Automatic Control Systems. Part III: Root Locus Technique

Transcription

1 PDH Coure E40 Automatic Control Sytem Part III: Root Locu Technique By Shih-Min Hu, Ph.D., P.E. Page of 30

2 PDH Coure E40 VI. Root Locu Deign Method Since the performance of a cloed-loop feedback ytem can be adjuted by changing one or more parameter, the location of the root of the characteritic equation hould be evaluated. In the deign and analyi of control ytem, control engineer often need to invetigate a ytem performance when one or more of it parameter vary over a given range. Referring to the tranfer function of a typical negative feedback cloed-loop ytem a dicued previouly, namely, () () C R () () H() G (6-) + G The characteritic equation of the cloed-loop ytem i obtained by etting the denominator polynomial of the tranfer function, C( ), to zero. Therefore, the root R() of the characteritic equation mut atify () H() 0 + G (6-) Let the open-loop tranfer function ( ) H( ) () H() G () () KN D N () () K G expre a m m [ + b + + b + b ] n + a n + + a m n + a n m (6-3) where and D are finite polynomial and K i the open-loop gain. Moreover, m i the degree of N () and n i the degree of D ( ). Equation (6-) can be replaced a follow () () KN + 0 (6-4) D Furthermore, Equation (6-4) can be repreented with () () K N D (6-5) The root locu i the locu of value of for which Equation (6-5) hold for ome poitive value of K. Namely, if K > 0, and () () K N D () () (6-6) N ( k + ) π (or odd multiple of 80 ) (6-7) D where k0, ±, ±, ±3,. It i worth mentioning that the portion of the root locu when K varie from - to 0 ( K < 0 ) i called Complementary Root Locu (CRL). For K > 0, the portion of Root Page of 30

3 PDH Coure E40 Locu i called Direct Root Locu, or imply, Root Locu (RL). Thi coure will only dicu the part that K > 0. If K doe not appear a a multiple factor of G ( ) H( ) a hown in (6-3), one may need to rearrange the function into the following form: () D() + KN() 0 F (6-8) Example 6-: Conider the open-loop tranfer function of a control ytem i () D ) D ( ) KN( ) K G() H() ) ( + )( ( 3 + K) + 5 ( + )( + ) K 0 K ( + ) ( + ) + () H() G + ( 3 + K) ( + )( + ) Find N and ( in term of +. Solution: The characteritic equation of the cloed-loop ytem i Then, ( ) ( )( ) Divide both ide by the term that do not contain K, namely, ( + )( + ) Therefore, N () and 3 () ( + )( + ) D + 5 The open-loop tranfer function of a cloed-loop ytem may ometime expre in term of the zero and pole. In thi cae, Equation (6-3) can be expreed a () H() G () () ( )( ) ( ) ( )( ) ( ) KN K + z + z + z m (6-9) D + p + p + p where the zero ( z, z,, z m ) and pole ( p, p,, pn ) of G H are real or in complex-conjugate pair. Example 6-: For the open-loop tranfer function of a given ytem () H() G n K( + ) ( + )( + 3) () () Page 3 of 30

4 PDH Coure E40 Locate the zero() and pole() on the -plane. Solution: There are one zero and two pole for the open-loop tranfer function, namely, and Zero at ( z ), Pole at ( p ), Pole at 3 ( p 3 ). Conventionally, one ue a for a pole and a ο for a zero. The pole-zero map for the given ytem can be located on the -plane a hown below. Example 6-3: Conider the characteritic equation of the cloed-loop ytem lited below, ( + )( + 3) + K( + ) 0 Locate the pole and zero of the open-loop tranfer function. Solution: Similar to Example 6-, one can divide both ide by ( + )( + 3) and obtain the characteritic equation a follow: Page 4 of 30

5 PDH Coure E40 K( + ) ( + )( + 3) + 0 Therefore, the open-loop tranfer function can be obtained, () H() G K( + ) ( + )( + 3). Now it i clear that there are one zero and three pole for the open-loop tranfer function, namely, and Zero at ( z ), Pole at 0 ( p 0 ), Pole at ( p ), Pole 3 at 3 ( p 3 3 ). The pole-zero map for the given ytem can be located on the -plane a hown below. Number of Branche: A one may expect, the number of branche of the Root Locu i the number of pole of the open-loop tranfer function of the given ytem. Conequently, the number Page 5 of 30

