PID tuning — Ziegler-Nichols, loop shaping: Exercises

Before we start, two symbols and three tools that the whole page leans on. We define the symbols on this page so it stands alone.

Formula The three tools you will keep using

Phase of a first-order factor. For a factor $(1 + j ω τ)$ , its phase is $arctan (ω τ)$ ; for $\frac{1}{1 + j ω τ}$ it is $- arctan (ω τ)$ . Why arctan? Because $1 + j ω τ$ is a point in the complex plane with horizontal part $1$ and vertical part $ω τ$ ; the angle it makes with the horizontal axis is the one whose tangent (opposite $ω τ$ over adjacent $1$ ) equals $ω τ$ — and "which angle has this tangent?" is exactly what $arctan$ answers. Branch caution: $arctan$ only ever returns an angle between $- 90°$ and $+ 90°$ , which is fine here because the real part is $+ 1$ (the point sits in the right half-plane). If you ever meet a factor whose real part goes negative (e.g. $(1 + j ω τ_{1}) (1 + j ω τ_{2})$ multiplied out, or a factor like $(- 1 + j ω τ)$ ), the true angle is past $\pm 90°$ and you must add or subtract $180°$ to the raw $arctan$ — the calculator's principal value silently folds it back. Every factor on this page keeps its real part positive, so plain $arctan$ is safe throughout.
Magnitude of a first-order factor. $∣1 + j ω τ ∣ = 1 + (ω τ)^{2}$ — the hypotenuse of that same little triangle.
A pure integrator $\frac{1}{j ω}$ . Magnitude $1/ ω$ , phase exactly $- 90°$ . Why exactly $- 90°$ ? Dividing by $j$ rotates any complex number a quarter turn clockwise; no arctan needed because the point sits straight down on the negative imaginary axis.

Closed-loop Ziegler–Nichols gives, for a full PID in the standard form, the rules $K_{p} = 0.6 K_{u}$ , $T_{i} = T_{u} /2$ , $T_{d} = T_{u} /8$ . A test found $K_{u} = 10$ and $T_{u} = 4$ s. Write down $K_{p}$ , $T_{i}$ , $T_{d}$ and convert to $K_{i}$ and $K_{d}$ .

Recall Solution 1.1

WHAT we do: read the table, then convert using $K_{i} = K_{p} / T_{i}$ and $K_{d} = K_{p} T_{d}$ (from the parent note).

$K_{p} = 0.6 \times 10 = 6$ .
$T_{i} = 4/2 = 2$ s $\Rightarrow K_{i} = K_{p} / T_{i} = 6/2 = 3$ .
$T_{d} = 4/8 = 0.5$ s $\Rightarrow K_{d} = K_{p} T_{d} = 6 \times 0.5 = 3$ .

WHY the conversion: the table speaks in times ( $T_{i}$ , $T_{d}$ ); your code wants gains ( $K_{i}$ , $K_{d}$ ). Multiplying/dividing by $K_{p}$ is the dictionary between the two forms.

Answer: $K_{p} = 6, K_{i} = 3, K_{d} = 3$ .

Problem 1.2

On a Bode plot the open loop $L (j ω)$ crosses $0$ dB (magnitude $= 1$ ) at $ω_{c} = 5$ rad/s, and there $∠ L = - 140°$ . State the phase margin. Is it inside the healthy $30°$ – $60°$ band?

Recall Solution 1.2

Definition (parent): $PM = 180° + ∠ L (j ω_{c})$ . $PM = 180° + (- 140°) = 40°.$ $40°$ lies inside $30°$ – $60°$ , so yes, healthy.

WHY this formula: instability happens when phase reaches $- 180°$ while gain is still $1$ . Phase margin measures the leftover angle you have before hitting that cliff. Here you can lose another $40°$ of phase before the loop sings.

Level 2 — Application

Problem 2.1

A plant $G (s) = \frac{1}{( s + 1 ) ^{3}}$ (the parent's test plant). Confirm the ultimate frequency, gain and period, then produce the ZN PID gains $K_{p}, K_{i}, K_{d}$ .

