The control systems exam: Bode, Nyquist, root locus

Control Systems (Controlli Automatici) is the exam in which many engineering students — mechanical, aerospace, electronics, automation — meet for the first time a way of reasoning completely different from that of Calculus or Physics: you don’t solve an equation to find a number, you reason about the shape of a frequency response to decide whether a system behaves well or blows up.

It is also an exam where it is very easy to “almost” understand: you know what a Bode diagram is, you know that the Nyquist criterion counts encirclements around a point, but put in front of a specific transfer function you no longer know where to start. Here I bring you the method I use with my students: what the exam really asks, where marks are lost, and three exercises solved down to the calculations.

What the exam asks

Almost all (introductory) Control Systems courses ask, on a given transfer function $G(s)$ (often open-loop, sometimes with a controller to design):

1. Analysis of the system: poles, zeros, gain, type (number of poles at the origin), open-loop stability.
2. Closed-loop stability with the Routh-Hurwitz criterion on the characteristic equation.
3. Asymptotic Bode diagram (magnitude and phase) drawn by hand, with reading of the phase margin and gain margin.
4. Nyquist criterion for closed-loop stability starting from the open-loop diagram.
5. Root locus as a gain $K$ varies: asymptotes, breakaway points, crossings of the imaginary axis.
6. Often, the design of a compensator (phase lead or lag) to meet specifications on phase margin or steady-state error.

The common thread is always the same: from the transfer function, with pen and paper, you have to reach saying yes, it is stable or no, it isn’t — and for which range of parameters.

Where marks are lost, in concrete terms

1. Confusing phase margin and gain margin. The phase margin is read at the gain crossover frequency ${\omega }_{gc}$ (where $|G(j\omega )|=1$, that is 0 dB): it is $180°+\phi ({\omega }_{gc})$. The gain margin is read at the phase crossover frequency ${\omega }_{pc}$ (where the phase is $-180°$): it is $-20{\log }_{10}|G(j{\omega }_{pc})|$. They are two different frequencies, read on two different axes, and swapping them is the most common mistake of the whole exam.

2. Forgetting the contribution of poles/zeros at the origin. A pole at the origin ($1/s$) already starts at $-90°$ of phase at all frequencies and contributes a slope of $-20$ dB/decade from the very start of the diagram — not from a break point. If the system is type 2 (two poles at the origin), you start at $-180°$ of phase: it is very easy to forget and get the whole reading of the phase margin wrong.

3. Getting the slope in dB/decade wrong. Each real pole adds $-20$ dB/decade after its break frequency, each real zero adds $+20$ dB/decade. A double pole (or a complex conjugate pair) adds $-40$. A sign error here flips the whole graph.

4. Routh: a table with a zero in the first column handled at random. If a zero appears in the first column (but the row is not entirely zero), you replace it with a small $\epsilon >0$ and carry on taking the limit; if instead the entire row is zero, the system has a symmetry of the roots with respect to the origin and you need the auxiliary polynomial. They are two different cases and must be recognised, not improvised.

5. Nyquist: forgetting the open-loop unstable poles. The criterion says $N=Z-P$: the number of clockwise encirclements of the polar diagram around $(-1,0)$ equals the number of zeros minus the number of open-loop unstable poles of the closed-loop system. If the open-loop $G(s)$ already has an unstable pole ($P=1$), you need counter-clockwise encirclements to stabilise, not the absence of encirclements: it is the point that confuses people most.

6. Root locus: forgetting that it starts at the poles and ends at the zeros (or at infinity). The locus for $K:0\to \infty$ always starts at the poles of $G(s)$ ($K=0$) and arrives at the finite zeros or, if there are not enough of them, goes to infinity along the asymptotes. Drawing a branch that “comes out of nowhere” is almost always a sign that something in the real-axis segment rule has been applied wrongly.

The method, step by step

1. Preliminary analysis

Write $G(s)$ in standard form, with the Bode gain isolated:

$N$ is the type of the system (number of poles at the origin): it determines the steady-state error for a step/ramp/parabola input and the starting slope of the Bode plot.

2. Routh-Hurwitz on the characteristic equation

Given the characteristic polynomial $a_n{s}^n+\cdots +a_{1}s+a_0$, build the Routh table starting from the even and odd coefficients on the first two rows. The system is stable (all roots with negative real part) if and only if all the elements of the first column have the same sign — no sign changes, no zeros.

3. Asymptotic Bode diagram

Break $G(j\omega )$ into its elementary factors and add up their contributions:

• Constant gain $K$: $20{\log }_{10}|K|$ dB, phase $0°$ (or $180°$ if $K<0$) — a horizontal line.
• Pole/zero at the origin ($1/s^N$ or $s^N$): slope $\mp 20N$ dB/decade over the whole graph, constant phase $\mp 90N°$.
• Simple real pole/zero ($1/(1+s/p)$ or $(1+s/z)$): flat asymptote up to the break frequency, then slope $\mp 20$ dB/decade; the phase varies continuously from $0°$ to $\mp 90°$, passing through $\mp 45°$ exactly at the break frequency (approximated as a linear variation between a decade before and a decade after, in the asymptotic plot).
• Complex conjugate pair: slope $\mp 40$ dB/decade after ${\omega }_n$, but watch for resonance if the damping $\zeta$ is low — the asymptote alone underestimates the peak.