6 PDH Coure E40 of branche equal to the order of the polynomial D ( ), n. All the branche tart at pole (K 0) and approach zero (K ) of the open-loop tranfer function. If there are more pole than zero ( n m > 0 ), then, n m branche go to. A mentioned earlier, thi coure only dicue the Root Locu for K > 0. Root Locu on the Real Axi: The Root Locu on a given ection of the Real Axi can be found in the ection only if the total number of pole and zero to the right of the ection i an odd number. For the ytem given in Example 6-, it Root Locu will be in the ection between and ince there i no pole or zero (total of, an even number) to the right of, but there i one pole (at ) to the right of. Similarly, one can expect there i no RL in the ection between and 3 ince there are one pole and one zero to the right of 3. Finally, one can expect RL i in the ection on the left of 3 ince there are a total of three pole/zero (two pole and one zero for thi ytem) to the right of thi ection. To ummarize the dicuion, Figure 6- how the ection where RL on the real axi. By analogy, Figure 6- how the ection where RL on the real axi for the ytem given in Example 6-3. Im[] RL on Real Axi -3 - RL on Real Axi - Re[] Fig. 6-. Section of real axi with Root Locu for the ytem given in Example 6-. Im[] -3 RL on Real Axi - - RL on Real Axi Re[] Fig. 6-. Section of real axi with Root Locu for the ytem given in Example 6-3. Page 6 of 30

7 PDH Coure E40 Aymptote() of Root Locu: The propertie of the Root Locu when K in the -plane are important to know. If n m > 0, n m branche are approaching to infinity in the -plane when K. In the other word, for large ditance from the origin in the -plane, the branche of Root Locu approach a et of traight-line aymptote. For intance, a ytem with n 3 (the number of pole) and m (the number of zero), then, a K increae, the Root Locu of the two pole would approach to the two zero while the third pole would approach to the negative infinite of the real axi. In thi cae, there i only one ( n m ) aymptote. Another intance, a ytem with n and m 0 ( pole and no zero), a K increae, the root locu of the pole would approach the poitive and the negative infinite of the imaginary axi. In thi cae, there are two ( n m ) aymptote. Thee aymptote from a point in the -plane on the real axi called the center of aymptote, or Centroid. The Centroid can be calculated a the um of pole and zero divided by n m, namely, n m pi zi pole zero i i α (6-0) n m n m The angle between the aymptote and the real axi can be calculated a follow: A π + kπ n m φ rad, k 0,,,, n m ( K > 0) (6-) Example 6-4: What i the Centroid of the ytem given in Example 6-3? Solution: There are three pole ( p 0, p and p 3 3 ) and one zero ( z ) for the given ytem. α ( ) ( ) 3 4 The Root Locu for the given ytem can be obtained with Matlab a hown below. A one can ee clearly, when K increae the pole at 0 approache to the zero at while the other two pole approach to the two aymptote with it Centroid at. The angle of aymptote are π and π radian, or 90 and 90, repectively. Page 7 of 30

8 PDH Coure E40 A one may expect, the pole and zero are not neceary located on the real axi only. For ome pole and zero, they may be complex conjugate pair. Angle of Departure (from Pole) and Angle of Arrival (at Zero): The angle of departure (arrival) of the RL at a pole (zero) of G() H() denote the behavior of the RL near that pole (zero). Equation (6-7) can be expanded a follow: G m () H() ( + zi ) ( + p j ) ( k + ) π i n j (6-) where k0, ±, ±, ±3,. Since the direction of travel of the Root Locu departing from the pole() and arriving to the zero(), the angle() of departure aociated with the pole() and the angle() of arrival aociated with the zero() are important. The angle of departure of the locu from a pole and the angle of arrival of the locu at a zero can be determined from the phae criterion. The angle of locu departure from a pole i the difference between the net angle due to all other pole and zero and the criterion angle of ( k + ) π, or imply π ( or 80 ), and imilarly for the locu angle of arrival at zero. The angle of departure (or arrival) i particularly of interet for complex pole (and zero) becaue the information i helpful in completing the Root Locu. p p θ dp To find the angle of departure for a particular pole, for intance, p, pick, then, the angle of departure for,, can be calculated a the angle of the openloop tranfer function, θ 80 + G p H pole at. Namely, p dp ( ) ( p ), ignoring the contribution from the Page 8 of 30

