Nyquist stability criterion — encirclements of −1: Question bank

Reminders you will keep needing:

$L (s) = G (s) H (s)$ is the open-loop transfer function; we plot $L (j ω)$ .
$Z = N + P$ : $Z$ = right-half-plane (RHP) closed-loop poles, $N$ = clockwise encirclements of $- 1$ , $P$ = RHP open-loop poles.
Stable $⟺ Z = 0 ⟺ N = - P$ .

True or false — justify

The Nyquist plot always encircles the origin when the loop is unstable

False. We count wraps of

== - 1 ==

, not the origin. Encirclements of the origin correspond to

1 + L

, but we draw

L

, so the danger point shifts left by one to

- 1

If a system has

P = 0

open-loop RHP poles, then any encirclement of

- 1

means instability

True for $P = 0$ . Then

Z = N

, so any

N \neq = 0

gives

Z \neq = 0

. The claim quietly assumes

P = 0

; with

P > 0

it collapses.

A counter-clockwise encirclement of

- 1

always helps stability

False in general. CCW gives

N < 0

, which lowers

Z = N + P

. It helps only when it is cancelling a positive

P

; for a

P = 0

plant a CCW wrap makes

Z < 0

, which is impossible and signals a plotting or counting error.

The Nyquist criterion tells you where the unstable poles are

False. It only counts them (

Z

). It answers "how many RHP closed-loop poles?" without locating any of them — that is precisely its power (no root-solving).

Adding gain

K

scales the whole Nyquist plot radially about the origin

True.

L \to K L

multiplies every point's distance from the origin by

K

while preserving its angle, so the curve stretches/shrinks about the origin — which changes whether it swallows the fixed point

- 1

A stable open-loop plant guarantees a stable closed loop

False.

P = 0

only means

Z = N

; if the plot wraps

- 1

(

N > 0

) the closed loop is unstable. Stable-open does not imply stable-closed.

An unstable open-loop plant can never be stabilised by feedback

False. Example 3 (

L = K / (s - 1)

P = 1

) is stabilised when the plot encircles

- 1

once CCW (

N = - 1

), giving

Z = 0

. Feedback can move RHP poles into the LHP.

The Nyquist contour is traversed counter-clockwise

False (in this vault's convention). It is traversed clockwise so that positive

N

counts clockwise image encirclements; the argument principle then reads

N = Z - P

directly.

L (- j ω)

is an independent curve you must compute separately

False.

L (- j ω) = \overline{L (j ω)}

, so the negative-frequency branch is just the mirror image about the real axis of the positive branch — no new computation.

For a strictly proper

L

, the infinite semicircle of the contour maps to a nonzero arc you must draw carefully

False. Strictly proper means

∣ L ∣ \to 0

∣ s ∣ \to \infty

, so the big semicircle collapses to the origin — a single point, not an arc.

Spot the error

L (0) = 1 > 0

, so the plot starts on the positive real axis and can never reach

- 1

; therefore stable."

Error: the starting point being positive says nothing about whether the curve later loops around

- 1

. Stability is about encirclement of

- 1

by the whole closed curve, not the sign of

L (0)

"The plot passes exactly through

- 1

, so

N = 1

and the system is unstable."

Error: passing through

- 1

is neither inside nor outside — it means a closed-loop pole sits on the imaginary axis (marginal).

N

is undefined there; the system is on the stability boundary, not cleanly unstable.

P =

number of poles of

L

in the RHP, and

L = K / (s (s + 1) (s + 2))

has a pole at

s = 0

, so

P = 1

Error:

s = 0

is on the imaginary axis, not in the open RHP. We indent the contour to the right so this pole stays outside; it is not counted, so

P = 0

"We got

N = - 2

and

P = 0

, so

Z = - 2

: two extra-stable poles."

Error:

Z

counts poles and cannot be negative.

Z = - 2

is impossible, so either the encirclement direction or the count was read wrong — recheck the traversal sense.

"The gain margin is how far the curve is from the origin."

Error: margins are distances from the critical point

- 1

, not the origin. Gain margin is measured where the plot crosses the negative real axis relative to

- 1

(see Bode Plot & Gain/Phase Margins).

"Both Nyquist and Routh–Hurwitz need the closed-loop characteristic polynomial, so they're the same tool."

