Outcome 3: Control System Stability Analysis PDF

OUTCOME 3: Analyze the relation between location of poles and stability, Investigate the stability analysis in time domain using Routh Hurwitz criterion, Construct the root locus diagrams and discuss the properties and applications. 3.1 Introduction : For a control system designer, the main concern...

OUTCOME 3: Analyze the relation between location of poles and stability, Investigate the stability analysis in time domain using Routh Hurwitz criterion, Construct the root locus diagrams and discuss the properties and applications. 3.1 Introduction : For a control system designer, the main concern is whether the system is stable or unstable. Because, Stability is a very important characteristic of the transient performance of a system. Any working system is designed considering its stability. Therefore, all instruments are stable with in a boundary of parameter variations. Thus the study of whether the system is stable or unstable is known as the stability analysis.The term stability refers to the stable working condition of a control system. In a stable system the response or output is predictable, finite and stable for a given input. 3.2 Concept of Stability: Stability can be illustrated by considering a circular object resting a different types of surfaces under the presence of gravitational force as shown in fig 3.1 Figure 3.1 Concepts of Stability A necessary and sufficient condition for a feedback system to be stable is that all the poles of the system transfer function have negative real parts. There are two types of stability analysis. 1. BIBO stability A system is said to a BIBO stable system if it produces a bounded output for a given bounded input. 2. Asymptotic stability In the absence of input, the output of the system goes to zero irrespective of the initial conditions. This is known as asymptotic stability. A stable system produces a bounded output for a given bounded input. The following figure shows the response of a stable system.For a system stability, four cases can be possible. 1|Page i. Absolute stable: A system is said to be absolute stable if it is stable for all values of the parameters. ii. Marginally or critically stable: a system is said to be a marginally stable system if the output is having sustained oscillation with constant magnitude. iii. Unstable: A system is unstable if the output is unbounded for a bounded input. iv. Conditionally stable: if the system output is stable for a limited range of variations of its parameters then the system is called as conditionally stable system. Figure 3.1 Concepts of Stability 2|Page  3.3 Effect of location of poles on stability Nature of Location Output Stability Roots/ Poles Real and exponentially decaying time Stable negative response. Complex Stable conjugate with negative real parts damped oscillations Real and Unstable positive exponentially increasing response. Complex Unstable conjugate with positive real parts growing oscillations. 3|Page  Repeated Unstable pair of poles on the imaginary axis. Non Marginally Repeated j stable pair of poles on the imaginary axis. Exercise Problems Determine if the systems and the following roots are stable, marginally stable and unstable. (a) – 3, – 6 (b) – 2, + 1 (c) – 1, 0 (d) – 1 + j, – 1 – j (e) – 1, 1, – j, j (f) – 2 + j3, – 2 – j3, – 2 (g) an integrator (h) a step input (i) x(t) = cos ωt (j) x (t) = e–t sin 4t 3.4 Methods for Calculating Stability in time domain 3.4.1 Routh Hurwitz Criterion : Routh Hurwitz criterion is an algebraic method of determining the stability of the linear time invariant system. This criterion provides information about whether the system is stable or not depending upon the position of the roots of the characteristic equation whether they lies in the left half of the s plane or right half of the s plane. This method can also be used to determine the limiting values of a variable parameter within which the system is stable and beyond which system would become unstable. The T.F.of any linear closed loop system can be represented as, 𝐶(𝑠) 𝑏0 𝑠 𝑚 + 𝑏1 𝑠 𝑚−1 + ⋯. +𝑏𝑚 = 𝑅(𝑠) 𝑎0 𝑠 𝑛 + 𝑎1 𝑠 𝑛−1 + ⋯. +𝑎𝑛 Where a & b are constants. To find the closed loop poles, equate denominator (F(s)) to 0. This equation is called as Characteristic Equation of the system. 