Tranfer Function and Impule Repone Solution of Selected Unolved Example. Tranfer Function Q.8 Solution : The -domain network i hown in the Fig... Applying VL to the two loop, R R R I () I () L I () L V() i.e. I () [R L] I () L V()... () I () R I () L I () L I () C 0 0 V() _ () L () C Fig.. I () I () R Uing () in (), C L R C LC I () L LC... () LCR C I () [R L] I () L V() LC ( LCRC) (R L) I () L LC I () V() V() LC LCR R R CR L C LR CL L C I () V() LC LC (R R ) (R R CL) R (-)
Principle of Control Sytem - Tranfer Function and Impule Repone.6 Laplace Tranform of Electrical Network Q. Solution : The -domian network i hown in the Fig.. (a). R C R RC R C Z () E () i E () i R C E () o Z () E () o R C R RC (a) Fig.. (b) T.F. E() o E() Z Z Z R RC R R RC RC R ( RC ) (R R ) R R (C C )
5 Block Diagram Repreentation of Control Sytem Solution of Selected Unolved Example 5. Rule for Block Diagram Reduction Q. Solution : Shifting take off point after G and eperating feedback path we get, G Parallel ( G ) G G R() G G G H H Minor loop H Minor loop R() G G G G GH G G H H R() G G GH G GH GH Serie Minor loop G G G G H G G GH GGH R() G G(G) GH GH GGHH GGH GGGHH (5-)
Principle of Control Sytem 5 - Block Diagram Repreentation of Control Sytem Q. Solution : No erie, parallel combination and no minor feedback loop exit. So take off point before the block of. 0 hifting 0 Parallel R() 0 8 4 Minor loop R() 0 0 8 4 R() 0 0 8 4 After implification 0 After implification (7) ( 0) ( ) R() 0 (7) (0)() 8 4
Principle of Control Sytem 5 - Block Diagram Repreentation of Control Sytem R() ( 7) ( ) ( ) ( 7) ( 4) ( 7) ( 8) ( ) ( ) ( 4) ( 7) ( 8) ( ) ( ) ( 4) ( 8) R() 5 7 4 0 9 5.4 Analyi of Multiple Input Multiple Output Sytem Q.4 Solution : A there are two input, conider each input eparately. Conider R(), auming Y() 0. R() G G R() GG H H Minor feedback loop R() G G G G H i.e. R() G G G G H Now conider Y() acting with R() 0. Now ign of ignal obtained from H i negative which mut be carried forward, though umming point at R() i removed, a R() 0, o we get, To carry forward negative ign G G H Y() Y() G G H ey Point While finding equivalent G, trace forward path from input umming point to output in direction of ignal. While finding equivalent H, trace the feedback path from output to input umming point in the direction of ignal. Now equivalent G G, tracing forward path from input umming point to output. Equivalent H G H tracing feedback path from output to input umming point. While ign of the final feedback i poitive at the input umming point.
Principle of Control Sytem 5-4 Block Diagram Repreentation of Control Sytem Y() G GH G G ( G H ) H itelf i negative Y() G G G H i.e. Y() G G G H Hence the net output i given by algebraically adding it two component, G G R() G Y() G G H Q.5 Solution : For C/R, conider R 0, hence umming point at R can be removed. R G G Parallel G G Minor loop G G 4 G GH C H H Fig. 5. (a) Serie G R G G G GH G 4 C R G C H H Fig. 5. (b) C R G GH where G (G G )G G G H 4 (G G )G G 4 C G H R (G G )G G H G H 4 (G G )G G 4 G H (G G )G G H 4
Principle of Control Sytem 5-5 Block Diagram Repreentation of Control Sytem For C/R, conider R 0 but negative ign of H mut be conidered while removing umming point at R. The parallel combination of G,G and minor loop of G and H can be ued directly. G G G GH R G (Forward path) G 4 eq C H H eq (Feedback path) Fig. 5. (c) G eq G 4 H eq (G G )G H ( G H ) C R G eq G H eq eq R G 4 (G G )G H G4 ( G H ) G eq H eq Fig. 5. (d) C C R G 4( G H ) G H (G G )G G H 4 Q.6 Solution : i) With N() 0 block diagram become Minor loop R() E() 4 0 () 0.5 Minor feedback loop 0 () 0 () 0.5 0 5 0 6 Aume output of econd umming point a X(), Hence E() R()... (i) 0 ( 4) X() 6... (ii)
Principle of Control Sytem 5-6 Block Diagram Repreentation of Control Sytem R() E() 4 0 6 4 R() E() X() 4 0 6 X() E() R()... (iii) 4 Subtituting value of X() and R() from (i) & (ii) in (iii) we get, 6 E() 0 ( 4) 4 E() 4 6 0( 4) ( 4) 4 E() ( 6 0) 0 ( 4) ( 7) ( 4) E() E() 0 ( 7) 6 0 when N() 0 ii) To find R(), we have to reduce block diagram olving minor feedback loop and hifting umming point to the left a hown earlier in (i). So referring to block diagram after thee two tep i.e. 4 R() E() 4 0 6
Principle of Control Sytem 5-7 Block Diagram Repreentation of Control Sytem Exchanging two umming point uing aociative law, R() Block diagram become, 4 Parallel of '' and 4 4 Minor loop G GH 0 ( 4) 6 0(4) 6 0(4) 6 R() 4 0 ( 4) 640 R() 7 0 ( 4) 4 6 40 0( 7) 6 40 iii) With R() 0 block diagram become, 4 0 () 0.5 N() The block of will not exit a R() 0. Similarly firt umming point will alo vanih but tudent hould note that negative ign of feedback mut be conidered a it i, though umming point get deleted. In general while deleting umming point, it i neceary to conider the ign of the different ignal at that umming point and hould not be diturbed. So introducing block of to conider negative ign. 4 0 () 0.5 N()
Principle of Control Sytem 5-8 Block Diagram Repreentation of Control Sytem N() 0 () 0.5 (4) Parallel Two block are in parallel, adding them with ign. 0 ( ) (4)0.5 N() After implification (.5 4) Removing umming point, a ign i poitive no need of adding a block. 0 () N() N() (.54) 0(.54) () Minor loop with G N() 0 (.5 4) ( ) 5 40 ( ) N() ( ) 6 40 5.5 Block Diagram from Sytem Equation Q. Solution : The network can be redrawn in -domain a,
Principle of Control Sytem 5-9 Block Diagram Repreentation of Control Sytem V a() Z () Z () A V () Z () Z 4() V () I () I () Z R R C CR Z R L Z R R C RC Z 4 R4 I () V () V () V a () Z a I () Z V() Z V a ()... () I I Z Z I () Z I ()... () V a () V () Z Z V () a Z V ()... () V () Z4 I ()... (4) Simulating each equation, the complete block diagram i, Shift V () I () Z Z V a() I () Z Z 4 V () Z Shift Z Z Interchange Minor feedback loop V () Z Z Z Z 4 V () Z Z Z Z 4 Z
Principle of Control Sytem 5-0 Block Diagram Repreentation of Control Sytem Minor feedback loop V () Z Z Z Z 4 Z 4 Z V () Z ZZ 4 Z Z4 Z 4 V () Z Z Z Z ZZ 4 Z (Z Z 4) V () ZZ Z Z Z 4 V () V() Z ZZ Z Z Z 4 Z Z Z Z Z Z Z Z Z Z Z Z 4 4 4 Z Z Z Z Z Z 4 Minor feedback loop 4 4 Subtituting the value of Z, Z, Z, Z 4 V () V() R C R R L R L R 4 R R C R 4 R R L R C... An.
