Worked examples — Nyquist stability criterion — encirclements of −1

This is the "grind through every case" page for the parent Nyquist note. If a symbol here feels unfamiliar, the parent note builds it from scratch — we assume you know that $Z=N+P$ and why $-1$ is the danger point. Our job now: hit every scenario class so no exam problem surprises you.

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The scenario matrix

Here $P$ = RHP open-loop poles, $N$ = clockwise encirclements of $-1$, $Z=N+P$ = RHP closed-loop poles (unstable iff $Z>0$).

Cell	$P$	Plot behaviour near $-1$	$N$	Result	Example
A	$0$	never wraps $-1$	$0$	stable	Ex 1
B	$0$, all poles LHP	high gain wraps $-1$ CW twice	$+2$	unstable	Ex 2
C	$0$ (integrator)	tangent — passes through $-1$	boundary	marginal	Ex 3
D	$1$	one counter-clockwise wrap	$-1$	stable	Ex 4 (high gain)
E	$1$	no wrap	$0$	unstable	Ex 4 (low gain)
F	$0$, pole at $s=0$	integrator → indent contour	(mechanics)	sets up B/C	Ex 5
G	$0$	limiting endpoints $\omega \to 0,\infty$; no finite crossing	$0$	stable (read the ends)	Ex 6
H	$0$	degenerate pure gain, single point	$0$	trivially stable	Ex 7
I	$0$ (integrator)	real-world GNC autopilot, no finite crossing	$0$	stable for all $K>0$	Ex 8
J	$0$ or $2$	same plot, one CW wrap	$+1$	depends on $P$	Ex 9

We now walk cells A→J.

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Example 2 uses the integrator mechanics from Example 5

Because Example 2 and Example 3 both use $L(s)=\frac{K}{s(s+1)(s+2)}$ (which has an integrator pole at $s=0$), we present Example 5 first is not necessary — but Example 2's Step 1 below is written to be self-contained: it derives $P=0$ on the spot and points to Ex 5 only for the picture.

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Example 1 — Cell A: $P=0$, no wrap → stable

Step 1 — Find $P$. Why this step? $Z=N+P$ needs $P$ first. Poles solve $(s+1)(s+2)=0\Rightarrow s=-1,-2$, both LHP. So $==P=0==$.

Step 2 — Evaluate the two endpoints of the curve. Why this step? The start ($\omega =0$) and end ($\omega \to \infty$) frame where the plot lives. $L(0)=\frac{2}{1\cdot 2}=1(\text{on the }+\text{real axis}).$ As $\omega \to \infty$, two poles each contribute $-90^{\circ }$ of phase, so phase $\to -180^{\circ }$ and magnitude $\to 0$: the plot dies into the origin from below.

Step 3 — Trace the shape (see figure). Why this step? We need the leftmost real-axis crossing to compare with $-1$. The curve starts at $+1$, swings through the fourth quadrant, and returns to $0$.

What to observe: the blue curve $L(j\omega )$ starts at $+1$ (green dot), loops clockwise through the lower half-plane, and dies at the origin. The red ✗ at $-1$ sits far to the left of every point on the blue curve — it is never surrounded.

Step 4 — Count $N$ and compute $Z$. The red $-1$ point is never enclosed, so $N=0$. Then $Z=N+P=0+0=0$ → stable.

Recall Verify

The closed-loop denominator is $1+L=1+\frac{2}{(s+1)(s+2)}=\frac{s^2+3s+4}{(s+1)(s+2)}$. Roots of $s^2+3s+4$: $s=\frac{-3\pm \sqrt{9-16}}{2}=-1.5\pm 1.32i$ — both LHP. $Z=0$ confirmed. ✓

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Example 2 — Cell B: $P=0$, gain too high → two CW wraps

Step 1 — Find $P$ (self-contained). Why this step? $Z=N+P$ needs $P$, and there is a pole exactly on the imaginary axis at $s=0$, so we must decide whether it counts. The Nyquist contour hugs the $j\omega$-axis but must enclose only the RHP. We deform it with a tiny semicircle that bulges into the RHP around $s=0$, so the pole is left just outside the enclosed region. Therefore the $s=0$ pole is not an RHP pole for our count: the enclosed poles are none of $\{0,-1,-2\}$, giving $==P=0==$. (Ex 5 draws this indentation; the algebra of $P=0$ needs nothing beyond this paragraph.)