4|Page  F(s) = a0 sn + a1 sn-1 + a2 sn-2 + ….. + an = 0. The RH criterion is based on arranging these coefficients of the characteristics equation in the form of an array which is called as Routh’s Array. Formation of Routh’s array: The coefficients of the characteristics equation are represented by array form as follows. Sn a0 a2 a4 a6 ………. Sn-1 a1 a3 a5 a7 Sn-2 b1 b2 b3 Sn-3 c1 c2 c3 - - - - - - - - S0 an Coefficients of first two rows are written directly from characteristics equation. From these two rows next rows can be obtained as follows. a1 a2 – a0 a3 a1 a4 – a0 a5 a1 a6 – a0 a7 b1 = , b2 = , b3 = a1 a1 a1 From 2nd & 3rd row , 4th row can be obtained as b1 a3 – a1 b2 b1 a5 – a1 b3 C1 = , C2 = b1 b1 This process is to be continued till the coefficient for s0 is obtained which will be an. From this array stability of system can be predicted. The Routh-Hurwitz criterion states that “The necessary and sufficient condition for stability is that all the terms in the first colum of Routh’s array must have same sign.if there are any sign changes exists then the system is unstable system”.The number of sign changes is equal to number of roots present in the Right Hand side of the s plane. 5|Page Example Problems: 1. Examine the stability of given equation using Routh’s method: 𝑠 3 + 9𝑠 2 + 26𝑠 + 24 = 0 𝑠3 1 26 𝑠2 9 24 𝑠1 70 0 = 23.33 3 𝑠0 24 Since there is no changes of the sign in the first column of the Routh table, so this system is stable. 2. Examine the stability of given equation using Routh’s method: s4 + 2s3 + 3s2 +4s +5=0 𝑠4 1 3 5 𝑠3 2 4 𝑠2 1 5 𝑠1 -6 𝑠0 5 There are two sign changes in the first column, so there 2 poles are at RHS plane. Hence the system is an unstable system. Exercise Problems: 1. By means of the Routh criterion, determine the stability of the system represented by the following characteristic equation. For the systems found to be unstable, determine the number of roots of the characteristic equation in the right-half of the S-plane. a) 9s5 -20 s4 + 10 s3 - s2 -9 s -10 = 0 b) s6 + 2s5 + 8 s4 + 15 s3 + 20 s2 + 16 s + 16 = 0 2.Determine the range of K for stability of unity feedback system whose open loop transfer function is given by 𝐾 𝐺(𝑆) = 𝑆(𝑆 + 1)(𝑆 + 2) Special Cases: Case 1: The first element in any row of the Routh array is zero while the rest of the row has at least one non-zero element. 1 Method: Modify the characteristic equation by replacing s by 𝑧 and rearrange the characteristic equation. Then form the Routh’s array from the new characteristics equation in x and examine the stability with this array. 6|Page Example 1: Examine stability of the following system given by s5 +2s4 +4s3 +8s2 +3s+1 using Routh-Hurwitz Stability criterion. Solution The Routh’s array is Here first element is zero and the other element is a non-zero. 1 So replace the term s by 𝑧.The characteristic equation is s5 +2s4 +4s3 +8s2 +3s+1 =0 becomes 1 2 4 8 3 + + + + +1=0 𝑍 5 𝑍 4 𝑍 3 𝑍 2 𝑍1 Z5 +3Z4 +8Z3 +4Z2 +2Z+1 =0 The new Routh’s array is There are two sign changes so the system is an unstable system. 2. Examine stability of the following system given by s5 +2s4 +3s3 +6s2 +2s+1 using Routh- Hurwitz Stability criterion. Case 2: All the elements in any one row of the Routh array are zero. Method: Step1: Form an auxiliary equation using the coefficients of a row just above the row of zeros. 2. Take the derivative of an auxiliary equation with respect to s. 3. Replace the rows of zeros with the coefficients of the derivative equation. 