Poles, Zeros & the Complex Plane
Poles, Zeros & the Complex Plane
In advanced mathematics and engineering, poles and zeros describe where a rational function blows up or vanishes. They are the roots of the denominator and numerator, respectively, and they live in the complex $s$-plane.
A rational function (also called a transfer function in engineering) has the form:
$$H(s) = \frac{N(s)}{D(s)} = \frac{b_m s^m + b_{m-1}s^{m-1} + \cdots + b_0}{a_n s^n + a_{n-1}s^{n-1} + \cdots + a_0}$$
where $s$ is a complex variable.
The zeros of $H(s)$ are the values of $s$ where the numerator $N(s) = 0$. At these points, $H(s) = 0$. Marked with ○ on the $s$-plane.
The poles of $H(s)$ are the values of $s$ where the denominator $D(s) = 0$. At these points, $H(s) \to \infty$. Marked with × on the $s$-plane.
- Write $H(s)$ as a ratio of two polynomials $N(s)/D(s)$.
- Zeros: Set $N(s) = 0$ and solve. Use factoring, the quadratic formula, or synthetic division.
- Poles: Set $D(s) = 0$ and solve the same way.
- Check for cancellations — if a factor appears in both $N(s)$ and $D(s)$, it cancels (removable singularity).
- Plot zeros as ○ and poles as × on the complex plane.
- Stability: A system is stable if all poles have negative real parts (left half of the $s$-plane).
- Frequency response: Near a zero, the output is attenuated. Near a pole, it is amplified.
- System behavior: Complex poles produce oscillations. Real poles produce exponential decay/growth.
- Order: The number of poles determines the system order and the number of energy-storage elements.
The complex $s$-plane has the real axis (horizontal, $\sigma$) and the imaginary axis (vertical, $j\omega$).
- Left half-plane (LHP): $\text{Re}(s) < 0$ — stable poles (decaying signals)
- Right half-plane (RHP): $\text{Re}(s) > 0$ — unstable poles (growing signals)
- Imaginary axis: $\text{Re}(s) = 0$ — sustained oscillation (marginally stable)
Find the poles and zeros of $H(s) = \dfrac{s + 2}{s + 5}$.
Zeros: $s + 2 = 0 \implies s = -2$
Poles: $s + 5 = 0 \implies s = -5$
Both are real and in the LHP → stable. The zero is at $-2$ and the pole is at $-5$.
Find the poles and zeros of $H(s) = \dfrac{2s}{s^2 + 4s + 13}$.
Zero: $2s = 0 \implies s = 0$
Poles: $s^2 + 4s + 13 = 0$
Quadratic formula: $s = \dfrac{-4 \pm \sqrt{16 - 52}}{2} = \dfrac{-4 \pm \sqrt{-36}}{2} = \dfrac{-4 \pm 6i}{2}$
$$s = -2 + 3i \quad \text{and} \quad s = -2 - 3i$$
Complex conjugate poles in the LHP → stable with oscillation.
Find all poles and zeros, check for cancellations, determine stability, and sketch the pole-zero plot for:
$$H(s) = \frac{s^3 + 5s^2 + 8s + 4}{s^4 + 6s^3 + 13s^2 + 12s + 4}$$
Step 1: Factor the numerator.
Try $s = -1$: $(-1)^3 + 5(1) + 8(-1) + 4 = -1 + 5 - 8 + 4 = 0$ ✓
Divide by $(s+1)$: $s^3+5s^2+8s+4 = (s+1)(s^2+4s+4) = (s+1)(s+2)^2$
Step 2: Factor the denominator.
Try $s = -1$: $1 - 6 + 13 - 12 + 4 = 0$ ✓
Divide by $(s+1)$: $s^4+6s^3+13s^2+12s+4 = (s+1)(s^3+5s^2+8s+4)$
But we just factored $s^3+5s^2+8s+4 = (s+1)(s+2)^2$, so:
$$D(s) = (s+1)^2(s+2)^2$$
Step 3: Write the factored form.
$$H(s) = \frac{(s+1)(s+2)^2}{(s+1)^2(s+2)^2}$$
Step 4: Cancel common factors.
Cancel $(s+2)^2$ and one $(s+1)$:
$$H(s) = \frac{1}{s+1}$$
Step 5: Identify remaining poles and zeros.
Zeros: None (numerator is constant)
Poles: $s = -1$ (simple pole)
Cancelled: $s = -2$ (double, removable), $s = -1$ (one copy cancelled)
Step 6: Stability.
The only pole is at $s = -1$, which has $\text{Re}(s) = -1 < 0$ → stable.
Step 7: Interpretation.
Despite starting as a 4th-order system, pole-zero cancellation reduces it to a simple first-order system with time constant $\tau = 1$ second.
Is $H(s) = \dfrac{s - 1}{s^2 - 4}$ stable?
Zeros: $s = 1$
Poles: $s^2 - 4 = (s-2)(s+2) = 0 \implies s = 2, \; s = -2$
Pole at $s = 2$ is in the RHP ($\text{Re}(s) = 2 > 0$) → unstable.
✨ Use Sliders mode to drag and watch poles move in real-time, or Presets for specific transfer functions.
Practice Problems
Show Answer Key
1. Zero: $s = -3$. Pole: $s = -7$.
2. $s^2+6s+9 = (s+3)^2$. Double pole at $s = -3$ (repeated root).
3. Zeros: $s = \pm 1$. Poles: $s = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i$. Yes, complex conjugate pair in LHP → stable.
4. $s^2+s-6 = (s+3)(s-2)$. Poles at $s = -3$ and $s = 2$. Pole in RHP → unstable.
5. $s^2+4s = s(s+4)$. Cancel $(s+4)$: $H(s) = 1/s$. Pole at $s = 0$ (integrator, marginally stable).
6. $s = \pm 2j$ (purely imaginary — on the imaginary axis). Marginally stable (sustained oscillation).
7. $|s| = \sqrt{1+4} = \sqrt{5} \approx 2.236$
8. Cancel $(s+1)$: $H(s) = \frac{s-3}{(s+2)(s+5)}$. Zero: $s = 3$. Poles: $s = -2, -5$.
9. $s = -3 - 4j$
10. $s = \frac{-2 \pm \sqrt{4-40}}{2} = -1 \pm 3i$. Real part $= -1 < 0$ → yes, LHP.
11. $N = (s+2)^2$, $D = s(s+2)^2$. Cancel $(s+2)^2$: $H(s) = 1/s$. Zero: none. Pole: $s = 0$. Cancelled: double at $s = -2$.
12. $H(s) = Ks/((s+5)(s+10))$. DC gain: $H(0) = 0$, so DC gain of $1$ isn't achievable. With steady-state gain at $\omega = 1$: typically $K = 50$ gives $H(s) = 50s/((s+5)(s+10))$.