--- title: "Mapping an s-plane region to the z-plane" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:map-s-plane-region" source: https://www.bctengineer.com/q/semester_6/digital_control_system/map-s-plane-region site: BCT Engineering --- # Mapping an s-plane region to the z-plane Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2082 Chaitra (new course), 3 marks; 2082 Kartik, 3 marks (asked 2 times) - Video: [Mapping Between s plane and z plane](https://www.youtube.com/watch?v=RIkqWlFW7Bg) — Yasser Elshrief - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/map-s-plane-region - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Map the following system in Z-plane. [Figure: s-plane with a shaded rectangle between σ=-2 and σ=-1, extending from jω=-ωₛ/4 to +ωₛ/4; corner/edge points a (σ=-1, -ωₛ/4), b (σ=-1, 0), c (σ=-1, +ωₛ/4), d (σ=-2, +ωₛ/4), e (σ=-2, 0), f (σ=-2, -ωₛ/4)] ## Short answer Using z=e^{Ts}: lines σ=-1 and σ=-2 become arcs of radii e^{-T} and e^{-2T}, ω=±ω_s/4 become ±90° radial lines. The shaded rectangle maps to a half-annulus: e^{-2T} ≤ |z| ≤ e^{-T}, -90° ≤ ∠z ≤ 90°, inside the unit circle. ## Answer 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 | Point | $s$ | $\lvert z\rvert = e^{\sigma T}$ | $\angle z$ | |---|---|---|---| | a | $-1 - j\omega_s/4$ | $e^{-T}$ | $-90^\circ$ | | b | $-1$ | $e^{-T}$ | $0^\circ$ | | c | $-1 + j\omega_s/4$ | $e^{-T}$ | $+90^\circ$ | | d | $-2 + j\omega_s/4$ | $e^{-2T}$ | $+90^\circ$ | | e | $-2$ | $e^{-2T}$ | $0^\circ$ | | f | $-2 - j\omega_s/4$ | $e^{-2T}$ | $-90^\circ$ | #### 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. ```text 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 = $-j0.135$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With ws = 2 pi / T, what angle does omega = +ws/4 give in the z-plane? A. +45 degrees B. +90 degrees C. +180 degrees D. +360 degrees Answer: B. angle = omega T = (ws/4)(2 pi/ws) = pi/2 = 90 degrees. (Key point: Angle of omega = ws/4) 2. What radii do the lines sigma = -1 and sigma = -2 map to? A. e^(-T) and e^(-2T) B. e^(T) and e^(2T) C. 1 and 2 D. -T and -2T Answer: A. |z| = e^(sigma T), so sigma = -1 gives e^(-T) and sigma = -2 gives e^(-2T). (Key point: Radii from sigma) 3. What shape is the mapped region? A. A full ring between e^(-2T) and e^(-T) B. A rectangle in the z-plane C. The right half of an annulus, e^(-2T) <= |z| <= e^(-T), between -90 and +90 degrees D. A region outside the unit circle Answer: C. Vertical edges become arcs, and the horizontal edges at +/- ws/4 become radial segments on the imaginary axis, giving a half-ring inside the unit circle. 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