Error: Routh works algebraically on the closed-loop polynomial's coefficients; Nyquist works graphically from the open-loop frequency response

L (j ω)

and never forms the closed-loop polynomial. Different inputs, same question.

Why questions

Why do we study

L

relative to

- 1

instead of

1 + L

relative to

0

They are the same map shifted by one. Plotting

L

directly (data we already have) and moving the critical point to

- 1

saves us from computing and drawing

1 + L

everywhere.

Why does a zero of

1 + L

inside the clockwise contour add

- 2 π

ar g (1 + L)

Near a simple zero

z_{0}

1 + L \approx a (s - z_{0})

; as

s

loops clockwise once around

z_{0}

, the factor

(s - z_{0})

rotates its angle by

- 2 π

, so the image winds once clockwise about the origin (see Argument Principle (Cauchy)).

Why does an RHP pole of

L

contribute

+ 2 π

(opposite sign to a zero)?

A pole means

1 + L

behaves like

1/ (s - p)

; the reciprocal reverses the sense of rotation, so a clockwise loop around

p

unwinds the image by

+ 2 π

. That opposite sign is exactly why

P

appears with a

+

Z = N + P

Why must we indent to the right around a pole on the

j ω

-axis, not the left?

Right-indenting keeps the axis pole outside the RHP region the contour encloses, so it is not miscounted as an unstable open-loop pole. Left-indenting would enclose it and corrupt

P

Why is

- 1

(not, say,

- 2

) the reinforcement point?

1 + L = 0

means

L = - 1

: the returned signal is flipped in sign and equal in magnitude, so it adds perfectly to itself. Any other point does not solve the characteristic equation.

Why can we determine stability without ever solving for the roots?

The argument principle converts counting RHP roots into counting how many times a drawn curve winds around a point — a geometric tally replaces algebra. This is the whole reason Nyquist exists alongside Root Locus.

Why does each pole of

L

add asymptotically

- 9 0^{\circ}

of phase at high frequency?

A first-order factor

1/ (s + p)

has phase

\to - 9 0^{\circ}

ω \to \infty

;

n

poles stack to

- 90 n^{\circ}

, which sets the direction the plot spirals into the origin.

Edge cases

If the plot grazes

- 1

tangentially without enclosing it, is the system stable?

It is marginally stable: a closed-loop pole lies on the imaginary axis. Encirclement count is ill-defined at that instant; any perturbation decides which side it falls to.

What is

N

if the Nyquist plot never crosses the negative real axis at all?

Then it cannot loop around

- 1

, so

N = 0

. Stability reduces to whether

P = 0

A plant has

P = 2

RHP poles; what

N

is required for closed-loop stability?

We need

Z = 0

, so

N = - P = - 2

: exactly two counter-clockwise encirclements of

- 1

. Anything else leaves

Z > 0

For a pure integrator loop

L = K / s

, why must we handle the

ω = 0

point specially?

L

blows up at

ω = 0

(pole on the axis). We indent right with a tiny semicircle; its image is a large clockwise arc sweeping around infinity, which must be included before counting

- 1

wraps.

ω \to \infty

for a strictly proper

L

, where does the curve end, and why does it matter for counting?

It ends at the origin. Because both

ω \to \pm \infty

branches meet cleanly at

0

(well away from

- 1

), the closure adds no spurious wraps — the encirclement count comes only from the finite part.

What happens to

Z = N + P

if you accidentally use the closed-loop poles as

P

You would be feeding the answer (

Z

-related data) into the count and double-counting.

P

must be open-loop RHP poles of

L

; mixing them up makes the criterion self-referential and wrong.

K \to 0

L = K / (s - 1)

(

P = 1

), is the loop stable?

No. The plot shrinks to a tiny circle near the origin that no longer encircles

- 1

, so

N = 0

and

Z = N + P = 1

— one RHP pole remains. The unstable plant needs enough gain (

K > 1

) to wrap

- 1

Connections

Argument Principle (Cauchy) — supplies the "why" behind every sign in these traps.
Bode Plot & Gain/Phase Margins — reframes "distance from $- 1$ " as margins.
Routh–Hurwitz Criterion — the algebraic cross-check when a plot is ambiguous.
Root Locus — shows the poles the encirclement count is silently tracking.
Feedback Control Basics — origin of $T = G / (1 + L)$ .
Stability Margins in GNC Loops — where these traps bite in real autopilots.