6 Signal Flow Graph Repreentation of Control Sytem Solution of Selected Unolved Example 6.5 Maon' Gain Formula Q. Solution : Number of forward path T.F. T T G G G G 4 uing Maon' gain Formula G G G G H H L G GH L GGH G 5 G 4 H H H T G G 5 Individual feedback loop, Loop L andl are non touching loop. L L L L L L L GH 4 L 4 GHH 5 4 [GGH GGH G4H G5HH ] [GGG 4HH ] Conider T, all loop are touching Conider T, all loop are touching (6-)
Principle of Control Sytem 6 - Signal Flow Graph Repreentation of Control Sytem T.F. T T G G G G4 G G5 G G H G G H G4 H G5 H H G G G4 H H R() G G G G 4 G G5 G G H G G H G H G H H G G G H H 4 5 4 Q. Solution : Number of forward path Maon' gain formula, T.F. T T T G H L G H G G G G 4 L GGGGH 4 H G 5 G 4 L GGH 5 4 H T G G G G 4, T G G Individual feedback loop are, L andl are two non touching loop. [L L L ][L L ] 5 4 [ G H G G G G4 H G5 G4 H ][G H G5 G4 H ] G H G G G G H G G H G G G H H 4 5 4 5 4 For T all loop are touching G 5 G G 4 L i non touching to T. H
Principle of Control Sytem 6 - Signal Flow Graph Repreentation of Control Sytem eliminating all loop gain and product from Conider T, [L ] (GH ) GH R() T T G G G G4 G5 G 4 ( G H ) R() G G G G 4 G4 G 5 ( G H ) G H G G G G4 H G5 G4 H G G5 G4 H H Q. Solution : Number of forward path T.F. T... Maon' gain formula G G G G G G 4 G 5 H H H 4 H 5 H L G H L G H L G GGH L 4 G4 H 4 L 5 G5 H 5 T G G G G 4 G5 Individual feedback loop, Combination of two non touching loop, i) L andl ii) L andl5 iii) L andl4 iv) L and L5 v) L andl 5 Combination of three non touching loop, i L, L andl5. [L L L L4 L 5] [LL LL5 LL4 LL5 LL 5][LLL 5] G H G H G G G H G4 H 4 G5 H5 G G H H G G4 H H4 G G5 H H5 G G5 H H 5 G G G G H H G G G H H H Now conidering T G G G G4 G5 All loop are touching to thi forward path hence, R() 5 5 5 5 T G G G G 4 G 5
Principle of Control Sytem 6-4 Signal Flow Graph Repreentation of Control Sytem R() G G G G4 G5 G H G H G G G H G H G H 4 4 5 5 G G H H G G H H G G H H G G H H G G G G H H G G G H H 4 4 5 5 5 5 5 5 5 H 5 Q.4 Solution : Number of forward path T G G4 G5 G 6, T G G 4 G 5 G 6, T G G G G Individual feedback loop are, 4 5 6 G G G 4 G 4 G 4 G H H H L G G H 4 L G G H 4 L G G H 4 [ L L L] No combination of non touching loop G G4 H G G4 H G G4 H Conider T G G4 G5 G 6 All loop are touching, T G G4 G5 G 6 All loop are touching, T G G4 G5 G 6 All loop are touching, T.F. T T T G G4 G5 G6 G G4 G5 G6 G G4 G5 G 6 R() G G G G G G G G G G G G G G H G G H G G H 4 5 6 4 5 6 4 5 6 4 4 4 Q.5 Solution : Number of forward path T G G G G 4, T G G G 6, T G G G G Individual feedback loop 5 G G G G G 4 H H
Principle of Control Sytem 6-5 Signal Flow Graph Repreentation of Control Sytem L G H L G G G G H No combination of nontouching loop 4 L L G H G G G G4 H Conider T G G G G4 All loop are touching T G G G6 All loop are touching T G G G G5 All loop are touching T.F. T T T G G G G4 G G G6 G G G G5 R() G G G G G G G G G G G G H G G G G H 4 6 5 4 Q.6 Solution : Number of forward path T G GG G4 Individual feedback loop G G G 4 H H 4 H L G H L G H4 L G H H 4 G 5 G G G 4 SELF LOOP L 4 G G G4 H L 5 G 5 Combination of two nontouching loop i) L andl ii) L and L iii) L andl5 iv) L and L5 Combination of three nontouching loop. i) L, L andl5 L L L L L L L L L L L L L L L L 4 5 5 5 5 GH G H 4 G4 H G G G4 H G 5 G G H H4 G G4 H H G H G5 G H 4 G5 G G G5 H H 4
Principle of Control Sytem 6-6 Signal Flow Graph Repreentation of Control Sytem All loop are touching to forward path T hence. R() T G G G G 4 R() G G G G4 G H G H4 G4 H G G G4 H G G G H H G G H H G H G G G H G G G H H 4 5 5 4 5 4 5 4 Q.7 Solution : R() t G t G H H R() G t G t H H T G G Individual loop are, G G G G H H L andl L G H L G G L G H i combination of two non-touching loop L L L L L R() G H G G G H G G H H T Conidering T, all loop are touching
Principle of Control Sytem 6-7 Signal Flow Graph Repreentation of Control Sytem R() G G G H G G G H G G H H Q.8 Solution : The ignal flow graph i how below repreenting each umming and take off point by eparate node. R() G G H H Forward path : T G G Individual feedback loop gain are, H G G G G H H L G H L G H H L G GH Nontouching loop : LL GGHH [L L L ] [LL ] All loop are touching to forward path T hence. R() T GG G H G H G G H G G H H By block diagram reduction, eliminate the two minor loop, R() G G R() GG G H G H ( G H )( GH ) H H R() GG ( G H ) ( G H ) GG H ( G H ) ( G H ) GG ( G H ) ( G H ) G G H R() GG G H G H G G H H G G G
Principle of Control Sytem 6-8 Signal Flow Graph Repreentation of Control Sytem 6.7 Application of Maon' Gain Formula to Electrical Network Q. Solution : Convert the given network in it laplace form and aume different loop current and node voltage a hown. C V C V i () R I I R V o () Writing down equation for I, V,I, Vo we get, I (V i V ) C (Vi V o) C V (I I ) R I (V V o ) C (V V o) C... (I)... (II)... (III) V o I R... (IV) Simulating above equation by ignal flow graph V i C I V For equation (I) C I R V I For equation (II) R V C I V o For equation (III) C I R V o For equation (IV) Combining we get ignal flow graph for given network. V C I R V C I R V o C R C
Principle of Control Sytem 6-9 Signal Flow Graph Repreentation of Control Sytem To find T.F. apply Maon' gain formula T T.F. Number of forward path T.F. T T C R C R R R C C Individual loop : L R C L R C L R C Out of three, L and L are nontouching L L L L L R C R C R C R C R C R C R C R C R C R C V o() T V() i All loop are touching to T, V o() V() i R C R C R C R C R C R C R C Q. Solution : Laplace Tranform of the given network i, R V R V i () I () R R 4 I () V o () Equation for different current and voltage are
