Part 3 · Chapter 13

Root Locus Analysis and Design

A performance specification is a region of the \(s\)-plane, a root locus is a set of curves drawn on that same plane, and design is the business of making the two meet — by choosing a gain when the curves already pass through the region, and by adding poles and zeros that bend them there when they do not.

Control Systems Prof. Mithun Mondal Reading time ≈ 45 min
i What you'll learn
  • How overshoot, settling time and peak time become a region of the \(s\)-plane — a wedge, a vertical line, a horizontal line.
  • How to pick the gain at the point where the locus enters that region, using the magnitude condition.
  • The dominance test that says when the two-pole formulas of Chapter 8 may be trusted on a higher-order loop.
  • Why an unreachable specification is a statement about the shape of the locus, not about the gain.
  • How an added zero pulls the locus left and an added pole pushes it right, argued from the centroid and the angle condition.
  • The angle-deficiency design of a lead compensator, and the lag compensator that multiplies \(K_v\) without moving the dominant poles.
  • How PD, PI and PID controllers appear on the locus as an added zero, an added pole–zero pair at the origin, and both together.
Section 13-1

What the Locus Is For

Chapter 12 built the locus. It never explained why one would want it. The answer is that a control specification and a root locus are objects of the same kind: both live in the \(s\)-plane. Chapter 8 showed that the damping ratio of a second-order pair is fixed by the angle its poles make with the negative real axis, the settling time by their distance from the imaginary axis, and the natural frequency by their distance from the origin. Every transient requirement therefore carves out a region of the plane in which the dominant poles must lie.

The locus, meanwhile, is the complete catalogue of pole positions the loop can be made to have by turning one knob. Design becomes a question of geometry: does the catalogue intersect the region?

🔑
The design question, stated geometrically
If the locus passes through the acceptable region, read off the gain at the crossing point. If it does not, no gain will work, and the pole–zero pattern itself must be altered.

These are the only two outcomes, and they divide this chapter in half. Sections 13-2 to 13-4 handle the first; Sections 13-5 to 13-9 handle the second.

Section 13-2

Specifications as Regions of the s-Plane

Take a dominant pair \(s = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}\) and translate each of Chapter 8's specifications in turn.

Overshoot. The percentage overshoot depends on \(\zeta\) alone. Since \(\cos\theta = \zeta\) where \(\theta\) is the angle measured from the negative real axis, a bound on overshoot is a bound on \(\theta\): the poles must lie inside a wedge straddling the negative real axis.

From overshoot to damping ratio
\[ \%\text{OS} = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}} \quad\Longleftrightarrow\quad \zeta = \frac{-\ln(\%\text{OS}/100)}{\sqrt{\pi^2 + \ln^2(\%\text{OS}/100)}} \]

Settling time. With the 2% criterion \(t_s = 4/(\zeta\omega_n)\), and \(\zeta\omega_n\) is exactly \(|\operatorname{Re}(s)|\). A bound \(t_s \le T\) is therefore \(\operatorname{Re}(s) \le -4/T\): the poles must lie to the left of a vertical line.

Peak time. \(t_p = \pi/\omega_d\), and \(\omega_d = |\operatorname{Im}(s)|\). A bound \(t_p \le T_p\) is \(|\operatorname{Im}(s)| \ge \pi/T_p\): the poles must lie above a horizontal line (and below its mirror image).

Natural frequency. \(\omega_n = |s|\), so a bound on \(\omega_n\) is a circle centred at the origin.

SpecificationConstraint on the poleBoundary in the \(s\)-plane
\(\%\text{OS} \le\) value\(\zeta \ge \zeta_{\min}\)Radial wedge, half-angle \(\cos^{-1}\zeta_{\min}\)
\(t_s \le T\) (2%)\(\zeta\omega_n \ge 4/T\)Vertical line \(\sigma = -4/T\)
\(t_p \le T_p\)\(\omega_d \ge \pi/T_p\)Horizontal lines \(\omega = \pm\pi/T_p\)
\(\omega_n \ge \Omega\)\(|s| \ge \Omega\)Circle of radius \(\Omega\)

Several specifications applied at once intersect their regions. The most common pairing — an overshoot limit together with a settling-time limit — produces a wedge truncated by a vertical line, sketched below.

