7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM


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1 ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s) to s Product of lengths of vectors drawn from zeros of G( s)h( s) to s (7.8) With reference to Fig. 7., we draw the constant damping ratio line θ cos (.77) ( s + 5) which meets the root locus of G (s) ( s+ + j)( s+ j) at s 5 + j5 and s 5 j5. The value of at the closed loop poles s is found as (see Fig. 7.) : L p. L p /L z (3.)(9.5)/5. s Lp p Imag axis of s z Lz Lp p s 8 Real axis of s Fig. 7.. Computation of at a specified rootlocation s on the locus diagram 7. STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM In order to find the roots of the characteristic equation graphically on the plane of the root, we shall present a step by step procedure incorporating the properties listed above by considering some illustrative examples. Example 7. Consider a control system with loop gain L(s) given by ( s + ) L(s) G(s)H(s) s( s+ )( s+ 3) We are interested in determining the locus of roots for positive values of in the range.
2 INTRODUCTION TO CONTROL ENGINEERING Step : Writing the characteristic equation using the loop gain. The characteristic equation for this example is given by + ( s + ) s( s+ )( s+ 3 ) (7.9) In general the characteristic equation should be written using the loop gain L(s) as F(s) + L(s) and the equation be rearranged, if required, such that the parameter of interest,, appears as the multiplying factor in the form: + L (s) (7.) Step : Write the polynomial in the factored form of poles and zeros as follows : F(s) + L(s) + m Π ( s+ zi ) i (7.) n Π ( s + p ) For this example, poles and zeros are already factored. There are 3 finite poles located at s p, s p, s p3 3 and one finite zero located at s z and two zeros at infinity. Step 3: Locate the poles and zeros on the splane on a graph paper. We note that Equation (7.) may be rewritten as j j Π (s + p j ) + Π (s + z i ). (7.) j n m i For this example, we have F(s) s(s + ) (s +3) + (s +) So, when, the roots of the characteristic equation are the poles of F(s), which was mentioned in the properties. Also, when, the roots of the characteristic equation are the zeros of F(s). The locus of the roots of the characteristic equation F(s), therefore, begins at the poles of F(s) and ends at the zeros of F(s) as increases from to infinity. So, the root locus starts from,, and 3 for and terminates at,, for (vide Fig. 7.7). Step : Determine the number of separate loci. In this example, there are three branches of the root loci, of which one terminates at the finite zero located at s z while two branches terminate at +. In general, with n finite poles and m finite zeros and n > m, we have N n m branches of the root locus approaching the N zeros at infinity. Step5: The root loci is symmetrical with respect to the horizontal real axis because the complex roots must occur as complex conjugate pairs. Step : Locate the segments of the real axis that are part of root loci. The section of the real axis will be part of the root locus if the count of real axis poles and zeros looking to the right is an odd number. So, the portion of the real axis lying between and 3 and between and are part of root loci.
3 ROOT LOCUS TECHNIQUE 3 Step 7: Asymptotes: In this example, the number of finite pole, n 3, and number of finite zero, m, so that n m branches of loci approach the two zeros at infinity asymptotically and makes angles θ k. Therefore, from Equation (7.) we find θ 9, for k and θ 7, for k. In general, if the number of finite zeros of F(s) are m and the number of finite poles are n, then N n m branches of root loci terminate at zeros at infinity as approaches infinity. These branches approach the zeros at infinity asymptotically and they meet the real axis at σ, which is computed, for this example, as: σ Σ real parts of poles of G( s) H( s) Σ real parts of zeros of G( s) H( s) (7.3) n m ( 3) ( ).5 3 Step 8: Determine the point at which the locus crosses the stability boundary  the imaginary axis for continuous system. The actual point at which the root locus crosses the imaginary axis is found with the help of Routh table (vide Section 5. in Chapter 5). For this example, the characteristic equation is given by F(s) s(s + ) (s + 3) + (s + ) s 3 + 5s + ( + )s + The Routh array becomes s 3 + s 5 s s o From the Routh table, the conditions of stability are: () >, implying > () 3 >, implying < 3 At 3, the roots lie on the imaginary axis and their values are obtained from the auxiliary equation 5s +, yielding a solution of s ± jω ± j. Step 9: Determine the breakaway point on the real axis (if any). For the present case, we can write Now s ( s + )( s + 3) F s + (s) (7.) 3 d s 3s s 3 ds ( s + ) Setting the above derivative to zero, we get s 8.57, s.587 and s 3.85 of which s 3.85 is the solution with a corresponding.8, whereas the other two solutions corresponds to negative values of, respectively, as can be verified by substitution in Equation (7.).