Step 2 — Derive $\mathrm{Im}\{L(j\omega )\}$ and find where it is zero. Why this step? A negative-real-axis crossing happens exactly where the imaginary part of $L$ is zero and the real part is negative — that is the phase $=-180^{\circ }$ condition, and its location decides if $-1$ is enclosed. Substitute $s=j\omega$ and expand the denominator: $s(s+1)(s+2){|}_{s=j\omega }=j\omega (j\omega +1)(j\omega +2)=j\omega [(j\omega {)}^2+3j\omega +2]=j\omega (2-{\omega }^2)+3j^2{\omega }^2.$ Collecting real and imaginary parts of the denominator: $D(j\omega )=\underset{\text{real}}{\underbrace{-3{\omega }^2}}+j\underset{\text{imag}}{\underbrace{\omega (2-{\omega }^2)}}.$ Then $L(j\omega )=\frac{K}{D}=\frac{K\overline{D}}{|D{|}^2}=\frac{K[-3{\omega }^2-j\omega (2-{\omega }^2)]}{|D{|}^2}.$ The imaginary part of $L$ vanishes when its numerator's imaginary part is zero: $-K\omega (2-{\omega }^2)=0\Rightarrow {\omega }^2=2(\text{taking }\omega >0).$

Step 3 — Evaluate the crossing magnitude. Why this step? We now know where ($\omega =\sqrt{2}$) the curve hits the real axis; we need how far left. At $\omega =\sqrt{2}$ the imaginary part is gone, so $L$ is purely the real part: $L(j\sqrt{2})=\frac{K(-3\cdot 2)}{|D{|}^2}.|D{|}^2=(-3\cdot 2{)}^2+0=36.$ $L(j\sqrt{2})=\frac{-6K}{36}=-\frac{K}{6}.$ So the crossing sits at $-K/6$ on the negative real axis (real part negative ✓ — a true $-180^{\circ }$ crossing).

Step 4 — Compare to $-1$ and count $N$ (see figure). For $K=12$: crossing at $-12/6=-2$, which is left of $-1$, so $-1$ is enclosed.

What to observe: the finite blue curve crosses the real axis at $-2$ (left of the red ✗ at $-1$). The huge orange arc is the image of the tiny right-indentation around $s=0$ (Ex 5): as $s$ sweeps $-90^{\circ }\to +90^{\circ }$ near $0$, $L\approx K/s$ traces a giant clockwise arc that, together with the finite lobe, wraps $-1$ twice clockwise. Follow the orange direction arrows: the point $-1$ is circled once by the big arc and once by the returning lobe → $N=+2$.

Step 5 — Compute $Z$. $Z=N+P=2+0=2$ → unstable, two RHP poles.

Recall Verify

$1+L=0\Rightarrow s(s+1)(s+2)+12=0\Rightarrow s^3+3s^2+2s+12=0$. Routh test: for $s^3+3s^2+2s+K$, the row entry is $(3\cdot 2-K)/3=(6-K)/3$; with $K=12$ this is $-2<0$ → two sign changes → 2 RHP roots. Matches $Z=2$. ✓

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Example 3 — Cell C: critical gain, tangent to $-1$

Step 1 — Set the crossing equal to $-1$. Why this step? Marginal stability = curve passes through the danger point. From Ex 2 Step 3 the crossing is at $-K/6$: $-\frac{K}{6}=-1\Rightarrow K=6.$

Step 2 — Interpret. Why this step? We translate the geometric threshold into stable/unstable ranges. For $K<6$ the crossing is right of $-1$ → $N=0$ → stable. For $K>6$ → $N=2$ → unstable. At $K=6$ the plot grazes $-1$: two closed-loop poles sit on the imaginary axis (sustained oscillation).

Recall Verify

At $K=6$: $s^3+3s^2+2s+6=(s+3)(s^2+2)=0\Rightarrow s=-3,\pm j\sqrt{2}$. Pure imaginary roots at $\pm j\sqrt{2}$ → marginal, oscillation frequency $\omega =\sqrt{2}$ (matches the crossing frequency from Ex 2). ✓

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Example 4 — Cells D & E: unstable plant, one CCW wrap needed

Step 1 — Confirm $P$. Why this step? The whole verdict hinges on $P$. Pole at $s=+1$ is in the RHP → $==P=1==$.

Step 2 — Write $L(j\omega )$ explicitly. Why this step? We need the actual curve to see the wrap. Multiply top and bottom by the conjugate: $L(j\omega )=\frac{K}{j\omega -1}=\frac{K(-1-j\omega )}{(-1{)}^2+{\omega }^2}=\frac{K(-1-j\omega )}{1+{\omega }^2}.$

Step 3 — Endpoints. Why this step? The two ends pin down the circle. At $\omega =0$: $L=-K$ (negative real axis). As $\omega \to \infty$: $L\to 0$. The full curve is a circle through $-K$ and the origin (see figure).

What to observe: the orange circle passes through $-K=-2$ (green dot) at $\omega =0$ and shrinks to the origin as $\omega \to \infty$. The direction arrow shows the counter-clockwise sweep, and the red ✗ at $-1$ sits inside the circle — so it is wrapped exactly once CCW.