σ ζ = 0.5 (θ = 60°) σ = −4/tₛ = −2 θ s_d = −2 + j3.46 acceptable region
Overshoot fixes the wedge; settling time truncates it with a vertical line
Read the corner. When both specifications are tight, the design target is the corner of the region — the point where the constant-\(\zeta\) ray meets the settling-time line. Meeting the corner exactly meets both specifications exactly; anything deeper into the shaded region beats both, at the price of more gain and more control effort.
Section 13-3

Choosing the Gain

Suppose the locus does cross the acceptable region. The crossing point \(s_d\) already satisfies the angle condition — that is what being on the locus means — so the only thing left to determine is the gain that puts a closed-loop pole exactly there. The magnitude condition of Chapter 12 answers it with a ruler.

🔑
Gain at a design point
\[ K = \frac{\text{product of distances from } s_d \text{ to all open-loop poles}}{\text{product of distances from } s_d \text{ to all open-loop zeros}} \]

Measured on a scale drawing this is accurate to a few percent, which is normally enough; computed from coordinates it is exact. If the system is low-order, matching coefficients against \(s^2+2\zeta\omega_n s+\omega_n^2\) gives the same answer without any geometry.

The chosen gain does more than fix the transient. It simultaneously fixes the static error constants of Chapter 9, since those are limits of \(KG(s)H(s)\) as \(s\to0\). Choosing \(K\) for damping therefore determines the steady-state error — it is not a second free choice. That single fact is the origin of most of the compensation in this chapter: one knob cannot independently satisfy two requirements.

Where the trade-off of Chapter 9 becomes visible. On a typical locus, moving up the branch increases \(K\) and lowers \(\zeta\). Larger \(K\) shrinks \(e_{ss}\) and grows the overshoot; smaller \(K\) does the reverse. The locus makes the conflict quantitative instead of merely qualitative: you can read the exact \(\zeta\) that a given \(K_v\) costs you.
Section 13-4

Dominance and the Second-Order Approximation

Everything in Section 13-2 came from second-order formulas, yet almost every real loop is third order or higher. The bridge is the notion of dominant poles. A closed-loop pole at \(-\sigma\) contributes a term \(e^{-\sigma t}\) to the response. If one real pole lies far to the left of a complex pair, its exponential decays long before theirs does, and its residue is small as well, because the residue carries the distance to that pole in the denominator. The response is then, to good accuracy, the response of the pair alone.

🔑
The five-times rule
A complex pair may be treated as dominant if every other closed-loop pole lies at least five times further from the imaginary axis, and no closed-loop zero lies near the pair.

At a factor of five the remote pole's transient has decayed to about \(e^{-5}\approx 0.7\%\) of its initial value in the time the dominant pair takes to decay to \(37\%\). The rule is a convention, not a theorem; a factor of ten leaves no doubt, a factor of three demands a check against the actual response.

The zero clause matters as much as the pole clause and is more often forgotten. A closed-loop zero close to the dominant pair does not decay away — it reshapes the residues, increasing overshoot and speeding the rise. A zero in the right half-plane produces an initial excursion in the wrong direction, which no pole placement can remove.

Where does a closed-loop zero come from? From the numerator of the closed-loop transfer function, which contains the zeros of the forward path. A compensator zero placed to bend the locus therefore reappears in the closed-loop numerator unless it is cancelled by a feedback element — a fact that quietly limits how aggressively lead compensation can be used.

Section 13-5

When Gain Alone Cannot Do It

Consider the loop \(G(s)=K/[s(s+4)]\) of Chapter 12. Its locus is the segment \([-4,0]\) together with the vertical line \(\operatorname{Re}(s)=-2\). Every closed-loop pole this system can ever have lies on that set. A settling-time specification \(t_s \le 1\ \text{s}\) demands \(\operatorname{Re}(s)\le-4\), and the vertical line sits at \(-2\). No gain reaches the region, because the region is not on the locus.