Recall Solution 2.1

Step A — find $ω_{u}$ (phase hits $- 180°$ ). Why phase first? The stability edge is defined by phase $= - 180°$ ; that equation has only $ω$ in it, so solve it alone. $∠ G (j ω) = - 3 arctan ω = - 180° \Rightarrow arctan ω_{u} = 60° \Rightarrow ω_{u} = tan 60° = 3 .$ Step B — magnitude there gives $K_{u}$ . $∣ G (j ω_{u}) ∣ = (ω_{u}^{2} + 1)^{- 3/2} = (3 + 1)^{- 3/2} = 4^{- 3/2} = \frac{1}{8}, K_{u} = \frac{1}{∣ G ∣} = 8.$ Why is $K_{u} = 1/∣ G ∣$ ? At the stability edge we crank up pure P gain until the open loop $L = K_{u} G$ has magnitude exactly $1$ at the frequency $ω_{u}$ where its phase is $- 180°$ — that is the "signal fed back inverted and equal in size reinforces itself forever" condition from the parent. Setting $∣ L (j ω_{u}) ∣ = K_{u} ∣ G (j ω_{u}) ∣ = 1$ and solving for $K_{u}$ gives $K_{u} = 1/∣ G (j ω_{u}) ∣$ : the ultimate gain is precisely the number that lifts the plant's shrunken magnitude back up to $1$ . Step C — period. $T_{u} = \frac{2 π}{ω _{u}} = \frac{2 π}{3} \approx 3.628$ s. Step D — ZN table.

$K_{p} = 0.6 \times 8 = 4.8$ .
$T_{i} = T_{u} /2 = 1.814 \Rightarrow K_{i} = 4.8/1.814 = 2.646$ .
$T_{d} = T_{u} /8 = 0.4535 \Rightarrow K_{d} = 4.8 \times 0.4535 = 2.177$ .

Answer: $K_{p} = 4.8, K_{i} \approx 2.65, K_{d} \approx 2.18$ — matching the parent's example.

Problem 2.2 (open-loop / reaction-curve method)

An open-loop step test on a heater fits a FOPDT model with static gain $K = 2$ , time constant $T = 10$ s, and dead time $L = 3$ s. Recall $R = K / T$ . Give the ZN open-loop PID parameters $K_{p}, T_{i}, T_{d}$ and then $K_{i}, K_{d}$ .

Recall Solution 2.2

Step A — slope parameter. $R = K / T = 2/10 = 0.2$ (per second). Step B — ZN open-loop formulas (parent). $K_{p} = \frac{1.2}{R L} = \frac{1.2}{0.2 \times 3} = \frac{1.2}{0.6} = 2.$ $T_{i} = 2 L = 6 s, T_{d} = 0.5 L = 1.5 s .$ Step C — gains. $K_{i} = K_{p} / T_{i} = 2/6 = 0.333$ , $K_{d} = K_{p} T_{d} = 2 \times 1.5 = 3$ .

WHY these ratios: the dead time $L$ is the villain — it is pure phase lag with no gain drop, so more $L$ forces a smaller $K_{p}$ (note $K_{p} \propto 1/ L$ ). The integral and derivative times are pinned to $L$ because for such plants the natural oscillation period is roughly $4 L$ .

Answer: $K_{p} = 2, T_{i} = 6, T_{d} = 1.5, K_{i} \approx 0.333, K_{d} = 3$ .

Level 3 — Analysis

Problem 3.1

For the same plant $G (s) = \frac{1}{( s + 1 ) ^{3}}$ , suppose we use pure P control with $K_{p} = 4$ (below $K_{u} = 8$ ). Find the gain margin in dB. Interpret it.

Recall Solution 3.1

WHAT is gain margin: at the frequency $ω_{180}$ where phase $= - 180°$ , how much more gain until $∣ L ∣ = 1$ . From 2.1, that frequency is $ω_{u} = 3$ and there $∣ G ∣ = 1/8$ . With $K_{p} = 4$ : $∣ L (j ω_{180}) ∣ = 4 \times \frac{1}{8} = \frac{1}{2}$ . $GM = \frac{1}{∣ L ( j ω _{180} ) ∣} = \frac{1}{0.5} = 2 ⟹ 20 lo g_{10} 2 = 6.02 dB .$ Interpretation: you may double the gain (from $4$ up to $8 = K_{u}$ ) before the loop reaches sustained oscillation — exactly consistent with the ultimate gain being $8$ . Where the " $6$ dB minimum" comes from: it is a design criterion, not a law of nature. The standard robustness rule of thumb (see Stability margins (gain & phase margin) and the Nyquist stability criterion) asks that gain margin be at least a factor of $2$ — i.e. $\geq 6.02$ dB — so that ordinary modelling errors and gain drift (component tolerances, temperature) cannot swallow your whole margin and tip the Nyquist curve past the $- 1$ point. So $K_{p} = 4$ sits exactly at that recommended lower bound: acceptable, but with no room to spare — the reason a designer would back off further.

Problem 3.2

Explain, using phase, why adding integral action $K_{i} / s$ to a loop that already had $PM = 45°$ reduces the phase margin — and estimate the loss if the integral corner $ω_{i} = K_{i} / K_{p}$ is placed at $ω_{c} /10$ .