9 PDH Coure E40 ( ) ( p + z ) + ( p + z ) + + ( p + z ) θ + ( p + p ) + + ( p + p ) 80 m dp n To find the angle of arrival for a particular zero, for intance, z, pick z, then, the angle of arrival for,, can be calculated a the angle of the open-loop z θ az tranfer function, θ 80 G z H z, ignoring the contribution from z. Namely, θ az az ( ) ( ) ( z + z ) + + ( z + z ) ( ( z + p ) + ( z + p ) + + ( z + p )) + 80 m n Example 6-5: Conider a cloed-loop ytem with an open-loop tranfer function which ha two zero at 3 + 4j pole at ( p ), 3 + () H() G ( i.e., z 3 4j) i.e., j ( i.e., p 3 j) K( + 3 4j)( j) ( + )( + 3 j)( + 3+ j) and 3 4j i.e., z 3 + 4j, and three, and 3 j i.e., p3 3 + j. Find ( ) ( ) the angle of departure for p, p and p 3, and the angle of arrival for z and z. Solution: The angle of departure from the pole at ( i.e., p ) Since there are three pole and two zero, the Root Locu would have three branche with two branche tarting from two of the complex pole and arriving at the two complex zero while the pole on the real axi, p, approache to the infinity of the negative Real Axi. Therefore, the angle of departure for p i 80. Or, it could be calculated a follow: + 3 4j j θ j j 80 ( ) ( ) ( ) dp ( ) ( ) ( 4j) + ( + 4j) ( θdp + ( j) + ( + j) ) θ + ( 45 ) ( ) 63 dp θdp which give the ame anwer. ( ) The angle of departure from the pole at 3 + j i.e., p 3 j : + j+ 3 4j j+ 3+ 4j 3+ j+ + θ j+ 3+ j 80 ( 3 ) ( ) ( ( ) dp ( )) ( j) + ( 6j) ( ( + j) + θdp + ( 4j) ) + 90 ( 35 + θ + 90 ) dp θdp The angle of departure from the pole at 3 j ( i.e., p3 3 + j) : + 3 j+ 3+ 4j ( 3 j ( j + 4j + θ ) ( 3 j+ 3 4j) ( ) ( + ) + ( 3 j+ 3 j) + θdp ) ( j) + ( j) ( ) ( ) dp 3 Page 9 of 30

10 PDH Coure E40 ( θ ) dp θdp 3 ( ) ( ) ( ( ) ( ) ( )) ( 8j) ( ( + 4j) + ( j) + ( 6j) ) The angle of arrival to the zero at 3 + 4j i.e., z 3 4j : j j 3 + 4j j+ 3 j j+ 3 + j 80 θaz θaz θaz ( ) ( 3 ) az ( ( ) ( ) ( )) ( 8j) + θaz ( ( 4j) + ( 6j) + ( j) ) ( 6.6 ) + ( 90 ) + ( 90 ) 6. 6 The angle of arrival to the zero at 3 4j i.e., z 3 + 4j : 4j+ 3 4j + θ 3 4j j+ 3 j + 3 4j+ 3 + j θaz Root Locu for the given ytem howing the angle of departure and arrival Page 0 of 30

11 PDH Coure E40 Angle of departure from 3 + j Angle of arrival at 3 + 4j Example 6-6: Conider a cloed-loop ytem with an open-loop tranfer function () H() G K( + ) ( + j)( + + j) What are the angle of departure from the pole at + j and the angle of arrival at the zero at? Solution: The angle of departure from the pole at + j ( i.e., p j) : ( + j+ ) ( θ + ( + j+ + j)) 80 dp ( θ + j ) ( + j) ( ) 80 θdp dp The angle of arrival at the zero i.e., z : + j j 80 θaz θaz θaz ( ) ( ( ) ( )) ( ( j) + ( + j) ) 80 + ( 35 ) Thoe angle can be eaily verified from the Root Locu hown below. Page of 30

12 PDH Coure E40 Croing of the Imaginary Axi: A dicued previouly, ome cloed-loop ytem would be table only with certain range of K, and that could be determined by Routh-Hurwitz Criterion. Alo, when a ytem i going untable from a table region, it Root Locu would interect with the Imaginary Axi (i.e., Root Locu traveling from the left-half-plane of -plane to the right-half-plane). The point where the Root Locu interect/croe the Imaginary Axi of the -plane, and the correponding value of K, may be determined by the mean of Routh-Hurwitz Criterion. Example 6-7: Conider a cloed-loop ytem with an open-loop tranfer function () H() K G ( + 3)( + + ) Where the Root Locu interect the Imaginary Axi and at what K value? Solution: K K The characteritic equation: ( )( ) 0 Routh Table Page of 30