This is a structural failure, not a tuning failure, and it is worth being blunt about the distinction. Raising \(K\) moves the poles along the locus. It cannot move the locus. Where the locus goes is decided entirely by the open-loop poles and zeros — by rules 1 to 9 of Chapter 12, none of which contains \(K\).

🔑
Why compensation exists
Gain selects a point on the locus; compensation selects the locus. If the required closed-loop poles are off the existing curve, the only remedy is to add open-loop poles and zeros that redraw it.

A cascade compensator \(G_c(s)\) in series with the plant does exactly that: the open-loop function becomes \(G_c(s)G(s)H(s)\), with a new pole–zero set and therefore a new locus.

The same argument applies to accuracy. If the gain that meets the transient specification gives too small a \(K_v\), no adjustment saves it — pushing \(K_v\) up means pushing \(K\) up, which slides the poles up the branch out of the acceptable wedge. Two conflicting demands need two design freedoms, and a compensator supplies the second.

Section 13-6

Reshaping the Locus: What a Pole or Zero Does

Before designing anything, it pays to know which direction each ingredient moves the curve. Two independent arguments give the same answer.

The centroid argument. Chapter 12 derived \(\sigma_a = \big(\sum p_i - \sum z_j\big)/(n-m)\). Zeros enter with a minus sign, so adding a zero at \(-z\) pulls the centroid to the left and simultaneously reduces \(n-m\), which widens the asymptote angles away from \(\pm90^\circ\) toward \(180^\circ\). Adding a pole does the reverse: the centroid moves right and the asymptotes narrow toward the imaginary axis.

The angle argument. At a test point \(s_d\) in the upper half-plane, an added zero at \(-z\) contributes \(+\phi\) to the angle sum, so a point that previously fell short of \(180^\circ\) can now reach it — the locus is pulled toward the zero. An added pole contributes \(-\theta\), and the locus is pushed away from it.

Added elementEffect on the locusEffect on performance
Zero in the left half-planeCentroid moves left; branches bend toward the zeroMore damping, faster response, better stability margin
Pole in the left half-planeCentroid moves right; branches bend toward the imaginary axisLess damping, lower critical gain, slower response
Pole–zero pair, pole nearer the origin (lag)Almost no change in shape; ratio \(z/p\) multiplies the low-frequency gainSteady-state error reduced, transient nearly untouched
Pole–zero pair, zero nearer the origin (lead)Net positive angle; branches pulled leftTransient improved, small loss of low-frequency gain
Every compensator is a pole–zero pair. A bare zero is not physically realisable — a differentiator has unbounded high-frequency gain and would amplify measurement noise without limit. Practical compensators therefore always pair the zero with a pole, and the whole art lies in choosing where to put the pole so that it collects the necessary realisability while spoiling as little as possible of what the zero achieved.
Section 13-7

Lead Compensation by Angle Deficiency

A lead compensator has its zero closer to the origin than its pole:

Cascade lead compensator
\[ G_c(s) = K_c\,\frac{s+z_c}{s+p_c}, \qquad 0 < z_c < p_c \]

At any test point in the upper half-plane the vector from the nearer zero subtends a larger angle than the vector from the more distant pole, so \(\angle G_c = \phi_z - \theta_p > 0\). The network supplies positive phase — hence "lead" — and by the angle argument of Section 13-6 it pulls the locus to the left.

The design method follows directly from the angle condition. The target point \(s_d\) is where we want a closed-loop pole. It will be a closed-loop pole precisely when the compensated open-loop function has angle \(-180^\circ\) there. The plant alone contributes \(\angle G(s_d)\), so the compensator must make up the difference.

🔑
Angle deficiency
\[ \angle G_c(s_d) \;=\; -180^\circ - \angle G(s_d)H(s_d) \]

If the required \(\angle G_c\) is positive the compensator must lead; if negative, it must lag. A deficiency beyond about \(60^\circ\) cannot be met by a single lead section without an impractically wide pole–zero spread, and two cascaded sections are used instead.

