Mapping an s-plane region to the z-plane

Use $z = e^{Ts}$ with $s = \sigma + j\omega$. Then $|z| = e^{\sigma T}$ and $\angle z = \omega T$. Since $\omega_s = 2\pi/T$, the frequency $\omega = \pm\omega_s/4$ gives the angle

$$\omega T = \pm\frac{\omega_s}{4}\cdot\frac{2\pi}{\omega_s} = \pm\frac{\pi}{2} = \pm90^\circ$$

Mapping of the corner and edge points

Mapping of the edges

• The line $\sigma = -1$ (a-b-c) maps to an arc of radius $e^{-T}$ from $-90^\circ$ to $+90^\circ$.
• The line $\sigma = -2$ (f-e-d) maps to an arc of radius $e^{-2T}$ from $-90^\circ$ to $+90^\circ$.
• The lines $\omega = \pm\omega_s/4$ (c-d and f-a) map to the radial segments along the positive and negative imaginary axes, between the two radii.

Mapped region

The shaded rectangle maps to the right half of an annulus (a half-ring) lying between the circles $|z| = e^{-2T}$ and $|z| = e^{-T}$, for $-90^\circ \le \angle z \le 90^\circ$. It lies entirely inside the unit circle.

          Im z
           |
           c        c: r=e^-T,  +90 deg
           |  .
           d    .   d: r=e^-2T, +90 deg
           | .   .
  ---------+--e---b------- Re z
           | .   .  e: e^-2T  b: e^-T
           f    .
           |  .     f: r=e^-2T, -90 deg
           a        a: r=e^-T,  -90 deg
           |
  Shaded: right half-ring, e^-2T <= |z| <= e^-T

Example with $T = 1$ s: the inner radius is $e^{-2} = 0.135$ and the outer radius is $e^{-1} = 0.368$. So b = 0.368, e = 0.135, c = $j0.368$, d = $j0.135$, a = $-j0.368$, f = $-j0.135$.