4. Stability margins

Both positive ⇒ the closed-loop system is stable (for minimum-phase systems, the most common case in the exam).

5. Nyquist criterion

$N$ = clockwise encirclements of the polar diagram around $(-1,0)$, $P$ = unstable poles of the open-loop $G(s)$, $Z$ = unstable closed-loop poles. For closed-loop stability you need $Z=0$, that is $N=-P$ (that is, $P$ counter-clockwise encirclements if $P>0$, none if $P=0$).

6. Compensator

A lead network adds positive phase near the crossover frequency to increase the phase margin; a lag network reduces the gain at high frequencies without touching the phase too much, to correct the steady-state error without destabilising. The choice depends on what is missing: phase (lead) or low-frequency gain (lag).

Three worked exercises

Exercise 1 — Routh-Hurwitz: for which $K$ is the system stable?

Open loop $G(s)=\frac{K}{s(s+2)(s+10)}$, unity feedback. The characteristic equation is:

Routh table:

The element of the $s^1$ row is the usual combination of the two rows above:

For stability the first column must be all positive: $12>0$ (always), $\frac{240-K}{12}>0\Rightarrow K<240$, and $K>0$. So:

For $K=240$ the $s^1$ row vanishes: it is the critical value at which the system has a pair of poles on the imaginary axis and starts to oscillate in a sustained way (limit of stability).

Exercise 2 — Bode: phase margin of $G(s)=\frac{10}{s(1+s/5)}$

Type 1 (one pole at the origin), one real pole at $\omega =5$ rad/s. Total phase: $\phi (\omega )=-90°-\arctan (\omega /5)$, always between $-90°$ (as $\omega \to 0$) and $-180°$ (as $\omega \to \infty$) — it never reaches $-180°$, so the gain margin is infinite (there is no phase crossover frequency).

For the phase margin we need ${\omega }_{gc}$. With the asymptotic plot: below $\omega =5$ the slope is $-20$ dB/decade and the asymptotic magnitude is $20{\log }_{10}(10/\omega )$, which vanishes at $\omega =10$; but $10>5$, so at that frequency you are already beyond the break, on the $-40$ dB/decade stretch. The asymptotic magnitude at $\omega =5$ is

From there the slope is $-40$ dB/decade:

Solving $|G(j\omega )|=1$ exactly (without the asymptotic approximation) gives ${\omega }_{gc}\approx 6.25$ rad/s — the asymptotic estimate overestimates a little, as is normal near a break, but it is already useful for getting your bearings before refining the calculation. With the exact value:

A phase margin of around 40° is the value typically sought in design (below 30° the system is close to practical instability, above 60° it is very conservative).

Exercise 3 — Root locus: breakaway point of $G(s)=\frac{K}{s(s+4)}$

Two real poles, at $s=0$ and $s=-4$, no finite zeros: the locus, for $K$ increasing from 0, starts at the two poles and moves towards each other on the real segment between $0$ and $-4$ (segment rule: an odd number of real poles/zeros to the right of the point), until they meet at a breakaway point and then leave vertically towards infinity (two branches, for two poles and zero finite zeros).

The characteristic equation is $s(s+4)+K=0$, that is $K=-s^2-4s$. The breakaway point is found where $dK/ds=0$:

which is, not by chance, the midpoint between the two poles (symmetry of the two-real-pole, no-zero case). The gain at that point:

At $K=4$ the two branches meet at $s=-2$ (a real double pole, critical damping) and for $K>4$ they become a complex conjugate pair: the system stays stable for every $K>0$ (both poles keep a negative real part), but the response goes from overdamped to underdamped exactly at $K=4$.

How we work in lessons

On Control Systems I always work by hand, with graph paper (or its digital equivalent on a graphics tablet): the asymptotic Bode plot has to be known how to draw quickly, because in the exam you don’t have MATLAB to check it. We start from simple transfer functions — one pole, one zero — and build fluency on the elementary contributions before summing them on complex cases.

A good part of the work is also about vocabulary: understanding why it is called phase margin, what “the system is minimum-phase” means, when the Nyquist criterion applies in its simplified version (the “margin criterion”, valid only for minimum-phase systems) and when the full version is needed. That is where the oral exam is played out, when there is one.

If your course also covers state-space representation (matrices $A,B,C,D$, controllability, observability, state feedback), we cover that as a separate module — ask me in the first lesson so we can plan it together.

Two foundations taken for granted here: the eigenvalues and matrices of the state-space representation are in linear algebra: rank, kernel and image, and the complex poles, magnitude and phase of a transfer function can be read with complex numbers explained properly. If instead you need to simulate the response numerically, you get there from the numerical analysis exam.

For the general picture of how I work on maths applied to engineering you will find the maths page; if your control course starts from notions of signals and systems that are not yet clear to you, also take a look at the physics page for the foundations of dynamics and differential equations.

Before booking, try the university exam mock tests: correct answer free, guided solution unlocked with your first lesson. If you have a specific exercise that doesn’t work out for you, write to me at contact for a free 30-minute introductory call.