Part 5 · Chapter 21

Lead, Lag, and Lead–Lag Compensator Design

A single gain moves the closed-loop poles only along a locus the plant has already fixed, so when accuracy and damping make demands that no point on that locus can satisfy at once, the cure is to change the locus itself — which is what a compensator does, by adding one zero and one pole and choosing whether they buy phase near crossover or gain near the origin.

Control Systems Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • Why a plant's root locus is fixed and what adding a pole–zero pair does to it.
  • The lead network, its parameter \(\alpha\), and the derivation of \(\sin\phi_m=(1-\alpha)/(1+\alpha)\) at \(\omega_m=1/T\sqrt{\alpha}\).
  • The complete Bode design procedure for a lead compensator, including why a safety allowance of \(5^\circ\) to \(12^\circ\) is always added.
  • The angle-deficiency method that places a lead compensator from a root-locus specification.
  • How a lag network multiplies the error constant by \(\beta\) while contributing almost no phase at crossover.
  • When one network cannot do the job, and how a lead–lag compensator splits the two tasks between its two halves.
  • The passive \(RC\) realisations, their loading and attenuation penalties, and how lead and lag relate to the PD and PI actions of Chapter 19.
Section 21-1

Why Gain Alone Is Not Enough

Chapter 12 built the root locus as the set of points satisfying the angle condition, and that set depends only on the open-loop poles and zeros — not on the gain. Turning the gain knob slides the closed-loop poles along the locus; it cannot move them off it. If the locus never passes through a region where the transient specification is met, no value of \(K\) will meet it.

The frequency-domain statement is the same fact in different clothing. Raising \(K\) lifts the whole magnitude curve without touching the phase curve, so the gain crossover slides to a higher frequency where the plant's phase is more negative, and the phase margin of Chapter 17 falls. Chapter 9 closed with the resulting deadlock: accuracy wants a large \(K\), damping wants a small one, and a plant of any complexity soon reaches a point where the two demands cannot both be met.

A compensator breaks the deadlock by adding dynamics of its own — in the classical case one zero and one pole, cascaded with the plant.

🔑
The two classical networks
\[ \underbrace{G_c(s)=K_c\alpha\,\frac{Ts+1}{\alpha Ts+1}}_{\text{lead},\;0<\alpha<1} \qquad\qquad \underbrace{G_c(s)=K_c\beta\,\frac{Ts+1}{\beta Ts+1}}_{\text{lag},\;\beta>1} \]

Both are one zero at \(-1/T\) and one pole. In a lead the pole lies further left than the zero, so the network contributes positive phase. In a lag the pole lies closer to the origin than the zero, so it contributes negative phase — and a useful attenuation at high frequency.

The two networks solve different problems. A lead adds phase where the phase margin is short, which improves damping and speed; it is a transient-response tool. A lag adds low-frequency gain without disturbing the phase near crossover, which improves the error constants of Chapter 9; it is a steady-state tool. Deciding which to use begins by asking which specification the uncompensated system fails.
Section 21-2

The Lead Network and Its Maximum Phase

Write the lead network in its two equivalent forms, one showing the corner frequencies, the other the pole and zero locations:

The lead network
\[ G_c(s)=K_c\alpha\,\frac{Ts+1}{\alpha Ts+1}=K_c\,\frac{s+1/T}{s+1/(\alpha T)},\qquad 0<\alpha<1 \]

The zero sits at \(-1/T\) and the pole at \(-1/(\alpha T)\); since \(\alpha<1\) the pole is a factor \(1/\alpha\) further from the origin. Between the two corner frequencies the magnitude rises at \(20\) dB/decade and the phase is positive. The phase contribution is the difference of two arctangents,

Phase of the lead network
\[ \phi(\omega)=\arctan(\omega T)-\arctan(\alpha\omega T) \]

which is zero at both extremes and therefore has a maximum somewhere between. Differentiating and setting the derivative to zero locates it.

Locating the maximum
\[ \frac{d\phi}{d\omega}=\frac{T}{1+\omega^2T^2}-\frac{\alpha T}{1+\alpha^2\omega^2T^2}=0 \]
\[ 1+\alpha^2\omega^2T^2=\alpha+\alpha\,\omega^2T^2 \;\Longrightarrow\; \alpha\,\omega^2T^2(\alpha-1)=\alpha-1 \;\Longrightarrow\; \omega^2T^2\alpha=1 \]
\[ \omega_m=\frac{1}{T\sqrt{\alpha}}=\sqrt{\frac{1}{T}\cdot\frac{1}{\alpha T}} \]

The maximum falls at the geometric mean of the two corner frequencies — the midpoint on a logarithmic axis, which is why the phase bump looks symmetric on a Bode plot. Substituting \(\omega_mT=1/\sqrt{\alpha}\) into the phase expression gives the peak value.