4. Complete the Routh array and examine the sign changes in the first column of the Routh’s Array to determine the stability 7|Page  Exercise: 1. Examine stability of the following system given by s6+2s5 +8s4 +12s3 +20s2 +16s+16 =0 using Routh-Hurwitz Stability criterion. 2. Examine stability of the following system given by s5+7s4+6s3+42s2+8s+56=0 using Routh-Hurwitz Stability criterion. Advantages of Routh-Hurwitz stability 1. Stability can be judged without solving the characteristic equation 2. Less calculation time 3. The number of roots in RHP can be found in case of unstable condition 4. Range of value of K for system stability can be calculated 5. Intersection point with the jw-axis can be calculated 6. Frequency of oscillation at steady-state is calculated Disadvantages of Routh-Hurwitz stability 1. It is valid for only real coefficient of the characteristic equation 2. Unable to give exact locations of closed-loop poles 3. Does not suggest methods for stabilizing an unstable system 4. Applicable only to the linear system 3.4.2 Root locus Technique The performance characteristics such as relative stability and transient response of a closed loop control system are directly related to the location of roots of the characteristics equation in the s plane.so it is necessary to determine how the roots migrate about the s plane as the parameters are varied. The root locus technique is a powerful tool for adjusting the location of closed loop poles to achieve the desired system performance by varying one or more system parameters. In addition to determining the stability of the system, the root locus can be used to design the damping ratio and natural frequency of a feedback system. Definition Root Locus is the plot of the roots of the characteristics equation of the closed loop system obtained when the system gain or loop gain K is varied from -∞ to ∞. Conditions of Root locus In general, the characteristic equation of closed loop system is given by 1+G(S) H(S) =0 Since ‘s’ is a complex variable ,G(S) H(S) is also a complex quantity.it can be split into equation by equating angle and phase magnitude on both sides to obtain the conditions of Rootlocus.They are Angle Condition and Phase Condition. 8|Page Angle Condition: 1+G(S) H(S) =0 G(S) H(S) = -1. Or G(S)H(S) = −1 + 𝑗0. Angle condition states that 𝐺(𝑆)𝐻(𝑆) = ±(2𝑞 + 1)180° where q =0,1,2 ….. = odd multiples of 180° If any value of ‘S’ which satisfies the angle condition then that point is considered to be on the root locus. Magnitude Condition: 1+G(S) H(S) =0 G(S) H(S) = -1. Or G(S)H(S) = −1 + 𝑗0. Magnitude condition can be stated as |𝑮(𝑺)𝑯(𝑺)| = |−𝟏 + 𝒋 𝟎| = 𝟏 But G(S)H(S) cannot be found if it contains unknown variable K. So using this condition it is not possible to check the existence of point on the rootlocus.But using magnitude condition ,we can find out the value of K for any point in s plane.If any point is on root locus it must satisfy the magnitude condition also. Example 𝑲 1. For the system with 𝑮(𝑺)𝑯(𝑺) = 𝑺(𝑺+𝟐).Find whether S= -0.5 lies on the root locus or not using angle condition. Angle condition states that ∠ 𝐺(𝑆)𝐻(𝑆) = ±(2𝑞 + 1)180° Where q =0, 1, 2 ….. = odd multiples of 180° ∠K + j0 ∠𝐺(𝑆)𝐻(𝑆)𝑠→−0.5 = ∠ − 0.5 + j0 ∗ (∠ − 0.5 + j0 + 2) ∠K+j0 = ∠−0.5+j0∗(∠−1.5+j0) 0° = = −180° 180°∗0° Thus it satisfies the angle condition.therefore S= -0.5 lies on the Root locus of the system. 9|Page  𝑲 2. For the system with 𝑮(𝑺)𝑯(𝑺) = 𝑺(𝑺+𝟓)(𝑺+𝟏𝟎).Find whether S= -2 lies on the root locus and also find the value of K using angle condition and magnitude condition. Angle condition states that ∠ 𝐺(𝑆)𝐻(𝑆) = ±(2𝑞 + 1)180° Where q =0, 1, 2 ….. = odd multiples of 180° ∠K + j0 ∠𝐺(𝑆)𝐻(𝑆)𝑠→−2 = −2∠ − 2 + 5 + j0 ∗ (∠ − 2 + 10 + 0j) ∠𝐾 + 𝑗0 = ∠ − 2 + 𝑗0 ∗ (∠3 + 𝑗0) ∗ (∠8 + 𝑗0) 0° = = −180° 180°∗0°∗0° ∴ s = -2 lies on the Root locus. To find the value of K Magnitude condition can be stated as |𝐺(𝑆)𝐻(𝑆)| = |−1 + 𝑗 0| = 1 𝐾 |𝐺(𝑆)𝐻(𝑆)| = |.