discretize a function using $z$-transform

Question

I would like to discretize the following continuous function using $z$-transform:

$$G(s)=\frac{s+1}{s^2+s+1}$$

The process I am using is to take the inverse Laplace transform of $\frac{G(s)}{s}$ and then take the z-transform of it. Finally I multiply it the result for $1-z^{1}$ to simulate a zero order hold.

The results I am obtaining are the following ones:

$$\mathcal{L} \{\frac{G(s)}{s}\}= F(t)=e^{-0.5}[\cos{\frac{\sqrt{3}t}{2}}+\frac{1}{{\sqrt{3}}} \sin{ \frac{\sqrt{3t}}{2}}]$$

$$Z\{F(t)\} = \frac{1-e^{-0.5t}z^{-1}\cos{\frac{\sqrt{3}T}{2}}}{1-2e^{-0.5t}z^{-1}\cos{\frac{\sqrt{3}T}{2}}+e^{-T}z^{-2}} + \frac{1-e^{-0.5t}z^{-1}\sin{\frac{\sqrt{3}T}{2}}}{1-2e^{-0.5t}z^{-1}\cos{\frac{\sqrt{3}T}{2}}+e^{-T}z^{-2}} $$

Definig $\alpha=e^{-0.5T}\cos \frac{\sqrt{3}T}{2}$, $\beta=\frac{e^{-0.5T}}{\sqrt{3}}e^{-0.5T}\cos \frac{\sqrt{3}T}{2}$ and by writing it in function of $z$ instead of $z^{-1}$ we have:

$$Z\{F(t)\} =\frac{ z^2+z(\alpha+\beta)}{z^2-2 \alpha z + e^{-T}}$$

Multiplying it by $ 1-z^{-1} $ we finally get:

$$\frac{z^2+z(1-\alpha- \beta)+ \alpha - \beta}{z^2-2 \alpha z + e^{-T}}$$

I already redone the calculation a number of times and I couldn't find any errors. However I know it is wrong because it is not matching the correct response to an impulse when I plot it in MATLAB.

I hope that it was clear enough. Sorry if I made any theoretical mistakes trying to explain the process, I am just a beginner at this subject.

EDIT: After taking the correct Laplace Transform presented below, the answer takes the form of: $$\frac{z(\beta - \alpha + 1)-\alpha - \beta + e^{-T}}{z^2-2 \alpha z + e^{-T}}$$

If anyone is interesd here are the plots I got from MATLAB corroborating the analysis:

The line is the step response of the continuous function, the dotted one is obtained via the c2d function in MATLAB and it is my benchmark. The points are the ones I get via my analysis showed at the top. As you can see everything fits perfectly. Thank you again for your support.

Answer 1

Here is the inverse Laplace transform

$$F(t) = 1+ \frac{{{\rm e}^{-\frac{t}{2}}}}{3}\, \left( \sqrt {3}\sin \left( \frac{\sqrt {3}t}{2} \right) -3\,\cos \left( \frac{\sqrt {3}t}{2} \right) \right) .$$

Added: The $z$-transform of $F(t)$ is given by

$$ {\frac {-z \left( -3\,z{{\rm e}^{1/2}}-3\,{{\rm e}^{1/2}} \right) \cos \left(\frac{\sqrt {3}}{2} \right) -z \left( \sqrt {3}{{\rm e}^{1/2}}z- \sqrt {3}{{\rm e}^{1/2}} \right) \sin\left( \frac{\sqrt {3}}{2} \right) -z \left( 3\,{{\rm e}^{1}}z+3 \right) }{6\,{{\rm e}^{1/2}}\cos \left( \frac{\sqrt {3}}{2} \right) {z}^{2}-3\,{{\rm e}^{1}}{z}^{3}-6\,z{{\rm e}^{1/2 }}\cos\left(\frac{\sqrt {3}}{2} \right) +3\,{z}^{2}{{\rm e}^{1}}-3\,z+3}}$$