The maximum phase lead
\[ \tan\phi_m=\frac{\omega_mT-\alpha\omega_mT}{1+\alpha\,\omega_m^2T^2}=\frac{(1-\alpha)/\sqrt{\alpha}}{1+1}=\frac{1-\alpha}{2\sqrt{\alpha}} \]
\[ \text{opposite}=1-\alpha,\quad \text{adjacent}=2\sqrt{\alpha},\quad \text{hypotenuse}=\sqrt{(1-\alpha)^2+4\alpha}=1+\alpha \]
🔑
Lead network design equations
\[ \sin\phi_m=\frac{1-\alpha}{1+\alpha} \qquad\Longleftrightarrow\qquad \alpha=\frac{1-\sin\phi_m}{1+\sin\phi_m} \]

at \(\omega_m=1/(T\sqrt{\alpha})\), where the magnitude of \(K_c\alpha(Ts+1)/(\alpha Ts+1)\) is \(K_c\sqrt{\alpha}\) — that is, \(10\log_{10}(1/\alpha)\) decibels above its low-frequency value. The phase and the gain are not independent: a network that supplies more phase necessarily supplies more gain at the same frequency, and this is the fact that drives the whole design procedure.

dB 0 +20 dB/dec 20 log(1/α) 1/T 1/αT log ω deg 0 φ_m ω_m = 1/(T√α)
The lead network: a gain step and a phase bump at the geometric mean
\(\alpha\)\(\phi_m\)High-frequency gain \(1/\alpha\)
0.5\(19.5^\circ\)2 (6 dB)
0.25\(36.9^\circ\)4 (12 dB)
0.1\(54.9^\circ\)10 (20 dB)
0.05\(64.8^\circ\)20 (26 dB)
0.01\(78.6^\circ\)100 (40 dB)

The last column is the reason a single lead network is rarely pushed past about \(60^\circ\). The factor \(1/\alpha\) is a high-frequency gain applied to everything the sensor delivers, and a sensor delivers noise as well as signal. Chapter 19 raised this objection against derivative action; the lead network is that same derivative action with the amplification deliberately limited, and \(\alpha\) is the limit. Taking \(\alpha\) below \(0.05\) trades an unreliable measurement for a marginal gain in phase. When more than \(60^\circ\) is genuinely needed, two lead sections in cascade — each supplying half — cost far less in noise than one section supplying all of it.

Section 21-3

Lead Design on the Bode Plot

The design problem is usually posed as a pair of specifications: an error constant, which fixes the low-frequency gain, and a phase margin, which fixes the behaviour at crossover. The procedure below satisfies the first exactly and the second by construction.

Its one subtlety is the reason \(\phi_m\) is not simply the shortfall in phase margin. Adding the lead network raises the magnitude curve near \(\omega_m\), which pushes the gain crossover to a higher frequency, where the plant contributes more phase lag than it did before. Some of the phase we just bought is therefore immediately spent. An allowance of \(5^\circ\) for a gently sloping plant, up to \(12^\circ\) for a steep one, covers the loss.

StepAction
1Choose the open-loop gain \(K\) so that the error-constant specification (\(K_p\), \(K_v\) or \(K_a\) from Chapter 9) is met exactly. Call the result \(G_1(s)\).
2Draw the Bode plot of \(G_1\) and read its phase margin \(\gamma_1\).
3Required lead \(\phi_m=\gamma_{\text{spec}}-\gamma_1+\varepsilon\), with \(\varepsilon=5^\circ\) to \(12^\circ\).
4\(\alpha=\dfrac{1-\sin\phi_m}{1+\sin\phi_m}\).
5Find \(\omega_m\), the frequency at which \(|G_1(j\omega)|=\sqrt{\alpha}\), i.e. \(10\log_{10}\alpha\) dB. This becomes the new gain crossover, because the network adds exactly \(10\log_{10}(1/\alpha)\) dB there.
6\(T=\dfrac{1}{\omega_m\sqrt{\alpha}}\); the zero is at \(-1/T\) and the pole at \(-1/(\alpha T)\).
7Set \(K_c=1/\alpha\) so that \(G_c(0)=K_c\alpha=1\) and the error constant of step 1 is untouched.
8Verify the phase margin of \(G_cG_1\). If it falls short, increase \(\varepsilon\) and repeat.
Step 5 is where the design actually happens. Choosing \(\omega_m\) to be the frequency at which the uncompensated magnitude is \(\sqrt{\alpha}\) guarantees that the compensated magnitude is exactly unity there — so the new crossover and the peak of the phase bump coincide. Every bit of phase the network can supply is delivered at the one frequency where phase margin is measured. A lead compensator that is centred anywhere else is simply wasting itself.
Section 21-4

Lead Design on the Root Locus

When the specification is given as overshoot and settling time rather than as a phase margin, Chapter 8 converts it into a pair of desired dominant poles \(s_d=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}\), and the design question becomes geometric: how must the locus be bent so that it passes through \(s_d\)?

