The Nyquist Plot

See whether the curve encircles −1

This is the Nyquist plot of the same open loop as in the F1, F2, and F3 units, L(jω) = K/(jω(jω + 1)(jω + 2)), drawn in the complex plane. The solid gold curve is where L(jω) passes as ω increases from 0.3 to 30, the faint gold curve is its mirror image for negative frequencies, and the red × is the −1 point. The curve crosses the negative real axis at −K/6, at ω = √2, where the phase reaches −180°. At the starting K = 8, that crossing is at −1.33, to the left of −1, so the curve encircles −1. Lower the gain K and the whole curve shrinks, pulling the crossing back to the right: for K < 6 it does not encircle −1 and the loop is stable, at K = 6 it passes exactly through −1 (marginal), and for K > 6 it encircles it again and the loop is unstable.

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 them in one picture. At each frequency ω, L(jω) is a single complex number with a magnitude and a phase, so you mark it as one point in the complex plane, and sweeping ω from 0 to infinity (and symmetrically through negative ω) makes the point trace a curve. The moment on the Bode plot when the phase is −180° while the magnitude is 1 is, here, the moment the curve touches exactly the −1 point. That is why every question about stability comes down to the relation between the curve and the single point −1.

Count the encirclements of −1

The Nyquist criterion is summed up in one equation, Z = N + P. 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 times the L(jω) curve encircles the −1 point clockwise. 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 the argument principle counts the roots inside. If the open loop is stable (P = 0), the stability condition is N = 0, that is, never encircling −1, and this example is such a case. For K < 6 the curve passes to the right of −1 without encircling it, so the loop is stable; for K > 6 it encircles −1 twice, giving Z = 2, two right-half-plane poles, and the loop is unstable.

Where Nyquist sees further

The strength of the Nyquist criterion shows when P is not 0, that is, in systems whose open loop is already unstable. For such systems the margins on a Bode plot alone cannot judge stability correctly, but the Nyquist criterion answers exactly through Z = N + P. 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), and the unstable plant of the D4 unit with gain k applied, L = k/(s − 1), is exactly such a case. Its curve is a circle with −k and 0 as the ends of a diameter, so for k > 1 it encircles −1 once counterclockwise, which is why the closed loop becomes stable for k > 1. The stability margins also reappear here: 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, expressed as a factor. Seeing how closely the curve brushes past −1 gives a quick sense of robustness.

Back to the first screen

On the first screen, you drew the same open loop L(jω) as in the F1, F2, and F3 units in the complex plane and looked at stability through its relation to the single point −1. The curve crossed the negative real axis at −K/6. With K below 6, that crossing was to the right of −1, so the curve did not encircle −1 and the loop was stable; at K = 6 it passed exactly through −1, the marginal state; and with K above 6 the crossing moved to the left of −1, encircling −1 twice, two right-half-plane poles and an unstable loop. The Routh test through the coefficients, the root locus through the path of the poles, the Bode plot through the frequency margins, and the Nyquist criterion through the encirclements of −1 all pointed to the same boundary of the same system, K = 6. Of them, only the Nyquist criterion can answer even when the open loop is already unstable.

So far, the Routh test, the root locus, the Bode plot, and the Nyquist plot have analyzed whether a given loop is stable and how robust it is. All of them used proportional control, adjusting only a single gain K. But a single proportional gain cannot bring the steady-state error for a step input to 0 unless the plant contains an integrator, and it cannot shape the transient response at will either. The next unit, PID Control, adds dynamics to the controller itself. The integral term keeps accumulating as long as error remains, wiping out steady-state error, and the derivative term reads changes ahead of time to add damping. By tuning the gains of the three terms, proportional, integral, and derivative, you move into designs that shape the poles and margins you have seen so far into the places you want.