Root Locus Technique

Unit-3

Time Response Analysis

Lecture 6

Root Locus Technique

• The root locus is the locus of the roots of the characteristics equation of a closed-loop system in the s-plane as a system parameter (usually the open loop gain K) is varied.

• It is a powerful method of analysis and design for stability and transient response of the control systems. Since the nature of the transient response is related to the location of the closed- loop poles, it is important to know how they move in the s-plane as the gain is varied.

• It gives information about the absolute stability as well as the relative stability of the system.

• Let us consider a system whose closed-loop transfer function is 𝐶(𝑠)

𝑅(𝑠) = 𝐾

𝑠² + 10𝑠 + 𝐾

• The table given below, shows the variation of pole location for different values of gain, K.

• The data of previous table is graphically displayed in Figure 1, which shows each pole and its gain.

Fig. 1

Fig. 2

• As the gain, K, increases in Table 8.1 and Fig. 1, the closed- loop pole, which is at -10 for K = 0, moves toward the right, and the closed-loop pole, which is at 0 for K = 0, moves toward the left. They meet at -5, break away from the real axis, and move into the complex plane. One closed-loop pole moves upward while the other moves downward.

• In Fig. 2, the individual closed-loop pole locations are removed and their paths are represented with solid lines. It is this representation of the paths of the closed-loop poles as the gain is varied that we call a root locus.

• The root locus shows the changes in the transient response as the gain, K, varies. First of all, the poles are real for gains less than 25. Thus, the system is overdamped. At a gain of 25, the poles are real and multiple and hence critically damped. For gains above 25, the system is underdamped.

Consider a single-loop feedback system shown in Figure 3.

The closed loop transfer function of this system is 𝐶(𝑠)

𝑅(𝑠) = 𝐺(𝑠)

1 + 𝐺 𝑠 𝐻(𝑠)

And its characteristic equation is^, 1 + 𝐺 𝑠 𝐻 𝑠 = 0

G(s)

H(s)

C(s) R(s)

+ -

Fig. 3

• The roots of the characteristics equation occur only for those values of ‘s’ for which

𝐺 𝑠 𝐻 𝑠 = −1

• Since s is a complex variable, the above equation is converted into two conditions:

𝐺 𝑠 𝐻 𝑠 = 1

𝑎𝑛𝑑 ∠𝐺 𝑠 𝐻 𝑠 = ± 2𝑞 + 1 𝜋, 𝑞 = 0, 1, 2, . . 𝑮 𝒔 𝑯 𝒔 = 𝟏 is the magnitude condition whereas,

∠𝑮 𝒔 𝑯 𝒔 = ± 𝟐𝒒 + 𝟏 𝝅 is the phase angle condition.

The roots of the characteristics equation 𝟏 + 𝑮 𝒔 𝑯 𝒔 = 𝟎 are those values of s for which both magnitude and phase angle condition are satisfied.

For the system shown in Fig. 3, the open loop transfer function is, 𝐺 𝑠 𝐻 𝑠 = 𝐾 𝑠 + 𝑧₁ 𝑠 + 𝑧₂ …

𝑠 + 𝑝₁ 𝑠 + 𝑝₂ . . = 𝐾 _𝑖=1^𝑚 (𝑠 + 𝑧_𝑖)

𝑗=1𝑚 (𝑠 + 𝑝_𝑗) As 𝑮 𝒔 𝑯 𝒔 = 1 and ∠𝑮 𝒔 𝑯 𝒔 = ± 𝟐𝒒 + 𝟏 𝝅,

𝐾 _𝑖=1^𝑚 (𝑠 + 𝑧_𝑖)

𝑗=1𝑚 (𝑠 + 𝑝_𝑗) = 1 ⇒ 𝐾 = ^𝑗=1

𝑚 (𝑠 + 𝑝_𝑗)

𝑖=1𝑚 (𝑠 + 𝑧_𝑖)

and _𝑖=1^𝑚 ∠ 𝑠 + 𝑧_𝑖 − _𝑗=1^𝑛 ∠ 𝑠 + 𝑝_𝑗 = ± 2𝑞 + 1 𝜋; 𝑞 = 0, 1, 2. .

