The s-Plane, Poles and Zeros

Move the pole, change the response

The upper plot is the s-plane, and the two gold × marks are a conjugate pole pair s = σ ± jω. The green-shaded left side is the stable left half-plane, the red-shaded right side is the right half-plane, and the vertical line between them is the imaginary axis. The lower plot draws the natural response e^(σt)cos(ωt) that these poles produce, from t = 0 to 6. You start at σ = 0.4, so the poles are in the right half and the response swings outward past the edge of the plot. Use the two sliders to bring the poles into the left half and watch the response settle to 0.

Infinity at a pole, 0 at a zero

A transfer function is written as a ratio of two polynomials, H(s) = N(s)/D(s). At an s where the denominator D(s) becomes 0, H shoots up to infinity; that point is called a pole and is marked × on the plane. At an s where the numerator N(s) becomes 0, H becomes 0; that point is a zero and is marked ○. The poles decide which modes (pieces of the form e^(σt)cos(ωt)) appear in the natural response, and the zeros change how much of each mode is mixed in, that is, the weights. The plot above shows only one pole pair and draws no zeros.

Real part is decay, imaginary is oscillation

A single pole s = σ + jω produces a natural response of the form e^(σt)cos(ωt). The real part σ is the decay rate of the exponential e^(σt): when σ is negative the response dies away, and the more negative σ is, the faster it dies away. The imaginary part ω is the angular frequency of the cosine, so the farther the pole moves up or down from the real axis, the faster the oscillation. Reading a pole's horizontal position on the plane therefore tells you how fast the response fades, and its vertical position tells you how fast it swings. Lower the ω slider to 0 and the two × marks merge on the real axis; the response then simply dies away or blows up, with no swinging at all.

Left half-plane means stable

If every pole of a system is to the left of the imaginary axis, in the left half-plane where the real part is negative, every mode of the natural response dies away and the system is BIBO stable. If even one pole is in the right half, that mode diverges and the system is unstable. A pole on the imaginary axis leaves an oscillation that never fades, which is marginal stability, but a bounded input at the same frequency resonates with it and drives the output up without limit, so it does not count as BIBO stable. This criterion takes the a < 0 stability condition from B4, extends it to complex poles, and condenses it into a single position on the s-plane. The poles decide stability; the zeros only change the shape of the response and cannot change stability.

Back to the first screen

On the first screen, dragging the poles left of the imaginary axis made the response die away to 0 (stable), pushing them right made the amplitude swell past the edge of the plot (unstable), and placing them on the axis made the response keep swinging at the same size (marginal). Raising ω to send the poles farther up and down from the real axis also made the oscillation faster. The real part, the pole's horizontal position, was the decay rate, and the imaginary part, its vertical position, was the angular frequency of the oscillation. How the system would move was already written in the location of one point on the s-plane.

If the poles and zeros fix nearly all of a system's properties, then the function H(s) that holds them both works like the system's ID card. In the next unit, The Transfer Function, you take the Laplace transform of a system's differential equation to set up H(s) = output/input, then multiply the transform of the input by H(s) to get the output. Convolution in the time domain becoming multiplication in the s-domain is exactly the property from C4. The poles and zeros are simply the roots you get when you factor the denominator and numerator of that H(s).