Exercises — Nyquist stability criterion — encirclements of −1

WHAT we do: find the poles, which are the values of $s$ that make the denominator zero. WHY: $P$ counts only the poles whose real part is positive.

Set $(s+3)(s+4)=0\Rightarrow s=-3$ or $s=-4$.

Both have negative real part, so both are in the left-half plane (LHP). None are in the RHP. $\boxed{P=0.}$

WHAT: plug straight into $Z=N+P$. WHY: that formula is the Nyquist criterion — no drawing needed once we know $N$ and $P$.

Two clockwise wraps means $N=+2$. With $P=0$: $Z=N+P=2+0=2.$ $Z\ne 0$, so the loop has 2 closed-loop poles in the RHP → unstable.

Step 1 — WHAT: evaluate at $\omega =0$. WHY: the point $L(0)$ is where the Nyquist curve starts on the real axis. $L(0)=\frac{4}{(1)(2)}=2.$ A positive real number: the curve starts at $+2$ on the positive real axis.

Step 2 — WHAT: phase at high frequency. WHY: each pole contributes $-90^{\circ }$ of phase in the limit, telling us where the curve ends. Two poles $\Rightarrow$ phase $\to 2\times (-90^{\circ })=-180^{\circ }$, and the magnitude $\to 0$ (denominator grows like ${\omega }^2$).

Step 3 — quadrants. Starting at $+2$ (phase $0^{\circ }$) and rotating to phase $-180^{\circ }$ as it shrinks toward the origin, the curve sweeps clockwise through the fourth quadrant (positive real, negative imaginary) and approaches the origin from the negative-real side.

What the figure shows (read even without the image): the solid blue curve starts on the positive real axis at $+2$ (marked in pale yellow), then bows downward and leftward into the fourth quadrant — its imaginary part goes negative, its real part shrinks — and it spirals into the origin. The pink $\times$ marks $-1$ on the negative real axis. Crucially, the blue curve stays entirely in the right half of the picture (real part $>-\frac{1}{2}$ or so at its leftmost) and never surrounds the pink $\times$. The dashed lighter-blue curve is the mirror image for $\omega <0$; together they form a closed loop that leaves $-1$ outside.

It never reaches out to $-1$, foreshadowing $N=0$.

WHAT: count wraps of the curve (from Ex. 2.1) around $-1$. WHY: $N$ is the only missing piece before $Z=N+P$.

The curve starts at $+2$, dips into the fourth quadrant, and slides into the origin — the point $-1$ sits on the negative real axis, outside the little loop the curve traces. The curve never surrounds it. $N=0.$ From Ex. 1.1-style factoring, poles at $-1,-2$ give $P=0$. Hence $Z=N+P=0+0=0\Rightarrow \text{stable.}$

Step 1 — WHAT: substitute $s=j\omega$ and find where the curve is purely real. WHY: the real-axis crossing is where the phase equals $-180^{\circ }$; that is exactly the point that can hit $-1$.

$L(j\omega )=\frac{K}{j\omega (j\omega +1)(j\omega +2)}.$ Multiply out the denominator: $j\omega (j\omega +1)(j\omega +2)=j\omega [(j\omega {)}^2+3j\omega +2]=j\omega (-{\omega }^2+3j\omega +2).$ Distribute the $j\omega$: $=j\omega (2-{\omega }^2)+j\omega \cdot 3j\omega =-3{\omega }^2+j\omega (2-{\omega }^2).$ So $L(j\omega )=\frac{K}{-3{\omega }^2+j\omega (2-{\omega }^2)}.$

Step 2 — WHAT: make the denominator purely real so $L$ is real. WHY: with a real numerator $K$, the curve crosses the real axis exactly when the denominator has zero imaginary part: $\omega (2-{\omega }^2)=0\Rightarrow {\omega }^2=2\Rightarrow \omega =\sqrt{2}.$ (We discard $\omega =0$, the origin pole.)

Step 3 — evaluate $L$ there. At ${\omega }^2=2$ the denominator is purely real: $-3{\omega }^2=-3(2)=-6$. So $L(j\sqrt{2})=\frac{K}{-6}=-\frac{K}{6}.$ This crossing lands on $-1$ when $-\frac{K}{6}=-1\Rightarrow \boxed{K=6.}$

WHAT: compare the crossing point $-K/6$ against $-1$. WHY: if the crossing is to the left of $-1$ (more negative), the curve now wraps around $-1$, switching $N$ from $0$ to $2$.