• Thus, a point in the s-plane is on the root locus for a particular value of gain K, if the angles of the zeros minus the angle of poles, which are drawn to the selected point on the s-plane, add up to (2q+1)180⁰.

• The value of gain K for which the angles add up to (2q+1)180⁰ found by dividing the product of the pole lengths by the product of the zero lengths.

• The value of K at any point s_o on the root locus can be found out as

𝐾 = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝ℎ𝑎𝑠𝑜𝑟 𝑙𝑒𝑛𝑔𝑡ℎ𝑠 𝑓𝑟𝑜𝑚 𝑠_𝑜 𝑡𝑜 𝑡ℎ𝑒 𝑜𝑝𝑒𝑛 − 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝ℎ𝑎𝑠𝑜𝑟 𝑙𝑒𝑛𝑔𝑡ℎ𝑠 𝑓𝑟𝑜𝑚 𝑠_𝑜 𝑡𝑜 𝑡ℎ𝑒 𝑜𝑝𝑒𝑛 − 𝑙𝑜𝑜𝑝 𝑧𝑒𝑟𝑜𝑠

• A function can also have infinite poles and zeros. If the function approaches infinity as s approaches infinity, then the function has a pole at infinity. If the function approaches zero as s approaches infinity, then the function has a zero at infinity.

• For example, the function G(s)= s has a pole at infinity, since G(s) approaches infinity as s approaches infinity. On the other hand, G(s)= 1/s has a zero at infinity, since G(s) approaches zero as s approaches infinity.

• Every function of s has an equal number of poles and zeros if we include the infinite poles and zeros as well as the finite poles and zeros.

Rules for drawing Root Locus

1. Starting and Ending Point: The root locus starts at open- loop poles and ends at open-loop zeros. At the open-loop poles, open loop gain (K) = 0 and at the open loop zeros, K = ∞.

2. Number of Branches: The number of branches of the root- locus is equal to the number of the closed-loop poles.

3. Real Axis Segments: Segments of the real axis having an odd number of real axis open-loop poles and zeros to there right are parts of the root locus.

4. Symmetry: The root locus is symmetrical about the real axis.

5. Angle of asymptotes: The root locus approaches straight lines as asymptotes as the locus approaches infinity. The angle of asymptotes with the real axis is given by,

𝜃_𝑞= ± (2𝑞 + 1)𝜋

𝑛 − 𝑚 ; 𝑞 = 0, 1, 2 … (𝑛 − 𝑚 − 1) where 𝑛 = number of open loop poles

𝑚 = number of open loop zeros

6. Centroid (𝝈): The asymptotes cross the real axis at a point known as centroid. It is determined as follows:

𝝈 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠 − 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑧𝑒𝑟𝑜𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠

7. Breakaway point: The closed loop poles move toward each other, as the value of gain K is increased from zero, and become coincident for the maximum value of K along the real axis.

As the value of K is increased above this value, the poles move into the complex plane. The point at which the poles move into the complex plane is called the breakaway point.

The breakaway points are obtained by solving the equation, 𝑑𝐾

𝑑𝑠 = 0

8. Angle of departure and Angle of arrival: The root locus starts at the open-loop poles and ends at the open-loop zeros.

In order to sketch the root locus more accurately, we have to calculate the root locus departure angle from the complex poles and the arrival angle to the complex zeros. Thus, angle of departure and arrival need to be calculated only when there are complex poles and zeros.

The angle of departure from an open loop pole is calculated as 𝜃_𝑑= ± 2𝑞 + 1 𝜋 + ϕ ; 𝑞 = 0, 1, 2 …

Where ϕ is the net angle contribution at this open loop pole, of all other open loop poles and zeros.

Similarly, The angle of arrival at an open loop zero is given as 𝜃_𝑎= ± 2𝑞 + 1 𝜋 − ϕ ; 𝑞 = 0, 1, 2 …

Where ϕ is the net angle contribution at this open loop zero, of all other open loop poles and zeros.