The angle fixes only the combination of \(z_c\) and \(p_c\) — there are infinitely many pairs that subtend the required angle at \(s_d\), and the freedom is used to serve a second objective. Three standard choices:

Choice of \(z_c\)RationaleCaution
Cancel a plant poleSimplest algebra; the compensated locus is that of a lower-order plantOnly for well-damped, accurately known, stable plant poles — never for a pole in the right half-plane
Directly below \(s_d\)Places the zero where it contributes \(90^\circ\), leaving a modest angle for the poleLeaves a closed-loop zero near the dominant pair
Bisector constructionMinimises \(p_c/z_c\), so the least high-frequency gain and least noise amplificationMore construction work; geometry must be drawn carefully

Whichever is chosen, the last step is always the same: apply the magnitude condition at \(s_d\) to the full compensated open-loop function and read off \(K\).

σ uncompensated: Re = −2 compensated: Re = −4 0 −4 −8 zero cancels the −4 pole new pole at −8 s_d = −4 + j6.93 (tₛ = 1 s) −2 + j3.46 (tₛ = 2 s) ζ = 0.5
Lead compensation moves the whole vertical branch from Re = −2 to Re = −4
Section 13-8

Lag Compensation and Steady-State Accuracy

The lag compensator solves the opposite problem. The transient specification is already met; what is unacceptable is the steady-state error, and Section 13-3 showed that the gain is no longer free to fix it. The trick is to raise the low-frequency gain without disturbing the angle sum near the dominant poles.

A lag compensator reverses the lead's ordering — the pole lies closer to the origin than the zero:

Cascade lag compensator and its effect on \(K_v\)
\[ G_c(s) = \frac{s+z_c}{s+p_c}, \qquad 0 < p_c < z_c \]
\[ K_v^{\text{new}} = \lim_{s\to0} s\,G_c(s)G(s) = \left(\lim_{s\to0}\frac{s+z_c}{s+p_c}\right)\lim_{s\to0}sG(s) = \frac{z_c}{p_c}\,K_v^{\text{old}} \]

The error constant is multiplied by the ratio \(z_c/p_c\) and by nothing else. Choose that ratio equal to the required improvement factor (with perhaps 10% to spare, since the gain has to be retuned slightly afterwards) and the accuracy problem is solved on paper.

The remaining question is whether the transient survives. Place both singularities very close to the origin — an order of magnitude nearer than the dominant poles — and the two vectors drawn from them to \(s_d\) are almost identical in direction. Their angles nearly cancel, leaving a net contribution of a degree or two, far too small to move the locus appreciably near \(s_d\).

σ s_d (dominant) −z_c −p_c φ θ φ − θ ≈ −1° while z_c / p_c = 10 (not to scale)
The lag dipole: a large gain ratio bought with a negligible angle
🔑
Lag design rule
Set \(z_c/p_c\) equal to the required increase in the error constant, and place \(z_c\) roughly one-tenth of \(|\operatorname{Re}(s_d)|\) from the origin so that the angle contribution stays within a couple of degrees.

The price is a slow closed-loop pole that appears between \(-z_c\) and \(-p_c\), paired with the compensator zero. The two nearly cancel, so the residue is small — but the time constant is long, and the step response carries a low-amplitude tail lasting tens of seconds. That tail is the real cost of lag compensation.

Lead and lag do different jobs. Lead moves the locus and therefore the transient; lag moves the gain and therefore the accuracy. Neither substitutes for the other, and a specification that constrains both — as most do — is met by cascading one of each, a lead–lag compensator. Design the lead first, against the transient specification; then design the lag against whatever error constant the lead-compensated loop happens to deliver.
Section 13-9

PD, PI and PID on the Locus

The three-term controller familiar from industrial practice is the same geometry under different names. Writing each form over a common denominator exposes its poles and zeros immediately.