The answer comes straight from the angle condition. Evaluate the open-loop transfer function at the desired point; if the angle is not \(-180^\circ\), the compensator must make up the difference.

🔑
Angle deficiency
\[ \phi_c=-180^\circ-\angle G(s_d) \pmod{360^\circ} \]

If \(\phi_c\) is positive the point lies to the left of the existing locus and a lead network is required; if it is negative the point lies to the right and a lag — or a reduction in gain — is called for. A single lead section can supply any \(\phi_c\) up to roughly \(60^\circ\); beyond that, use two.

Knowing how much angle is needed does not by itself fix the pole and the zero: infinitely many pairs supply the same angle at \(s_d\). Three choices are standard. The first is cancellation — put the compensator zero on top of an existing plant pole, which removes that pole from the picture and leaves only the compensator pole to place. It is the simplest choice and the one that gives the cleanest algebra, but it must never be used on an unstable or very lightly damped plant pole, because an imperfect cancellation leaves that pole in the closed-loop response where it is no longer controllable by the gain.

The second is the bisector rule. Draw the line from \(s_d\) to the origin and the horizontal line through \(s_d\); bisect the angle between them, then place the zero and the pole on the real axis at \(\pm\phi_c/2\) from that bisector, as seen from \(s_d\). This construction can be shown to maximise \(\alpha\) — that is, to place the pole and zero as close together as possible for the required angle — which minimises the high-frequency gain \(1/\alpha\) and therefore the noise penalty.

The third is simply to fix the zero somewhere convenient, just to the left of the plant's slowest pole, and solve for the pole from the angle condition. Whichever placement is used, the gain follows from the magnitude condition \(|G_c(s_d)G(s_d)|=1\), and the design is not finished until two checks have been made: that the remaining closed-loop poles are far enough left for \(s_d\) to dominate, and that the error constant has not been ruined — a lead network multiplies \(K_v\) by \(\alpha\) at fixed \(K_c\), which is a reduction.

σ 0 −2 s_d wanted ζω_n fixed at 1
Uncompensated: the branches cannot reach s_d
σ 0 −4 s_d reached ζω_n now 2
Lead compensated: the asymptote moves left
Section 21-5

The Lag Network

Reverse the ordering of pole and zero and the network changes character completely.

The lag network
\[ G_c(s)=K_c\beta\,\frac{Ts+1}{\beta Ts+1}=K_c\,\frac{s+1/T}{s+1/(\beta T)},\qquad \beta>1 \]

Now the pole at \(-1/(\beta T)\) is nearer the origin than the zero at \(-1/T\). The magnitude starts at \(K_c\beta\) at low frequency, falls at \(-20\) dB/decade between the two corners, and settles at \(K_c\) — a factor \(\beta\) lower. The phase is negative throughout, with a minimum of \(-\phi_m\) at the geometric mean, where \(\sin\phi_m=(\beta-1)/(\beta+1)\) by the same algebra as before.

Negative phase is exactly what a stability-limited loop does not need, so the lag network is not used for its phase at all. It is used for the ratio \(\beta\) between its low- and high-frequency gains, and the design consists of arranging for the phase penalty to be paid somewhere harmless.

dB Kcβ −20 dB/dec 20 log β 1/βT 1/T log ω deg 0 −φ_m crossover is placed out here
The lag network: a gain reduction bought with a phase dip left behind

The trick that makes this possible is to place both corner frequencies well below the intended gain crossover. Put the zero one decade below the new crossover \(\omega_2\), so \(1/T=\omega_2/10\). At \(\omega_2\) the zero contributes \(\arctan 10=84.3^\circ\) and the pole contributes very nearly \(90^\circ\) of lag, so the network's net phase there is only about \(-5.7^\circ\) — an amount easily covered by the same safety allowance \(\varepsilon\) used in the lead procedure. Meanwhile the full attenuation \(1/\beta\) is already in effect at \(\omega_2\), because \(\omega_2\) lies a decade above the upper corner.

🔑
What a lag network buys
\[ \text{low-frequency gain}\;\times\;\beta \quad\text{while}\quad |G_c(j\omega_2)|\approx K_c,\;\; \angle G_c(j\omega_2)\approx-6^\circ \]

Equivalently: keep the low-frequency gain fixed and the lag attenuates the loop near crossover by \(\beta\), dragging the crossover down to a frequency where the plant's own phase lag is smaller. Either reading is correct — the network improves the error constant at fixed margin, or the margin at fixed error constant.

