$\dfrac{C(z)}{R(z)}=\dfrac{G(z)}{1+H(z)G(z)}$$

So let's do that: $$ \require{cancel} \dfrac{\dfrac{z^{-1}}{1-e^{-3}z^{-1}}}{1+\dfrac{1}{1-z^{-1}}\dfrac{z^{-1}}{1-e^{-3}z^{-1}}} = \frac{z^{-1}(1-z^{-1})}{(1-z^{-1})(1-e^{-3}z^{-1}) + z^{-1}}\\ = \frac{z^{-1}(1-z^{-1})}{1 - z^{-1} - e^{-3}z^{-1} + e^{-3}z^{-2} + z^{-1}} = \frac{z^{-1}(1-z^{-1})}{1 \cancel{- z^{-1}} - e^{-3}z^{-1} + e^{-3}z^{-2} \cancel{+ z^{-1}}}\\ = \frac{z^{-1}(1-z^{-1})}{1 - e^{-3}z^{-1} + e^{-3}z^{-2} } $$ so it looks like either the solution you have is wrong, or that $G(z)$ is not right.

So, let's see what could be wrong. Either $$ z \not = e^{sT}\tag{1} $$ or $$ s \not = 1 - z^{-1}\tag{2} $$ or $$ G(z) \not = \dfrac{1}{1-e^{-3}z^{-1}} \cdot {\cal T}_{s \rightarrow z}(e^{-s})\tag{3} $$ where ${\cal T}_{s \rightarrow z}$ is the $s$ to $z$ conversion.

Comparing (1) and (2), it looks like (1) is assuming the starred transform and (2) is assuming the impulse invariance method --- which can't both be right.

  • $\begingroup$ That $G(z)$ is not right. That's actually my question, learning how to compute the Z-transform of $G(s)$ because the one I provided is the result of my wrong computation. $\endgroup$ – Airman01 Apr 27 '16 at 17:41
  • $\begingroup$ @Airman01 Any comment on my update? $\endgroup$ – Peter K. Apr 27 '16 at 18:10
  • $\begingroup$ Hmmm, thank for taking your time answering :) I'll ask my professor, to ask him how he got the $G(z)$ transformation. And I'll post the solution, to compare it. I'm a bit lost with this topic... Thank you anyway. $\endgroup$ – Airman01 Apr 28 '16 at 7:19

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