Recall Solution 3.2

WHAT integral does to phase: the factor $\frac{1}{j ω}$ contributes exactly $- 90°$ ; but a real PI controller $K_{p} (1 + \frac{1}{T _{i} s})$ contributes only $- arctan (ω_{i} / ω)$ at frequency $ω$ , because far above its corner $ω_{i}$ the integral term becomes negligible and phase returns toward $0°$ . Estimate at $ω = ω_{c}$ with $ω_{i} = ω_{c} /10$ : $Δ ϕ = - arctan (\frac{ω _{i}}{ω _{c}}) = - arctan (0.1) = - 5.71°.$ New phase margin $\approx 45° - 5.7° = 39.3°$ . WHY it must be a loss: integral only ever adds lag (never lead), so it can only pull phase down at crossover; the design trick is to push its corner far below $ω_{c}$ so that lag has almost decayed by the time you reach $ω_{c}$ — hence the " $ω_{i} \approx ω_{c} /10$ " rule keeps the loss to a tolerable few degrees.

Level 4 — Synthesis

Problem 4.1 (lead design by phase margin)

Plant $G (s) = \frac{1}{s ( s + 1 )}$ . Design a PD controller $C (s) = K_{p} (1 + T_{d} s)$ so that the gain crossover is at $ω_{c} = 2$ rad/s with phase margin $PM = 45°$ . Find $T_{d}$ and $K_{p}$ .

Figure s01 (below) is the geometric heart of Step C. It draws the single complex number $1 + j T_{d} ω_{c}$ as a right triangle: adjacent side $= 1$ (black), opposite side $= T_{d} ω_{c}$ (black), hypotenuse the red arrow. Look at the red arrow — the angle it makes above the horizontal axis IS the phase lead the PD zero donates, and reading "opposite over adjacent" off this triangle is exactly why $tan (lead) = T_{d} ω_{c}$ .

Figure — PID tuning — Ziegler-Nichols, loop shaping

Recall Solution 4.1

Step A — plant phase at $ω_{c} = 2$ . $G$ has an integrator ( $- 90°$ flat) and one lag $(s + 1)$ ( $- arctan ω$ ). $∠ G (j 2) = - 90° - arctan 2 = - 90° - 63.43° = - 153.43°.$ Step B — how much phase must the controller supply? We want the total open-loop phase at $ω_{c}$ to be $- 135°$ , because $PM = 180° + ∠ L = 45°$ means $∠ L = - 135°$ . $ϕ_{lead} = ∠ L - ∠ G = - 135° - (- 153.43°) = + 18.43°.$ Step C — a PD zero supplies lead $arctan (T_{d} ω_{c})$ (this is the triangle in figure s01). Why arctan again? The factor $1 + j T_{d} ω$ is the little triangle with vertical side $T_{d} ω$ ; its angle above the axis is $arctan (T_{d} ω)$ — positive, i.e. phase lead. $arctan (T_{d} \cdot 2) = 18.43° \Rightarrow T_{d} \cdot 2 = tan 18.43° = 0.3333 \Rightarrow T_{d} = 0.1667.$ Step D — set magnitude to 1 at $ω_{c}$ (that is what "crossover" means). $∣ G (j 2) ∣ = \frac{1}{2 2 ^{2} + 1} = \frac{1}{2 5} = 0.2236, ∣1 + j T_{d} \cdot 2∣ = 1 + 0.333 3^{2} = 1.0541.$ $∣ L (j 2) ∣ = K_{p} \cdot 0.2236 \cdot 1.0541 = 1 \Rightarrow K_{p} = \frac{1}{0.2236 \times 1.0541} = 4.243.$

Answer: $T_{d} \approx 0.167$ s, $K_{p} \approx 4.24$ . Matches the parent example.

Figure s02 (below) makes Step B visible as a phase budget. Each black bar is one factor's phase contribution at $ω_{c} = 2$ ; they stack. Watch the red "total" bar land exactly on the red dotted $- 135°$ target line — that is the phase-margin condition being met, and the dashed $- 180°$ line above which we must never let the red bar reach is the instability cliff.

Problem 4.2 (build the full PID from that lead design)

Take the $K_{p} = 4.24$ , $T_{d} = 0.167$ from 4.1 and add integral action with corner $ω_{i} = ω_{c} /10 = 0.2$ rad/s (the rule of thumb). Give $K_{i}$ , and state by how much this trims the phase margin.

Recall Solution 4.2

Step A — $K_{i}$ from the corner. $ω_{i} = K_{i} / K_{p} \Rightarrow K_{i} = ω_{i} K_{p} = 0.2 \times 4.243 = 0.8485$ . Step B — phase penalty at $ω_{c}$ (as in 3.2): $Δ ϕ = - arctan (ω_{i} / ω_{c}) = - arctan (0.2/2) = - arctan (0.1) = - 5.71°.$ So the phase margin drops from $45°$ to about $39.3°$ — still healthy. Answer: $K_{i} \approx 0.849$ , PM $\approx 39.3°$ . (In practice you would nudge $T_{d}$ up slightly to recover the lost $5.7°$ .)