Principle of Control Sytem 6-0 Signal Flow Graph Repreentation of Control Sytem S.F.G. (I) I R I V V V V i i... (I) R I R V R I S.F.G. (II)V (I I ) R... (II) V R I S.F.G. (III) I V V R R V o... (III) R o I R 4 V o S.F.G. (IV) V o I R4... (IV) Total S.F.G. i a hown in the following figure. V i R I R V R I R 4 V o Ue Maon' gain formula to find V o Vi Vo T Vi Number of forward path T Individual feedback loop are, T R R 4 R R L and L L R R, L R R are nontouching, L R 4 R L L L L L R R R R R 4 R R R R R R R R 4 all loop are touching to T R R R R R R R R 4 R R 4 R R,
Principle of Control Sytem 6 - Signal Flow Graph Repreentation of Control Sytem V V o i T V V o i R R4 R R R R R R R R R R 4 4 Q. Solution : The Laplace domain repreentation of the given network i hown below. I R () I () ( I I ) V () V () i V () i I () C ( I I) R C I () V () o V () o ( I I) The variou branch current are hown, V i () V () I () C Then, I () C V i () C V () V()V i o() R... () and alo, R V() R V () i o... () V() [I () I ()]R RI () RI () I () V o() V () C... () C V o() C V () From thi obtain the equation for V o() a I () equation i already obtained. Note : Write the eparate equation for eparate branch and each element mut be conidered atleat once. V o () V() i C I ()... (4)
Principle of Control Sytem 6 - Signal Flow Graph Repreentation of Control Sytem Hence the ignal flow graph i, R R R V () i C I () R V () I () C V () o C The forward path gain are, T CR T T CR R R The variou loop gain are, L CR L L CR R R The loop L andl are non-touching L L Hence ytem determinant i, [L L L ] [LL ] CR CR CR C R CR For T, all loop touching to T For T, L a L i non touching to T (CR) For T, all loop touching to T According to Maon gain formula, V o() T T T V() i CR ( CR) CR CR CR CR CR CRCR CR CR C R CR V o() V() i C R CR C R CR
Principle of Control Sytem 6 - Signal Flow Graph Repreentation of Control Sytem Q.4 Solution : Laplace tranform of the given network. R V i () R I() C V o () I() V i V R o V o () I() R C S.F.G. for equation () : S.F.G. for equation () : R R I V i V o I C V o... ()... () R Complete S.F.G. i : V i R I R C V o R Ue Maon' gain formula. Number of forward path T R R R C C R C Individual loop, L R C C R R C RC ( R C) [L ] R C A L i touching to T, R C V o() T R C V() i ( R C) R C V o() V() i R C R CR C RC C(R R ) R C R
7 Time Domain Analyi of Control Sytem Solution of Selected Unolved Example 7.8 Analyi of TYPE 0, and Sytem Q. Solution : Expre the given open loop tranfer function in time contant form. G()H() () 5 4 Now, p Lim G()H() 0 Lim G()H() v 0.. 0 ( ) ( ) () (5) (4) 5 4 () 5 4 0 a Lim G()H() 0 0 Now input i, r(t) tt t t The input i combination of three tandard input. A, tep of A, ramp of A, parabolic input of Note that parabolic input mut be expreed a A t. a) For tep of the error i, A e 0 p TECHNICAL PUBLICATIONS (7-) - An up thrut for knowledge
Principle of Control Sytem 7 - Time Domain Analyi of Control Sytem b) For ramp of the error i, e A v c) For parabolic of, the error i e A a 0 Hence teady tate error i, e e e e 0 Q. Solution : From the ytem hown we can write, G() ( ), H() The input i r(t) 0. t i.e. ramp of magnitude 0.. For ramp input v control the error. v Lim 0 e A v 0. Maximum e allowed i 0.005 0.005 0. G()H() Lim 0. ( ) 0. 0.005 0 For any value of greater than 0, e will be le than 0.005. Hence the range of value of for e 0.005 i, 0 < Q. Solution : G() H() ( ) ( 0.4 ) i) e Lim R() 0 G() H(), r(t) 4t, R() 4
Principle of Control Sytem 7 - Time Domain Analyi of Control Sytem 4 Lim 0 ( ) ( 0.4 ) 4 0 4 given Lim 0 4 ( ) ( 0.4 ) e ii) Now for the ame input e 4 and it i deired to have e 0. 0. 4 0 iii) Now input i r(t) 6t R() 6 e Lim 0 6 ( ) ( 0.4) Lim 0 ( ) ( 0.4) Lim 6 0 ( ) ( 0.4) 0 6, 0 given e 6 0 0.6 Q.4 Solution : Solving internal feedback loop we get, 0 ( 4) G() 0 6 ( 4) 4 0 60 0 [6 4] R() 0 (4) 0 (4) 6 Fig. 7.
Principle of Control Sytem 7-4 Time Domain Analyi of Control Sytem G()H() 0 6 4 6 Comparing with tandard form ( T ) ( T )... j ( T ) ( T )... where j type of ytem. i.e. j, Sytem i TYPE ytem Error coefficient : p 0 Lim 0 Lim 6 0 4 6 v a a 0 Lim 0 G()H() Lim 6 0 4 6 Lim 0 G()H( ) Q.5 Solution : G() 0. 00 ( ) ( ) G()H() 00( 0.) 0.0 ( ) ( ) Lim 0 00 0.0 0. ( 0) ( ) ( 0.5) b 0 6 0 6 4 0 6 ( 0.)00, H() 0.0 ( ) ( ) 0.( 0) ( ) ( 0.5) i) For unit tep input p Lim Lim G()H() 0 0 e 0 p 0.( 0) ( ) ( 0.5) ii) For unit ramp input v Lim Lim G()H() 0 0 e 5 v 0. 0.( 0) ( ) ( 0.5) 0.
Principle of Control Sytem 7-5 Time Domain Analyi of Control Sytem Q.6 Solution : When ytem i not in the imple cloed loop form then we can not apply the error coefficient. In uch cae, we have to ue final value theorem i.e. e Lim t e(t) Lim 0 E() In the ytem given E() when R() 0 Now let u find out, o that for unit tep diturbance we can calculate and hence T() E(). When R 0, umming point at R() can be removed and block of `' i to be added to conider ign of the ignal at that umming point. T 0 0 0. 00 0. Shifting umming point to right. Fig. 7. T 0 0 0. 00 0. 0 0. Fig. 7. Combining the two umming point and redrawing the diagram. T() 00 ( 0. ) 00 ( 0. ) Fig. 7.4
Principle of Control Sytem 7-6 Time Domain Analyi of Control Sytem Negative ign of 00 0. can be taken out to change ign of the ignal at the umming point from poitive to negative. T() 00 0. 00 ( 0. ) T() 00 00 ( 0. ) Fig. 7.5 Fig. 7.6 T() 00 00 (00 ) ( 0.) 00 ( 0.) (0. 00 0000) For T(), 00 ( 0.) (0. 00 0000) 00 ( 0.) Now E() (0. 00 0000) e Lim 0 E() e Lim 0 00 ( 0.) (0. 00 0000) Steady tate error 0.0 00 0000 Q.7 Solution : G()H() ( ) ( 4) ( 0) Given G()H() Type ytem, ( ) ( 4) 0 ( 0. 0. ) 0.( ) ( 4) ( 0. 0. )
Principle of Control Sytem 7-7 Time Domain Analyi of Control Sytem p Lim 0 v Lim 0 G() H() G() H() a Lim 0 G() H() 0. Q.8 Solution : G() ( ) ( ) ( 0.4) The ytem i Type- ytem. For an input of r(t) t, calculate v. e A v Lim G() Lim 0 0 v But e 0 % 0. (given) ( ) ( ) ( 0.4) Aforr(t) t 0. i.e. 0 7.7 Derivation of Time Domain Specification Q.9 Solution : R() 0 ( ) ( 4) 0 ( ) ( 4) 0 5 4 ey Point Now though T.F. i not in tandard form, denominator alway reflect n and n from middle term and the lat term repectively. comparing 5 4 with n n n 4 n 4.8989 rad/ec. n 5 0.50 d n 4.9 rad/ec. Now for c(t) we can ue tandard expreion for in tandard form. So writing R()
