CR · s-domain

Poles in the s-Plane and Stability

The Laplace transform turns differentiation into multiplication by s and compresses a circuit’s behavior into a single point, a pole. Learn how where that point sits in the plane decides whether the response dies away or blows up.

Which side of the imaginary axis is the pole on?

Slide the pole’s real part σ. On the left the response fades away; on the right it grows without bound. Find the boundary between stable and unstable, the imaginary axis σ = 0.

Pole real part σσ = -0.80

Response regime

Stable · decays

Far from the boundary

The pole is the blueprint of the time response

Under the Laplace transform, the derivative d/dt becomes multiplication by s. So a circuit’s differential equation turns into an algebraic equation in s, and the values of s that make its denominator zero are the poles. Each pole s = σ + jω gives rise to one piece of the response in the time domain, e^(σt)cos(ωt). The whole circuit’s natural behavior is the sum of such pieces.

The real part decides life or death

In the response piece e^(σt)cos(ωt), the cosine merely swings between −1 and 1; what grows or shrinks the amplitude is the e^(σt) in front. If σ is negative this envelope converges to zero and the response fades (stable); if σ is positive the envelope grows without bound and it diverges (unstable). If σ is exactly zero the envelope stays at one, so the oscillation neither shrinks nor grows — the boundary state.

The imaginary part is the ringing rate

The pole’s imaginary part ω sets how fast the response oscillates. A large ω rings quickly; ω = 0 means no oscillation, only an exponential. So from the location (σ, ω) of a pole pair alone you can read at a glance whether the circuit is stable (is it on the left?) and how much it rings (how far from the imaginary axis?). The steady-state jω response is the special case of placing the pole on the imaginary axis.

Observes = σ + jω → |resp| ∝ e^σt

The real part forms the envelope.

Chooseσ < 0 → e^σt ?

Left half-plane fades to zero.

Fill in → σ = ?

The imaginary axis is the stability edge.

On your own → ?

The imaginary part is the ringing rate.

Back to the first screen

With the pole on the left, the time response oscillated yet steadily died down; moving it to the right, the same oscillation grew out of control. On the imaginary axis σ = 0 the oscillation neither shrank nor grew — the boundary. The big question of a circuit’s stability was compressed into one thing: which side of the plane the pole is on.

The Laplace transform turns the derivative d/dt into multiplication by s, solving a circuit algebraically. The poles s = σ + jω that make the denominator zero set the time response e^σtcos(ωt). The real part σ decides stability: σ < 0 (left half-plane) fades and is stable, σ = 0 (imaginary axis) is the boundary, σ > 0 (right half-plane) diverges and is unstable. The imaginary part ω is the ringing rate.