9. Intersection with the imaginary axis: The point of intersection of the root locus branches with the imaginary axis and the critical value of K can be determined by the use of Routh Hurwitz criterion.

Example 1: Draw the root locus for with the following open loop transfer function:

𝐺 𝑠 𝐻 𝑠 = 𝐾

𝑠(𝑠 + 1)(𝑠 + 3)

Solution: There are three open loop poles at s = 0,s = -1 and s= -3.

There are no finite open loop zeros which means all the three zeros are at infinity. Therefore, there are three asymptotes.

The three branches of the root locus start at the open loop poles s = 0, 1 and -3 (where K = 0) and terminate at the open loop zeros at infinity (where K = ∞).

Step 1: Calculate angle of asymptotes:

Since, n = 3 and m = 0 in the given problem, therefore 𝜃_𝑞= ± (2𝑞 + 1)𝜋

3 − 0 ; 𝑞 = 0, 1, 2

Putting q = 0,1, 2 in the given expression, the angles are obtained as 𝜃₀ = 60°, 𝜃₁ = 180°, 𝜃₂ = 300°

Step 2: Calculate centroid i.e. point of intersection of the asymptotes on the real axis:

𝝈 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠 − 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑧𝑒𝑟𝑜𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠

= 0 − 1 − 3 − (0)

3 − 0 = −1.33

Step 3: Calculate breakaway point:

The breakaway points are obtained by the solution of ^𝑑𝐾

𝑑𝑠 = 0.

According to the magnitude condition,

𝐺 𝑠 𝐻 𝑠 = 1

⇒ 𝐾

𝑠(𝑠 + 1)(𝑠 + 3) = 1 ⇒ 𝐾 = 𝑠 𝑠 + 1 𝑠 + 3 = 𝑠³ + 4𝑠² + 3𝑠 𝑑𝐾

𝑑𝑠 = 𝑑

𝑑𝑠 𝑠³ + 4𝑠² + 3𝑠 = 3𝑠² + 8𝑠 + 3 = 0 ⇒ 𝑠 = −0.451 𝑜𝑟 − 2.28 According to rule 3, as discussed previously, the root locus exists on the real axis between s = 0 to s = -1 and from s = -3 to −∞.

The actual breakaway point , from the two points which are obtained, is s = -0.451. The second point which is obtained i.e. s = -2.28 can be ignored as no root locus exists there.

Step 4: Intersection with the imaginary axis:

It is found by applying the Routh Hurwitz criterion.

The characteristic equation of the system is, 𝑠³ + 4𝑠² + 3𝑠 + 𝐾 = 0 The Routh table can be formed as follows

𝑠³ 1 3

𝑠² 4 K

𝑠¹ 12 − 𝐾

4

𝑠⁰ K

For all the roots of the characteristic equation to lie to the left of the imaginary axis, the following conditions should be satisfied.

𝐾 > 0 𝑎𝑛𝑑 12 − 𝐾

4 > 0 ⇒ 𝐾 < 12.

The value of K for which the roots lie on the jω axis is given by, 12 − 𝐾

4 = 0 ⇒ 𝐾 = 12

For K = 12, all the coefficients of s¹ row will be zero. The auxiliary equation formed from the coefficients of the s² row is,

4𝑠² + 𝐾 = 0

For K =12, the roots of the above equation lie on the jω axis.

4𝑠² + 12 = 0 ⇒ 𝑠² = −3 ⇒ 𝑠 = ±𝑗 3 = ±𝑗1.73.

Thus, the root locus will intersect the jω axis at s = ±𝑗1.73.

The complete root locus for the given system is shown on Fig. 4.

Fig. 4

Example 2: Draw the root locus for with the following open loop transfer function:

𝐺 𝑠 𝐻 𝑠 = 𝐾

𝑠(𝑠 + 4)(𝑠² + 4𝑠 + 20)

Solution: There are four open loop poles at s = 0,s = -4 and s= -2+j4, s = -2-j4. Therefore, n = 4

There are no finite open loop zeros therefore m = 0.