Principle of Control Sytem 7-8 Time Domain Analyi of Control Sytem R() 0 4 4 5 4 For the bracket term ue tandard expreion, and then c(t) can be obtained by multiplying thi expreion by contant 0 4. c(t) 0 4 e tan n t in ( d t ) radian.0 radian 0 c(t).68 e.5 t in (4.9 t.0) 4 Q.0 Solution : d y dt To find T.F. condition. Sytem differential equation i, 4 dy dt Y() X() 8y 8x, take Laplace tranform from above equation and neglect initial Y() 4 Y() 8 Y() 8 X() Y() [ 4 8] 8 X() T.F. Y() X() 8 4 8 Comparing thi with tandard T.F. of econd order ytem n n n n 8 n.8 rad/ec. n 4 0.7067 d n.8 (0.7067).00 rad/ec. T p Time for peak overhoot.57 ec..00 d
Principle of Control Sytem 7-9 Time Domain Analyi of Control Sytem % M p e 00 e 0.706 (0.706) 00 4.% T Settling time 4 n t c(t) e n in ( t ) d 4 0.7067.8 ec. where tan c(t) e (0.7067) 0.7067.87 t 45 4 rad in t 4 c(t).4 t e in t 4 n.8 rad/ec. T p.57 ec. d.00 rad/ec. % M p 4.% T ec. 0.7067 Q. Solution : Aume T L a zero for [I] i) To calculate M, p G() 6 0.5 0.9 H() R() G() G()H() 6 ( 0.5 0.9) 6 (0.5 0.9) 40 6 40 Comparing the characteritic equation with, n 40 i.e. n 6.45 n 6 0.474 / Now, M p e 00 8.4 % n n ii) For unit ramp input, v Lim 0 G()H() Lim 0 6.. 6.67 ( 0.5 0.9)
Principle of Control Sytem 7-0 Time Domain Analyi of Control Sytem e v 0.5 rad II) For T L i Nm, aume R() a zero. With R() zero, the ytem get modified a T () L E() 6 0.5 0.9 Fig. 7.7 From the Fig. 7.7, we can write. G() 0.5 and H() 6 0.9 G() T L () G() H() Negative ign a ign of T L () applied i negative and G()H() a ign of the feedback i poitive. A block of i to continue ign of, though R() 0. ( 0.5 0.9) T L ().( 6) 0.5 0.9 6 ( 0.5 0.9 ) T () L C Lim [0.5 0.9 6] 0 Lim 0. [ 0.5 0.9 6] 6 0.66 Q. Solution : R() T() R() hence a Finding partial fraction, we get A A a ( a)
Principle of Control Sytem 7 - Time Domain Analyi of Control Sytem where A a and A a a a a Taking invere Laplace tranform, c(t) a a e at The teady tate of c(t) i Lim t c(t) a While lope at t 0 i.0 0. dc(t) i.e. dt t 0 a ( a) e at t 0 and a / Q. Solution : The repone of firt order ytem to the tep input i given a c(t) t T A(te ) Where A Final repone value T Time contant Now c(t) 0.98 A for t 60 ec. 0.98 A A 60 T ( e ) 0.98 e 60 t 60 e T 0.0 60 T.9 T 5. ec. Now, i Input temperature 0 Output temperature
Principle of Control Sytem 7 - Time Domain Analyi of Control Sytem For firt order ytem, 0 i () () T 5. 0 i 5. Now error () () i 0 E ( ) i() i() 5. i () 5. 5. i () ( 5. ) But input temperature i varying a a rate of 0º C/min i.e. 6 ºC/ec. i (t) t 6 i () E() 6 6 e Lim 0.556 5. ( 5. ) Lim E() 0 6 5. ( 5. ) 5. 6 Q.4 Solution : Cloed loop T.F. Comparing with tandard form G() G(), H() ( T) ( T) n n n T T T T n T n T n T T T T
Principle of Control Sytem 7 - Time Domain Analyi of Control Sytem Now for M p 0.6, Let 0.6 e Solving, 0.60 i.e. 0.5 For M p 0., Let 0. e Solving, 0.4559 i.e..6094 For 0.60, and 0.4559, 0.60 T T T 0.4559 From equation () T T.0 0.60... ()... () T 9.74 i.e. T 8.74 From equation () T.0967 0.4559 T.08 i.e. T 0.08 T T 0.08 4. Q.5 Solution : To prove that ytem i overdamped mean to prove > and not dependent on the value of R and C. R V V () () R o So firt to find it C.L.T.F. ue ignal V() i flow graph method. I () V i V V i() /C /C V o () R I () I () V() (I I ) C I () V V Fig. 7.8 o R V o () I C
Principle of Control Sytem 7-4 Time Domain Analyi of Control Sytem Signal flow graph i hown in the following Fig. 7.9. V i /R I C V /R I C V o L L L T /R C Fig. 7.9 C R, L L L CR /R L andl i combination of nontouching loop. L L C R Uing Maon' gain formula, L L L L L a all loop are touching to T, T C R R() CR C R C R C R CR CR C R Comparing with tandard form, n n C R CR rad/ec. CR C R n CR CR CR.5 A > ytem i overdamped and i independent of R and C value. For thi, two reitance mut be equal, and two capacitor value mut be equal. Q.6 G() Solution : R() G()H() (J ) ( m) J m (J ) J R() m J J
Principle of Control Sytem 7-5 Time Domain Analyi of Control Sytem Comparing denominator with 0 n n n m J n m J n J i.e. J J m J m For M p 5%, we can calculate. 5 00 e i.e. ln 0.5 Solving, 0.407 Now T p d n n.769 rad/ec. Now teady tate of ytem i 5 unit. Now teady tate i defined a, C Lim t c(t) i.e. n (0.407) Uing final value theorem we can ay, C Lim Lim c(t) t 0 From C.L.T.F. R() J J J C Lim 0 m Lim 0 J m J J J 5 i.e. m 0. m J m J and R() / 5 (given)
Principle of Control Sytem 7-6 Time Domain Analyi of Control Sytem Now n m J hence.769 0. J J 0.06784 and J m 0.9059 i.e. 0.47 0.06784 0. Q.7 Solution : From the ytem hown, G() and H() T G() R() G()H() ( T) Comparing denominator with nn T n n... () and n T T... () Now M p 0% 0. e ln (0.) Solving for, 0.4559 T p ec. d n n 0. 4559 n.7648 rad/ec Subtituting (4) and () in () and ().47 and T0.566... ()... (4) Q.8 Solution : The T.F. of the ytem i, R()
Principle of Control Sytem 7-7 Time Domain Analyi of Control Sytem Now a numerator ha term of '' tandard expreion of c(t) cannot be ued, though ytem i econd order. Hence ue partial fraction method ubtituting R(). R(), a unit tep input ( ) A B ( ) C ( ) Partial fraction A() BC() A AABC C AC 0, ABC, A C, B Taking invere Laplace tranform, c(t) te t e t ( ) ( ) Q.9 Solution : G() A, H() ( P) R() A ( P) A ( P) A n P A n n n A i.e. A and P i.e. n n P A 00 % M p e 00 i. e. 0 e Solving for, 0.59 T 4 n 4 i.e. n 4 4 0.59.697 radec n A.86 and P A
Principle of Control Sytem 7-8 Time Domain Analyi of Control Sytem 7. Senitivity of a Control Sytem Q.6 Solution : From given ignal flow graph, T U()Q()G(), T M()G() L Q()G(),, T() Y R U Q G M G Q G T() G UQ M Q G Senitivity of T() with repect to G() i, ST G TT GG G T T G ST G T G QG UQ M UQ M Q G Q G UQ UQ G M MQG UQ G MQG Q G UQ M Q G G UQ M G UQ M Q G Q G S G T Q G... Required enitivity It doe not depend on U() or M().