13 PDH Coure E b c d 8 6 b K b 5 5 b 5 K 0 5 K K K c d c b b c K To have a table ytem, K > 0 and K > 0. Therefore, 0 < K < 8.6. When K8.6, the Root Locu interect the Imaginary Axi which can be obtained from the equation tarting with term in the Routh Table, namely, b + b 0 with K ( j.0954)( + j.0954) 0 Therefore, the two branche of the Root Locu interect the following two conjugate point at ±.0954 rad ec. The Root Locu for the given ytem i hown below. Page 3 of 30

14 PDH Coure E40 Breakaway Point: A Breakaway Point i a point on the Real Axi where two or more branche of the Root Locu depart from or arrive at the Real Axi. Figure 6-3 and 6-4 how two branche leaving and coming onto the Real Axi, repectively. Im[] Breakaway Point Re[] Fig Breakaway Point with two branche leaving the Real Axi. Page 4 of 30

15 PDH Coure E40 Im[] Breakaway Point Re[] Fig Breakaway Point with two branche coming onto the Real Axi. The Breakaway Point on the Real Axi can be evaluated graphically or analytically. Conider a unity feedback cloed-loop ytem with an open-loop tranfer function () H() G K ( + )( + 4) The Root Locu i hown in Figure 6-5, and the Breakaway Point i at 3. Fig Root Locu of ytem with open-loop tranfer function G() H() K. ( + )( + 4) Page 5 of 30

16 PDH Coure E40 The Breakaway Point() of the Root Locu of + G( ) H( ) + KG ( ) H( ) 0, or K (6-3) G() H() mut atify dk 0 (6-4) d For the given ytem, dk d or d K ( + )( + 4). Therefore, G () H () ( ) ( + 6) 0 d 3 i the Breakaway Point, ame a obtained graphically from Figure 6-5. It i important to point out that the condition for the Breakaway Point given by Equation (6-4) i neceary but not ufficient condition. In other word, all breakaway Point() mut atify Equation (6-4), but not all olution of Equation (6-4) are Breakaway Point. Example 6-8: For the ame ytem a given in Example 6-7, find the Breakaway Point. Solution: K The open-loop tranfer function i G () H(). ( + 3)( + + ) Therefore, the K can be expreed a K G () H () ( + 3)( + + ) 4 3 ( ) dk d d d 4 3 ( + 3)( + + ) ( ) Therefore,.886, j and j However, the Breakaway Point i located only at.886 where the Root Locu on that ection of the Real Axi, a hown in the Root Locu for Example 6-7. Example 6-9: For the ame ytem a given in Example 6-6, find the Breakaway Point. Solution: Page 6 of 30

17 PDH Coure E40 The open-loop tranfer function i G() H() Therefore, the K can be expreed a K G K( + ) ( + j)( + + j) ( + ) K. ( + + ) () () ( ) ( ) ( ) H ( )( ) ) ( )( ) ( )( ) ( + + ) 0 dk d d d ( + ) + ( + ) ( + ) + ( + )( + ) ( + ) ( + ) ( )( ) + 0 ( + + ) Therefore, 3.44 and However, the Breakaway Point i located only at 3.44 where the Root Locu on that ection of the Real Axi, a hown in the Root Locu for Example 6-6. Symmetry of the Root Locu: Another important property of the Root Locu i that a Root Locu i ymmetrical with repect to the Real Axi of the -plane. In general, the Root Locu i ymmetrical with repect to the axe of ymmetry of the pole and zero of G ( ) H( ). One hould oberve uch a property from the plot given throughout in thi coure. Page 7 of 30