| = 1 𝑆(𝑆 + 5)(𝑆 + 10) Sub s = -2 , |𝐾| =1 |−2||3||8| 𝑲 = 𝟒𝟖 3.5 Root locus Plot The plot representing the root locus is always drawn in the s plane. ‘s’ is a complex quantity ,expressed as 𝜎 + 𝑗𝑤.where 𝜎 represents the real axis and jw represents the imaginary axis. To construct the root locus, open loop poles and open loop zero are necessary. Consider the general equation for open loop system 𝐾(𝑠 + 𝑍1 )(𝑠 + 𝑍2 ) … ….. 𝐺(𝑠) = 𝑠 𝑛 (𝑠 + 𝑃1 )(𝑠 + 𝑃2 ) … … Where K is the open loop system gain Z1, Z2… are the open loop zeros. P1,P2 …. are the open loop poles. n is the number of poles lies at origin. 10 | P a g e 3.5.1 Terms in Root Locus 1. Characteristic Equation Related to Root Locus Technique: 1 + G(s)H(s) = 0 is known as characteristic equation. Now on differentiating the characteristic equation and on equating dk/ds equals to zero, we can get break away points. 2. Break away Points : Suppose two root loci which start from pole and moves in opposite direction collide with each other such that after collision they start moving in different directions in the symmetrical way. Or the break away points at which multiple roots of the characteristic equation 1 + G(s)H(s)= 0 occur. The value of K is maximum at the points where the branches of root loci break away. Break away points may be real, imaginary or complex. 3. Break in Point : Condition of break in to be there on the plot is written below : Root locus must be present between two adjacent zeros on the real axis. 4. Centre of Gravity : It is also known centroid and is defined as the point on the plot from where all the asymptotes start. Mathematically, it is calculated by the difference of summation of poles and zeros in the transfer function when divided by the difference of total number of poles and total number of zeros. Centre of gravity is always real & it is denoted by σA. Where, N is number of poles and M is number of zeros. 5. Asymptotes of Root Loci : Asymptote originates from the center of gravity or centroid and goes to infinity at definite some angle. Asymptotes provide direction to the root locus when they depart break away points. 6. Angle of Asymptotes : Asymptotes makes some angle with the real axis and this angle can be calculated from the given formula, Where, p = 0, 1, 2....... (N-M-1) N is the total number of poles M is the total number of zeros. 11 | P a g e  7. Intersection of Root Locus with the Imaginary Axis: In order to find out the point of intersection root locus with imaginary axis, we have to use Routh Hurwitz criterion. First, we find the auxiliary equation then the corresponding value of K will give the value of the point of intersection. 8. Gain Margin: We define gain margin as a by which the design value of the gain factor can be multiplied before the system becomes unstable. Mathematically it is given by the formula 9. Phase Margin: Phase margin can be calculated from the given formula: 3.5.3 Rules for constructing Root locus. Rule 1: The root locus is symmetric about the real axis Rule 2: At the open-loop poles, k = 0 and at the open-loop zeros, k = ∞ As the open-loop gain K is varied from zero to infinity, each branch of the root locus originates from an open-loop pole where K = 0 and terminates on an open-loop zero or zero at infinity where K = ∞. The number of branches of root locus terminating on infinity equals the number of open- loop poles minus open-loop zeros. If n  number of open-loop poles; and m number of open-loop zeros and n > m , the open loop transfer function has (n - m) zeros at infinity and (n – m) branches of the root locus terminates on these zeros. Rule 3: Segments of the real axis having an odd number of real axis open-loop poles and zeros to their right are parts of the root locus. Rule 4: The (n-m) branches of the root locus which go to infinity travel along straight line