• $K=3$: crossing at $-3/6=-0.5$. That is to the right of $-1$ → $-1$ is not enclosed → $N=0\Rightarrow Z=0$ → stable.
• $K=6$: crossing exactly at $-1$ → curve passes through the danger point → marginally stable (a pole sits on the imaginary axis; $Z=0$ but on the boundary).
• $K=12$: crossing at $-12/6=-2$, to the left of $-1$ → $-1$ now enclosed twice clockwise → $N=2\Rightarrow Z=2$ → unstable (2 RHP closed-loop poles).

What the figure shows: three coloured dots sit on the negative real axis at $-0.5$ (blue, $K=3$), $-1$ (yellow, $K=6$), and $-2$ (pink, $K=12$). The chalk-white $\times$ at $-1$ is the fixed danger point. As $K$ grows, the crossing dot slides steadily leftward; the moment it passes the $\times$ (between $K=6$ and above) the curve begins to surround $-1$ and encirclements switch on.

WHAT: solve $Z=N+P$ for $N$ with $Z=0,P=1$. WHY: stability is defined by $Z=0$; everything else is bookkeeping. $0=N+1\Rightarrow N=-1.$ A negative $N$ means one counter-clockwise encirclement of $-1$.

What that looks like: the Nyquist curve must loop around the danger point once, spinning counter-clockwise. This is the surprising heart of Nyquist for unstable plants — the encirclement is not a warning sign here, it is the cure: it accounts for the RHP open-loop pole being cancelled out by the feedback.

Step 1 — WHAT: rationalise $L(j\omega )$. WHY: to see the geometry we separate real and imaginary parts. $L(j\omega )=\frac{K}{j\omega -1}=\frac{K(-1-j\omega )}{(j\omega -1)(-1-j\omega )}=\frac{K(-1-j\omega )}{1+{\omega }^2}.$ So $\mathrm{Re}L=\frac{-K}{1+{\omega }^2},\mathrm{Im}L=\frac{-K\omega }{1+{\omega }^2}.$

Step 2 — key points.

• At $\omega =0$: $L=-K$ (on the negative real axis).
• As $\omega \to \infty$: $L\to 0$.

The full curve (with its mirror image for negative $\omega$) is a circle through $-K$ and the origin, sitting in the left half of the plane.

What the figure shows: two circles are drawn, both passing through the origin. The blue circle ($K=0.6$) reaches only to $-0.6$ on the real axis — the yellow $\times$ at $-1$ lies outside it. The pink circle ($K=1.6$) reaches to $-1.6$, so the yellow $\times$ at $-1$ now sits inside it. Inside = encircled = $N=-1$.

Step 3 — when is $-1$ inside the circle? The circle spans from the origin ($0$) to $-K$ on the real axis. The point $-1$ lies inside exactly when $K>1.$ Then the curve encircles $-1$ once counter-clockwise, $N=-1$, giving $Z=N+P=-1+1=0$ → stable.

If $K<1$, the point $-1$ is outside the circle: $N=0$, so $Z=0+1=1$ → unstable. $\boxed{\text{Stable for }K>1.}$

Part (a) — WHAT: count RHP poles. WHY: we need $P$ for $Z=N+P$. Denominator $s^2(s+10)$: poles at $s=0$ (double, on the imaginary axis — indented to the right per the definition above, so not counted as RHP) and $s=-10$ (LHP). Therefore $P=0.$

Part (b) — WHAT: find every real-axis crossing. WHY: the curve can only touch $-1$ where it is purely real; if it never becomes real at a finite negative value except a computable one, that value determines whether $-1$ is crossed. Substitute $s=j\omega$: $L(j\omega )=\frac{K(1+j\omega )}{(j\omega {)}^2(j\omega +10)}=\frac{K(1+j\omega )}{-{\omega }^2(10+j\omega )}.$ Multiply top and bottom by the conjugate $(10-j\omega )$: $L(j\omega )=\frac{K(1+j\omega )(10-j\omega )}{-{\omega }^2(100+{\omega }^2)}.$ Expand the numerator (using the zero-factor identity): $(1+j\omega )(10-j\omega )=10-j\omega +10j\omega +{\omega }^2=(10+{\omega }^2)+j9\omega .$ So $\mathrm{Re}L=\frac{-K(10+{\omega }^2)}{{\omega }^2(100+{\omega }^2)},\mathrm{Im}L=\frac{-9K\omega }{{\omega }^2(100+{\omega }^2)}=\frac{-9K}{\omega (100+{\omega }^2)}.$ A real-axis crossing needs $\mathrm{Im}L=0$. But $\mathrm{Im}L=\frac{-9K}{\omega (100+{\omega }^2)}$ is never zero for any finite $\omega >0$ (numerator $-9K\ne 0$). Therefore $\boxed{\text{there is NO finite frequency at which the plot crosses the real axis, hence it never crosses }-1\text{ for any }K.}$

Part (c) — WHAT: determine $N$ by tracking the quadrants and the indent arc. WHY: with no finite real-axis crossing, $N$ is fixed by the phase sweep, so we walk the whole locus.