The three terms as pole–zero patterns
\[ \text{PD:}\quad K_p + K_d s = K_d\left(s + \frac{K_p}{K_d}\right) \qquad\Longrightarrow\qquad \text{one zero at } -K_p/K_d \]
\[ \text{PI:}\quad K_p + \frac{K_i}{s} = K_p\,\frac{s + K_i/K_p}{s} \qquad\Longrightarrow\qquad \text{a pole at the origin and a zero at } -K_i/K_p \]
\[ \text{PID:}\quad K_p + \frac{K_i}{s} + K_d s = K_d\,\frac{s^2 + (K_p/K_d)s + K_i/K_d}{s} \qquad\Longrightarrow\qquad \text{two zeros and a pole at the origin} \]

Proportional–derivative action adds a single left-half-plane zero, which by Section 13-6 pulls the locus left: more damping and a faster response. The catch is that \(K_d s\) has gain proportional to frequency, so measurement noise is amplified without limit and the actuator is asked for impulsive effort. The lead compensator is the realisable version — the same zero, with a far-off pole added to roll the high-frequency gain off.

Proportional–integral action adds a pole at the origin, raising the system type by one. By the table of Chapter 9 that drives one error to zero outright: a type-0 plant becomes type 1 and tracks a step with no error at all. The accompanying zero, placed close to the origin, keeps the added pole from wrecking the angle sum near the dominant poles. The lag compensator is the realisable and gentler version: its pole sits at \(-p_c\) rather than exactly at the origin, so the error is made small rather than zero, but the integrator windup and the extra \(90^\circ\) of lag at low frequency are avoided.

PID is simply both at once, and mirrors the lead–lag structure: the two zeros do the transient work while the pole at the origin does the accuracy work.

ControllerAdds to the open loopRealisable counterpartImproves
PDOne zeroLead compensatorTransient: damping, speed
PIPole at origin + zeroLag compensatorSteady state: error, system type
PIDTwo zeros + pole at originLead–lag compensatorBoth, at the cost of three parameters
Section 13-10

Worked Examples

1 Turning specifications into a target point

Problem. A unity-feedback design must satisfy \(\%\text{OS} \le 16.3\%\) and \(t_s \le 2\ \text{s}\) on the 2% criterion. Describe the acceptable region and give the pole location that meets both specifications exactly.

Solution. Convert the overshoot bound first:

Working
\[ \zeta = \frac{-\ln(0.163)}{\sqrt{\pi^2+\ln^2(0.163)}} = \frac{1.814}{\sqrt{9.870+3.291}} = \frac{1.814}{3.628} = 0.500 \]
\[ \theta = \cos^{-1}(0.5) = 60^\circ, \qquad \zeta\omega_n \ge \frac{4}{t_s} = \frac{4}{2} = 2 \]

The region is the wedge within \(60^\circ\) of the negative real axis, truncated by the vertical line \(\sigma=-2\) — exactly the shaded area of Section 13-2. The corner of that region is

Target point
\[ s_d = -2 \pm j\,2\tan 60^\circ = -2 \pm j3.464, \qquad \omega_n = |s_d| = 4 \]
2 Reading the gain off the locus

Problem. For \(G(s)=\dfrac{K}{s(s+10)}\) with unity feedback, find the gain that gives \(\zeta=0.6\), and state the resulting settling time, overshoot and velocity error constant.

Solution. The locus is the segment \([-10,0]\) with a breakaway at \(-5\), after which it is the vertical line \(\operatorname{Re}(s)=-5\). For \(\zeta = 0.6\) the constant-damping ray makes \(\theta=\cos^{-1}0.6 = 53.13^\circ\) with the negative real axis, so it meets that line at

Design point and gain
\[ s_d = -5 + j\,5\tan 53.13^\circ = -5 + j6.667 \]
\[ K = |s_d - 0|\cdot|s_d+10| = |{-5}+j6.667|\cdot|5+j6.667| = 8.333 \times 8.333 = 69.4 \]