asymptotes whose angles are given by (2q  1) θq = ± ; q = 0,1,2,… (n-m-1) ( n  m) Rule 5: The asymptote cross the real axis at a point known as centroid sum of real parts of poles - sum of real parts of zeros Centroid = σ = number of poles - number of zeros 12 | P a g e  dK Rule 6: The breakaway point of the root locus are the solution of = 0 ds Rule 7: The intersection points of root locus branches with an imaginary axis can be determined by using the Routh criterion. Or by letting s=jw in the characteristic equation and equating the real part and imaginary part to zero. Rule 8: The open loop gain K at any point s=sa on the root locus is given by ∏ni=1|sa + pi | K= m ∏i=1|sa + zi | product of vector length from open loop poles to the point s =product of the vector length from open loop zeros to the pointsa a 3.5.4 Procedure for constructing Root locus Step 1: Locate the poles and zeros of G(S).H(S) on the S plane.The root locus branch start from open loop poles and terminates at zeros. Step 2: Determine the Root locus on the real axis. Step 3: Determine the Asymptote and centroid Step 4: Find the break away points and break in points. Step 5: Find the intersection point or crossing point. Example Problems: K 1. Draw the root locus plot for G(s). H(s) = s(s  1)(s  3) Solution: Step 1: Determine the number of open-loop poles and zeros Number of open-loop poles n=3 Number of open-loop zeros m=0 Open-loop poles: s=0, s= -1 and s= -3 Step 2: Determine parts of the root-locus on the real axis There are 3 poles on the real axis. 13 | P a g e  i. Choose a test point in the real axis between s=0 to -1.To the right of this point the total number of poles and zeros is one ,which is an odd number.Hence the real axis between s=0 to -1 will be a part of Root locus. ii. Choose a test point in the real axis between s= -1 to -3.To the right of this point the total number of poles and zeros is two ,which is an even number.Hence the real axis between s=-1 to -3 will not be a part of Root locus. iii. Choose a test point in the real axis between s=-3 to −∞.To the right of this point the total number of poles and zeros is three ,which is an odd number.Hence the real axis between s=-3 to −∞ will be a part of Root locus. Step 3: Find Asymptote and Centroid (2q  1) q   ; q = 0,1,2 (n  m)  3 5 θ0 = 3 = 60 ; ο θ1 = 3 = 180ο ; θ2 = 3 = 300ο ; with the real axis. The point of intersection of the asymptotes on the real axis is called centroid σ. sum of real parts of poles - sum of real parts of zeros Centroid = σ = number of poles - number of zeros (0  1  3)  (0) σ = = -1.33 (3  0) Step 4: Determine breakaway and break-in point 𝐾 𝐶(𝑆) 𝑆(𝑆+1)(𝑠+3) The closed loop transfer function = 𝐾 𝑅(𝑆) 1+ 𝑆(𝑆+1)(𝑆+3) 𝐾 = 𝑆(𝑆+1)(𝑆+3)+𝐾 The characteristic equation is S(S+1) (S+3) +K = 0 S(S2+4S+3)+K = 0 S3+4S2+3S+K = 0 ∴ K = - S3- 4S2- 3S On differentiating with respect to K 14 | P a g e 𝑑𝐾 = −3𝑆 2 − 8𝑆 − 3 𝑑𝑆 𝑑𝐾 Put =0 𝑑𝑆 −3𝑆 2 − 8𝑆 − 3 = 0  8  82  (4  3  3)  8  28 ∴s = = = - 0.451 or - 2.28. (2  3) 6 For K: When S = - 0.451, the value of K is given by K =−[(−0.451)3 + 4(−0.451)2 + 3(−0.451)] =0.6327 Since K is positive and real for S = - 0.451, this point is the actual breakaway point. When S =-2.28 ,the value of K is given by K =−[(−2.28)3 + 4(−2.28)2 + 3(−2.28)] = −2.1 Since K is negative and real for S =-2.28, this point is not actual breakaway point. Step 5: Determine points on the root-locus crossing imaginary axis The characteristic equation is S3+4S2+3S+K = 0 Put s =jw (𝑗𝑤)3 + 4(𝑗𝑤)2 + 3𝑗𝑤 + 𝐾 = 0 −𝑗𝑤 3 − 4𝑤 2 + 3𝑗𝑤 + 𝐾 = 0 Equate real and imaginary part to zero −𝑤 3 + 3𝑤 = 0 & −4𝑤 2 + 𝐾 = 0 ∴ −𝑤 2 = −3𝑤 ; −4𝑤 2 = −𝐾 𝑤 = √3 = ±1.732 ; 𝐾 = 4𝑋3 = 12 15 | P a g e The crossing point of root locus is ±1.732. The value of K at this point is 12.This is the limiting value of K for stability. 