8 Stability Analyi of Control Sytem Solution of Selected Unolved Example 8.7 Special Cae of Routh' Criterion Q.4 Solution : 6 4 5 5 6 0 4 4 0 0 0 0 0 Row of zero A() 4 4 0 i.e. 4 0 da() d 4 4 A () 0 da () d 4 0 6 4 5 5 6 0 4 4 0 4 4 0 0 0 0 0 0 0 0 Row of zero again (8-)
Principle of Control Sytem 8 - Stability Analyi of Control Sytem 6 4 5 5 6 0 4 4 0 4 4 0 0 0 0 4 0 0 0 0 0 0 0 No ign change, hence no root i located in R.H.S. of -plane. A row of zero occur, ytem may be marginally table or untable. To examine that find the root of firt j auxiliary equation. A() 4 0 4 4 j,,, j,, 4 j 0 The root of A () 0 are the root of A() 0. So do not olve econd auxiliary equation. Predict the tability from the nature of root of firt auxiliary equation. A there are repeated root on imaginary axi, ytem i untable. j Fig. 8. 8.9 Marginal and Frequency of Sutained Ocillation Q.6 Solution : H() G H 0 65 ( ) ( 6 5) 0 i.e. 4 0 5 4 00 0 Routh' array i, 4 5 00 For tability, 00 > 0... from 0 0 4 0 40.6 00 504.4 0 00 0 40.6 i.e. >00 504.4 0 00 0... from 404.4 0 i.e. 40.44 0 00
Principle of Control Sytem 8 - Stability Analyi of Control Sytem For tability, 00 40.44 For utained ocillation, mar for which row of become row of zero. mar 40.44 For thi mar, A 40.6 00 0 00 40.44 00.4 40.6 40.6 j.5 j Hence the frequency of ocillation i.5 rad / ec. Q.7 Solution : The characteritic equation i G()H() 0 ( 0) ( 0) 0 i.e. ( ) 0 00 0 ( ) Routh' array i, 0 From row, >0 > 00 From 0 row, 00>0, >0 0()00>0 0 00 0 From row, a>0 i.e. 0 ( ) 00 > 0 i.e. > 4.667 0 00 0 > 40 For 4.667 < <, ytem i table. mar 4.667 which make row of a row of zero. A() ( ) 00 0 i.e. 00 ( ) 00 4.667 40 (4.667 ) i.e. j.8 Frequency of ocillation.8 rad/ec. Q.8 Solution : Refer example 8.9. for the procedure and verify the anwer : 0<<4.667, mar 4.667,.04 rad/ec.
0 Introduction to Stability Analyi in Frequency Domain Solution of Selected Unolved Example 0.4 Conceptual Approach to Frequency Repone Q. Solution : T() T(j ) 0 ( ) ( ) ( 5) 0 ( j ) ( j ) (5 j ) Now the teady tate repone i given by, c(t) A M in t ) where A amplitude, input frequency, phae angle of tranfer function M Magnitude of tranfer function 0 4 M T(j ) 5 and T (j ) tan tan tan 5 A 5 a x(t) i 5 in (t ) Steady tate repone can be obtained a, c(t) 50 4 5 50 4 5 in ( t ) in t tan tan tan 5 (0-)
Principle of Control Sytem 0 - Introduction to Stability Analyi in Frequency Domain 0.6 B. W. (Bandwidth) Q. Solution : For the equivalent econd order ytem M r and r n Now M r We are given M r 6 ( ) 6 6 4 6 4 6 0 i.e. 4 4 ( ) Hence 6 56 64 6 9 0.5 0.4 0.0669 or 0.9 and 0.588 or 0.966 ey Point A ytem with relative damping factor greater than 0.707 doe not exhibit any peak in the frequency repone hence the acceptable anwer i only 0.0669. Alo r n i.e. n (. 0 588) n. rad ec The tranfer function of equivalent econd order ytem i, M() n (.) n n (0.588) (.) (.) 0.87.668 0.87 Rie time, T r d where tan In thi cae, tan (. 0 588) 0. 558 radian tan 0. 9659. radian 0. 588 Alo d n. ( 0. 588). rad/ec
Principle of Control Sytem 0 - Introduction to Stability Analyi in Frequency Domain Rie time, T r. 0.59 ec. Time to peak, T p d.00 ec. 4 Settling time, T (For % error band) 4 n 0.588. 4.795 ec T.596 ec (for 5 % error band) Period of ocillation, oc..0 ec. d. n Number of ocillation during ettling N T T oc. 4795. 0..7 Peak overhoot, M p e Since 0.588 (. 0 588) 0.847 Overhoot e 0.847 0.409 Thu the ytem exhibit 4. % overhoot. Sketche of frequency repone and etimated time repone are hown. M r M c(t) 0.7 Total.7 ocillation before it ettle 0 (rad/ec) 0 4 5 6 7 t(ec) Fig. 0. Frequency repone Fig. 0. Step repone (etimated) Q. Solution : M p 6. % and T p 5
Principle of Control Sytem 0-4 Introduction to Stability Analyi in Frequency Domain Now M p e 00 6. e 00 ln(0.6). 80 i.e. 0.5 0.5 And T p d 5 d 5 n n 0 rad/ec For inuoidal input, i) r n 0 ( 0. 5) 7.07 rad/ec ii) M r 0. 5 ( 0. 5).547 Q.4 Solution : G() ( a), H() i) R() G() G()H() ( a) ( a) a Comparing denominater with, n n n, n...() n a, a...() Now, M r.04 Solving, 0.60... cannot be more than 0.707. While r n.55
Principle of Control Sytem 0-5 Introduction to Stability Analyi in Frequency Domain n.07 rad/ec From equation (), 485.86 From equation (), a 6.548 ii) B.W. 4 4 n 4 4.07 (. 0 60) 4(. 0 60) 4(. 0 60) T 5.08 rad/ec 4 4 0. 60. 07 n 0.06 ec Q.5 Solution : G() ( T), H() R() G() G() H() ( T) ( T) T T T T Comparing thi with n n n, n T n T n T i.e. T Now M p 5 % for unit tep input 5 e 00 i.e. ln 0.5 Solving, 0.406 In frequency domain, r 8 rad/ec n and 0.406 8 n (0.406) n 9.744 rad/ec n i.e. T n T... ()
Principle of Control Sytem 0-6 Introduction to Stability Analyi in Frequency Domain T i.e. 4T T n From equation (), ubtituting in equation (), 4 T T n... () T 0.06 i.e. T 0.7.07 M r 0.406 (0.406).54 Q.6 Solution : For the econd order ytem, M p 0 % i.e. 0 e 00 ln0. i.e..05 Solving, 0.59 T r 0. d 0.98 n where tan 0.98 rad.0 i.e. n 0. 0.59 n 7.58 rad/ec From thee value, M r and B.W. can be calculated a, M r 0.59 0.59.0487 B.W. 4 4 n 4 7.58 0.59 40.59 40. 59 4.686
Bode Plot Solution of Selected Unolved Example.7 Calculation of G.M. and P.M. from Bode Plot Q.6 Solution : Step : G() i in the time contant form. Step : In thi example, i unknown. Draw the magnitude plot without. The factor are, i) Pole at the origin (). The traight line of lop 0 db/dec paing through the interection point of and 0 db. ii) Simple pole, 05., T 0.5, C T. The traight line of lop 0 db/dec for. iii) Simple pole, T 0. 0., C 5. T The traight line of lope 0 db/dec for >5. Range of 0< < C () < < C (5) 5< < Reultant lope 0 db/dec 0040dB/dec 40060dB/dec Note : The effect of i to hift the entire magnitude plot upward or downward by 0 Log db. So find the hift required in magnitude plot without to match the required pecification. Step : Phae angle table, G(j) j( 0. 5j)( 0. j) ( - )
Principle of Control Sytem - Bode Plot j tan 0.5 tan 0.0 R 0. 90º 5.7º.9º 98º 90º 45º.8º 56.8º 5 90º 68.9º 45º 0.9º 0 90º 78.69º 6.4º.º 90º 90º 90º 70º Step 4 : Sketch the Bode plot. From the plot pc. rad/ec. Step to find for given G.M. : ) Draw the horizontal line below 0 db at a ditance of given G.M., till it interect vertical line of pc. Thi i point A. ) Draw the vertical line from pc, till it interect magnitude plot without. Thi i point A. ) The point A mut be at A to match given G.M. Hence A to A required i.e. contribution by which i 0 Log db. i the hift 4) Upward hift mut be taken poitive and downward hift negative. 0 Log Shift (AA ) In thi example, 0 Log 6 db... G.M. 6 db... G.M. 6 db Step to find for given P.M. : ) Draw the hoizontal line above 80º line at a ditance of given P.M., till it interect phae angle plot. Thi i point B. ) Draw the vertical line from B, till it interect the magnitude plot without. Thi i point C. ) Thi C mut be on 0 db line at C, a the vertical line through point B mut be gc line for given P.M. 4) Thu ditance of C from 0 db line i the hift required to atify given P.M. Thi mut be 0 Log db. 5) Upward hift ( C C) mut be taken poitive while downward hift ( C C) mut be taken negative. 0 Log Shift ( CC)
Principle of Control Sytem - Bode Plot In thi example, 0 Log 6 db... for P.M. 5º... for P.M. 5º Note : For ame value of, both pecification are matched. But thi may not be the cae, all the time. (See Fig.. on next page). M R 60 40 0 0dB 6dB 0º 40º R 90º 0º 50º 80º 0º 40º 70º SEMI-LOG PAPER (5 CYCLES X /0) 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 Scale : On Y axi unit 0 db, 0º 0 db/dec 40 db/dec C' C A' AA' hift for G.M. 6 db C'Chift for P.M. 5º A 60 db/dec (/) Without 5º B 0. pc. 0 00 000 Log Fig..
