Nyquist stability criterion — encirclements of −1

WHAT is the Nyquist criterion?

WHY does $-1$ matter and WHY encirclements?

Why $-1$

The characteristic equation is $1+L(s)=0$ , i.e. $L(s)=-1$ . So a closed-loop pole sits wherever the open-loop $L$ equals $-1$ . Studying how $L$ behaves relative to $-1$ is the same as studying $1+L$ relative to $0$ .

Why encirclements — the Argument Principle (derive it)

Formula Cauchy's Argument Principle (first principles)

Take a function $F(s)$ and a closed contour $C$ in the $s$ -plane traversed clockwise. Consider $\frac{1}{2\pi i}\oint_C \frac{F'(s)}{F(s)}\,ds.$ Why this integrand? Because $\dfrac{F'}{F}=\dfrac{d}{ds}\ln F$ , and $\ln F = \ln|F| + i\arg F$ . Integrating the derivative around a closed loop, the $\ln|F|$ part returns to its start (net zero), leaving only the change in argument (angle): $\oint_C \frac{F'}{F}\,ds = i\,\Delta_C(\arg F).$ If $F$ has a zero of order 1 at $z_0$ , near it $F\approx a(s-z_0)$ so $\arg F$ gains $-2\pi$ as $s$ loops clockwise around it. A pole gains $+2\pi$ . Summing over all zeros $Z$ and poles $P$ enclosed: $\frac{1}{2\pi i}\oint_C \frac{F'}{F}\,ds = Z - P.$ Geometrically: the image curve $F(C)$ winds around the origin $(Z-P)$ times clockwise.

Now set $F(s)=1+L(s)$ and let $C$ be the Nyquist contour — the entire imaginary axis $s=j\omega$ closed off by an infinite semicircle in the RHP (clockwise). This contour encloses the whole RHP.

Zeros of $F$ enclosed $=Z$ (unstable closed-loop poles — what we want to count).
Poles of $F$ enclosed = poles of $L$ in RHP $=P$ .
Clockwise encirclements of the origin by $1+L$ $= N$ .

But $1+L$ encircling $0$ is identical to $L$ encircling $-1$ (just shift left by 1). Hence: $N = Z - P \;\;\Rightarrow\;\; \boxed{Z = N + P}.$

That's the whole criterion — falling straight out of counting angle changes.

Figure — Nyquist stability criterion — encirclements of −1

HOW to read a Nyquist plot (procedure)

Sketch $L(j\omega)$ for $\omega:0^+\to\infty$ .
Reflect for $\omega:-\infty\to 0^-$ (mirror image about the real axis, since $L(-j\omega)=\overline{L(j\omega)}$ ).
Close the contour (map the big semicircle — usually shrinks to the origin for strictly proper $L$ ).
Count net clockwise encirclements $N$ of $-1$ .
Find $P$ from the poles of $L$ .
Compute $Z=N+P$ . Stable iff $Z=0$ .

Worked example Example 2 — Gain pushes it unstable

$L(s)=\dfrac{K}{s(s+1)(s+2)}$ with a pole at the origin.

$P=0$ (poles at $0,-1,-2$ ; the origin pole is on the contour — we indent to the right around it, keeping it outside, so it isn't counted as RHP).
The indentation contributes a large clockwise arc; the finite plot crosses the negative real axis at some point $-a K$ .
Why a critical gain? As $K$ grows, that crossing moves left. When it passes through $-1$ , encirclements switch $0\to 2$ .
Routh gives the crossing at $K=6$ . For $K<6$ : $N=0\Rightarrow Z=0$ stable. For $K>6$ : $N=2\Rightarrow Z=2$ unstable (2 RHP poles).
Why this step matters: encirclements literally count the poles that crossed into the RHP.

Worked example Example 3 — Unstable open-loop plant made stable

$L(s)=\dfrac{K}{s-1}$ (an RHP pole at $s=+1$ , so $P=1$ ).

For stability we need $Z=0\Rightarrow N=Z-P=-1$ : we need one counter-clockwise encirclement of $-1$ .
$L(j\omega)=\dfrac{K}{j\omega-1}=\dfrac{K(-1-j\omega)}{1+\omega^2}$ . At $\omega=0$ : $L=-K$ . As $\omega\to\infty$ : $\to 0$ .
The plot is a circle passing through $-K$ on the negative real axis. If $K>1$ , the point $-1$ lies inside and is encircled once counter-clockwise ⇒ $N=-1$ , $Z=0$ → stable. If $K<1$ , no encirclement ⇒ $Z=1$ → unstable.
Why this feels backwards but is right: an unstable plant needs feedback strong enough to wrap the curve around $-1$ the correct way — the encirclement is doing the stabilizing bookkeeping.