• Small $\omega \to 0^{+}$: $\mathrm{Re}L\to \frac{-K\cdot 10}{{\omega }^2\cdot 100}=\frac{-K}{10{\omega }^2}\to -\infty$ and $\mathrm{Im}L\to \frac{-9K}{\omega \cdot 100}\to -\infty$. The curve comes in from the third quadrant (both parts negative), far out near $\text{phase}=-180^{\circ }$ tilting slightly below. Total phase at DC: two integrators give $-180^{\circ }$, the zero and the $-10$ pole are negligible → phase $\approx -180^{\circ }$.
• Large $\omega \to \infty$: magnitude $\to 0$; phase $=\underset{\text{zero }(s+1)}{\underbrace{+90^{\circ }}}\underset{s^2}{\underbrace{-180^{\circ }}}\underset{(s+10)}{\underbrace{-90^{\circ }}}=-180^{\circ }$. So it also ends approaching the origin at phase $-180^{\circ }$, i.e. along the negative real axis, from the second/third quadrant boundary.
• In between: since $\mathrm{Im}L<0$ for all $\omega >0$ (it equals $\frac{-9K}{\omega (100+{\omega }^2)}$, always negative), the entire $\omega >0$ branch lives in the lower half-plane (third quadrant, real part also negative throughout because $\mathrm{Re}L<0$ for all $\omega$). It never rises above the real axis, so it cannot loop around $-1$.
• The indent at $s=0$: the tiny right-half semicircle around the double pole maps to a large clockwise arc of radius $\to \infty$ sweeping through $+360^{\circ }$ (because $s^2$ doubles the angle) out at infinity, closing the contour on the right side — away from $-1$.
• Mirror for $\omega <0$: reflection about the real axis puts that branch in the upper half-plane.

Putting it together: both finite branches hug the lower/upper left region without ever encircling the finite point $-1$; the closing arc sits at infinity on the right. The net winding of $-1$ is therefore $N=0\text{for every }K>0.$ Hence $Z=N+P=0+0=0\Rightarrow \boxed{\text{stable for every }K>0.}$ Lesson: the zero at $s=-1$ injects $+90^{\circ }$ of phase lead (defined above), which cancels one integrator's $-90^{\circ }$ before the crossing frequency could form, so the imaginary part never returns to zero and the curve never wraps $-1$. A well-placed zero (lead compensation) is what makes a two-integrator GNC loop usable at all. See Stability Margins in GNC Loops and Bode Plot & Gain/Phase Margins.

Step 1 — WHAT: form the characteristic polynomial. WHY: closed-loop poles are the roots of $1+L=0$, i.e. of $s(s+1)(s+2)+K=0$. $s(s+1)(s+2)+K=s^3+3s^2+2s+K=0.$

Step 2 — WHAT: build the Routh array. WHY: Routh–Hurwitz counts RHP roots from sign changes in the first column; marginal stability is when a first-column entry hits zero.

The $s^1$ entry is $\frac{6-K}{3}$.

Step 3 — marginal condition. It vanishes when $6-K=0$: $\boxed{K=6.}$ This exactly matches the Nyquist real-axis crossing found in Exercise 3.1 — two independent methods, one answer. For $K<6$ the first column entries $1,3,\frac{6-K}{3},K$ are all positive (no sign change → $Z=0$, stable); for $K>6$ the third entry $\frac{6-K}{3}$ turns negative while $K>0$ stays positive, giving two sign changes (positive→negative→positive) → $Z=2$, unstable. This matches the $N=2$ from Nyquist and is the Root Locus crossing of the imaginary axis in disguise.

Critical gain of $\frac{K}{s(s+1)(s+2)}$? ::: $K=6$. Stable range for $\frac{K}{s-1}$? ::: $K>1$ (needs $N=-1$ since $P=1$). Real-axis crossing of $\frac{K}{s(s+1)(s+2)}$ occurs at which $\omega$? ::: $\omega =\sqrt{2}$. For $\frac{K}{(s+1)(s+2)}$, is the loop ever unstable for $K>0$? ::: No — $P=0$ and the curve never reaches $-1$, so $Z=0$ always. What does a zero at $-1$ do for a double integrator? ::: Adds $+90^{\circ }$ phase lead, keeps $\mathrm{Im}L$ one-signed so the plot never wraps $-1$ → stable for all $K>0$.