4 Asymptotes INTRODUCTION TO CONTROL ENGINEERING 8 3 Root Locus 3 Asymptotes Imag part of s Real part of s ( s + ) Fig. 7.7 Root loci of G(s)H(s) s ( s + )( s + 3) Step : Determine the angles of departure from complex poles and angles of arrival at complex zeros of the root locus. This rule is not applicable for this example, since all the open loop poles and zeros are real. Step : We can determine the parameter value s at a specific root s s on the locus using the magnitude requirement of relation (7.). The magnitude requirement at s s is s n Π ( s + p ) j m s Π ( s + z ) i The parameter s can also be computed graphically as follows: s i j (7.5) s n Π L j m Π L i i j (7.) where L j and L i are vector lengths of the search point s s measured from pole p j and zero z i. The complete locus, based on the above data, may be drawn as shown in Fig It is to be noted that all the steps discussed above may not be executed for a given problem. For instance, step was not needed for drawing the root loci in Fig. 7.7 but will be necessary in the next example. Example 7. We consider the loop gain G(s)H(s) ss ( + )( s+ + j)( s+ j) and follow the step by step procedure to obtain the root locus diagram for < <.
5 ROOT LOCUS TECHNIQUE 5 Step : Characteristic equation: The characteristic equation is given by + ss ( + )( s+ + j)( s+ j ) Step : Poles and zeros of loop gain : There are four finite poles for the loop gain located at p, p, p 3 + j and s j and no finite zero such that n, m and N n m. Step 3: The poles are located on a graph paper as shown in Fig Root locus Asymptotes 3 p 3 Breakaway point at s.9 for 7. Asymptotes Imaginary axis crossing at s j.58 for. Imag part of s Asymptotes p p 5 p Asymptotes Imaginary axis crossing at s j.58 for. 3.5 Real part of s Fig Illustration of the steps for root locus plot for Example 7. Step : Number of branches of the root loci: Since, maximum (n, m), there will be branches of root loci and all four will terminate at zeros located at infinity. Step 5: The root loci is symmetrical with respect to the horizontal real axis because the complex roots must occur as complex conjugate pairs. Step : Real axis root locus: For this example, the portion of the real axis lying between and is part of root loci. Step 7: Asymptotes: The angles of the asymptotes are found from the relation θ k ( k + ) 8 ( k + ) 8 n m Therefore, we have θ 5, θ 35, θ 5 and θ 3 35 corresponding to k,,, and 3 respectively. The asymptotes intersect on the real axis at σ (see Fig. 7.8) and are found as σ.5
6 INTRODUCTION TO CONTROL ENGINEERING Step 8: Points at which the locus crosses the imaginary axis. For this example, the characteristic equation is given by F(s) s + s 3 + 8s + s + The Routh table for the above characteristic equation is shown below: where a (9/3) s 8 s 3 s /3 s a s For stability the following conditions must be satisfied: () > () <. The roots will lie on the imaginary axis for. and the imaginary axis roots are found from the auxiliary equation (/3)s +. The values of s are found to be s m jω m j.58. Step 9: Breakaway points: From step 8 the characteristic equation is rearranged as: (s + s 3 +8s + s) F (s) Therefore, we find d ( s 3 8s ds 5s + ), and by setting this derivative to zero, we get the following values of breakaway points: s.938, s j.73 and s.789 j.73 The value of s.938 is the only permissible solution for the breakaway point lying on the real axis and the corresponding value of is found to be 7. by substituting s.938 in the characteristic equation. The other solutions for the breakaway points are not acceptable as they produce complex values of. Step : The angles of departure from complex poles. By taking the search point s s very close to the complex pole p 3 + j and following the procedure illustrated in Fig. 7.3, will produce an angle of departure θ 3 of 5. Similarly, by taking a search point very close to the complex pole p j, the angle of departure θ is found to be 35 (see Fig. 7.8). Using the above information we can draw the complete root loci diagram shown in Fig.7.9.
7 ROOT LOCUS TECHNIQUE 7 Root Locus Imag part of s Real part of s Fig. 7.9 Root locus diagram for G(s)H(s) s ( s + )( s + + j)( s + j) Example 7.3 We now consider the following transfer function of a system to illustrate the root loci plot F(s) s(s + ) (s + 8s + 8) + (s + 5) We divide the above equation by the factors not involving to get + ( s + 5) s( s+ )( s + 8s + 8) Step : The equation above is the characteristic equation. Step : The finite poles are at p, p, p 3 + j8 and p j8 and one finite zero is located at z 5. So we have n, m and N n m 3. Step 3: The poles and the zero are located on a graph paper as shown in Fig. 7.. Step : Since, maximum (n, m), there will be branches of root loci and three will terminate at 3 zeros located at infinity and one branch will terminate at z 5. Step 5: The root loci is symmetrical with respect to the horizontal real axis. Step : The portion of the real axis lying between and is part of root loci. Step 7: The angles of the asymptotes are found from the relation: θ k ( k + ) 8 ( k + ) 8 n m Therefore, we have θ 5, θ 35, θ 5 and θ 3 35, corresponding to k,,, and 3 respectively. The asymptotes intersect on the real axis at σ given by σ ( 5 ).7
8 8 INTRODUCTION TO CONTROL ENGINEERING Step 8: For this example, the characteristic equation is simplified as: F(z) s + s 3 + 9s + ( + )s + 5 Therefore, the RouthHurwitz array is obtained as : where A s 9 5 s 3 + s 8 5 s A s So, we have the following conditions for stability: () < 8 () 8 < (3) > For finding the values of for which the loci crosses the imaginary axis in the splane, the inequality in condition () be replaced by equality to get the positive values of 3.55 for which the root loci will cross the imaginary axis. Putting this value of in the auxiliary equation corresponding to the s row in the RouthHurwitz table we get: s This gives the frequencies at which the root loci crosses the imaginary axis as s ± jω ± j7.7 Step 9: The breakaway(breakin) points are given as the solution of 3 d ( 3s + s + s + 9s+ 8) ds ( s + 5) which are s.89, s 7., s j5. and s.399 j5. The solutions s and s satisfy the phase conditions for positive values of and the corresponding values of are 8. and 5 respectively, whereas s 3 and s do not satisfy the phase conditions for real values of and are rejected. We note that the point on the root locus corresponding to 5 is a breakin point between a zero at and a finite zero at 5. Step : The angles of departure from complex poles p 3 and p are found as was illustrated in Fig These are given by: θ 3 7. and θ 85. Using the above information, the complete root loci may be drawn for > and is shown in Fig. 7..
