Electronic – How an equation in s domain is converted into z domian using zero hold method

control system

I am just reading a research paper on matrix converter and i came through transfer function conversion from s domain into z domain. which i am unable to understand. how the equation 3.4 is transferred into 3.5 using zero hold method. can anyone explain step by step? attached is the screenshot.

equation 3.4 and 3.5

Best Answer

The Laplace TF of a ZOH is usually taken as \$\small \dfrac{1-e^{-sT}}{s}\$, where \$\small T\$ is the sampling increment. Multiply this by \$\small G_p(s)\$, giving $$\small G^*_p(s)=(1-e^{-sT}) \frac{G_p(s)}{s}$$

Since \$\small e^{-sT}\$ and \$\small z^{-1}\$ represent pure time delays of \$\small T\$ in the s and z domains, respectively, we may write:

$$ \small 1-e^{-sT} \rightarrow \dfrac{z-1}{z}$$

The z-transform of the remaining term, \$\small \dfrac{G_p(s)}{s}\$, may be obtained from the tables {http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html}

Working (very) approximately from your equations to check the result, the original Laplace TF may be simplified to: $$\small G_p(s)\approx\dfrac{ \omega_n^2}{s^2+\omega_n^2}$$

with \$\small \zeta=0.017\approx0\$; \$\small \omega_n=7\times 10^3rad/s\$; \$\small \omega_n\:T=0.547rad\$ \$\small\approx30^0\$.

Hence, after z-transforming \$\small \dfrac{G_p(s)}{s}\$ and multiplying by \$\small \dfrac{z-1}{z}\$, the approximate z-TF is: $$\small G^*_p(z)\approx\frac{0.13z+0.13}{z^2-1.73z+1}$$

Which compares favourably with the answer you give.