Principle of Control Sytem - 4 Bode Plot Q.7 Solution : Step : Arrange G()H() in time contant form G()H() ( ) () (0) 0 0 ( ) ( 0.5) ( 0.) Step : Factor i) i unknown. 0 ii) Simple pole, iii) Simple pole, iv)simple pole,, T, C 0.5 T, T 0.5 0.5, C T, T 0. 0., C 0 T Step : Magnitude plot analyi. i) No pole at the origin, i unknown o tarting Bode plot i nothing but 0 db line 0 which will change it lope at C 0.5. ii) At C 0.5, imple pole occur which contribute a 0 db/dec line. Thi line will continue upto C. iii) At C, imple pole occur which contribute a 0 db/dec individually and hence reultant lope will become 0 0 40 db/dec from C onward till next corner frequency C 0. iv)a C 0, imple pole occur which contribute a 0 db/decade individually and hence reultant lope will become 40 0 60 db/dec from C 0 onward till a there i no other factor preent. Step 4 : Phae angle plot 0 G(j) H(j) ( j)( j05. )( 0. j) Though ( 0 ) i unknown, it i not going to contribute any phae angle. Hence phae angle plot i independent of value of ( 0 ) and hence pc i true, reultant pecification from thi phae angle plot which i independent of value of ( 0 ). j tan, 05. j tan 05., 0.j tan 0.,
Principle of Control Sytem - 5 Bode Plot tan tan 05. tan 0. R 6.4 6.56 5.7 95.7º 75.96 45..6º 5 84.8 68.9 6.5 79.0º 0 87. 78.69 45 0.8º 90 90 90 70º Step 5 : Sketch the Bode plot, a hown in the Fig.. 4 5 6 789 4 5 6 789 4 5 6 789 4 5 6 789 pc 60 60 db/dec 40 40 40 db/dec 0 60 0 db/dec B AB Upward hift required to meet P.M.0 8dB 0dB o 0 Shifted magnitude plot to meet P.M. 0 o A 0 40 0 Magnitude plot without 90 o Fig.. SOLUTION For P.M. 0 (/0) 7.94 i.e. 58.86 0 o o 50 o P.M. 0 o 80 o 0 o 40 o Log 0. gc C 0.5 C pc 4.9 0 00 000 C Required. for P.M. 0 o
Principle of Control Sytem - 6 Bode Plot pc 4.9 rad/ec For P.M. 0º, 0 P.M. 80 G(j) H(j) gc G(j) H(j) 50º gc Without effect of ( 0 ), gc cannot be determined. So now adjut effect of ( 0 ) i.e. hifting plot upward till it interect 0 db line at uch a frequency where G(j) H(j) i.e. gc P.M. 0. To get `.8 a gc, magnitude plot i required to be hifted upward from A to B a hown in plot which i the effect of ( 0 ) i.e. 0 log ( 0 ). Shift AB 8 db upward. 0 log ( 0 ) 8 ( 0 ) 7.94 58.86 to meet P.M. a 0º Q.8 Solution : Step : Arrange G()H() in the time contant form. The factor of ( 0 0) are ( 9.) ( 0.68) ey Point A the factor of quadratic are real, they mut be treated a eparate imple pole. So quadratic factor no longer remain quadratic in nature. G()H() 000 0. ( 0. ) 4( 4 ) ( 9.) ( 0.68) 000 0. ( 0. ) 4 ( 4 ) 9. ( 9. ) 0.68 ( 0.68 ) Step : Factor i) 7.5 7.5(.) ( 0.5) ( 0.04) (.466) ii) Simple zero, (.), T., C T 0. iii) Simple pole, iv) Simple pole, v) Simple pole,. 466, T.466, C 0.68 T 0.5, T 0.5, 4 C 05., T 0.04 4 0.04, C4 9. T 4
Principle of Control Sytem - 7 Bode Plot Step : Magnitude plot i) Contribution by 7.5 i 0 log 7.5 7.5 db. So tarting magnitude plot a 0 i traight line of lope 0 db. Thi will continue till C 0.. ii) At C 0., imple zero occur which contribute a 0 db/dec individually and hence reultant lope will become 0 0 0 db/dec from C 0. onward till next corner frequency C 0.68. iii) At C 0.68, imple pole occur which contribute a 0 db/decade individually and hence reultant lope will become 0 0 0 db/dec from C 0.68 onward till next corner frequency C 4. iv) At C 4, imple pole occur which contribute a 0 db/dec individually and hence reultant lope will become 0 0 0 db/dec from C 4 onward till next corner frequency C4 9.. v) At C4 9. imple pole occur which contribute a 0 db/dec individually and hence reultant lope will become 0 0 40 db/dec from C4 9. onward till a there i no other factor preent. Step 4 : Phae angle plot : 7.5(. j) G(j) H(j) ( 0. 5j)( 0. 04j)(. 466j) 7.5 j0 0. j tan. 05. j tan 05., 0. 04 j tan 04,.. 466 j tan. 466 Phae angle table : tan. tan 05. tan 0. 04 tan. 466 R 0. 8.4.40 0.948 8.40 8.45 7.8 4.06.947 55.70.6 0 85.7 68.9 8.77 86.09 87.5 0 89. 78.69 4.4 88.04.8 50 89.65 85.4 59.5 89. 44.5
Principle of Control Sytem - 8 Bode Plot 00 89.8 87.7 7.6 89.6 6.0 90 90 90 90 80 Step 5 : Sketch the Bode plot With 000, P.M. 40º, gc 45 rad/ec, G.M. db and pc ytem table. For P.M. 0º, gc hould be 64 a hown in plot. So plot mut be hifted upward from A to B to get P.M. 0º. Thi hift AB 7 db upward. So effect of contant mut be (7.5 due to 7 db) 4.5 db required So 0 log 4.5 i.e. 6.78. But /66.67, by adjuting G()H() in time contant form. New 6.78 66.67 4474.65 for P.M. 0º. (See Fig..) 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9 60 40 0 db/dec 0 db/dec 0 db/dec 0 db/dec gc 40 db/dec 0dB 0 0 0 o 0 o 0 o 60 o 90 o 0 o 50 o 80 o For P.M. 0º o 40 P.M. 0. 0 gc 45 00 000 o New gc 64 required for P.M. 0 Fig.. B A For7.5 Reultant SOLUTION With 000 gc 45rad/ec pc o P.M. 40 G.M. STABLE To get P.M. 0 additional hift required to be introduced i A to B i.e. 7 db by increaing. 4475.65 to get o P.M. 0 o P.M. 0 o Log
