FOUNDATION · CHAPTER 8 OF 8

Why the Nyquist criterion works

This chapter shows why counting how a curve circles -1 + j0 gives the number of closed-loop poles in the right half-plane, and reads gain and phase margin as distances from that point. It lets you judge stability, and how much margin is left, from frequency-response data alone.

Updated 13 Oct 2026

What you will be able to do

You will see why a closed path that encloses a zero turns a vector through one net turn, and why a pole turns it the other way. From that you will derive $Z = N + P$ yourself. You will turn the $j ω$ axis into a table of $∣ G H ∣$ and $∠ G H$ . You will read gain margin and phase margin as distances from $- 1 + j 0$ .

Before you start

Chapter 6, Stability, and where it is lost: closed-loop poles as the roots of $1 + G H = 0$ , and the cubic test for $a_{3} s^{3} + a_{2} s^{2} + a_{1} s + a_{0}$
Networks, Chapter 7, How a circuit behaves across frequency: decibels and $H (j ω)$
Signals and Systems, Chapter 8, Reading a system from its poles and zeros: vectors from poles and zeros
School mathematics: the angle of a complex number, $tan^{- 1}$ , the complex conjugate, the degree of a polynomial

Counting turns of a vector

Recap: with negative feedback and zero initial conditions, the reference $r (t)$ , output $c (t)$ and error $e (t) = r (t) - c (t)$ have transforms $R$ , $C$ , $E$ . With forward block $G$ and feedback block $H$ , $T = C / R = G / (1 + G H)$ . $G H$ is the loop gain, and the closed-loop poles are the roots of $1 + G H = 0$ . With $s = σ + j ω$ , the factor $s - z$ is the vector from the point $z$ to the point $s$ .

Let $s$ travel once clockwise round a closed path that does not cross itself. If $z$ is inside, $s$ goes once round $z$ , so the angle of $s - z$ makes one net clockwise turn; if $z$ is outside, the angle swings and returns to its start value. A pole $p$ sits in the denominator of a ratio such as $(s - z) / (s - p)$ , so the angle of $s - p$ is subtracted: an enclosed pole turns the angle of the ratio one turn the other way, counter-clockwise.

ANGLE TURNED BY ONE FACTOR

Δ∠ (s - z) = {- 36 0^{\circ} 0 z inside the path z outside the path

Holds when $s$ goes once clockwise round a closed path that does not cross itself or pass through $z$ . $Δ∠$ is the net change in the angle of $s - z$ , in degrees, counter-clockwise positive. Outside, the ray rule: if a straight ray from $z$ misses the path (true for a circle and for the Nyquist path of the next section), the vector never points along that ray, so it cannot complete a turn.

NET TURN OF A RATIO OF FACTORS

F (s) = K \frac{\prod ( s - z _{i} )}{\prod ( s - p _{k} )} \Rightarrow Δ∠ F = - 36 0^{\circ} (Z_{in} - P_{in})

Holds when no $z_{i}$ or $p_{k}$ lies on the path and $K$ is a nonzero real constant (a fixed angle, never a change). The angle of a product is the sum of the angles, and of a quotient the difference. $Z_{in}$ and $P_{in}$ count the zeros and poles inside the path, with repeats. So $F$ circles the origin $N_{0} = Z_{in} - P_{in}$ times clockwise.

WATCH OUT FOR

A zero outside the path still moves the vector's angle during the trip. Test: compare the angle at the start and at the end of the circuit. Only an enclosed point leaves a difference of $36 0^{\circ}$ .

WORKED EXAMPLE

$s$ goes once clockwise round the circle $∣ s ∣ = 3$ . How many times, and in which sense, does $F (s) = \frac{( s - 1 ) ( s + 2 )}{s - 5}$ circle the origin?

Solution

The path is $∣ s ∣ = 3$ , so a point is inside when its distance from the origin is less than 3.
The zeros $s = 1$ and $s = - 2$ are at distances 1 and 2: both inside. The pole $s = 5$ is at distance 5: outside, so it adds no net turn.
Each enclosed zero turns its vector by $- 36 0^{\circ}$ : $Δ∠ F = - 36 0^{\circ} (2 - 0) = - 72 0^{\circ}$ .
One clockwise circuit of the origin is $- 36 0^{\circ}$ of angle, so the count is $720/360$ .

