SC · Control

The Nyquist Plot

Stability becomes a relationship to a single point, −1. Draw the open loop L(jω) in the complex plane along the frequency ω and it forms a curve, and how many times that curve encircles the −1 point tells the number of right-half-plane poles of the closed loop. Move the gain K and watch the curve pass to the right of −1, through −1, and then encircle it on the left.

See whether the curve encircles −1

This is the Nyquist plot of the same open loop L(jω) = K/(jω(jω+1)(jω+2)) as F1, F2, and F3, drawn in the complex plane. A point moves along the frequency ω and traces the gold curve. The red × is the −1 point. Where this curve crosses the negative real axis is at −K/6, at the ω=√2 where the phase reaches −180°. Raise the gain K and the whole curve grows, pushing that crossing to the left. If the crossing is to the right of −1 (K<6) the curve does not encircle −1, so stable; through −1 exactly (K=6) is marginal; past it to the left (K>6) the curve encircles −1, unstable.

complex plane (× = −1, curve = L(jω))

gain K8.0

Relation to −1

K = 8.0 · crossing = -1.33 · GM = 0.75×

N = 2 · Z = 2 (P = 0)

The curve crosses the negative real axis to the left of −1 (at −K/6 < −1) and encircles the −1 point. In the Nyquist criterion this encirclement is counted as a right-half-plane pole of the closed loop. The full curve encircles −1 twice, so there are two right-half-plane poles, the same instability as F1’s two Routh sign changes and F2’s complex pole pair.

Unstable (−1 encircled)

Drawing the loop in the complex plane

The Bode plot drew the magnitude and phase of the open loop L(jω) separately against frequency. The Nyquist plot combines the two into one picture. At each frequency ω, L(jω) is a single complex number with a magnitude and a phase, so you plot it as one point in the complex plane. Sweep ω from zero to infinity (and symmetrically to negative ω) and the point traces a curve. The instant where the Bode phase is −180° and the magnitude is 1 is, here, the instant the curve touches exactly the −1 point. So every question of stability gathers into the relation between this one point, −1, and the curve.

ObserveNyquist plot = L(jω) in the complex plane

Plot L(jω) as points in the complex plane.

ChooseZ = N + P : count encirclements of ?

Z = N + P, count encirclements of −1.

Fill inopen-loop stable: stable ⟺ encirclements = ?

If open-loop stable, no encirclement means stable.

On your owncurve passes through −1 at K = ?

The curve hits −1 at K=6.

Count the encirclements of −1

The Nyquist criterion is summed up in one equation: Z = N + P. Here Z is the number of right-half-plane poles of the closed loop (the value we want to be 0), P is the number of right-half-plane poles of the open loop, and N is the number of clockwise encirclements of the −1 point by the L(jω) curve. The closed-loop poles are the roots of 1 + L(s) = 0, and the L curve encircling −1 is the same as the 1 + L curve encircling the origin, so by the argument principle the roots inside are counted. If the open loop is stable (P = 0), the stability condition is simply N = 0: never encircle −1. This example is such a case. With K < 6 the curve passes to the right of −1 and does not encircle it, so stable; with K > 6 it encircles −1 twice, giving Z = 2, two right-half-plane poles, unstable.

Where Nyquist sees further

Nyquist’s power shows when P is not zero, that is for a system whose open loop is already unstable. For such a system the plain margins of a Bode plot cannot judge stability correctly, but Nyquist answers exactly with Z = N + P. For example, if the open loop has one right-half-plane pole (P = 1), then for the closed loop to be stable (Z = 0) the curve must encircle −1 exactly once counterclockwise (N = −1). The D4 situation of stabilizing an unstable plant with feedback, Nyquist explains by the direction and number of encirclements of −1. And the stability margins reappear here too: the phase margin is the angle from where the curve meets the unit circle to the negative real axis, and the gain margin is the distance from where the curve meets the negative real axis to −1. In the end, how closely the curve grazes −1 tells the robustness at a glance.

Back to the first screen

Drawing the same open loop L(jω) as F1, F2, and F3 in the complex plane, we saw stability as a single relation to the −1 point. The curve crossed the negative real axis at −K/6. When K was below 6 that crossing was to the right of −1, so the curve did not encircle −1 and it was stable; at K = 6 it passed exactly through −1, marginal; when K was above 6 the crossing went left of −1 and the curve encircled −1 twice, two right-half-plane poles, unstable. Routh by coefficients, the root locus by the path of the poles, the Bode plot by the frequency margins, and now Nyquist by the count of encirclements of −1, all point to the same boundary K = 6 of the same system, a fourth way of seeing. And Nyquist alone can answer even when the open loop is already unstable.

The Nyquist plot is the curve drawn by the open loop L(jω) in the complex plane along the frequency ω. The Nyquist criterion: Z = N + P — Z = closed-loop RHP poles, P = open-loop RHP poles, N = clockwise encirclements of −1 by the curve. The closed-loop poles are the roots of 1+L=0, so encircling −1 = 1+L encircling the origin (the argument principle). If the open loop is stable (P=0), no encirclement of −1 (N=0) means stable. This example L=K/(s(s+1)(s+2)) crosses the negative real axis at −K/6, so K<6 is right of −1 (stable), K=6 passes through −1 (marginal, GM=0), K>6 encircles −1 twice (2 RHP poles, unstable). The same boundary K=6 as F1, F2, F3, which Nyquist generalizes to the case P≠0.

On to the next unit

So far, with Routh, the root locus, the Bode plot, and Nyquist, we analyzed whether a given loop is stable and how robust it is. All of it was proportional control, adjusting only the single gain K. But proportional alone can neither drive the steady-state error to zero nor shape the transient at will. The next unit, PID control, adds dynamics to the controller itself. The integral term keeps accumulating as long as an error remains and erases the steady-state error, while the derivative term reads the change ahead of time and adds damping. By tuning the gains of the proportional, integral, and derivative terms, we step into designing the poles and margins we have been seeing directly into the places we want.