Principle of Control Sytem - 9 Bode Plot Q.9 Solution : The open loop tranfer function i G() ( 0. )( 00. ) The variou factor are : i) Contant i unknown ii) One pole at the origin. The magnitude plot i the traight line of lope 0 db/decade paing through interection point of and 0 db line. iii) The imple pole, ( 0. ) C 0. 5 Magnitude plot i traight line of 0 db upto 5 and traight line of lope 0 db/decade after 5. iv) The imple pole, ( 00. ) C 50 00. Magnitude plot i traight line of 0 db upto 50 and traight line of lope 0 db/decade after 50. The reultant lope table i, Range of Slope tart < < 5 5< < 50 0 db/decade 40 db/decade 50 < < 60 db/decade G(j) j( 0. j)( 00. j) The reultant phae angle plot i,
Principle of Control Sytem - 0 Bode Plot for 90 for 0.. tan 0 for 00.. tan 00 R 0.5 90º 5.7º 0.57º 96º 90º.º.4º 0º 5 90º 45º 5.7º 4º 50 90º 84º 45º 9º 5 90º 78º 6.5º 94º From the Bode plot hown in the Fig..4. (See Fig..4 on next page.) pc 6 rad/ec Now to find limiting value of, gc pc So point A hould at point B to get gc pc. So hift required from A to B i the contribution by, which i allowed till ytem doe not reach on the verge of intability. 0 log hift AB 0 log 5 db 56. Hence range of for tability i, Q.0 0 < < 56. Solution : G()H() ( 0.)( 0.5)( 0.00) The G()H() i in time contant form. The variou factor are, i) i unknown. ii) One pole at origin, traight line of lope 0 db /dec paing through interection of and0db. iii), imple pole, T 0.5 0.5, C 4 05. Straight line of lope 0 db/dec for 4. iv), imple pole, T 0. 0., C 0 0. v) Straight line of lope 0 db/dec for 0., imple pole, T 0.00 0. 00, C 000 0. 00
Principle of Control Sytem - Bode Plot 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 60 40 0 db/dec pc 6 Scale On Y axi : unit 0 db, 0º 0 0dB B 0 SHIFT A B 5dB R 40 db/dec A 90-60 db/dec 0 50 80 0 40 70 0. 0 00 000 Log Fig..4
Principle of Control Sytem - Bode Plot Straight line of lope 0 db/dec for 000 (See Fig..5 on next page.) Range of Reultant lope 0 C () 4 0 db/dec 4 C ( 0) 0040dB/dec 0 C ( 000) 40060dB/dec 000 60080dB/dec Phae angle table : G(j)H(j) j ( 0.5j )( 0.j )( 0.00j ) j tan 05. tan 0. tan 0. 00 R 0.4 90º 5.7º.9º 0.0º 98º 4 90º 45º.8º 0.º 57.0º 0 90º 68.9º 45º 0.57º 0.76º 00 90º 87.7º 84.8º 5.7º 67.69º 90º 90º 90º 90º 60º Draw the Bode plot without. A P.M. 40º, draw the horizontal line at 40º above 80º line till it interect phae angle plot at point C. Draw vertical line from C till it interect magnitude plot at point D. The point D mut be on 0 db line to have P.M. 40º a the frequency correponding to point D mut be gc. The hift required from D till 0 db line i the contribution by. Shift DE 0 Log 8dB(Upward hence poitive).58...for P.M. 40º Then draw magnitude plot parallel to that without and then find the point of interection of pc line with new magnitude plot. Thi point i A. The correponding gain margin i A to B. G.M. db Q. 80000 Solution : G()H() ( )( 50)( 00) G()H() 80000 50 00 50 00
Principle of Control Sytem - Bode Plot 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 M in db R Scale: On y axi: unit 0 db, 0º 80 60 40 gc.5 0 E B 0dB G.M. db A 0dB/dec DE Upward hift to get P.M. 40º D 0 40dB/dec R 90º (/) New magnitude plot 60 db/dec C 0º 50º Fig..5 P.M. 40º 80º 0º 80 db/dec 40º Without 70º 00º 0. 0 00 000 Log pc 6.
Principle of Control Sytem - 4 Bode Plot 4 ( 0.5)( 0.0)( 0.005) The variou factor are, i) 4 i.e. 0 log 4.04 db, line parallel to log axi.... Time contant form ii), one pole at origin, traight line of lope 0 db/dec paing through interection of and 0 db. iii), imple pole, T 0.5 05., C T Straight line at lope 0 db/dec for. iv), imple pole, T 0.0 00., 50 C T Straight line of lope 0 db/dec for 50 v), imple pole, T 0.005 0. 005, 00 C T Straight line of lope 0 db/dec for 00 Range of Reultant lope 0 C () 0 db/ dec C ( 50) 0040dB/dec 50 C ( 00) 40060dB/dec 00 60080dB/dec. Phae angle table : G(j)H(j) 4 j ( 0.5j )( 0.0j )( 0.005j ) j tan 05. tan 00. tan 0. 005 R 0. 90º 5. 7º 0. 9º 005. º 969. º 90º 45º 9. º 057. º 7. 86º 50 90º 87.º 7 45º 4. 0º 67. º 0 90º 84.º.º 8 57. º 0. 7º 90º 90º 90º 90º 60º The Bode plot i hown in the Fig..6. From Bode plot, (See Fig..6 on next page.)
Principle of Control Sytem - 5 Bode Plot SEMI-LOG PAPER (5 CYCLES X /0) 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 M in db R 80 Scale: Onyaxi:unit0dB,0º pc 0 0dB/dec gc 60 40 0 4,dB B 0dB G.M. 0dB A 40 db/dec 0 R P.M. º Fig..6 gc rad/ec, pc 0 rad /ec. G.M. 0dB, P.M. º 90º (/) 0º 60 db/dec C 50º D 80º 0º 80 db/dec 40º rad/ec 0 rad/ec G.M. 0 db P. M. º Sytem Stable gc pc 70º 00º 0. 0 00 000 Log gc pc
Principle of Control Sytem - 6 Bode Plot Q. Solution : G() 0 ( 0. ) ( 0.5 ) Step : G() i in time contant form. Step : Factor are, i) 0 i.e. 0 log 6.0 db, line parallel to log axi. ii) One pole at origin, traight line of lope 0 db/dec paing through interection point of and 0 db. iii) Simple pole,, T 0.5 0.5, C T traight line of lope 0 db/dec for. iv) Simple zero, ( 0. ), T 0., C T 5 traight line of lope 0 db/dec for 5. Frequency range Reultant lope 0 C () 0 db/dec C (5) 0040dB/dec 5 4000dB/dec Step : Phae angle table 0 ( 0. j ) G(j ) j ( 0.5 j ) j tan 05. tan 0. R 0. 90º.86º.4º 9.7º 90º 45º.8º.9º 5 90º 68.9º 45º.9º 0 90º 84.8º 75.96º 98.º 90º 90º 90º 90º Step 4 : The bode plot i hown in the Fig..7. (See Fig..7 on next page.) Step 5 : From the bode plot, gc 9 rad/ec pc rad/ec G.M. db P.M. 70º The ytem i abolutely table in nature.