Level 5 — Mastery

Problem 5.1 (judge two designs)

For $G (s) = \frac{1}{( s + 1 ) ^{3}}$ you have two candidate PID controllers:

A (raw ZN): $K_{p} = 4.8, K_{i} = 2.65, K_{d} = 2.18$ (from 2.1).
B (detuned): the same $K_{p}$ reduced by $30%$ , integral and derivative times unchanged.

Compute B's gains, and argue in one line which you would ship for a position servo that must not overshoot much.

Recall Solution 5.1

Step A — B's $K_{p}$ . $K_{p}^{B} = 0.7 \times 4.8 = 3.36$ . Step B — keep the times fixed ( $T_{i} = 1.814$ , $T_{d} = 0.4535$ ), so: $K_{i}^{B} = K_{p}^{B} / T_{i} = 3.36/1.814 = 1.852, K_{d}^{B} = K_{p}^{B} T_{d} = 3.36 \times 0.4535 = 1.524.$ Step C — judgement. Raw ZN aims at quarter-amplitude decay $\approx 25%$ overshoot — too bouncy for a position servo. Detuning $K_{p}$ by $30%$ pulls the crossover down, buys phase margin, and cuts overshoot at the cost of a little speed. Ship B.

Answer: $K_{p}^{B} = 3.36, K_{i}^{B} \approx 1.85, K_{d}^{B} \approx 1.52$ .

Problem 5.2 (mastery synthesis: full loop-shaping check)

Combine everything. For the lead-plus-integral controller from 4.2 ( $K_{p} = 4.243$ , $T_{d} = 0.1667$ , $ω_{i} = 0.2$ ) on $G (s) = \frac{1}{s ( s + 1 )}$ , verify the final open-loop phase at $ω_{c} = 2$ and confirm the resulting phase margin numerically.

Recall Solution 5.2

Assemble the phases at $ω_{c} = 2$ (everything superposes):

Plant integrator: $- 90°$ .
Plant lag $(s + 1)$ : $- arctan 2 = - 63.43°$ .
Controller lead zero $(1 + T_{d} s)$ : $+ arctan (0.1667 \times 2) = + arctan (0.3333) = + 18.43°$ .
Controller integral lag: $- arctan (ω_{i} / ω_{c}) = - arctan (0.1) = - 5.71°$ . $∠ L (j 2) = - 90 - 63.43 + 18.43 - 5.71 = - 140.71°.$ $PM = 180° + (- 140.71°) = 39.29°.$ Interpretation: exactly the $\approx 39.3°$ predicted in 4.2. The design is self-consistent — the loss from integral shows up precisely where we expected, and the loop is safely stable with a fast $ω_{c} = 2$ rad/s. To recover the full $45°$ , raise $T_{d}$ so the lead becomes $18.43° + 5.71° \approx 24°$ .

Answer: $∠ L (j 2) \approx - 140.7°$ , $PM \approx 39.3°$ .

Recall Final self-check (hide and answer)

ZN closed loop measured $K_{u} = 8$ , $T_{u} = 3.63$ on $(s + 1)^{- 3}$ — where did $3$ come from? ::: From $arctan ω_{u} = 60°$ , so $ω_{u} = tan 60° = 3$ , the frequency where phase hits $- 180°$ .
Why is the ultimate gain the reciprocal of $∣ G (j ω_{u}) ∣$ ? ::: Because at the edge $∣ L ∣ = K_{u} ∣ G ∣ = 1$ , so $K_{u} = 1/∣ G (j ω_{u}) ∣$ lifts the plant's magnitude back to $1$ .
Why does a PD zero raise phase margin? ::: It contributes $+ arctan (T_{d} ω)$ of lead near crossover, pushing $∠ L$ away from $- 180°$ .
What does a $6.02$ dB gain margin literally allow, and why is it the minimum? ::: Doubling the loop gain before oscillation; $6$ dB (factor $2$ ) is the standard robustness criterion so gain drift cannot erase the margin.
Why place $ω_{i} = ω_{c} /10$ ? ::: So the integral's lag $- arctan (ω_{i} / ω_{c}) \approx - 5.7°$ costs almost no phase margin while still boosting low-frequency gain.

Related vault notes: PID controller basics · Bode plot & frequency response · Stability margins (gain & phase margin) · Nyquist stability criterion · Integral windup and anti-windup · Root locus method · State-space control & LQR.

Exercises — PID tuning — Ziegler-Nichols, loop shaping

Level 1 — Recognition

Problem 1.1