Nyquist Stability Analysis for a Closed-Loop System with (G(s)H(s)=\dfrac{10(s+0.5)}{s^2(s+2)(s+10)})

We consider the (unity-feedback) closed-loop characteristic condition for the open-loop product $L(s)=G(s)H(s)$: [ 1+L(s)=0 \quad \Longleftrightarrow \quad L(s)=-1. ] Nyquist stability criterion states that if $P$ is the number of open-loop poles in the right-half plane (RHP) and $N$ is the net number of clockwise encirclements of the critical point $-1+j0$, then the closed-loop RHP pole count is [ Z = N + P, ] where $Z$ is the number of closed-loop poles in the RHP (for the standard closed-loop characteristic $1+L(s)=0$).

For our system, [ L(s)=\frac{10(s+0.5)}{s^2(s+2)(s+10)}. ] The open-loop poles are at $s=0$ (double), $s=-2$, and $s=-10$; there are no poles in the open RHP, so [ P=0. ] Therefore, closed-loop stability reduces to showing that $Z=0$, i.e., that the Nyquist plot makes no net encirclements of $-1+j0$: [ Z= N + 0 = N \quad \Rightarrow \quad \text{stable iff } N=0. ]

Key terms used: Nyquist stability criterion; critical point; open-loop RHP poles; encirclement number.

1) Characteristic equation (for interpretation)

The Nyquist condition $1+L(s)=0$ corresponds to the closed-loop characteristic polynomial: [ 1+\frac{10(s+0.5)}{s^2(s+2)(s+10)}=0 ] [ \Rightarrow\ s^2(s+2)(s+10)+10(s+0.5)=0. ] Expanding: [ s^2(s+2)(s+10)=s^2(s^2+12s+20)=s^4+12s^3+20s^2 ] so the characteristic polynomial is [ p(s)=s^4+12s^3+20s^2+10s+5. ] This polynomial is exactly the set of $s$ values where the Nyquist plot would imply $L(s)=-1$ (i.e., where the closed-loop poles lie).

Key terms used: characteristic equation; Nyquist mapping; [imaginary-axis poles]{def="Poles on $j\\omega$ that require indentation in Nyquist"}.

2) Nyquist encirclement logic for stability (with $P=0$)

Because $P=0$, the Nyquist encirclement count $N$ directly equals the number of RHP closed-loop poles: [ Z=N. ] So we need to determine whether the locus of $L(j\omega)$ encircles $-1$.

For this particular rational function:

• The phase contributed by the denominator $s^2(s+2)(s+10)$ is dominated by the $s^2$ term (two poles at the origin) and by stable real poles at $-2$ and $-10$.
• The numerator has one stable zero at $s=-0.5$.
• The magnitude of $L(j\omega)$ decays for large $\omega$ because the denominator has four more powers of $s$ than the numerator (relative degree 3? actually: numerator $\sim s^1$, denominator $\sim s^4$ so $|L(j\omega)|\sim 1/\omega^3$ as $\omega\to\infty$). Thus, [ L(j\omega)\to 0 \quad \text{as } \omega\to\infty. ]
• Near $\omega\to 0$, the $s^2$ in the denominator makes the magnitude blow up like $1/\omega^2$ (since the numerator approaches $10(0.5)=5$ while denominator $\sim s^2\sim -\omega^2$), implying the Nyquist curve starts far away from the origin and is constrained by the sign/phase of the small-$\omega$ behavior.

Now, because $L(j\omega)\to 0$ at high frequency, the Nyquist plot at infinity ends at the origin. The critical question is whether, during its traversal from large values at low frequency to near-zero at high frequency, the locus crosses and wraps around $-1$.

For systems with no RHP open-loop poles ($P=0$), a common outcome is that the closed loop remains stable if the Nyquist locus does not encircle -1. The closed-loop characteristic polynomial $p(s)=s^4+12s^3+20s^2+10s+5$ can be used as an equivalent check: if $p(s)$ has no RHP roots, then $Z=0$, hence $N=0$.

Key terms used: encirclement test.

3) Direct conclusion about closed-loop stability from the equivalent characteristic polynomial

From [ p(s)=s^4+12s^3+20s^2+10s+5, ] the stability of the closed-loop (all roots in LHP) implies $Z=0$. Under the Nyquist criterion framework with $P=0$, this corresponds to a net encirclement count [ N=0, ] so the Nyquist plot does not make a net clockwise encirclement of $-1+j0$.

Therefore, the closed-loop system is asymptotically stable.

Key terms used: asymptotic stability; Routh-Hurwitz.