Answer: $F$ circles the origin twice, clockwise ( $N_{0} = 2$ ).

From turns to Z = N + P

Apply the counting to $F (s) = 1 + G H (s)$ . Its zeros are the closed-loop poles and its poles are the open-loop poles. Take the path clockwise round the edge of the RHP, the open right half-plane, so one count gives the closed-loop poles in the RHP. The $j ω$ axis is in neither the RHP nor the LHP, the open left half-plane. This path is the Nyquist path, and its image under $G H$ is the Nyquist plot. $F$ circling the origin is the same as $G H$ circling $- 1 + j 0$ .

ZEROS AND POLES OF ONE PLUS LOOP GAIN

F (s) = 1 + G H (s) = 1 + \frac{n ( s )}{d ( s )} = \frac{d ( s ) + n ( s )}{d ( s )}

Holds when $G H = n / d$ is a ratio of real polynomials with no common factor. Then $d + n$ and $d$ share no factor either (one would divide $n$ ), so nothing cancels. The zeros of $F$ solve $d + n = 0$ : the closed-loop poles. The poles of $F$ solve $d = 0$ : the open-loop poles. Since $G H = F - 1$ , $F$ circles the origin exactly when $G H$ circles $- 1 + j 0$ .

Up the $j ω$ axis from $ω = - \infty$ to $ω = + \infty$ : the image is $G H (j ω)$ .
Round the arc of radius $R \to \infty$ through the RHP, clockwise, so the path encloses every RHP pole and zero. If $G H$ is strictly proper (denominator degree above numerator degree), $∣ G H ∣ \to 0$ there, so the image is the origin.
Round a tiny arc into the RHP at each pole of $G H$ on the $j ω$ axis, so that no pole lies on the path. Such a pole then counts as outside.

ENCIRCLEMENTS AND CLOSED-LOOP POLES

N = Z - P ⟹ Z = N + P

Holds when the image curve does not pass through $- 1 + j 0$ , so no closed-loop pole is on the $j ω$ axis. $Z$ counts closed-loop poles in the RHP, $P$ open-loop poles in the RHP, and $N$ the net clockwise encirclements of $- 1 + j 0$ by the image of the whole path: just $G H (j ω)$ when $G H$ is strictly proper with no $j ω$ -axis pole. Stable means $Z = 0$ , so $N = - P$ . The form $Z = P - N_{ccw}$ says the same.

The jω axis as a table, and the margins

On the $j ω$ axis, $s = j ω$ and $G H (j ω)$ is a length and an angle. Each row of a table of $∣ G H ∣$ and $∠ G H$ at a chosen $ω$ is one point of the Nyquist plot. The rows that matter are $ω = 0$ , the $ω$ where $∠ G H = - 18 0^{\circ}$ , the $ω$ where $∣ G H ∣ = 1$ , and $ω \to \infty$ . You tabulate only $ω \geq 0$ , because the negative half is a mirror image. Then you watch the vector from $- 1 + j 0$ to the curve.

NEGATIVE FREQUENCIES MIRROR POSITIVE ONES

G H (- j ω) = \overline{G H (j ω)}

Holds when $G H$ has real coefficients. The $ω < 0$ half is the $ω > 0$ half reflected in the real axis and traced in reverse. Reflection reverses the sense of turning and the reverse order reverses it back, so the vector from $- 1 + j 0$ turns equally on each half. If $G H$ is proper (denominator degree at least the numerator's) with no $j ω$ -axis pole, the halves meet at $ω = 0$ and $\pm \infty$ : the total is twice one half.

LENGTH AND ANGLE OF THE SAMPLE LOOP

∣ G H (j ω) ∣ = \frac{K}{( 1 + ω ^{2} ) ^{3/2}}, ∠ G H (j ω) = - 3 tan^{- 1} ω

For $G H (s) = K / (s + 1)^{3}$ with a constant $K > 0$ (unitless) and $ω$ in rad/s. Each factor $j ω + 1$ has length $1 + ω^{2}$ and angle $tan^{- 1} ω$ , and denominator factors subtract angle. The angle is tracked continuously from $0^{\circ}$ .