Check by coefficients: \(s^2+10s+69.4\) gives \(\omega_n=\sqrt{69.4}=8.33\) and \(\zeta = 10/(2\times8.33) = 0.60\). The performance follows:

Resulting performance
\[ t_s = \frac{4}{5} = 0.8\ \text{s},\qquad \%\text{OS} = 100e^{-\pi(0.6)/\sqrt{1-0.36}} = 9.5\%,\qquad K_v = \frac{69.4}{10} = 6.94 \]

The unit-ramp error is \(1/K_v = 0.144\). Notice that this number was never chosen — it fell out of the gain that the damping specification demanded, which is the point made in Section 13-3.

3 Is the pair really dominant?

Problem. Chapter 12 found that \(K=8.30\) places the closed-loop poles of \(K/[s(s+2)(s+4)]\) at \(-0.667\pm j1.155\) and \(-4.667\). Test the second-order approximation and compare its predictions with the true response.

Solution. Apply the five-times rule to the remote pole:

Dominance test
\[ \frac{|-4.667|}{|-0.667|} = 7.0 \;\ge\; 5 \quad\checkmark \]

The closed-loop numerator is the constant \(K\), so there is no closed-loop zero to worry about either. Both clauses pass, and the second-order formulas may be used with \(\zeta = 0.5\) and \(\omega_n = 1.333\):

Prediction
\[ \%\text{OS} \approx 16.3\%, \qquad t_s \approx \frac{4}{0.667} = 6.0\ \text{s}, \qquad t_p \approx \frac{\pi}{1.155} = 2.72\ \text{s} \]

Expanding the exact step response by partial fractions gives

Exact response
\[ c(t) = 1 - 0.103\,e^{-4.667t} + 1.294\,e^{-0.667t}\cos\!\big(1.155t + 133.9^\circ\big) \]

whose peak is \(1.155\) at \(t \approx 3.0\ \text{s}\). The true overshoot is therefore \(15.5\%\) against a predicted \(16.3\%\), and the true peak time \(3.0\ \text{s}\) against \(2.72\ \text{s}\): the remote pole slightly slows the rise and slightly shaves the peak. Errors of that size are entirely acceptable for design, which is what the five-times rule is meant to guarantee.

4 Lead design by angle deficiency

Problem. The plant is \(G(s)=\dfrac{K}{s(s+4)}\) with unity feedback. Design a cascade compensator giving \(\zeta = 0.5\) and \(t_s = 1\ \text{s}\), and find the required gain.

Solution. With gain alone the locus reaches \(\zeta=0.5\) at \(-2\pm j3.464\), where \(t_s = 4/2 = 2\ \text{s}\) — twice too slow, and unreachable because the branch is the fixed line \(\operatorname{Re}=-2\). The specification demands \(\zeta\omega_n = 4/1 = 4\), so

Target
\[ s_d = -4 + j\,4\tan60^\circ = -4 + j6.928 \qquad (\omega_n = 8) \]

Evaluate the plant angle there. The vector from the pole at \(0\) is \(-4+j6.928\), at \(120^\circ\); the vector from the pole at \(-4\) is \(j6.928\), at \(90^\circ\):

Angle deficiency
\[ \angle G(s_d) = -(120^\circ + 90^\circ) = -210^\circ \]
\[ \angle G_c(s_d) = -180^\circ - (-210^\circ) = +30^\circ \]

A positive requirement, so a lead section is called for. Take the cancellation choice and place the zero at \(-4\), on top of the plant pole; it contributes \(90^\circ\), leaving the compensator pole to contribute \(90^\circ - 30^\circ = 60^\circ\). If that pole is at \(-p_c\), the vector to \(s_d\) is \((p_c-4) + j6.928\), so

Placing the compensator pole
\[ \tan 60^\circ = \frac{6.928}{p_c-4} \;\Longrightarrow\; p_c - 4 = \frac{6.928}{1.732} = 4 \;\Longrightarrow\; p_c = 8 \]