A lag improves the margin by making the loop slower. This is the essential difference from a lead, which improves the margin by adding phase and, in doing so, makes the loop faster. Bandwidth rises with lead compensation and falls with lag compensation. When a specification asks for both better accuracy and a faster response, no single network can deliver it, and Section 21-7 is the answer.
Section 21-6

Lag Design in Both Domains

The frequency-domain procedure mirrors the lead procedure but reads the Bode plot in the opposite order: the phase curve is consulted first to choose the new crossover, and the magnitude curve then tells us how much attenuation is needed there.

StepAction
1Set \(K\) from the error-constant specification, giving \(G_1(s)\).
2On the phase curve of \(G_1\), find \(\omega_2\) where \(\angle G_1=-180^\circ+\gamma_{\text{spec}}+\varepsilon\), with \(\varepsilon=5^\circ\) to \(12^\circ\). This will be the new gain crossover.
3Read \(|G_1(j\omega_2)|\). The lag must remove exactly this much: \(\beta=|G_1(j\omega_2)|\), or \(20\log_{10}\beta=|G_1(j\omega_2)|_{\mathrm{dB}}\).
4Place the zero a decade below: \(1/T=\omega_2/10\).
5Place the pole at \(1/(\beta T)\).
6Set \(K_c=1/\beta\) so that \(G_c(0)=K_c\beta=1\) and the error constant survives.
7Verify the phase margin; the \(-6^\circ\) contributed by the network is already covered by \(\varepsilon\).

On the root locus the lag plays a different-looking but identical role. Suppose the uncompensated locus already passes through an acceptable \(s_d\) — the transient specification is met — but the gain there gives an error constant that is too small by a factor \(\beta\). Add a zero at \(-z\) and a pole at \(-z/\beta\), both very close to the origin, with \(z\) chosen small compared with \(|s_d|\).

The lag dipole and the error constant
\[ G_c(s)=\frac{s+z}{s+z/\beta}\;\Longrightarrow\; G_c(0)=\frac{z}{z/\beta}=\beta,\qquad G_c(s_d)\approx 1 \]

The pair is called a dipole. Seen from a point \(s_d\) far away, the pole and the zero are almost at the same place: their angles nearly cancel and their magnitudes nearly cancel, so the locus in the region of interest is essentially unchanged and the gain required at \(s_d\) is essentially unchanged. Seen from the origin, where \(s\) is comparable with \(z\), they are far apart, and the ratio of their distances is exactly \(\beta\). The error constant is multiplied by \(\beta\) for almost nothing.

"Almost" deserves a word. The dipole adds a slow closed-loop pole near \(s=-z\), close to the compensator zero but not on it. Its residue is small, so it contributes a small-amplitude term to the step response — but that term decays with time constant \(1/z\), which is deliberately long. The result is a long, low tail on the settling response, familiar to anyone who has watched a lag-compensated servo creep into its final position. Making \(z\) smaller reduces the tail's amplitude and lengthens its duration; there is no setting that removes it.

Section 21-7

Lead–Lag Compensation

Specifications frequently demand both things at once: a large error constant and a fast, well-damped transient. A lag alone would meet the first and slow the system down; a lead alone would meet the second and might even reduce the error constant. Cascading the two gives a network with two zeros and two poles that does both jobs, each in its own frequency band.

The lead–lag compensator
\[ G_c(s)=K_c\underbrace{\frac{T_1s+1}{(T_1/\gamma)s+1}}_{\text{lead},\;\gamma>1}\cdot\underbrace{\frac{T_2s+1}{\beta T_2s+1}}_{\text{lag},\;\beta>1},\qquad \frac{1}{T_2}\ll\frac{1}{T_1} \]

The condition on the corner frequencies is what keeps the two halves from interfering: the lag section does all its work one or two decades below crossover, where its \(-6^\circ\) is harmless, while the lead section does all of its work at crossover, where its phase bump is worth the most. Read the magnitude curve from left to right and the compensator lifts the low frequencies by \(\beta\), drops back down, then lifts again by \(\gamma\) around crossover.

The design is done sequentially, and the order matters. Start with the lead half, because it is the lead that determines where the crossover ends up: choose the dominant poles or the phase margin, design the lead section against the plant alone, and compute the error constant the lead-compensated loop achieves. Then take the ratio of the required error constant to the achieved one — that ratio is \(\beta\) — and design a lag dipole to supply it, placed far enough below the new crossover to leave the lead's work undisturbed. A single verification pass at the end confirms both specifications.

🔑
Division of labour
\[ \beta=\frac{K_v^{\text{required}}}{K_v^{\text{after lead}}} \]

The lead half sets the crossover frequency and the phase margin; the lag half then multiplies the error constant by whatever factor is still missing. Neither half is asked to do the other's work, which is why a lead–lag design usually needs less phase from its lead section than a lead-only design of the same plant would.

