Root Locus Method¶

Possible Exam Questions¶

Syllabus Focus¶

• Root locus method
• Closed-loop pole movement with gain variation
• Stability from root locus
• Root locus construction rules
• Transient-response interpretation from pole locations

1. Introduction to Root Locus¶

The root locus is the path traced by the roots of the closed-loop characteristic equation as a system parameter, usually gain \(K\), varies from \(0\) to \(\infty\).

For a negative-feedback system:

The characteristic equation is:

The root locus shows how the closed-loop poles move in the \(s\)-plane as \(K\) changes.

Importance¶

Root locus helps determine:

• Stability range of gain \(K\)
• Transient response as gain changes
• Dominant closed-loop pole locations
• Damping ratio and natural frequency
• Effect of adding poles and zeros
• Controller or compensator design

Basic Idea¶

Open-loop poles and zeros determine the possible paths of closed-loop poles.

As \(K\) changes:

• Closed-loop poles start at open-loop poles when \(K=0\).
• Closed-loop poles end at open-loop zeros or infinity when \(K\to\infty\).

2. Characteristic Equation and Root Locus Conditions¶

For negative feedback:

Therefore:

Since \(-1\) has magnitude 1 and angle odd multiples of \(180^\circ\):

These are the two root locus conditions.

Angle Condition¶

A point \(s=s_0\) lies on the root locus if:

where:

• \(z_i\) = open-loop zeros
• \(p_i\) = open-loop poles

Magnitude Condition¶

Once a point is known to lie on root locus, the gain \(K\) at that point is:

assuming open-loop transfer function is:

3. Root Locus Construction Rules¶

Consider:

Let:

• \(n\) = number of open-loop poles
• \(m\) = number of open-loop zeros

Rule 1 - Number of Branches¶

The number of root locus branches equals the number of open-loop poles.

Rule 2 - Starting and Ending Points¶

• Root locus starts at open-loop poles when \(K=0\).
• Root locus ends at open-loop zeros when \(K\to\infty\).
• If \(n>m\), then \(n-m\) branches end at infinity.

Rule 3 - Symmetry¶

Root locus is symmetric about the real axis because complex roots occur in conjugate pairs.

Rule 4 - Real-Axis Segments¶

A point on the real axis lies on the root locus if the total number of open-loop poles and zeros to its right is odd.

Rule 5 - Asymptotes¶

If \(n>m\), then \(n-m\) branches go to infinity along asymptotes.

Number of asymptotes:

Angles of asymptotes:

Centroid of asymptotes:

where sums are algebraic sums of pole and zero locations.

Rule 6 - Breakaway and Break-In Points¶

Breakaway points occur where branches leave the real axis.

Break-in points occur where branches enter the real axis.

From characteristic equation:

Solve for \(K\) as a function of \(s\):

Break points are found from:

Only points that lie on valid real-axis root locus segments are accepted.

Rule 7 - Angle of Departure from Complex Poles¶

For a complex pole \(p_k\), angle of departure is:

Rule 8 - Angle of Arrival at Complex Zeros¶

For a complex zero \(z_k\), angle of arrival is:

Rule 9 - Imaginary-Axis Crossing¶

The points where root locus crosses the imaginary axis are found using the Routh-Hurwitz criterion on the characteristic equation.

At crossing, substitute \(s=j\omega\) or use the auxiliary equation from the Routh array.

Rule 10 - Gain at Any Point¶

Use magnitude condition:

4. Root Locus and Stability¶

A continuous-time closed-loop system is stable if all closed-loop poles lie in the left half of the \(s\)-plane.

Root locus shows the closed-loop pole locations for all values of gain \(K\).

Stability from Root Locus¶

Gain Range¶

To find stable gain range:

1. Write characteristic equation \(1+KG(s)H(s)=0\).
2. Apply Routh-Hurwitz criterion.
3. Find range of \(K\) for which all first-column elements are positive.
4. Imaginary-axis crossing gain matches root locus crossing.

5. Routh-Hurwitz Criterion for Root Locus Support¶

For characteristic equation:

construct the Routh array. The system is stable if all elements in the first column have the same sign and none are zero.

Routh Array Structure¶

For a fourth-order equation:

Routh table:

where:

Interpretation¶

Number of sign changes in the first column equals the number of roots in the right half-plane.

Special cases:

• Zero in first column: replace by small \(\epsilon\) and continue.
• Entire row zero: use auxiliary equation from the row above and differentiate it.

