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 single points, its poles. This unit shows that where such a point sits in the plane decides whether the response dies away or blows up.

Which side of the imaginary axis is the pole on?

On the left is the complex plane with a pair of poles marked, and on the right is the time response those poles produce. Change the pole's real part σ: to the left of the imaginary axis the response rings and gradually dies down, while to the right it grows out of control. 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. The circuit's differential equation therefore turns into an algebraic equation in s, and the values of s that make its denominator 0 are the poles. Each pole, s = σ + jω, gives rise to one piece of response in the time domain, e^(σt)cos(ωt). The natural behavior of the whole circuit is the sum of such pieces.

The real part decides life or death

In the response piece e^(σt)cos(ωt), the cos merely swings between −1 and 1; what grows or shrinks the size is the e^(σt) in front. If σ is negative this envelope converges to 0 and the response dies away (stable); if σ is positive the envelope grows without bound and the response diverges (unstable). If σ is exactly 0 the envelope stays constant at 1, a boundary state in which the oscillation neither shrinks nor grows.

The imaginary part is the ringing rate

The imaginary part of the pole, ω, sets how fast the response swings. A large ω means fast oscillation, and an ω of 0 means the response changes only exponentially, with no oscillation. So from the position (σ, ω) of a pole pair alone, you can read at a glance whether the circuit is stable, that is, whether the poles lie left of the imaginary axis, and how much it rings. This plot fixes ω at 6 rad/s and moves only σ. The jω response of the sinusoidal steady state is the special case of bringing a pole onto the imaginary axis.

Observes = σ + jω → |resp| ∝ e^σt

The real part forms the envelope.

Chooseσ < 0 → e^σt ?

In the left half-plane it dies away to 0.

Fill in → σ = ?

The imaginary axis is the stability boundary.

On your own → ?

The imaginary part is the pace of the oscillation.

Back to the first screen

With the poles on the left, the time response rang yet gradually died down; moved to the right, the same oscillation grew out of control. On the imaginary axis σ = 0, it sat at the boundary where the oscillation neither shrank nor grew. The big question of whether a circuit is stable came down to a single thing: which side of the plane its poles lie on.

The Laplace transform turns the derivative d/dt into multiplication by s, so a circuit can be solved with algebra. The poles s = σ + jω that make the denominator 0 set the time response e^σtcos(ωt), and among them the real part σ decides stability. With σ < 0 (left half-plane) the response dies away and is stable; σ = 0 (imaginary axis) is the boundary; σ > 0 (right half-plane) diverges and is unstable. The imaginary part ω sets the pace of the oscillation.