From S to Z plane transformation

This article cannot start without a short exposé of the s‑plane. It is a map of all possible complex frequencies s = σ + jω. The horizontal axis (σ) tells you whether a signal grows or decays over time. The vertical axis (jω) tells you how fast it oscillates — its frequency.

A pole, marked with ×, is a point where the system naturally resonates. If the pole lies in the left half‑plane (σ negative), that resonance dies out on its own — a stable system. If the pole lies in the right half‑plane (σ positive), the resonance grows without bound — unstable. If the pole sits exactly on the imaginary axis (σ = 0), the system oscillates forever at a constant amplitude, like an ideal oscillator.

Because real systems have real coefficients, complex poles always appear in conjugate pairs — one above the real axis, one below. That's why one sees them as symmetric pairs, like the green ×s in the figure. Real poles (those on the horizontal axis) stand alone — like the red × in the figure.

Zeros, marked with ○, are points where the transfer function becomes zero. They do not cause instability. Instead, they shape the frequency response — cutting off certain frequencies or boosting others. If poles are the engine of the system, zeros are the sculptor.

Understanding the s‑plane starts with an observation: left means stable, right means unstable. The rest — damping, natural frequency, response shape — follows directly from where the poles sit on this map.

The mapping from the s-plane (continuous-time) to the z-plane (discrete-time) is a fundamental concept in digital signal processing. It enables the transformation of analog systems into their discrete-time equivalents while preserving stability — meaning that a stable analog system always maps to a stable digital system, and an unstable one maps to an unstable one.

Unlike the Laplace transform in the s-domain which represents continuous systems, the Z-transform (z-domain) represents discrete systems. The relationship that bridges these two domains is:

$z = e^{sT}$

where T is the sampling period.

The z-plane is the discrete-time counterpart of the s-plane. Here, stability is determined by the unit circle — the circle of radius 1 centered at the origin. If a pole lies inside the unit circle ($|z| < 1$), the system is stable. If it lies outside ($|z| > 1$), the system is unstable. Poles exactly on the unit circle ($|z| = 1$) represent oscillatory behavior — marginally stable.

The transformation $z = e^{sT}$ has a crucial property: the left half of the s-plane ($σ < 0$) maps to the inside of the unit circle in the z-plane. The right half ($σ > 0$) maps to the outside. And the imaginary axis ($σ = 0$) maps onto the unit circle itself.

This is why the green ×s (stable poles in the left half-plane) end up inside the dotted circle in the z-plane, while the red × (unstable pole in the right half-plane) ends up outside. The transformation preserves stability — and that is the beauty of the z-transform.

Just like in the s-plane, complex poles in the z-plane appear in conjugate pairs for real systems. Their angular position (argument) determines the frequency of oscillation, and their radial distance from the origin determines the rate of decay or growth.

Figure: Mapping of poles from s-plane to z-plane (z = e^{sT}, T = 1)

This figure shows how pole locations in the continuous-time domain (s-plane) map into the discrete-time domain (z-plane), which determines system stability:

• Green × (left half-plane, σ = -0.4, ω = ±0.6): Complex conjugate pole pair with negative real part. Maps inside the unit circle (|z| ≈ 0.67). → Stable system (exponentially decaying impulse response).
• Red × (right half-plane, σ = +0.3, ω = 0): Real pole with positive real part. Maps outside the unit circle (z ≈ 1.35). → Unstable system (exponentially growing impulse response).
• Dotted circle: Unit circle (|z| = 1) – the stability boundary in the z-plane.
• Solid/dashed gray lines: Real and imaginary axes in both domains.

Key insight: The left half of the s-plane (σ < 0) maps inside the unit circle → stability. The right half (σ > 0) maps outside → instability. The imaginary axis (σ = 0) maps onto the unit circle → marginal stability.

Mapping the s‑plane to the z‑plane

The core relationship that maps the s-plane to the z-plane is:

$z = e^{sT}$

where $T$ is the sampling period — the time between two consecutive samples. It is the reciprocal of the sampling frequency: $T = 1 / F_s$, where $F_s$ is measured in samples per second (Hz).

Substituting $s = \sigma + j\omega$ into $z = e^{sT}$ gives $z = e^{(\sigma + j\omega)T} = e^{\sigma T} \cdot e^{j\omega T}$.

This factorization separates the transformation into two independent parts:

• $e^{\sigma T}$ — a real number that determines the radius of the pole in the z-plane. The radius is the distance from the pole to the origin $(0, 0)$ of the z-plane. As the phase changes, this radius defines a circular trajectory around the origin.
• $e^{j\omega T}$ — a complex number on the unit circle that determines the angle (argument) of the pole. The angle is measured from the positive real axis (the horizontal axis) to the vector connecting the origin to the pole. A positive angle means counter‑clockwise rotation; a negative angle means clockwise rotation.