Common mistakes (Steel-man + fix)

Active recall

Recall Test yourself (hide the answers)

What equation defines closed-loop poles? → $1+L(s)=0$ .
Why the point $-1$ ? → because $L=-1$ solves that equation.
State the count formula. → $Z=N+P$ .
Which principle underlies it? → Cauchy's argument principle.
Stable condition in terms of $N,P$ ? → $N=-P$ (so $Z=0$ ).

Recall Feynman: explain to a 12-year-old

Imagine shouting into a canyon. If the echo comes back weaker, the sound dies — safe. But if the echo comes back just as loud and pushing the same way you pushed, it stacks up and gets louder and louder — that's instability. The Nyquist plot is a treasure-map of how the echo behaves at every pitch of your voice. There's a "danger X" on the map called $-1$ . We walk the whole map's edge and count how many times the drawn line loops around that X, and which way it spins. That count tells us whether the echo will explode — and we never had to actually listen to it blow up.

Connections

Argument Principle (Cauchy) — the mathematical engine.
Bode Plot & Gain/Phase Margins — margins are distances from $-1$ along the plot.
Routh–Hurwitz Criterion — algebraic alternative for counting RHP roots.
Root Locus — tracks closed-loop poles vs gain $K$ .
Feedback Control Basics — where $T=G/(1+L)$ comes from.
Stability Margins in GNC Loops — application to attitude/autopilot loops.

Nyquist: what point do we count encirclements of?

The point

-1+0j

Nyquist count formula?

Z=N+P

, with

Z

=RHP closed-loop poles,

N

=CW encirclements of

-1

P

=RHP open-loop poles.

Why is

-1

special?

Characteristic eq

1+L=0

⇒

L=-1

; closed-loop poles occur where open-loop

L

equals

-1

Which theorem gives Nyquist?

Cauchy's argument principle: image of a closed contour winds around origin

Z-P

times.

Stability condition in

N,P

terms?

N=-P

so that

Z=0

P=0

and plot doesn't encircle

-1

, stable?

Yes:

N=0\Rightarrow Z=0

Unstable plant

P=1

needs what?

One counter-clockwise encirclement of

-1

(

N=-1

Why plot

L

not

1+L

1+L

around

0

equals

L

around

-1

; we plot

L

, so critical point shifts to

-1

How handle a pole on the

j\omega

-axis?

Indent contour with a small semicircle to the right (keep pole outside RHP).

What does a CW encirclement of

-1

signify (P=0)?

A closed-loop pole crossed into the RHP → instability.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, feedback control me sabse important sawaal hai: system stable rahega ya blow up ho jayega? Nyquist criterion isko bina roots solve kiye batata hai. Idea simple hai — open-loop transfer function $L(s)$ ki value har frequency par nikaalo, aur uska graph (Nyquist plot) banao complex plane me. Ab ek special point hai: $-1$ . Ye point isliye khaas hai kyunki characteristic equation $1+L=0$ hoti hai, matlab jahan $L=-1$ , wahi closed-loop pole banta hai. Toh $-1$ ke aas-paas kya ho raha hai, wahi asli khel hai.

Formula yaad rakho: $Z = N + P$ . Yahan $P$ = open-loop ke RHP (right-half plane) poles ki ginti, $N$ = plot kitni baar $-1$ ko clockwise ghera (encircle kiya), aur $Z$ = closed-loop ke unstable poles. Stable tabhi jab $Z=0$ . Agar plant already stable hai ( $P=0$ ), toh bas dhyan rakho ki plot $-1$ ko na ghere — $N=0$ chahiye. Lekin agar plant hi unstable hai ( $P=1$ ), toh ulta chahiye: ek counter-clockwise encirclement, taki $Z$ zero ho jaye. Ye thoda ulta lagta hai par yahi correct bookkeeping hai.

Iske peeche maths hai Cauchy ka argument principle — jab tum ek closed loop ke around chalte ho, toh function ka angle kitna ghoomta hai wahi $Z-P$ deta hai. Bas $1+L$ ka origin ke around ghoomna, wahi $L$ ka $-1$ ke around ghoomna hai (bas left shift). GNC me — jaise satellite attitude control ya autopilot loops — yahi criterion se gain kitna badha sakte ho, gain margin aur phase margin sab decide hote hain. $-1$ se plot kitni door hai, utni safety margin.

Nyquist stability criterion — encirclements of −1

WHAT is the Nyquist criterion?

WHY does $-1$ matter and WHY encirclements?

Why $-1$

Why encirclements — the Argument Principle (derive it)

HOW to read a Nyquist plot (procedure)

Common mistakes (Steel-man + fix)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Guidance, Navigation & Control (GNC)

Connections