6. Root Locus and Transient Response¶

Closed-loop pole locations determine transient response.

For dominant complex poles:

Natural frequency:

Damping ratio:

Settling time:

Peak time:

Percent overshoot:

Constant Damping Ratio Lines¶

Lines of constant damping ratio are radial lines from the origin.

If the angle from the negative real axis is \(\theta\):

Constant Natural Frequency Circles¶

Lines of constant natural frequency are circles centered at the origin:

Constant Settling Time Lines¶

Settling time depends on real part \(\sigma\):

Thus constant settling time lines are vertical lines in the left half-plane.

7. Effect of Adding Poles and Zeros¶

Adding an Open-Loop Pole¶

Adding a pole tends to pull the root locus toward the right half-plane.

Effects:

• Decreases relative stability
• Slows response
• Increases overshoot
• Can make system unstable

Adding an Open-Loop Zero¶

Adding a zero tends to pull the root locus toward the left half-plane.

Effects:

• Improves relative stability
• Speeds up response
• Can reduce settling time
• Often used in lead compensation

Compensator Interpretation¶

8. Root Locus Design Procedure¶

Typical design steps:

1. Write open-loop transfer function \(G(s)H(s)\).
2. Plot open-loop poles and zeros.
3. Sketch root locus using construction rules.
4. Translate time-domain specifications into desired pole region.
5. Choose desired dominant pole location on the root locus.
6. Use magnitude condition to calculate gain \(K\).
7. Verify remaining poles are non-dominant.
8. If root locus does not pass through desired point, add compensator.

Desired Pole from Specifications¶

If percent overshoot is specified, find \(\zeta\):

If settling time is specified:

Desired dominant poles are:

or:

9. Solved Examples¶

NTC 2081 PYQ - Parameterized Four-Pole Root Locus¶

The source paper gives:

Its open-loop poles are:

There are four branches and no finite zeros. Therefore the four asymptote angles are:

and the centroid is:

The real-axis locus lies between the two real poles \(-\sigma\) and \(0\). From

the breakaway candidates satisfy:

Only a real root lying in \((-\sigma,0)\) is a valid breakaway point. For the upper complex pole, the angle of departure is:

The lower-pole departure angle is its complex-conjugate reflection.

Example 1 - Basic Root Locus Data¶

Q. For \(G(s)H(s)=\frac{K}{s(s+2)(s+4)}\), find number of branches, real-axis segments, asymptote angles, and centroid.

Solution:

Open-loop poles: \(0,-2,-4\)

Open-loop zeros: none

Number of branches:

Real-axis segments exist where the number of poles/zeros to the right is odd:

• \((-\infty,-4)\): 3 to the right → root locus exists
• \((-4,-2)\): 2 to the right → no root locus
• \((-2,0)\): 1 to the right → root locus exists
• \((0,\infty)\): 0 to the right → no root locus

Asymptotes:

Centroid:

Example 2 - Breakaway Point¶

Q. For \(G(s)H(s)=\frac{K}{s(s+4)}\), find the breakaway point.

Solution:

Characteristic equation:

Breakaway point:

This point lies between poles at \(0\) and \(-4\), which is a valid root-locus segment.

Example 3 - Stability Range Using Routh¶

Q. Find range of \(K\) for stability if characteristic equation is:

Solution:

Routh array:

For stability, first column must be positive:

For positive gain:

Example 4 - Gain at Desired Pole¶

Q. For \(G(s)H(s)=\frac{K}{s(s+2)}\), find gain \(K\) when closed-loop pole is at \(s=-1+j2\).

Solution:

Use magnitude condition:

At \(s=-1+j2\):

10. Quick Revision Table¶

Model Answer — Root-Locus Definition and Construction Rules [10 marks]¶

Exam-ready answer

For a negative-feedback LTI system, write the characteristic equation as

\[ \boxed{1+K L_0(s)=0}, \qquad K\ge0, \]

where \(L_0(s)=G(s)H(s)\) excludes the variable real gain \(K\). The root locus is the set of all closed-loop pole positions in the \(s\)-plane as \(K\) varies from \(0\) to \(\infty\). The horizontal axis is \(\sigma\) in s\(^{-1}\) and the vertical axis is \(j\omega\) in rad/s.