18 PDH Coure E40 VII. Guideline for Sketching a Root Locu The detail of the baic propertie of the Root Locu of the cloed-loop ytem have been dicued in the previou ection. In thi ection, a et of tep-by-tep guideline for ketching a Root Locu i ummarized, and a detailed tep-by-tep example i preented to illutrate how to ue thoe guideline to ketch a Root Locu. Rule for the Contruction of the Root Locu: ) Number of Branche: n, where n i the number of pole. ) Direction of Travel: A K, the Root Locu approache zero of open-loop tranfer function. If n m > 0 (m i the number of zero), then, n m branche go to. 3) Root Locu on the Real Axi: If the number of pole plu zero on the right hand ide i an odd number, Root Locu i on the real axi. On the other word, draw Root Locu on the Real Axi to the left of an odd number of pole and zero. 4) Number of Aymptote, Angle of Aymptote and Centroid: Number of aymptote: Angle of aymptote: Centroid: n m The angle() between the real axi and the aymptote() can be calculated a A π + kπ n m φ, k 0,,,,n m ( K > 0) pi z pole zero i i α n m n m Table 6- how the angle() of aymptote() for n m,, 3 and 4, numerically (in radian) and graphically. n m i Page 8 of 30

19 PDH Coure E40 Table 6-. Graphical repreentation of Aymptote (for n-m,, 3 & 4). n m φ Graphical Repreentation of Aymptote A π π π, 3 π, 3 π, π 3 π π 3 π 3 4 π, 4 3π 5π,, 4 4 7π 4 π 4 π 4 6) Angle of Departure and Angle of Arrival: Angle of departure, : θ dp ( ) ( p + z ) + ( p + z ) + + ( p + z ) θ + ( p + p ) + + ( p + p ) 80 θ az m dp Angle of Arrival, θ : az ( z + z ) + + ( z + z ) ( ( z + p ) + ( z + p ) + + ( z + p )) + 80 m 7) Croing of the Imaginary Axi: The location of the Root Locu interect the Imaginary Axi can be obtained from the equation tarting with term in the Routh Table, namely, b + b 0 with K equal the maximum value of the table range obtained by Routh-Hurwitz Criterion. 8) Breakaway Point(): The breakaway point() of the Root Locu of dk + G() H() + KG () H( ) 0, or K mut atify 0. G H d () () 9) Root Locu i ymmetry with repect to the Real Axi. n n Page 9 of 30

20 PDH Coure E40 Example 6-0: A cloed-loop negative feedback ytem ha an open-loop tranfer function of () H() G ( + 0) K ( ) Sketch the Root Locu for K > 0. Solution: There are three pole and one zero, i.e., n3 and m. Therefore, the Root Locu ha three branche (n3), one branch tarting from one of the pole and approaching to the zero while the other two branche tarting from the two ( n m 3 ) remaining pole and approaching to their Aymptote ( n m ) going to infinite when K increae. The pole are located at 0, 3 + j and 3 j while the zero i located at 0. The Root Locu i on the Real Axi between 0 and 0 ince the number of pole() i an odd number on the right hand ide of the zero at 0. The angle of Aymptote can be calculated a follow π π π + kπ, for k 0 φ A 3 n m π + π 3π π, for k 3 The Centroid can be obtained by pi z pole zero i i α n m n m n m i ( 0 + ( 3 j) + ( 3+ j) ) ( 0) 3 A Centroid with a poitive value implie two branche of the Root Locu would travel from the left-half-plane to the right-half-plane of the -plane. Therefore, thoe branche would cro the Imaginary Axi and the location can be obtained a follow: 3 The characteritic equation: ( ) + K( + 0) ( 3 + K) + 0K 0 Routh Table b c 6 b 3+ K 0K ( 3 + K) 0K 78 4K 6 6 c 0K To have a table ytem, 0 and 78 K > 4 K > 0. Therefore, 0 < K < 9.5. Page 0 of 30