The term $e^{j\omega T}$ is handled by Euler's identity: $e^{j\omega T} = \cos(\omega T) + j \sin(\omega T)$.

Putting both parts together, a pole at $s = \sigma + j\omega$ maps to a pole in the z-plane with:

• Radius: $r = e^{\sigma T}$ (distance from the origin)
• Angle: $\theta = \omega T$ in radians (angle from the positive real axis)

Let us walk through the actual calculation for the two poles shown in the figure. Assume a sampling period of $T = 1$ second.

Example 1 — Stable complex pole pair:
The pole is at $s = -0.4 + j0.6$.

We start with the mapping formula:

$z = e^{sT} = e^{(-0.4 + j0.6) \cdot 1} = e^{-0.4} \cdot e^{j0.6}$

Using Euler's identity: $e^{j0.6} = \cos(0.6) + j \sin(0.6)$

Therefore:

$z = e^{-0.4} \cdot \bigl(\cos(0.6) + j \sin(0.6)\bigr)$

Now substitute the numerical values:
$e^{-0.4} \approx 0.6703$
$\cos(0.6) \approx 0.8253$
$\sin(0.6) \approx 0.5646$

$z \approx 0.6703 \cdot (0.8253 + j0.5646) = 0.553 + j0.378$

From this complex number we extract:
Radius: $|z| = \sqrt{0.553^2 + 0.378^2} = 0.6703$ (distance from the origin)
Angle: $\theta = \arctan\left(\dfrac{0.378}{0.553}\right) \approx 0.6$ radians (angle from the positive real axis)

The conjugate pole at $s = -0.4 - j0.6$ gives $z \approx 0.553 - j0.378$ — symmetric below the real axis, with the same radius $0.6703$ and angle $\theta = -0.6$ radians.

Example 2 — Unstable real pole:
The pole is at $s = 0.3 + j0$.

$z = e^{(0.3 + j0) \cdot 1} = e^{0.3} \cdot e^{j0} = e^{0.3} \cdot 1 = e^{0.3}$

Numerically: $e^{0.3} \approx 1.3499$

Thus $z \approx 1.3499 + j0$ — on the positive real axis, at distance $1.3499$ from the origin, outside the unit circle. Its angle is $0$ radians (pointing along the positive real axis).

Summary of the mapping:

• Radius: $r = e^{\sigma T}$ — determined only by the real part $\sigma$ and the sampling period $T$. This is the distance from the pole to the origin of the z-plane.
• Angle: $\theta = \omega T$ (in radians) — determined only by the imaginary part $\omega$ and the sampling period $T$. This is the angle measured from the positive real axis to the vector pointing to the pole.
• Stability condition: $r < 1$ ⇔ $e^{\sigma T} < 1$ ⇔ $\sigma < 0$ — exactly the condition for a stable continuous-time system (poles in the left half-plane map to poles inside the unit circle).

This step-by-step calculation shows why the green poles (with $\sigma = -0.4$) end up inside the unit circle and the red pole (with $\sigma = +0.3$) ends up outside. The transformation preserves stability in a very transparent way.

Exercises

For deeper understanding of the material covered in this article, it is strongly recommended that you work through the following exercises. Each exercise presents a different pole location in the s-plane. Apply the mapping $z = e^{sT}$ with $T = 0.1$ seconds ($F_s = 10$ Hz).

Exercise 1
$s = -0.5 + j1$
Calculate the radius $r$, the angle $\theta$, and the mapped pole $z = a + jb$.

Exercise 2
$s = -0.3 + j0.8$
Calculate the radius $r$, the angle $\theta$, and the mapped pole $z = a + jb$.

Exercise 3
$s = +0.2 + j0.5$
Calculate the radius $r$, the angle $\theta$, and the mapped pole $z = a + jb$. Is this system stable?

Exercise 4
$s = -0.6 + j0$ (real pole)
Calculate the radius $r$ and the mapped pole $z$. Is this system stable?

Exercise 5
$s = 0 + j1.2$ (pole on the imaginary axis)
Calculate the radius $r$, the angle $\theta$, and the mapped pole $z$. Where does this pole land in the z-plane?

Answers: Solutions to these exercises are provided on the next page.

Reference: For further reading, see Bob Meddins, Introduction to Digital Signal Processing, Butterworth-Heinemann, 2000.