A control system has one open-loop right-half-plane pole (P = 1). Its Nyquist plot makes exactly one counter-clockwise encirclement of −1. How many closed-loop RHP poles does this system have?
AZ = 2 — counter-clockwise encirclements add instability
BZ = 0 — the closed-loop system is stable
CZ = 1 — the encirclement count equals the closed-loop pole count
DCannot be determined without counting all encirclements
By N = Z − P: one counter-clockwise encirclement means N = −1 (counter-clockwise encirclements are negative by convention). So Z = N + P = −1 + 1 = 0. The closed-loop system has no RHP poles and is stable. This illustrates the non-intuitive result that when the open-loop system has unstable poles (P > 0), the Nyquist plot *must* encircle −1 counter-clockwise for stability — zero encirclements would give Z = P = 1, meaning one unstable closed-loop pole. Stability requires counter-clockwise encirclements to cancel the open-loop instability.
Question 2 Multiple Choice
Why is the critical point −1+0j rather than the origin when applying Cauchy's argument principle to closed-loop stability?
ABy engineering convention, −1 is always the gain crossover point for stable systems
BMapping G(s)H(s) instead of 1+G(s)H(s) shifts the reference by −1, so encirclements of the origin in F(s) correspond to encirclements of −1 in the G(s)H(s) plot
CThe −1 point corresponds to the location of all open-loop poles for typical plants
DCauchy's theorem requires the critical point to be on the negative real axis
Cauchy's argument principle counts encirclements of the *origin* by F(s) = 1+G(s)H(s). The zeros of F(s) are the closed-loop poles (roots of 1+GH = 0). But instead of mapping F(s), we map G(s)H(s) directly — a plot that is already easy to construct from Bode data. Shifting by −1: F(s) = 0 ↔ G(s)H(s) = −1. So counting encirclements of the origin by F(s) is equivalent to counting encirclements of −1 by G(s)H(s). The critical point moves from the origin to −1 solely because we substituted G(s)H(s) for the original function F(s) = 1+G(s)H(s).
Question 3 True / False
For an open-loop stable system (P = 0), zero encirclements of −1 is sufficient for closed-loop stability, but any number of counter-clockwise encirclements is also acceptable.
TTrue
FFalse
Answer: False
When P = 0, stability requires Z = 0 closed-loop RHP poles. From N = Z − P = Z − 0 = Z, we need N = 0: exactly zero net encirclements. Counter-clockwise encirclements correspond to N < 0, which would mean Z = N < 0 — a mathematical impossibility (you can't have negative poles). Any clockwise or counter-clockwise net encirclements when P = 0 indicate instability. The Nyquist curve must pass to the right of −1 at the −180° crossing (or equivalently, −1 must lie outside the curve) for an open-loop stable system to be closed-loop stable.
Question 4 True / False
An open-loop unstable system with two RHP poles (P = 2) requires exactly two counter-clockwise encirclements of −1 for closed-loop stability.
TTrue
FFalse
Answer: True
For stability, Z = 0 (no closed-loop RHP poles). From N = Z − P, we need N = 0 − 2 = −2. A negative N means 2 counter-clockwise encirclements. This is the counterintuitive result: for an open-loop unstable plant, the Nyquist plot of the closed-loop stabilizing controller must encircle −1 counter-clockwise exactly P times. Too many or too few counter-clockwise encirclements both result in closed-loop instability. This is why Bode analysis (which only checks whether the curve crosses −1) can give wrong stability conclusions for open-loop unstable plants.
Question 5 Short Answer
Why does the Nyquist criterion require counter-clockwise encirclements of −1 for closed-loop stability when the open-loop system has right-half-plane poles?
Think about your answer, then reveal below.
Model answer: The relation N = Z − P connects encirclements (N), closed-loop RHP poles (Z), and open-loop RHP poles (P). For a stable closed-loop system we need Z = 0, so N must equal −P. A negative N means P counter-clockwise encirclements. The physical intuition: the open-loop plant already has P unstable poles that 'want to' push the closed-loop system unstable. The feedback loop must 'compensate' for each of these by generating one counter-clockwise encirclement of the critical point — each CCW encirclement cancels one unit of open-loop instability in Cauchy's counting. If the Nyquist curve instead makes fewer than P CCW encirclements, the closed-loop system inherits some of the open-loop instability (Z > 0).
This is why Nyquist extends beyond Bode analysis. Bode analysis only checks whether the phase is above −180° at gain crossover — a rule derived implicitly assuming P = 0. When P > 0, the Bode rules predict instability (the curve must cross −1), but a well-designed feedback controller can still be stable because those crossings are CCW rather than CW. The Nyquist criterion correctly counts signed encirclements and handles this case; Bode's approximate rules do not.