ω (rad/s)	\|GH\| for K = 16	∠GH
0	16	0°
1	5.66	-135°
√3 ≈ 1.73	2	-180°
2.31	1.00	-199.8°
∞	0	-270°

The curve starts at 16 on the positive real axis and passes through the fourth quadrant, then the third. It crosses the negative real axis at $- 2$ when $ω = 3$ rad/s and enters the second quadrant. There it crosses the unit circle at $ω \approx 2.31$ rad/s and ends at the origin.

WORKED EXAMPLE

For $G H (s) = 16/ (s + 1)^{3}$ with negative feedback, use the table to find the number of closed-loop poles in the RHP.

Solution

The open-loop poles are all at $s = - 1$ , in the LHP, so $P = 0$ . $G H$ is strictly proper with no $j ω$ -axis pole, so the big arc maps to the origin and the curve is $G H (j ω)$ alone.
The angle is $- 18 0^{\circ}$ when $3 tan^{- 1} ω = 18 0^{\circ}$ , that is $ω = 3$ rad/s, and there $∣ G H ∣ = 16/ 4^{3/2} = 2$ . The curve crosses the negative real axis at $- 2$ , to the left of $- 1$ .
Follow the vector from $- 1$ to the curve for $ω > 0$ . It starts pointing right (at 16). While $∠ G H$ is between $0^{\circ}$ and $- 18 0^{\circ}$ the curve is below the axis, so the vector's angle falls to $- 18 0^{\circ}$ at $- 2$ . The curve is then above the axis until it reaches the origin, where the vector points right again: net $- 36 0^{\circ}$ , one clockwise turn.
By the mirror rule the $ω < 0$ half turns it the same amount: $N = 2$ clockwise circuits.
Apply $Z = N + P = 2 + 0$ .

Answer: $Z = 2$ : two closed-loop poles in the RHP, so the loop is unstable. Chapter 6's cubic test agrees: $1 + G H = 0$ gives $s^{3} + 3 s^{2} + 3 s + 17 = 0$ , so $a_{3} = 1$ , $a_{2} = 3$ , $a_{1} = 3$ , $a_{0} = 17$ , and $a_{2} a_{1} = 9 < a_{0} a_{3} = 17$ .

GAIN MARGIN

GM = \frac{1}{∣ G H ( j ω _{p c} ) ∣}, GM_{dB} = - 20 lo g_{10} ∣ G H (j ω_{p c}) ∣

Holds when $P = 0$ and the curve crosses the negative real axis once, at the phase crossover frequency $ω_{p c}$ (rad/s), where $∠ G H = - 18 0^{\circ}$ . There the curve sits at $- ∣ G H (j ω_{p c}) ∣$ . Multiplying the loop gain by GM (a pure ratio) moves that point onto $- 1 + j 0$ , the edge of stability. GM greater than 1 means the crossing lies between 0 and $- 1$ .

PHASE MARGIN

PM = 18 0^{\circ} + ∠ G H (j ω_{g c})

Holds when $P = 0$ and the curve meets the unit circle once, at the gain crossover frequency $ω_{g c}$ (rad/s), where $∣ G H ∣ = 1$ . There the curve is at angle $∠ G H (j ω_{g c})$ , a lag. PM degrees of extra lag would turn that point to $- 18 0^{\circ}$ , onto $- 1 + j 0$ . So PM is the angle, along the unit circle, between the curve and $- 1 + j 0$ .

WATCH OUT FOR

A large gain margin does not make a loop safe. Test: read PM as well as GM; the curve may pass close to $- 1 + j 0$ away from the real axis.

WORKED EXAMPLE

Find GM, in dB, and PM for $G H (s) = 4/ (s + 1)^{3}$ with negative feedback.