Step 4 — Where is $-1$? Why this step? Enclosure of $-1$ is decided by comparing $-K$ with $-1$. If $K>1$, then $-K<-1$, so $-1$ lies inside the circle → encircled once, and the traverse direction makes it counter-clockwise: $N=-1$. Then $Z=N+P=-1+1=0$ → stable (Cell D). If $K<1$, then $-1$ is outside → $N=0$, $Z=0+1=1$ → unstable (Cell E).

Recall Verify

$1+L=0\Rightarrow s-1+K=0\Rightarrow s=1-K$. Stable iff $s<0$ iff $K>1$. Boundary at $K=1$ (pole at origin). Matches the geometric threshold. ✓

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Example 5 — Cell F: integrator, indent the contour

Step 1 — Indent to the right. Why this step? The Nyquist contour must enclose only the RHP. We deform it with a tiny semicircle of radius $\epsilon \to 0$ bulging into the RHP around $s=0$, keeping the pole outside the enclosed region. Hence it is not counted in $P$: $P=0$.

Step 2 — Map the small semicircle. Why this step? We must know what that tiny detour becomes on the $L$-plot, because it contributes the big arc that helps wrap $-1$. On it $s=\epsilon e^{j\varphi }$ with $\varphi$ running $-90^{\circ }\to +90^{\circ }$. Then $L\approx \frac{K}{s}=\frac{K}{\epsilon }{e}^{-j\varphi }$ — a large arc of radius $K/\epsilon \to \infty$, sweeping $+90^{\circ }\to -90^{\circ }$ (clockwise). See figure.

What to observe: the blue contour runs up the $j\omega$-axis but makes a small right-hand bump (green) around the red ✗ pole at $s=0$, keeping it outside. In the $L$-plane this tiny bump balloons into the giant clockwise arc seen in Ex 2's figure.

Step 3 — Consequence. Why this step? This closes the plot consistently so the encirclement count is well-defined. This giant clockwise arc is what closes the plot at infinity and lets us count wraps of $-1$ consistently (used in Ex 2 & 3).

Recall Verify

With the indentation, the enclosed RHP contains poles $-1,-2$? No — those are LHP. Contains $s=0$? No — excluded by the bump. So $P=0$, confirming the count used in Ex 2 ($Z=2$) and Ex 3 (boundary at $K=6$). ✓ (both cross-checked by Routh above.)

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Example 6 — Cell G: reading the limiting endpoints

Step 1 — $\omega =0$ endpoint. Why this step? The start of the curve fixes one anchor. $L(0)=\frac{5}{1}=5$ on the $+$real axis.

Step 2 — $\omega \to \infty$ endpoint. Why this step? The end fixes the other anchor. Magnitude $\to 0$, phase $\to -180^{\circ }$: approaches origin from the negative-real side.

Step 3 — Derive $\mathrm{Im}\{L(j\omega )\}$ and solve for a crossing. Why this step? A finite negative-real crossing requires $\mathrm{Im}\{L\}=0$ at a finite $\omega$; let us test that algebraically rather than guess. Expand: $(j\omega +1{)}^2=(1-{\omega }^2)+2j\omega ,L(j\omega )=\frac{5}{(1-{\omega }^2)+2j\omega }=\frac{5[(1-{\omega }^2)-2j\omega ]}{(1-{\omega }^2{)}^2+4{\omega }^2}.$ So $\mathrm{Im}\{L(j\omega )\}=\frac{-10\omega }{(1-{\omega }^2{)}^2+4{\omega }^2}.$ This is zero only at $\omega =0$ (where $L=+5$, on the positive axis) and in the limit $\omega \to \infty$ (where $|L|\to 0$). There is no finite $\omega >0$ giving a negative-real crossing. As a spot-check, at $\omega =1$: $L(j)=\frac{5}{2j}=-2.5j$ — purely imaginary, not on the real axis.

Step 4 — Conclusion. Why this step? We convert "no finite crossing" into an encirclement count. Since the curve only touches the negative real axis in the limit at the origin, it can never enclose $-1$ for this $L$: $P=0$, $N=0$, $Z=0$ → unconditionally stable.

Recall Verify

$1+L=0\Rightarrow (s+1{)}^2+5=0\Rightarrow s=-1\pm j\sqrt{5}$ — both LHP. $Z=0$. And $|L(j)|=5/|(j+1{)}^2|=5/2=2.5$. ✓

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Example 7 — Cell H: degenerate pure-gain plant

Step 1 — Poles? Why this step? $P$ comes from poles, and here there are none → $P=0$. The Nyquist "plot" is a single point at $K$ (same for all $\omega$).