Hence \(G_c(s) = \dfrac{s+4}{s+8}\), and the compensated open loop collapses to \(K/[s(s+8)]\) — a vertical branch at \(\operatorname{Re}=-4\) passing exactly through \(s_d\). The magnitude condition gives the gain:

Gain
\[ K = |s_d|\cdot|s_d+8| = |-4+j6.928|\cdot|4+j6.928| = 8 \times 8 = 64 \]

Verify: \(s^2+8s+64\) has \(\omega_n = 8\) and \(\zeta = 8/16 = 0.5\), so \(t_s = 4/4 = 1\ \text{s}\) as required. A bonus arrives free of charge — \(K_v\) rises from \(16/4 = 4\) to \(64/8 = 8\), because the lead raised the gain more than it lowered the low-frequency gain ratio.

5 Lag design for a ramp-error specification

Problem. The same plant \(G(s)=\dfrac{K}{s(s+4)}\) is operated at \(K=16\), giving \(\zeta=0.5\) and \(t_s = 2\ \text{s}\), which is acceptable. The unit-ramp error, however, must not exceed \(0.025\). Design a lag compensator.

Solution. The present accuracy is

Present and required accuracy
\[ K_v = \lim_{s\to0}\frac{16}{s+4} = 4, \qquad e_{ss} = \frac{1}{K_v} = 0.25 \]
\[ \text{required } K_v \ge \frac{1}{0.025} = 40 \quad\Longrightarrow\quad \frac{z_c}{p_c} \ge 10 \]

Raising \(K\) tenfold would do it numerically but would drive \(\zeta\) from \(0.5\) to \(2/\sqrt{160} = 0.158\) — an overshoot near \(60\%\). Instead insert a lag with ratio \(10\), placed close to the origin: \(G_c(s) = \dfrac{s+0.1}{s+0.01}\). Confirm the new error constant:

New error constant
\[ K_v^{\text{new}} = \lim_{s\to0} s\cdot\frac{16(s+0.1)}{s(s+4)(s+0.01)} = \frac{16 \times 0.1}{4 \times 0.01} = 40 \quad\checkmark \]

Now check the damage at the dominant pole \(s_d = -2+j3.464\):

Angle contribution of the dipole
\[ \angle(s_d+0.1) = 180^\circ - \tan^{-1}\!\frac{3.464}{1.9} = 118.74^\circ \]
\[ \angle(s_d+0.01) = 180^\circ - \tan^{-1}\!\frac{3.464}{1.99} = 119.87^\circ \]
\[ \angle G_c(s_d) = 118.74^\circ - 119.87^\circ = -1.13^\circ \]

Just over one degree of lag, which shifts the dominant pair by a fraction of a percent — recoverable with a small retune of \(K\). The compensated loop also acquires a slow closed-loop pole just left of \(-0.1\), nearly cancelled by the compensator zero; its residue is small but its time constant is around ten seconds, so the step response carries a faint tail long after the visible transient has finished.

6 Lead and lag together

Problem. Take the lead-compensated loop of Example 4, \(L(s)=64/[s(s+8)]\), which meets \(\zeta=0.5\) and \(t_s=1\ \text{s}\). A further requirement arrives: the unit-ramp error must not exceed \(0.0125\). Complete the design.

Solution. The lead-compensated loop has \(K_v = 64/8 = 8\), so \(e_{ss}=0.125\) — a factor of ten too large. The transient is already correct and must not be disturbed, so the second stage must be a lag of ratio \(10\). Reuse \(G_c^{\text{lag}}(s) = \dfrac{s+0.1}{s+0.01}\):

Check the error constant
\[ K_v^{\text{new}} = \frac{64 \times 0.1}{8 \times 0.01} = 80, \qquad e_{ss} = \frac{1}{80} = 0.0125 \quad\checkmark \]

And the angle cost at \(s_d = -4+j6.928\):

Angle contribution
\[ \angle G_c^{\text{lag}}(s_d) = \left(180^\circ - \tan^{-1}\!\frac{6.928}{3.9}\right) - \left(180^\circ - \tan^{-1}\!\frac{6.928}{3.99}\right) = 119.37^\circ - 119.94^\circ = -0.57^\circ \]