Rearranging gives \(KL_0(s)=-1=1\angle(2q+1)180^\circ\). Therefore a point belongs to the positive-\(K\) locus only if

\[ \boxed{\angle L_0(s)=(2q+1)180^\circ} \]

and its gain is

\[ \boxed{K=\frac1{|L_0(s)|}}. \]

If \(L_0=\prod_{i=1}^m(s-z_i)/\prod_{i=1}^n(s-p_i)\), use these construction rules:

1. There are \(n\) branches, one for each open-loop pole.
2. At \(K=0\) branches start at \(p_i\). As \(K\to\infty\), \(m\) end at finite zeros \(z_i\) and \(n-m\) end at infinity.
3. With real coefficients, the locus is symmetric about the real axis.
4. A real-axis point is on the positive-\(K\) locus if the number of real poles and zeros to its right is odd.
5. The \(n-m\) asymptotes toward infinite zeros have

\[ \boxed{\theta_q=\frac{(2q+1)180^\circ}{n-m}}, \quad q=0,1,\ldots,n-m-1, \]

and intersect at

\[ \boxed{\sigma_a=\frac{\sum p_i-\sum z_i}{n-m}}. \]

1. On a real segment, write \(K(s)=-1/L_0(s)\) and solve \(dK/ds=0\) for breakaway or break-in candidates. Retain only points on a valid locus segment with \(K\ge0\).
2. From a complex pole \(p_k\), the departure angle is

\[ \boxed{\phi_d=180^\circ +\sum_i\angle(p_k-z_i) -\sum_{i\ne k}\angle(p_k-p_i)\pmod{360^\circ}}. \]

1. At a complex zero \(z_k\), the arrival angle is found by applying the same angle condition while excluding that zero:

\[ \boxed{\phi_a=180^\circ -\sum_{i\ne k}\angle(z_k-z_i) +\sum_i\angle(z_k-p_i)\pmod{360^\circ}}. \]

1. Find imaginary-axis crossings and the associated \(K\) by Routh-Hurwitz or by substituting \(s=j\omega\) into the characteristic equation and equating real and imaginary parts.
2. At any proposed point, use the magnitude condition, equivalently \(K=\prod|s-p_i|/\prod|s-z_i|\), and mark branch arrows from poles toward zeros as \(K\) increases.

Worked check: for \(KL_0=K/[s(s+2)(s+4)]\), poles are \(0,-2,-4\), there are no zeros, real segments are \((-2,0)\) and \((-\infty,-4)\), the centroid is \(-2\), and asymptote angles are \(60^\circ,180^\circ,300^\circ\). Since

\[ K=-s(s+2)(s+4), \quad \frac{dK}{ds}=-(3s^2+12s+8), \]

the candidates are \(s=-2\pm2\sqrt3/3\). Only \(s=-0.845\) lies on a valid segment; \(s=-3.155\) lies in \((-4,-2)\) and must be rejected.

Root locus directly relates gain to stability, damping and speed, but it is based on the modeled pole-zero set, usually varies one scalar gain, and becomes cumbersome for delays or high-order uncertain plants. It guides compensator placement; final gain and robustness should be checked by Routh and frequency-response margins.

Practice target: 18–20 minutes; derive angle/magnitude conditions, list all ten rules, and validate one candidate rather than accepting every dK/ds root.

Model Answer — NTC 2081 Parameterized Root-Locus Construction [10 marks]¶

Exam-ready answer

The available NTC 2081 source states

\[ \boxed{G(s)H(s)=\frac{K}{s(s+\sigma)(s^2+4s+13)}} \]

and does not assign a numerical value to \(\sigma\). Assume only \(\sigma>0\) so \(-\sigma\) is a left-half-plane real pole. Negative feedback gives \(1+G(s)H(s)=0\) and \(K\ge0\).

Part A — Root-locus construction and breakaway [6 marks]¶

The open-loop poles are

\[ \boxed{0,\quad-\sigma,\quad-2+j3,\quad-2-j3}, \]

with no finite zeros. Hence four branches start at these poles and all four end at infinity. The locus is conjugate-symmetric. On the real axis the only segment is between the two real poles, namely the open interval joining \(-\sigma\) and \(0\), because exactly one real pole lies to the right of each point there.