9 ROOT LOCUS TECHNIQUE p 3 Root locus Imag axis 5 z.7 p p p Real axis Fig. 7. Root loci of F(s) s(s + ) (s + 8s + 8) + (s + 5) Example 7. We now consider another loop transfer function for illustrating the root locus plot. Let G(s)H(s) ss ( + )( s+ j3)( s+ + j3) Step : The characteristic equation is given by: + s( s+ )( s+ j3)( s+ + j3 ) Step : There are four finite poles for the loop gain located at s p, s p, s p3 + j3 and s p j3 and no finite zero. So we have n, m and N n m. Step 3: The poles are located on a graph paper as shown in Fig. 7. Step : Since, maximum (n, m), there will be branches of root loci and all four will terminate at zeros located at infinity. Step 5: The root loci is symmetrical with respect to the horizontal real axis. Step : The portion of the real axis lying between and is part of root loci. Step 7: The angles of the asymptotes are found from the relation (7.). Therefore, we have θ 5, θ 35, θ 5, and θ 3 35, corresponding to k,, and 3 respectively. The asymptotes intersect on the real axis at σ given by σ
10 INTRODUCTION TO CONTROL ENGINEERING Step 8: Points at which the locus crosses the imaginary axis in the splane. For this example, the characteristic equation is given by F(z) s + s 3 + s + s + Therefore, the RouthHurwitz array is where A 8 9 s s 3 s 9 s A s From the first column of the RouthHurwitz Table, we find the following conditions for stability: () > () < 5 For finding the values of for which the loci crosses the imaginary axis in the splane, the inequality in condition () be replaced by equality to get value of 5 for which the root loci will cross the imaginary axis. Putting this value of in the auxiliary equation corresponding to the s row in the RouthHurwitz table we get: 9s + 5 This gives the frequencies at which the root loci crosses the imaginary axis as s ± jω ± j.3 Step 9: The breakaway points are given as the solution of d (s 3 + s ds + 8s + ) which are s, s + j and s 3 j All the above three values of s are breakaway points since they satisfy the phase conditions. The corresponding values of are 9 and 5 respectively. Step : The angles of departure from complex poles s 3 and s are found as before. These are given by: θ 3 9 and θ 9 Using the above information we draw the complete root loci shown in Fig. 7..
11 ROOT LOCUS TECHNIQUE Root locus 5 5 Imag axis Real axis Fig. 7. Root locus of G(s)H(s) s( s + )( s + j3)( s + + j3) 7.5 ROOT LOCUS DESIGN USING GRAPHICAL INTERFACE IN MATLAB The root locus is a powerful design tool of control systems in classical approach. Since, by varying the forward path gain, the location of the closed loop poles can be varied, it affects the overshoot, rise time, settling time as well as steady state error. However, we can not set all the above performance measures to specified values simultaneously by varying only a single parameter. The ability to change along with the freedom of introducing compensator poles and zeros (vide Chapter 8) in the structure of control systems, greatly enhances the capability of a control engineer to meet more than one design specifications simultaneously. The design of compensators using root locus technique has been taken up in Section 8.3. of Chapter 8. In the example below, we shall discuss the use of rltool in MATLAB for the selection of to meet a prescribed constraint on overshoot, settling time, and natural frequency of a closed loop system from its root locus diagram. Example 7.5 Draw the root locus diagram of a plant with loop gain G(s)H(s) given below, and set the value of so that the step response exhibits an overshoot less than 5%, when the loop is closed. G(s)H(s) (s + ) / s(s + ) (s + 8); In the MATLAB command window enter >> gh zpk([ ], [ 8],); Enter >> rltool, The GUI SISO Design for System Feedback config opens with the following feedback configuration: + F C G H Fig. 7. Feedback configuration in MATLAB GUI rltool
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