A single passive network can realise the whole compensator if the two ratios are linked, \(\gamma=\beta\); this is the classical lag–lead network of the analogue era, and its restriction is the price of using one \(RC\) section instead of two. With an operational amplifier, or in software, the two ratios are independent and the more general form above applies.

You have seen this controller before. A lead is a PD action whose derivative is filtered — the pole at \(-1/(\alpha T)\) is precisely the filter \(N\) of Section 19-8, with \(N=1/\alpha\). A lag is a PI action whose integrator has been moved off the origin to \(-1/(\beta T)\), trading exact zero error for a finite but \(\beta\)-fold improvement. And a lead–lag is a PID with both refinements applied. The classical networks and the PID family are the same three ideas expressed in the languages of two different industries: servomechanisms and process control.
Section 21-8

Realisation and Comparison

Both networks are two-resistor, one-capacitor circuits, which is why they dominated control hardware for forty years before digital implementation made the question moot. Their transfer functions follow from a voltage divider with one frequency-dependent arm.

e_i R1 C1 R2 e_o T = R1C1, α = R2/(R1+R2)
Passive lead network
e_i R1 R2 C e_o T = R2C, β = (R1+R2)/R2
Passive lag network
Deriving the lead network from the divider
\[ \frac{E_o}{E_i}=\frac{R_2}{R_2+\dfrac{R_1}{1+R_1C_1s}}=\frac{R_2(1+R_1C_1s)}{R_1+R_2+R_1R_2C_1s}=\frac{R_2}{R_1+R_2}\cdot\frac{1+R_1C_1s}{1+\dfrac{R_1R_2}{R_1+R_2}C_1s} \]
\[ \Longrightarrow\quad T=R_1C_1,\qquad \alpha=\frac{R_2}{R_1+R_2}<1 \]

The lag network is the same exercise with the capacitor moved into the lower arm: \(E_o/E_i=(1+R_2Cs)/\bigl(1+(R_1+R_2)Cs\bigr)\), giving \(T=R_2C\) and \(\beta=(R_1+R_2)/R_2\).

Two practical points follow from these expressions. The passive lead has a DC gain of \(\alpha\), an attenuation that must be made up by an amplifier of gain \(1/\alpha\) somewhere in the loop — the \(K_c=1/\alpha\) of the design procedure is not a bookkeeping device but a real amplifier. The passive lag, by contrast, has unity DC gain and needs no such amplifier, which is one reason lag networks were popular in the era of expensive amplification. And because both are passive dividers, cascading two of them directly makes the second load the first and changes both transfer functions; a buffer between the sections, or a single active realisation, is required.

PropertyLeadLagLead–lag
Pole–zero orderzero then polepole then zeroboth pairs
Phase at crossover\(+\phi_m\) addedabout \(-6^\circ\)\(+\phi_m\) added
Bandwidthincreasedreducedincreased
Rise / settling timefasterslowerfaster
Error constantsslightly reducedmultiplied by \(\beta\)multiplied by \(\beta\)
Noise sensitivityworse by \(1/\alpha\)improvedworse by \(1/\alpha\)
PID counterpartfiltered PDapproximate PIapproximate PID
Use it whenmargin or speed is shortaccuracy is shortboth are short
Section 21-9

Worked Examples

1 Reading a lead network

Problem. For \(G_c(s)=\dfrac{1+0.2s}{1+0.05s}\), find \(\alpha\), the maximum phase lead, the frequency at which it occurs, and the gain there.

Solution. Compare with \(K_c\alpha(Ts+1)/(\alpha Ts+1)\): the zero corner gives \(T=0.2\), and the pole corner gives \(\alpha T=0.05\).

Working
\[ \alpha=\frac{0.05}{0.2}=0.25,\qquad \sin\phi_m=\frac{1-0.25}{1+0.25}=0.6\;\Longrightarrow\;\phi_m=36.87^\circ \]
\[ \omega_m=\frac{1}{T\sqrt{\alpha}}=\frac{1}{0.2\times 0.5}=10\ \mathrm{rad/s},\qquad |G_c(j10)|=\frac{1}{\sqrt{\alpha}}=2\;(6.02\ \text{dB}) \]

Checking directly: \(G_c(j10)=(1+j2)/(1+j0.5)\), whose magnitude is \(2.236/1.118=2\) and whose phase is \(63.43^\circ-26.57^\circ=36.87^\circ\). The network's DC gain is unity, so \(K_c=1/\alpha=4\).

2 Lead design from a phase-margin specification

Problem. For \(G(s)=\dfrac{K}{s(s+2)}\) with unity feedback, design a lead compensator giving \(K_v=20\ \mathrm{s^{-1}}\) and a phase margin of at least \(50^\circ\).