With \(n-m=4\), the asymptote angles and centroid are

\[ \boxed{\theta_q=45^\circ,135^\circ,225^\circ,315^\circ}, \]

\[ \boxed{\sigma_a =\frac{[0-\sigma+(-2+j3)+(-2-j3)]-0}{4} =-\frac{\sigma+4}{4}}. \]

From the characteristic equation,

\[ K(s)=-s(s+\sigma)(s^2+4s+13). \]

The breakaway candidates satisfy \(dK/ds=0\):

\[ \boxed{4s^3+3(4+\sigma)s^2 +2(13+4\sigma)s+13\sigma=0}. \]

Only a real root lying between \(-\sigma\) and \(0\) and producing \(K>0\) is a valid breakaway. Because its coordinate varies with the unspecified parameter, no unique decimal value can be reported.

Part B — Angle of departure [4 marks]¶

For the upper pole \(p_+=-2+j3\), measure all vectors from the other poles to \(p_+\). Their angles are

\[ \angle(p_+-0)=\operatorname{atan2}(3,-2)=123.69^\circ, \]

\[ \angle[p_+-(-\sigma)]=\operatorname{atan2}(3,\sigma-2), \qquad \angle[p_+-(-2-j3)]=90^\circ. \]

There are no zero angles. Applying the root-locus angle condition while excluding \(p_+\) gives

\[ \boxed{\phi_{d,+}=180^\circ- \left[123.69^\circ+ \operatorname{atan2}(3,\sigma-2)+90^\circ\right] \pmod{360^\circ}}. \]

The lower-pole departure is the reflection across the real axis:

\[ \boxed{\phi_{d,-}=-\phi_{d,+}\pmod{360^\circ}}. \]

Consistency check: the conjugate pole sum is \(-4\), so the centroid numerator must be \(-(\sigma+4)\), and conjugate symmetry requires opposite departure angles. Once a numerical \(\sigma\) is supplied, solve the cubic, reject invalid candidates with the real-axis and \(K>0\) tests, evaluate the angle, and use \(K=|s(s+\sigma)(s^2+4s+13)|\) to label the sketch.

The symbolic result is the most specific answer justified by the source. Choosing a convenient \(\sigma\) would change the pole order, centroid, breakaway and departure angle and would therefore be technically dishonest. Root locus also represents the stated linear model only; unmodeled delay and parameter uncertainty still require robustness checks.

Practice target: 20 minutes; allocate about 12 minutes to the 6-mark construction/breakaway and 8 minutes to the 4-mark departure-angle geometry.

Model Answer — Root Locus, Stability Range, and Marginal Gain [5 marks]¶

Exam-ready answer

For negative feedback with characteristic equation

\[ \boxed{1+KG(s)H(s)=0}, \qquad K\ge0, \]

the points on the root locus are the closed-loop poles for each \(K\). A continuous-time rational system is asymptotically/BIBO stable when all branches for that gain lie strictly in the left half-plane. Their distance from the imaginary axis indicates relative stability: poles farther left generally decay faster, while a complex pole angle gives damping ratio \(\zeta=-\operatorname{Re}(p)/|p|\).

The gain at which a conjugate pair reaches \(s=\pm j\omega_c\) is the marginal-stability gain. Two standard procedures are:

1. Form the polynomial \(D(s)+KN(s)=0\), substitute \(s=j\omega\), and solve simultaneous real and imaginary equations for \(K\) and \(\omega\).
2. Construct the Routh array. Require every first-column term to have the same sign for stability. Set the row that reaches zero to its boundary value; the auxiliary polynomial from the row above gives \(\omega_c\).

Worked check: for

\[ G(s)H(s)=\frac{K}{s(s+2)(s+4)}, \]

the characteristic polynomial is

\[ s^3+6s^2+8s+K=0. \]

Its Routh first column is

\[ 1,\quad6,\quad\frac{48-K}{6},\quad K. \]

Therefore

\[ \boxed{0<K<48\quad\text{is stable}}. \]

At \(K=48\), the \(s^1\) row vanishes. The auxiliary equation from the \(s^2\) row is \(6s^2+48=0\), so

\[ \boxed{s=\pm j\sqrt8=\pm j2\sqrt2\ \text{rad/s}}, \qquad \boxed{K_{\rm marginal}=48}. \]

Indeed, the polynomial factors as \((s+6)(s^2+8)\), confirming one LHP pole and a simple imaginary pair. For \(K>48\), branches enter the RHP and the system is unstable. The statement “below marginal gain is stable” is not universal without checking all Routh inequalities; systems with RHP open-loop poles or multiple crossings can have several stable/unstable intervals. Root locus gives the geometry, while Routh gives an exact gain range for a polynomial model.

Practice target: 9 minutes; state the LHP condition, build the short Routh array, and verify the crossing with the auxiliary polynomial.