Converting from S-Domain Transfer Function to Z-Domain Transfer Function

9.2k Views Asked by At

I have the following transfer function in the S-Domain:

$$G(s)=\frac{a}{s+a}$$

I want to determine a transfer function in the Z-Domain that does a good job of modelling this system for a discrete time approximation.

Should I use the bilinear transformation:

$$s=\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}$$

$$G(z)=G(\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}})$$

the impulse invariant transformation

$$G(z)=Z(L^{-1}(G(s))_{t=nT})$$

the step invariant transformation:

$$G(z)=(1-z^{-1})Z(L^{-1}(\frac{G(s)}{s})_{t=nT})$$

or something else? (In the above, $Z$ is for Z-Transform and $L$ is for Laplace transform, I couldn't figure out how to do the fancy symbols)

What are the instances when I should use each type of transformation?