Step 2 — Encircle $-1$? Why this step? A single point has no winding. It cannot wrap around anything → $N=0$ → $Z=0$: always "stable" in the encirclement sense.

Step 3 — Check the actual closed-loop pole. Why this step? Encirclement bookkeeping should agree with the algebra. $1+L=1+K=0\Rightarrow$ there is a closed-loop pole only if $K=-1$ (then $1+K=0$ identically — degenerate). For $K\ne -1$ the loop gain is just $\frac{G}{1+K}$, a static number — no pole to destabilize.

Step 4 — Edge case $K=-1$. Why this step? Every case must be covered, including the degenerate singular one. Denominator $1+L=0$ everywhere: the transfer function is undefined (division by zero at all $s$). This is the degenerate "sits exactly on $-1$" limit.

Recall Verify

$1+K=0\Rightarrow K=-1$ is the only special value. For any $K=2$, closed-loop static gain $=G/(1+2)=G/3$, finite → stable. ✓

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Example 8 — Cell I: real-world GNC autopilot

Step 1 — Poles. Why this step? Fix $P$ before anything else. $s=0$ (indent, not counted) and $s=-4$. $P=0$.

Step 2 — Phase at finite $\omega$. Why this step? A finite negative-real crossing requires phase $=-180^{\circ }$ at finite $\omega$; let us test it. $\angle L(j\omega )=-90^{\circ }-\arctan \frac{\omega }{4}.$ This reaches $-180^{\circ }$ only as $\omega \to \infty$, where $|L|\to 0$. So the plot's negative-real crossing is at the origin — never at a finite point left of it.

Step 3 — Conclusion. Why this step? Turn "no finite crossing" into the count. $-1$ is never enclosed for any $K>0$: $N=0$, $Z=0$. The loop is stable for all positive gain — a classic type-1 second-order shape.

Step 4 — Sanity via characteristic eq. Why this step? Cross-check the encirclement verdict with the algebra. $s^2+4s+K=0$ has roots $s=-2\pm \sqrt{4-K}$; real part $=-2<0$ for all $K>0$ (complex when $K>4$, still stable). So the "largest stable $K$" is unbounded — the design limit comes from margins/noise, not encirclement (see Stability Margins in GNC Loops and Bode Plot & Gain/Phase Margins).

Recall Verify

$s^2+4s+K$: sum of roots $=-4<0$, product $=K>0$ ⇒ both roots LHP for every $K>0$ (Routh: all coefficients positive, second-order ⇒ stable). ✓

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Example 9 — Cell J: exam twist — same plot, two different $P$

Step 1 — Apply $Z=N+P$ for student A. Why this step? Plug the given $N$ and each claimed $P$ into the master formula. $Z=1+0=1$ → one unstable pole → unstable.

Step 2 — Apply for student B. Why this step? Same formula, different $P$. $Z=1+2=3$ → three unstable poles → very unstable.

Step 3 — Twist resolved. Why this step? Show that the number of unstable poles, not just the yes/no, depends on $P$. Neither reading is stable here, but $Z$ differs. To make a $P=2$ plant stable you'd need $N=-2$ (two CCW wraps). A single CW wrap ($N=+1$) can never stabilize any plant.

Step 4 — General rule. Why this step? Extract the reusable law. Stable requires $N=-P$. Only $N\le 0$ (net CCW) can rescue an unstable plant; $N>0$ always adds instability.

Recall Verify

$Z_A=N+P=1+0=1$; $Z_B=1+2=3$. Stability requires $Z=0$; neither is $0$. ✓

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Recall

Recall Quick self-test

First thing to compute in any Nyquist problem? ::: $P$ — the number of RHP open-loop poles. Ex 2 critical gain of $K/[s(s+1)(s+2)]$? ::: $K=6$ (crossing at $-K/6=-1$). Ex 4: threshold gain for $K/(s-1)$? ::: $K=1$; stable for $K>1$. Why is $L=K/(s+1{)}^2$ always stable? ::: Its $\mathrm{Im}\{L\}=0$ only at $\omega =0$ (positive axis) and $\omega \to \infty$ (magnitude zero) — the curve never crosses the negative real axis, so $-1$ is never enclosed. Same plot ($N=+1$), does $P$ change the answer? ::: Yes — $Z=N+P$, so $Z$ (and thus how unstable) depends on $P$.

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Connections

• Argument Principle (Cauchy) — why counting wraps counts poles.
• Routh–Hurwitz Criterion — the algebra behind every "Verify" here.
• Root Locus — watch the closed-loop poles move as $K$ grows (Ex 2, 3).
• Bode Plot & Gain/Phase Margins — Ex 8's "how much $K$" question lives here.
• Feedback Control Basics — origin of $T=G/(1+L)$.
• Stability Margins in GNC Loops — the applied autopilot design (Ex 8).