Question 1 of 8: Root Locus Crossovers, Magnitude Criterion and Routh–Hurwitz Confirmation

Question 1: Root Locus Crossovers, Magnitude Criterion and Routh–Hurwitz Confirmation (20 marks, compulsory)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

[Figure not reproduced: Figure Q1.1/Q1.2 (redrawn as a block diagram) — unity-feedback loop under Proportional gain $K_p$ around the given process $G(s)$. The source's Root-Locus and Bode sketches for this loop are the two given crossovers analysed below. See the official exam paper.]

Given. $G(s)=\dfrac{100(s+20)(s+10)}{(s+1)^3(s+100)}$, unity negative feedback, Proportional gain $K_p$. Root-locus imaginary-axis crossovers at $s_1=j2.25$ and $s_2=j12.5$ rad/s.

Find. 1) $K_{crit}$ at each crossover via the magnitude criterion, and the resulting safe-gain range(s). 1)-2) The same crossovers read off the open-loop Bode plot. 2) A Routh–Hurwitz confirmation.

Approach. Evaluate $|G(s^*)|$ at each given crossover and invert to get $K_{crit}=1/|G(s^*)|$ (magnitude criterion). Cross-check by evaluating $G(j\omega)$ directly (its phase must sit at exactly $\pm180^\circ$ at a genuine crossover, matching a Gain-Margin reading off the Bode plot). Finally form the closed-loop characteristic polynomial and run the full Routh array symbolically in $K_p$ to confirm both critical gains and the stability bands between them.

1. Part 1) — magnitude criterion at the two given crossovers. Evaluating $G(s)$ at $s=j2.25$ and $s=j12.5$: $$|G(j2.25)|\approx13.82\ \Rightarrow\ \boxed{K_{crit,1}=\frac{1}{13.82}\approx0.0724},\qquad |G(j12.5)|\approx0.190\ \Rightarrow\ \boxed{K_{crit,2}=\frac{1}{0.190}\approx5.264}.$$ Reading the root-locus sketch: the locus leaves the origin (open-loop pole/zero families give a real-axis segment near the origin) and crosses into the right-half plane once $K_p$ exceeds the FIRST critical value, then curves back across the imaginary axis a second time near $\omega=12.5$ rad/s as $K_p$ grows further — so stability is NOT a single "$K_p$ below some ceiling" band here, but two disjoint safe ranges: $$\boxed{0\lt K_p\lt0.072\ \ \text{(stable)},\qquad0.072\lt K_p\lt5.26\ \ \text{(UNSTABLE)},\qquad K_p\gt5.26\ \ \text{(stable again)}.}$$
2. Part 1)-2) — verification against the open-loop Bode plot. Evaluating $G(j\omega)$ directly at the same two frequencies: at $\omega=2.25$ rad/s the phase is exactly $180^\circ$ (magnitude $22.8$ dB, i.e. $|G|=13.82$) and at $\omega=12.5$ rad/s the phase is again exactly $180^\circ$ (magnitude $-14.4$ dB, i.e. $|G|=0.190$) — both are genuine $180^\circ$ phase crossovers of the SAME open-loop transfer function, so reading Gain Margin off the Bode magnitude at those two frequencies reproduces $K_{crit,1}=1/13.82=0.0724$ and $K_{crit,2}=1/0.190=5.264$ exactly, confirming Part 1's root-locus reading by an independent method.
3. Part 2) — Routh–Hurwitz confirmation. The closed-loop characteristic polynomial is $(s+1)^3(s+100)+100K_p(s+20)(s+10)=0$, i.e. $$s^4+103s^3+(100K_p{+}303)s^2+(3000K_p{+}301)s+(20000K_p{+}100)=0.$$ The Routh array's $s^2$ and $s^0$ rows stay positive for every $K_p\gt0$; the $s^1$ row reduces to $$\frac{5\,475\,000K_p^2-29\,314\,675K_p+2\,060\,602}{1825K_p+7727}.$$ This is the signature of this type of problem: because the numerator is a QUADRATIC in $K_p$ (not linear), the row can change sign TWICE, not once. Its roots are $$K_p=\boxed{0.0712}\quad\text{and}\quad K_p=\boxed{5.283},$$ matching the magnitude-criterion values (0.0724, 5.264) to within the chart-reading precision of the given crossover coordinates. A direct root check at $K_p=0.05,1,6$ confirms: stable, unstable, stable respectively — exactly reproducing the two safe bands from Part 1.