The four branches of the root locus start at the open loop poles s = 0, -4, -2+j4 and -2-j4 (where K = 0) and terminate at the open loop zeros at infinity (where K = ∞).

Step 1: Calculate angle of asymptotes:

Since, n = 4 and m = 0 in the given problem, therefore 𝜃_𝑞= ± (2𝑞 + 1)𝜋

4 − 0 ; 𝑞 = 0, 1, 2,3

Putting q = 0,1, 2,3 in the given expression, the angles are obtained as 𝜃₀ = 𝜋

4 , 𝜃₁ = 3𝜋

4 , 𝜃₂ = 5𝜋

4 , 𝜃₃ = 7𝜋 4

Step 2: Calculate centroid i.e. point of intersection of the asymptotes on the real axis:

𝝈 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠 − 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑧𝑒𝑟𝑜𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠

= 0 − 4 − 2 − 2 − (0)

4 − 0 = −2

Step 3: Calculate breakaway point:

The breakaway points are obtained by the solution of ^𝑑𝐾

𝑑𝑠 = 0.

According to the magnitude condition,

𝐺 𝑠 𝐻 𝑠 = 1

⇒ 𝐾

𝑠(𝑠 + 4)(𝑠² + 4𝑠 + 20) = 1 ⇒ 𝐾 = 𝑠 𝑠 + 4 𝑠² + 4𝑠 + 20

= 𝑠³ + 4𝑠² + 3𝑠 𝑑𝐾

𝑑𝑠 = −(4𝑠³ + 24𝑠² + 72𝑠 + 80) = 0 ⇒ 𝑠³ + 6𝑠² + 18𝑠 + 20 = 0

𝑠 + 2 𝑠² + 4𝑠 + 10 = 0 ⇒ 𝑠 + 2 𝑠 + 2 + 𝑗2.45 𝑠 + 2 − 𝑗2.45 = 0 There is one breakaway one point at s= -2 and there are two complex breakaway points at s = 2+j2.45 and s = 2-j2.45.

According to rule 3, as discussed previously, the root locus exists on the real axis between s = 0 to s = -4.

Step 4: Intersection with the imaginary axis:

It is found by applying the Routh Hurwitz criterion.

The characteristic equation of the system is,

𝑠⁴ + 8𝑠³ + 36𝑠² + 80𝑠 + 𝐾 = 0 The Routh table can be formed as follows

𝑠⁴ 1 36 K

𝑠³ 8 80

𝑠² 26 K

𝑠¹ 2080 − 8𝐾 26

𝑠⁰ K

For all the roots of the characteristic equation to lie to the left of the imaginary axis, the following conditions should be satisfied.

𝐾 > 0 𝑎𝑛𝑑 2080 − 8𝐾

26 > 0 ⇒ 𝐾 < 260.

The value of K for which the roots lie on the jω axis is given by, 2080 − 8𝐾

26 = 0 ⇒ 𝐾 = 260

For K = 260, all the coefficients of s¹ row will be zero. The auxiliary equation formed from the coefficients of the s² row is,

26𝑠² + 𝐾 = 0

For K =260, the roots of the above equation lie on the jω axis.

26𝑠² + 260 = 0 ⇒ 𝑠² = −10 ⇒ 𝑠 = ±𝑗 10 = ±𝑗1.73.

Thus, the root locus will intersect the jω axis at s = ±𝑗 10.

Step 5: Find angle of departure:

The angle of departure from the complex pole s = -2+j4 is given by, 𝜃_𝑑 = ± 2𝑞 + 1 𝜋 + ϕ

ϕ = − 𝜃₁ + 𝜃₂ + 𝜃₃ = − 117° + 90° + 63° = −270°

𝜃_𝑑 = 180° − 270° = −90°

Because of symmetry, the angle of departure from the complex pole at s = -2-j4 is +90°.

The complete root locus is shown in Fig. 5.

Fig. 5

References

1. Control Systems Engineering by Norman S. Nise.

2. Control Systems by A. Anand Kumar.

3. Control Systems by M. Gopal

4. NPTEL Lectures on Control Engineering

(https://nptel.ac.in/courses/108/106/108106098/)