Solution. Step 1 fixes the gain: \(K_v=\lim_{s\to0}sG=K/2=20\), so \(K=40\) and \(G_1(s)=40/\bigl(s(s+2)\bigr)\).

Step 2 finds the uncompensated crossover from \(|G_1|=1\):

Uncompensated margin
\[ \omega\sqrt{\omega^2+4}=40\;\Longrightarrow\;\omega^4+4\omega^2-1600=0\;\Longrightarrow\;\omega_{gc}=6.17\ \mathrm{rad/s} \]
\[ \angle G_1=-90^\circ-\arctan\frac{6.17}{2}=-162.0^\circ\;\Longrightarrow\;\gamma_1=18.0^\circ \]

Step 3: the shortfall is \(32^\circ\); adding a \(6^\circ\) allowance for the crossover shift gives \(\phi_m=38^\circ\). Steps 4 and 5:

Working
\[ \alpha=\frac{1-\sin 38^\circ}{1+\sin 38^\circ}=\frac{0.3843}{1.6157}=0.238,\qquad \sqrt{\alpha}=0.4877 \]
\[ |G_1(j\omega_m)|=0.4877\;\Longrightarrow\;\omega\sqrt{\omega^2+4}=82.0\;\Longrightarrow\;\omega_m=8.95\ \mathrm{rad/s} \]
\[ T=\frac{1}{\omega_m\sqrt{\alpha}}=\frac{1}{8.95\times0.4877}=0.229,\qquad \frac{1}{T}=4.36,\quad\frac{1}{\alpha T}=18.3 \]

With \(K_c=1/\alpha=4.20\) the compensator is

Result
\[ G_c(s)=\frac{0.229s+1}{0.0545s+1}=4.20\,\frac{s+4.36}{s+18.3} \]

Verification at the new crossover \(\omega=8.95\): the plant contributes \(-90^\circ-\arctan(4.47)=-167.4^\circ\) and the network contributes \(+38^\circ\), giving a phase margin of \(50.6^\circ\). Since \(G_c(0)=1\), \(K_v\) remains \(20\). Both specifications are met.

3 Lead design on the root locus

Problem. The same plant \(G(s)=K/\bigl(s(s+2)\bigr)\) must have dominant closed-loop poles with \(\zeta=0.5\) and \(\omega_n=4\). Design a lead compensator and find the resulting \(K_v\).

Solution. The desired poles are \(s_d=-2\pm j3.464\). Without compensation the characteristic equation is \(s^2+2s+K\), whose roots always have real part \(-1\); no gain can produce a real part of \(-2\), so compensation is unavoidable. Compute the angle at \(s_d\):

Angle deficiency
\[ \angle\frac{1}{s_d}=-120^\circ,\qquad \angle\frac{1}{s_d+2}=-90^\circ,\qquad \angle G(s_d)=-210^\circ \]
\[ \phi_c=-180^\circ-(-210^\circ)=+30^\circ \]

Choose the compensator zero to cancel the plant pole at \(-2\). Its angle is \(+90^\circ\), so the compensator pole at \(-p\) must satisfy

Placing the pole
\[ 120^\circ+90^\circ+\theta_p-90^\circ=180^\circ\;\Longrightarrow\;\theta_p=60^\circ \]
\[ \tan 60^\circ=\frac{3.464}{p-2}\;\Longrightarrow\;p-2=\frac{3.464}{1.732}=2\;\Longrightarrow\;p=4 \]

So \(G_c(s)=K_c(s+2)/(s+4)\), and the open loop collapses to \(K_cK/\bigl(s(s+4)\bigr)\) with characteristic equation \(s^2+4s+K_cK\). Matching \(s^2+2\zeta\omega_ns+\omega_n^2=s^2+4s+16\) gives \(K_cK=16\), and the closed-loop poles are exactly \(-2\pm j3.464\). The velocity constant is

Error constant
\[ K_v=\lim_{s\to0}s\,G_c(s)G(s)=\frac{16}{4}=4\ \mathrm{s^{-1}} \]

The transient specification is met exactly, but \(K_v=4\) is modest. If the specification also demanded \(K_v=20\), a lag section would have to be added — which is Example 6.

4 Lag design rescuing an unstable loop

Problem. For \(G(s)=\dfrac{K}{s(s+1)(0.5s+1)}\), design a lag compensator giving \(K_v=5\ \mathrm{s^{-1}}\) and a phase margin of at least \(40^\circ\).

Solution. \(K_v=K=5\), so \(G_1(s)=5/\bigl(s(s+1)(0.5s+1)\bigr)\). Check the uncompensated loop first: the characteristic equation \(s^3+3s^2+2s+10=0\) has a Routh \(s^1\) entry of \((6-10)/3=-1.33\), so the closed loop is unstable. Its gain crossover is at \(1.80\) rad/s where the phase is \(-193^\circ\) — a phase margin of \(-13^\circ\).

