Poles and zeros
A constant and a handful of points of the plane contain the whole circuit: the frequency response, the step response and the stability all come from reading where those points are.
01The two numbers that define a circuit
A lumped-parameter transfer function is always a quotient of polynomials, and every polynomial can be factored. So:
That factorization says something powerful: a constant and a handful of points of the plane contain all the information about the circuit. The frequency response, the step response, the stability and even the settling time come from reading where those points are.
The real part of a pole tells how fast that component of the response dies out: a pole at \( \sigma \) contributes \( e^{\sigma t} \). The imaginary part tells at what frequency it oscillates. A real pole gives an exponential; a conjugate pair, a damped sinusoid.
02The frequency response, read with a ruler
To obtain the frequency response, \( H \) is evaluated on the imaginary axis, that is, at \( s = j\omega \). Taking the modulus and phase of the factorization:
Each factor \( j\omega - p_k \) is the vector that goes from the pole to the operating point on the axis. The magnitude of the response is the product of the distances to the zeros divided by the product of the distances to the poles. With a ruler and a protractor on the diagram you can estimate the Bode plot without computing anything.
From this come the two rules that an electronics engineer uses all the time: bringing a pole closer to the imaginary axis raises a peak at that frequency, because the denominator distance becomes small; and a zero on the axis produces a null, because the numerator vanishes.
On the left, the \( s \)-plane with the poles (×) and zeros (○), and the vectors going to the operating point. On the right, the magnitude in decibels and the phase. Click or tap the plane to move the pole pair.
03Stability: a single line divides the plane
The natural response of a system is the sum of the terms \( e^{p_k t} \), one per pole. Whether it dies out or not depends only on the sign of the real part:
| Where the pole is | What it contributes to the response | Result |
|---|---|---|
| \( \operatorname{Re}(p) \lt 0 \) | \( e^{\sigma t} \) with \( \sigma \lt 0 \) | Dies out · stable |
| \( \operatorname{Re}(p) = 0 \), simple | Oscillation of constant amplitude | Marginally stable |
| \( \operatorname{Re}(p) = 0 \), double | \( t\cos\omega t \), grows | Unstable |
| \( \operatorname{Re}(p) \gt 0 \) | \( e^{\sigma t} \) with \( \sigma \gt 0 \) | Grows without bound · unstable |
Zeros do not affect stability: they can be anywhere. They do affect the shape of the response, and a zero in the right half-plane —what is called non-minimum phase— makes the output start off in the wrong direction before correcting. This is what happens in a boost converter and is one of the reasons they are hard to compensate.
A sinusoidal oscillator needs a pair of poles exactly on the imaginary axis. Since that cannot be sustained with real components, they are designed slightly to the right and an amplitude control is added —a small lamp, a JFET, a limiter— that pushes them toward the axis when the amplitude grows. The circuit settles by itself onto the line.
04The cases you have to recognize by heart
| Function | Poles and zeros | What it does |
|---|---|---|
| \( \dfrac{1}{s + \omega_0} \) | One real pole | Low-pass, −20 dB/dec, −45° at \( \omega_0 \) |
| \( \dfrac{s}{s + \omega_0} \) | A zero at the origin | First-order high-pass |
| \( \dfrac{\omega_0^2}{s^2 + \frac{\omega_0}{Q}s + \omega_0^2} \) | Conjugate pair | Low-pass with a peak if \( Q \gt 0.707 \) |
| \( \dfrac{s^2 + \omega_0^2}{s^2 + \frac{\omega_0}{Q}s + \omega_0^2} \) | Zeros on the axis | Notch: exact null at \( \omega_0 \) |
| \( \dfrac{s - a}{s + a} \) | Zero and pole in mirror image | All-pass: flat magnitude, only the phase changes |
The all-pass deserves attention because it is counterintuitive: it has a zero in the right half-plane and yet is perfectly stable, because the pole is on the left. Its magnitude is 1 at all frequencies and all it does is delay. It is used to equalize delays and to correct phase without touching the amplitude.
05In the lab
Build a series RLC with variable \( R \) and measure the frequency response. Locate the peak, measure its height relative to the passband and solve for \( Q \). Compare with \( Q = \frac{1}{R}\sqrt{L/C} \) and locate the poles in the plane.
Design a notch at 50 Hz with \( Q = 10 \), compute the positions of its two zeros and two poles, and verify in the lab how many decibels of attenuation it gives at 50 Hz and how much it affects 45 Hz and 55 Hz.
Build a first-order all-pass with an operational amplifier and measure magnitude and phase between \( 0.1\,a \) and \( 10\,a \). Check that the magnitude stays flat while the phase sweeps 180°.
06Common mistakes
- Believing that a zero in the right half-plane makes the system unstable. Stability is decided by the poles, nothing else.
- Forgetting the zeros at infinity. If the denominator has a higher degree than the numerator, there are zeros at infinity, and that fixes the final slope of the Bode plot.
- Confusing \( \omega_0 \) with the frequency of the peak. The peak is at \( \omega_0\sqrt{1 - 1/(2Q^2)} \), slightly below.
- Drawing a conjugate pair without its mirror partner. With real coefficients, complex poles always come in pairs.
- Canceling an unstable pole with a zero. On paper they cancel; in the real circuit the cancellation is never exact and the unstable mode is still there.
07Self-assessment
What is a pole and what is a zero?
A zero is a value of \( s \) where \( H \) vanishes; a pole, where it becomes infinite. They are the roots of the numerator and of the denominator.
How do you estimate \( |H(j\omega)| \) by looking at the diagram?
Product of the distances to the zeros divided by the product of the distances to the poles, all times \( |K| \).
A pole at \( -0.1 \pm j10 \): what response does it give?
An oscillation of 10 rad/s that decays with a time constant of 10 s. It is a high \( Q \), around 50: in frequency it will show a pronounced peak.
Why does the all-pass have constant magnitude?
Because the zero and the pole are mirror images across the imaginary axis: for each \( j\omega \) the two distances are equal and the quotient is 1.
What does bringing a pole closer to the imaginary axis do to the circuit?
It raises a peak at that frequency and makes the time response take longer to die out. On the axis, it oscillates without damping.
What is a non-minimum-phase system?
One with zeros in the right half-plane. It is stable, but its output starts off in the opposite direction before heading to the final value.
08Further reading
- Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1998. The Laplace chapter, with the relationship between pole location and time response.
- Adel S. Sedra and Kenneth C. Smith. Microelectronic Circuits. 7th ed. (Spanish edition, “Circuitos microelectrónicos”), Oxford University Press, 2016. The filters chapter: each classic response with its pole-zero diagram alongside.
- Katsuhiko Ogata. Modern Control Engineering. 5th ed. (Spanish edition, “Ingeniería de control moderna”), Pearson, 2010. The root locus, which is the study of how the poles move as a gain is varied.