Mapping from s-plane to z-plane

The s-plane and z-plane are linked by the sampling relation $z = e^{Ts}$, where $T$ is the sampling period. The inverse relation is $s = \dfrac{1}{T}\ln z$.

Mapping $s \to z$

Put $s = \sigma + j\omega$:

$$z = e^{T(\sigma + j\omega)} = e^{\sigma T}e^{j\omega T} \;\Rightarrow\; |z| = e^{\sigma T}, \quad \angle z = \omega T$$

As $\omega$ goes from $-\omega_s/2$ to $\omega_s/2$ (with $\omega_s = 2\pi/T$), the angle $\omega T$ goes from $-\pi$ to $\pi$, covering the circle once. This band is the primary strip. Each complementary strip above or below it maps onto the same z-plane again. So the mapping is many-to-one, and frequencies differing by $\omega_s$ cannot be told apart (aliasing).

   s-plane (primary strip)       z-plane
   jw
   | ws/2 ---------             .---.
   |   LHP  | RHP   ==>       /  in  \  out
 --+--------+---> sigma      | (LHP)  |
   |        |                 \      /
   | -ws/2 ---------            '---'  |z|=1

Mapping $z \to s$

$$s = \frac{1}{T}\ln z = \frac{1}{T}\left[\ln|z| + j\angle z\right]$$

This gives the principal value in the primary strip.

Example: $T = 0.1$ s and $s = -2 + j10$. Then $z = e^{-0.2}\angle 1\text{ rad} = 0.819\angle57.3^\circ$. Going back: $\sigma = \ln(0.819)/0.1 = -2$ and $\omega = 1/0.1 = 10$ rad/s.

Use

The left half-plane maps to the inside of the unit circle. So a discrete system is stable when all its closed-loop poles lie inside $|z| = 1$. Specifications such as $\zeta$, $\omega_n$ and settling time can also be carried from the s-plane to the z-plane for design.