Step 2: find where the phase equals \(-180^\circ+40^\circ+12^\circ=-128^\circ\).

Choosing the new crossover
\[ 90^\circ+\arctan\omega_2+\arctan 0.5\omega_2=128^\circ\;\Longrightarrow\;\arctan\omega_2+\arctan0.5\omega_2=38^\circ \]
\[ \omega_2=0.465\ \mathrm{rad/s} \]

Step 3: the magnitude there must be removed entirely.

Working
\[ |G_1(j0.465)|=\frac{5}{0.465\times1.103\times1.027}=9.50\;\;(19.6\ \text{dB})\;\Longrightarrow\;\beta\approx10 \]
\[ \frac{1}{T}=\frac{\omega_2}{10}=0.0465\approx0.05\;\Longrightarrow\;T=20,\qquad \beta T=200 \]
\[ G_c(s)=\frac{20s+1}{200s+1}=0.1\,\frac{s+0.05}{s+0.005} \]

Verification: with the network's factor of \(0.1\) in place, the new crossover is where \(|G_1|=10\), namely \(\omega=0.446\) rad/s. There the plant contributes \(-126.6^\circ\) and the network \(\arctan(8.91)-\arctan(89.1)=-5.8^\circ\), giving a phase margin of \(47.7^\circ\). A loop that was unstable now has a comfortable margin and the full \(K_v=5\), at the cost of a bandwidth reduced from \(1.80\) to \(0.45\) rad/s.

5 A lag dipole for the error constant

Problem. For \(G(s)=\dfrac{K}{s(s+1)(s+2)}\) the gain is set so that the dominant poles have \(\zeta=0.5\). Find that gain and the resulting \(K_v\), then design a lag network raising \(K_v\) to at least \(5\) without disturbing the transient.

Solution. Write the dominant pair as \(-0.5\omega_n\pm j0.866\omega_n\) and the third root as \(-p\). Matching \(s^3+3s^2+2s+K\) coefficient by coefficient:

Locating the operating point
\[ \omega_n+p=3,\qquad \omega_n^2+p\,\omega_n=2\;\Longrightarrow\;3\omega_n=2\;\Longrightarrow\;\omega_n=0.667,\;p=2.333 \]
\[ K=\omega_n^2\,p=0.4444\times2.333=1.037,\qquad s_d=-0.333\pm j0.577 \]
\[ K_v=\frac{K}{2}=0.519\ \mathrm{s^{-1}} \]

The required boost is \(5/0.519=9.6\), so take \(\beta=10\). Place the dipole far inside \(|s_d|=0.667\): zero at \(-0.02\), pole at \(-0.002\). Its angular contribution at \(s_d\) is

Checking the disturbance to the locus
\[ \angle(s_d+0.02)-\angle(s_d+0.002)=118.5^\circ-119.9^\circ=-1.4^\circ \]

Well under the \(5^\circ\) usually tolerated, so the dominant poles are essentially unmoved and the gain \(K=1.037\) is retained. The compensated loop has \(K_v\approx 10\times0.519=5.19\ \mathrm{s^{-1}}\). The price is a closed-loop pole near \(-0.02\), which adds a small tail decaying with a time constant of about \(50\) s.

6 A lead–lag design

Problem. For \(G(s)=\dfrac{K}{s(s+1)(s+2)}\), design a compensator giving dominant poles with \(\zeta=0.5\) and \(\omega_n=1.5\), together with \(K_v=10\ \mathrm{s^{-1}}\).

Solution. The desired poles are \(s_d=-0.75\pm j1.299\). Example 5 showed that \(\zeta=0.5\) alone gives \(\omega_n=0.667\); asking for \(\omega_n=1.5\) as well as a twenty-fold increase in \(K_v\) puts both specifications out of reach of gain adjustment. Design the lead half first.

Angle deficiency at s_d
\[ 120^\circ+79.1^\circ+46.1^\circ=245.2^\circ\;\Longrightarrow\;\phi_c=245.2^\circ-180^\circ=65.2^\circ \]

Cancel the plant pole at \(-1\) with the lead zero. That zero contributes \(79.1^\circ\), so the lead pole at \(-p_1\) must contribute

Placing the lead pole
\[ 245.2^\circ-79.1^\circ+\theta_p=180^\circ\;\Longrightarrow\;\theta_p=13.9^\circ \]
\[ p_1-0.75=\frac{1.299}{\tan13.9^\circ}=\frac{1.299}{0.2475}=5.25\;\Longrightarrow\;p_1=6 \]