21 PDH Coure E40 When K9.5, the Root Locu interect the Imaginary Axi which can be obtained from the equation tarting with term in the Routh Table, namely, 6 + 0K 0 with K ( j5.7009)( + j5.7009) 0 Therefore, the two branche of the Root Locu interect the following two conjugate point at ± rad ec with K9.5. The angle of departure from the pole at 0 (i.e., p 0 ): ( dp ) ( θdp + 3 j j ) ( θ ( 46.3 )) ( + 0) θ + ( j) + ( j) 80 0 ( ) ( ) ( ) dp 80 θdp ( 46.3 ) The angle of departure from the pole at 3 + j ( i.e., p 3 j) : ( dp ) ( 3 + j + θ + 4j ) ( + j+ 0) ( 3+ j+ 0) + θ + ( 3 + j+ 3 + j) 80 3 ( + j) ( ) ( ) 80 θdp 7 dp The angle of departure from the pole at 3 j ( i.e., p3 3 + j) : ( + θdp ) ( 3 j + 4j + θ ) ( j+ 0) ( 3 j+ 0) + ( 3 j+ 3 j) ( j) ( ) ( ) 80 θdp 7 dp 3 ( 46.3 ) ( 90 ) The angle of arrival at the zero 0 θaz θaz ( i.e., z 0) ( ( 0 + 0) + ( j) + ( j) ) 80 ( ( 0) + ( 7 j) + ( 7 + j) ) 80 θaz ( 80 ) + ( 64. ) With all the calculation, one might be able to conclude that there i no Breakaway Point for the given ytem. One might puruit evaluating the Breakaway Point anyway and hould conclude the ame thing. Page of 30

22 PDH Coure E40 K ( ) ( + 0) ( ) ( + 0) ( )( + 0) ( )( ) ) 3 d ( 3 3 )( 0 ) ( )( 0 ) ( ) d ( )( + 0) ( + 0) ( + 0) ( + 0) ( + 0) dk d 3 ( ) Therefore,.969± j and However, there i no Root Locu on the Real Axi at So, no Breakaway Point exit a expected. The Root Locu for the given ytem i ketched below uing the information obtained from following the calculation provided in the Guideline Im[] K Croing jω Axi K j Aymptote K 0-3+j K 0 RL on Real Axi -0-3 K 0-3-j j Centroid K 0 0 -j π π Re[] Aymptote K j Croing jω Axi K The Root Locu obtained by uing MATLAB i hown below to verify the ketched Root Locu. Page of 30

23 PDH Coure E40 Root Locu of ytem with open-loop tranfer function () H() ( + 0) K G. ( ) To how the angle of departure and arrival and the location where the two branche croing the Imaginary Axi, the Root Locu of the ame ytem with a modified cale i hown below. Page 3 of 30

24 PDH Coure E40 Root Locu with a better cale to how the angle of departure and arrival and the location where two branche croing the Imaginary Axi. Alo, a one can eaily oberved that the Root Locu i ymmetrical to the Real Axi. Page 4 of 30

25 PDH Coure E40 Reference: [] Gene F. Franklin, J. David Powell and Abba Emami-Naeini, Feedback Control of Dynamic Sytem nd Edition, Addion Weley, 99 [] Benjamin C. Kuo, Automatic Control Sytem 5 th Edition, Prentice-Hall, 987 [3] Richard C. Dorf, Modern Control Sytem 6 th Edition, Addion Weley, 99 [4] John A. Camara, Practice Problem for the Electrical and Computer Engineering PE Exam 6 th Edition, Profeional Publication, 00 [5] NCEES, Fundamental of Engineering Supplied-Reference Handbook 6 th Edition, 003 [6] Merle C. Potter, FE/EIT Electrical Dicipline-Specific Review for the FE/EIT Exam 5 th Edition, Great Lake Pre, 00 [7] Merle C. Potter, Principle & Practice of Electrical Engineering t Edition, Great Lake Pre, 998 [8] Joeph J. DiStefano, III, Allen R. Stubberud, and Ivan J. William, Feedback and Control Sytem nd Edition, McGraw-Hill, 990 Page 5 of 30

26 PDH Coure E40 Appendix A How to ue MATLAB to obtain the Root Locu Example 6-5: >> num[ 6 5]; >> den[ 7 9 3]; >> [z,p,k]tfzp(num,den); >> z z i i >> P p i i >> pzmap(p,z) Page 6 of 30

27 PDH Coure E40 >> rlocu(num,den) To change cale of the Real Axi and Imaginary Axi to how the angle of departure and arrival >> v[ ]; >> axi(v) Page 7 of 30

28 PDH Coure E40 >> v[ ]; >> axi(v) Example 6-6: >> num[ ]; >> den[ ]; >> rlocu(num,den) Page 8 of 30

29 PDH Coure E40 Example 6-0: >> num[ 0]; >> den[ 6 3 0]; >> [p,z]pzmap(num,den) p i i z -0 >> pzmap(p,z) >> v[- -6 6]; >> axi(v) >> rlocu(num,den) Page 9 of 30

30 PDH Coure E40 >> v[- -6 6]; >> axi(v) Page 30 of 30