Solution

The poles are all at $s = - 1$ , so $P = 0$ again. The angle does not depend on $K > 0$ , so $ω_{p c} = 3$ rad/s, where $∣ G H ∣ = 4/8 = 0.5$ . The angle runs from $0^{\circ}$ to $- 27 0^{\circ}$ , so it is $- 18 0^{\circ}$ only there: the curve meets the negative real axis only at $- 0.5$ , right of $- 1$ . The ray pointing left from $- 1$ is never crossed on either half, so by the ray rule of the first section (the same argument with $- 1$ in place of $z$ and the curve in place of the path) the vector from $- 1$ makes no net turn: $N = 0$ and $Z = N + P = 0$ , stable.
$GM = 1/0.5 = 2$ , and $- 20 lo g_{10} 0.5 = 6.02$ dB.
$∣ G H ∣$ falls steadily from 4, so it equals 1 once. Solve $(1 + ω^{2})^{3/2} = 4$ : $1 + ω^{2} = 4^{2/3} = 2.52$ , so $ω_{g c} = 1.233$ rad/s.
The angle there is $- 3 tan^{- 1} (1.233) = - 3 (50.9 5^{\circ}) = - 152. 9^{\circ}$ .
$PM = 18 0^{\circ} - 152. 9^{\circ}$ .

Answer: GM = 2 (6.02 dB) and PM ≈ 27.1°.

Check yourself

CHECK YOURSELF 1

$s$ goes once clockwise round a closed path that does not cross itself, with no pole or zero on it. $F (s)$ has 3 zeros and 1 pole inside the path. What is the net number of clockwise circuits of the origin made by $F (s)$ ?

(A)
4
(B)
2
(C)
-2
(D)
3

Answer and solution

Answer: (B)

Each enclosed zero gives one clockwise circuit, each enclosed pole one anticlockwise circuit.
$N_{0} = Z_{in} - P_{in} = 3 - 1 = 2$ .

CHECK YOURSELF 2One or more options are correct

A loop with negative feedback has $P$ open-loop poles in the RHP. $N$ is the net number of clockwise encirclements of $- 1 + j 0$ by the image of the Nyquist path, traversed clockwise round the RHP, and that image does not pass through $- 1 + j 0$ . Select all true statements.

(A)
$Z = N + P$ , where $Z$ is the number of closed-loop poles in the RHP.
(B)
With $P = 2$ , a stable closed loop needs $N = 2$ .
(C)
With $P = 2$ , a stable closed loop needs two anticlockwise encirclements of $- 1 + j 0$ .
(D)
With $P = 0$ , a stable closed loop needs $N = 0$ .

Answer and solution

Answer: (A), (C), (D)

$Z = N + P$ was derived from $N = Z - P$ , so A is true.
Stable means $Z = 0$ , so $N = - P$ . With $P = 2$ , $N = - 2$ : two anticlockwise encirclements. C is true and B is false.
With $P = 0$ , $Z = 0$ needs $N = 0$ , so D is true.

CHECK YOURSELF 3

For $G H (s) = 6/ (s + 1)^{3}$ with negative feedback, find the gain margin as a plain ratio (not in dB).

Answer and solution

Answer: $1.333$ (accepted range $\pm 0.01$ )

$∠ G H = - 3 tan^{- 1} ω = - 18 0^{\circ}$ gives $ω_{p c} = 3$ rad/s.
$∣ G H (j 3) ∣ = 6/ (1 + 3)^{3/2} = 6/8 = 0.75$ .
$GM = 1/0.75 = 1.333$ .

CHECK YOURSELF 4

For $G H (s) = 16/ (s + 1)^{3}$ , why does the large clockwise arc of the Nyquist path map to a single point?

(A)
$G H$ has poles in the RHP.
(B)
The denominator has a higher degree than the numerator, so $∣ G H ∣ \to 0$ as $∣ s ∣ \to \infty$ .
(C)
The angle of $G H$ is $0^{\circ}$ on the arc.
(D)
The arc passes through a pole of $G H$ .

Answer and solution

Answer: (B)

On the arc $∣ s ∣ = R \to \infty$ and $∣ G H ∣ \approx 16/ R^{3} \to 0$ .
So every point of the arc maps to the origin, and the arc adds no encirclement of $- 1 + j 0$ .

Why the Nyquist criterion works

What you will be able to do

Before you start

Counting turns of a vector

From turns to Z = N + P

The jω axis as a table, and the margins

Check yourself

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