The lead-compensated loop is \(K/\bigl(s(s+2)(s+6)\bigr)\), whose characteristic equation \(s^3+8s^2+12s+K\) passes through \(s_d\) when \(K=14.6\). That gives \(K_v=14.6/12=1.22\), short of \(10\) by a factor

Sizing the lag half
\[ \beta=\frac{10}{1.22}=8.2 \]

Place the lag zero at \(-0.05\) and its pole at \(-0.05/8.2=-0.0061\). The pair contributes \(118.3^\circ-119.8^\circ=-1.5^\circ\) at \(s_d\) and reduces the magnitude there by \(1.4\%\), so the gain is raised slightly to \(K_c=14.8\). The complete compensator is

Result
\[ G_c(s)=14.8\,\frac{(s+1)(s+0.05)}{(s+6)(s+0.0061)} \]
\[ K_v=14.8\times\frac{1\times0.05}{6\times0.0061}\times\frac{1}{2}=10.1\ \mathrm{s^{-1}} \]

The lead half moved the dominant poles from \(\omega_n=0.667\) to \(\omega_n=1.5\) at the same damping; the lag half then multiplied the velocity constant by \(8.2\). Neither section interferes with the other, because their corner frequencies are two decades apart.

Review

Chapter Summary

Why compensate

Gain moves poles along a fixed locus; a compensator changes the locus itself.

Lead network

\(\sin\phi_m=(1-\alpha)/(1+\alpha)\) at \(\omega_m=1/T\sqrt{\alpha}\), with gain \(1/\sqrt{\alpha}\) there.

Bode lead procedure

Set \(K\), find the shortfall, add \(5^\circ\)–\(12^\circ\), get \(\alpha\), centre \(\omega_m\) on the new crossover.

Angle deficiency

\(\phi_c=-180^\circ-\angle G(s_d)\); positive means a lead, negative means a lag.

Lag network

A dipole near the origin multiplies the error constant by \(\beta\) for about \(6^\circ\) of phase.

Lead–lag

Lead first to set the crossover, lag second to supply the missing \(\beta\) — a filtered PID in disguise.

Practice

Practice Problems

Decide first which specification the uncompensated system fails — margin or accuracy — because that decides the network. Assume unity feedback throughout. Difficulty rises down the list.

  1. For \(G_c(s)=\dfrac{1+0.5s}{1+0.1s}\), find \(\alpha\), \(\phi_m\), \(\omega_m\) and the gain in decibels at \(\omega_m\).
  2. What value of \(\alpha\) is needed for a maximum phase lead of \(45^\circ\), and what high-frequency gain does that imply?
  3. A lag network has \(\beta=8\) and its zero at \(-0.4\). Where is its pole, and what is its phase contribution at \(\omega=4\) rad/s?
  4. For \(G(s)=\dfrac{K}{s(s+4)}\), find \(K\) for \(K_v=10\) and the resulting phase margin.
  5. Continuing Problem 4, design a lead compensator that raises the phase margin to \(45^\circ\) while keeping \(K_v=10\).
  6. For \(G(s)=\dfrac{K}{s(s+1)}\), design a lead compensator giving dominant poles with \(\zeta=0.707\) and \(\omega_n=2\), using cancellation of the plant pole.
  7. Explain why a lead compensator reduces the velocity error constant when \(K_c\) is held fixed, and how the design procedure compensates for that.
  8. A system meets its transient specification with \(K_v=2\) but requires \(K_v=25\). Design a suitable lag dipole and state the resulting settling-tail time constant.
  9. Show that placing the lag zero one decade below the new gain crossover limits the network's phase contribution there to about \(-6^\circ\), whatever the value of \(\beta\).
  10. For \(G(s)=\dfrac{K}{s(s+2)(s+8)}\), design a lead–lag compensator for \(\zeta=0.5\), \(\omega_n=3\) and \(K_v=15\).
  11. Two lead sections, each supplying \(30^\circ\), are cascaded. Compare the total high-frequency gain with that of a single section supplying \(60^\circ\), and explain the practical significance.
  12. A plant has a lightly damped pole pair at \(-0.1\pm j5\). Explain why cancelling it with compensator zeros is a poor design choice, and what the root locus of the imperfectly cancelled system looks like.
Tip: in an examination the lead and lag procedures differ in exactly one respect — which Bode curve you read first. For a lead you read the magnitude curve to find \(\omega_m\), because \(\alpha\) has already been fixed by the phase shortfall. For a lag you read the phase curve to find \(\omega_2\), because the phase specification fixes the frequency and the magnitude there then fixes \(\beta\). Get that ordering right and every remaining step is substitution into \(\alpha=(1-\sin\phi_m)/(1+\sin\phi_m)\) or \(1/T=\omega_2/10\). On the root locus, the equivalent discipline is to compute the angle deficiency before touching anything else: its sign alone tells you which network you are about to design.