2. Sketch the root locus of the system whose open loop transfer function is 𝐾 𝐺(𝑠) = 𝑠(𝑠 + 2)(𝑠 + 4) Find the value of K so that the damping ratio is 0.5. Solution: Step 1: Determine the number of open-loop poles and zeros Number of open-loop poles n=3 Number of open-loop zeros m=0 Open-loop poles: s=0, s=-2 and s=-4 Step 2: Determine parts of the root-locus on the real axis There are 3 poles on the real axis. iv. Choose a test point in the real axis between s=0 to -2.To the right of this point the total number of poles and zeros is one ,which is an odd number.Hence the real axis between s=0 to -2 will be a part of Root locus. 16 | P a g e  v. Choose a test point in the real axis between s= -2 to -4.To the right of this point the total number of poles and zeros is two ,which is an even number.Hence the real axis between s=-2 to -4 will not be a part of Root locus. vi. Choose a test point in the real axis between s=-4 to −∞.To the right of this point the total number of poles and zeros is three ,which is an odd number.Hence the real axis between s=-4 to −∞ will be a part of Root locus. Step 3: Find Asymptote and Centroid (2q  1) q   ; q = 0,1,2 (n  m)  3 5 θ0 = 3 = 60ο ; θ1 = 3 = 180ο ; θ2 = 3 = 300ο ; with the real axis. The point of intersection of the asymptotes on the real axis is called centroid σ. sum of real parts of poles - sum of real parts of zeros Centroid = σ = number of poles - number of zeros (0  2  4)  (0) σ = = -2 (3  0) Step 4: Determine breakaway and break-in point 𝐾 𝐶(𝑆) 𝑆(𝑆+2)(𝑠+4) The closed loop transfer function = 𝐾 𝑅(𝑆) 1+𝑆(𝑆+2)(𝑆+4) 𝐾 = 𝑆(𝑆+2)(𝑆+4)+𝐾 The characteristic equation is S(S+2) (S+4) +K = 0 S(S2+4S+8)+K = 0 S3+6S2+8S+K = 0 ∴ K = - S3- 6S2- 8S On differentiating with respect to K 𝑑𝐾 = −3𝑆 2 − 12𝑆 − 8 𝑑𝑆 𝑑𝐾 Put =0 𝑑𝑆 17 | P a g e −3𝑆 2 − 12𝑆 − 8 = 0 −12 ± √122 − 4𝑋3𝑋8 ∴𝑆= = −0.845 𝑜𝑟 − 3.154. 2𝑋3 For K: When S =-0.845, the value of K is given by K =−[(−0.845)3 + 6(−0.845)2 + 8(−0.845)] = 3.08 Since K is positive and real for S =-0.845, this point is the actual breakaway point. When S =-3.154 ,the value of K is given by K =−[(−3.154)3 + 6(−3.154)2 + 8(−3.154)] = −3.08 Since K is negative and real for S =-3.154, this point is not actual breakaway point. Step 5: Determine points on the root-locus crossing imaginary axis The characteristic equation is S(S+2) (S+4) +K = 0 S3+6S2+8S+K = 0 Put s =jw (𝑗𝑤)3 + 6(𝑗𝑤)2 + 8𝑗𝑤 + 𝐾 = 0 −𝑗𝑤 3 − 6𝑤 2 + 8𝑗𝑤 + 𝐾 = 0 Equate real and imaginary part to zero −𝑤 3 + 8𝑤 = 0 & −6𝑤 2 + 𝐾 = 0 ∴ −𝑤 2 = −8𝑤 ; −6𝑤 2 = −𝐾 𝑤 = √8 = ±2.8 ; 𝐾 = 6𝑋8 = 48 The crossing point of root locus is ±2.8. The value of K at this point is 48.This is the limiting value of K for stability. To find the value of K corresponding to δ =0.5. Given δ = 0.5. Let α = cos−1 𝛿 = cos−1 0.5 = 60° Draw a line OP such that the angle between the line OP and negative real axis is 60°.The meeting point of the line OP and the root locus gives the dominate pole sd. 18 | P a g e Let Ksd be the value of K corresponding to the point s= sd 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑣𝑒𝑐𝑡𝑜𝑟 𝑙𝑒𝑛𝑔𝑡ℎ 𝑓𝑟𝑜𝑚 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑠 𝑎 𝑙1 𝑋𝑙2 𝑋𝑙3 Ksd =𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑙𝑒𝑛𝑔𝑡ℎ 𝑓𝑟𝑜𝑚 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑧𝑒𝑟𝑜𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 = = 7.96 = 8 𝑎 1 Exercise Problems 1. Draw the Root locus diagram for the open loop system K G(s) = s(s + 1)(s + 2) 2. Draw the Root locus diagram for the open loop system K G(s) = s(s + 3)(s + 6) Applications of Root locus: i. The Root Locus Plot technique can be applied to determine the dynamic response of the system. ii. This method associates itself with the transient response of the system and is particularly useful in the investigation of stability characteristics of the system. iii. It can also be used to determine the stability boundaries of the system. iv. Selection of suitable parameters may be made using the root locus analysis. 19 | P a g e