Barely half a degree — smaller than in Example 5, because the dominant poles are further from the origin and the dipole subtends an even narrower angle. The finished cascade compensator is

Lead–lag compensator
\[ G_c(s) = \underbrace{\frac{s+4}{s+8}}_{\text{lead}}\;\cdot\;\underbrace{\frac{s+0.1}{s+0.01}}_{\text{lag}}, \qquad K = 64 \]

The lead placed the dominant poles; the lag multiplied the error constant by ten while contributing under a degree at those poles. Two specifications, two design freedoms — the arrangement Section 13-5 argued was unavoidable.

Review

Chapter Summary

Specs are regions

Overshoot gives a wedge, settling time a vertical line, peak time a horizontal one.

Gain from geometry

\(K=\prod|s_d-p_i|/\prod|s_d-z_j|\) at the point where the locus enters the region.

Dominance

Other poles at least five times further left, and no nearby closed-loop zero.

One knob, one spec

The gain that fixes \(\zeta\) also fixes \(K_p, K_v, K_a\) — they are not independent.

Zeros pull left

Added zeros move \(\sigma_a\) left and add angle; added poles do the reverse.

Lead

\(\angle G_c(s_d) = -180^\circ - \angle G(s_d)\) fixes the zero–pole pair; the magnitude condition fixes \(K\).

Lag

\(z_c/p_c\) multiplies the error constant; near the origin it costs about a degree.

PID

PD is a zero, PI a pole–zero pair at the origin; lead and lag are their realisable forms.

Practice

Problems

Assume unity feedback throughout. Translate every specification into a point or region of the \(s\)-plane before touching any algebra; difficulty rises down the list.

  1. Convert \(\%\text{OS} \le 10\%\) into a bound on \(\zeta\) and on the wedge half-angle. Then add \(t_p \le 1\ \text{s}\) and describe the resulting region.
  2. For \(G(s)=\dfrac{K}{s(s+10)}\), find the gain giving \(\zeta=0.6\) and the corresponding \(t_s\), \(\%\text{OS}\) and unit-ramp error.
  3. For \(G(s)=\dfrac{K}{s(s+2)(s+4)}\) with \(K=8.30\), the closed-loop poles are \(-0.667\pm j1.155\) and \(-4.667\). Justify the second-order approximation and estimate \(t_s\) and \(t_p\).
  4. Explain, in terms of the shape of the locus rather than the value of \(K\), why \(G(s)=\dfrac{K}{s(s+4)}\) can never achieve \(t_s \le 1\ \text{s}\) under proportional control.
  5. Design a cascade lead compensator for \(G(s)=\dfrac{K}{s(s+2)}\) that places the dominant poles at \(\zeta=0.5\), \(\omega_n=6\). Use the pole-cancellation choice for the zero and find \(K\).
  6. A loop meets its transient specification with \(K_v = 5\), but requires \(K_v \ge 50\). Propose a lag compensator and evaluate its angle contribution at the dominant pole \(s_d = -1 + j1.7\).
  7. A PD controller \(K_p + K_d s\) is used with a plant whose design point is \(s_d = -2+j3\). Show that the controller contributes a single zero at \(-K_p/K_d\), and find the ratio \(K_p/K_d\) that makes that zero contribute \(+40^\circ\) at \(s_d\).
  8. For \(G(s)=\dfrac{K}{s(s+4)(s+10)}\), determine the gain that gives \(\zeta = 0.5\) for the dominant pair, verify dominance, and state whether a lag stage of ratio 10 could be added without violating the transient specification.
Tip: work in a fixed order — specification to region, region to target point \(s_d\), plant angle at \(s_d\), deficiency, compensator, and only then the gain. Most errors in compensator design come from computing a gain before the angle condition has been satisfied, which produces a confident number for a point that is not on the locus at all. Check the angle first; the magnitude condition is meaningless until it holds.