Principle of Control Sytem - 7 Bode Plot M in db R 60 40 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 4 5 6 7 89 0dB/dec 40 db/dec gc 9 Scale On Y axi unit 0 db, 5º 0, 6 db 0dB 0dB/dec 0 40 R 75º (/) 90º 05º 0º 5º 50º P.M. 70º 9 rad/ec G.M. db gc pc P.M. 70º Sytem abolutely table 65º 80º 0. 0 00 000 Log Fig..7
Principle of Control Sytem - 8 Bode Plot.0 Calculation of Tranfer Function from Magnitude Plot Q. Solution : Starting lope 0 db o no pole or zero at origin. At each corner frequency lope change. Firt change in lope i at, C. Change in lope i econd lope initial lope 0 00dB/dec. i.e. factor i imple zero with C i.e. T C Factor i ( T ) i.e. ( ) Now next change in lope i at C Change in lope 0 0 0dB/dec o factor i imple pole. Now i not known. Let u write equation for that line, y mxci.e. magnitude in db m Log C m 0 db/dec Now at, magnitude in db 0 db Subtituting 0 0 Log C C 0 Equation i magnitude in db 0 Log Now at, Magnitude in db 5 db given Subtituting 5 0 Log 5.6 rad/ec C. T C 5.6 0.77 Factor i T 077. Next change in lope i at. C. change i 0 00dB/dec Factor i imple pole. To find write equation for that line having lope 0 db/dec. mag in db m Log C mag in db 0 Log C.
Principle of Control Sytem - 9 Bode Plot At 000, mag in db 0 Subtituting 0 0 Log 000 C C 60 db Equation i, mag in db 0 Log 60 At, mag.in db 5 0 Log 60 0Log 45 77.8 C T 0.056 C Factor i 0.0056 The lat change in lope i at 000 C4 Change i 0(0) 0 db/dec o factor i imple zero C4 000 T 4 C4 0.00 Factor ( T4 ) ( 0.00) The total tranfer function i product of all of them ( ) ( 0.00 ) G()H() ( 0.77 ) ( 0.0056 ) Q. Solution : Aume that the magnitude at 40i0dB. Now the equation for the firt traight line of lope 40 db/dec i, y 40 log C... () At 40, y 0 db 0 40 log 40 C C 64.08 The equation of econd traight line of lope 0 db/dec i, y 0 log C... () At 40, y 0 db 0 0 log 40 C C.04
Principle of Control Sytem - 0 Bode Plot i) To find, ubtitute y0inequation () 0 0 log.04 5.04 log 400 rad/ec ii) To find A, ubtitute 0 in equation () A 40 log0 64.08 A 4.08 db iii) Now the equation of third traight line of lope 40 db/dec i, y 40 log C At 000, the y can obtained by ubtituting in equation () y 0 log 000.04 y 7.9589 db Thu A 7.9589 db iv) The equation of third line now i, y 40 log C... () At 000, y 7.9589 7.9589 40 log000 C C 9.04 So to find, ubtitute y40dbinequation (), 40 40 log 9.04 000 rad/ec Q. Solution : The lope 6 db/octave i ame a 0 db/decade while the lope db/octave i ame a 40 db/decade. Hence the tarting lope i 0 db/decade hence there i one pole at the origin. Now thi line of lope 0 db/decade would have interected 0 db line at 0. Hence at it magnitude i 0 db/dec. Thi indicate that at the magnitude plot ha experienced a hift of 0 db. Thi i the contribution by. 0 Log 0 Log 0 The firt change in lope i at C rad/ec. The change i of 0 db/dec due to which reultant lope become db/dec after C 5. Hence the correponding factor i imple zero.
Principle of Control Sytem - Bode Plot T C 5 0. Factor ( T ) ( 0. ) The next change in lope i at C 0 rad/ec. The change i of 40 db/dec. Hence the correponding factor i imple pole with order. So repeated imple pole i the factor. T C 0 0.0 Factor ( T ) ( 0.0 ) Hence the tranfer function i the product of all the factor. G()H() 0 ( 0. ) ( 0.0 )
Polar and Nyquit Plot Solution of Selected Unolved Example.5 Stability Determination from Polar Plot Q.4 Solution : For drawing the rough ketch of polar plot we note that, G(j)H(j) 00 4 8 and 90 tan tan tan 4 8 Giving gain and phae at 0.,, 0 a given below 0. 0 GH 5.6.9 0.00808 95.0º 7.7º 88º At 0., GH At, GH At 0, GH Similarly we calculate and write in the above table. It i clear that it i difficult to plot the polar plot to cale, however a rough ketch can be made. Thi i hown. To determine tability we mut find x. Thi can be done provided, we know the frequency at which 80. Thi can be eaily be achieved by trying more value of greater than. 00 0. 4 8 00 5 4 8 00 64. 5.6 00 7.55.9 00 0 0 0 64 0 64 0.00808 0 x 90º Fig.. A rough ketch of polar plot ( - )
Principle of Control Sytem - Polar and Nyquit Plot At, 90 45 6.56 4 75.5 At., 90 47. 7 8. 8 5. 4 8. 9 Obviouly the phae cro over frequency i very nearly. rad/ec Hence GH pc x 00... 4. 8 00..97 4.565 8.96 00 47.7 0.40 Since x <, the ytem i table and GM x.477 0. 40 Moreover GM in db 0 log.477 7.87 db Similarly by trial we find gc, the gain cro over frequency. For example at. 00 GH. 544. 744. 6544. 00.. 476. 8. 08 00.05 94. 4 Thu gc i around.05 rad/ec And at gc i, 05. 05. 90 tan tan tan 4 05. 8 8.5 Thi give PM 80 ( 8.5 ) 4.5 For the above problem Gain cro over frequency.05 rad/ec phae cro over frequency. rad/ec Gain Margin GM.477 or 7.87 db Phae Margin 4.5 The ytem i table. Verification of G.M. by ROUTH S CRITERION. We can find critical value of by ROUTH S criterion. G()H() 0 give ( ) ( 4) ( 8) 0
Principle of Control Sytem - Polar and Nyquit Plot or ( ) ( 4) ( 8) 0 or ( 6 8) ( 8) 0 or ( 4 56 64) 0 ROUTH'S TABLE 4 4 56 64 0 4 56 4 64 5.4 89.6 4 5.4 0 Thu for tability 89.6 4 >0 or 4 < 89.6 or < 4.97 critical 4.97 and GM critical actual 4.97.5 or 7.4 db 00 The anwer agree very cloely with what we determined by Polar Plot ( db i accepted uing Bode or M- plot). Incidentally when 4.47, row in the Routh Table i all zero and the auxiliary equation formed by row i 5.4 0 or j 5.4 j 4.97 5.4. rad/ec Thi i cloe to pc. Q.5 Solution : G(j)H(j) j( j)( 0. j) We can find pc by trial. However for thi problem there i a impler method. GH ( ) ( 0. ) 90 tan ( ) tan ( 0. )
Principle of Control Sytem - 4 Polar and Nyquit Plot At pc, 80 80 90 tan ( ) tan ( 0. ) 90 tan ( ) tan ( 0. ) 90 where tan and tan 0. tan tan Hence tan 90 tan ( ) tan tan The right hand ide i infinite only if tan tan Thi give 0. 0 o 0. 0 or 5 or 5.6 rad/ec Thu pc.6 rad/ec Then GH pc x. 6 (. 6) ( 0. 6) x.6 4.58.046 x 0. 447 The ytem ocillate if x or 0.497 Q.6 Solution : G (j ) j a Since G(j) will be plotted a a vector, we find the x and y component of the vector. x In thi cae G(j) xjy,where x and y are obtained by 0 rationaliation. a j ( a j) G(j) G(j ) a j a j a y a j a a Fig.. (a) Thu x a a and a... (ii) We now eliminate... (i)
Principle of Control Sytem - 5 Polar and Nyquit Plot Dividing equation (ii) by equation (i), y x a a and ya x Subtituting in equation (i), we have a a x a ( ya x) y a a x cro multiplying, a x y a a x Multiplying throughout by x a x y a ax Dividing by a, we have x y a x We bring thi to the form (x x ) (y y ) R which give centre (x,y ) and radiu R From equation (iv), x a x y 0... (Iii)... (iv) or x a x y a 0 a or x y a a... (v) Thi i the equation of the circle. With centre a,0 and radiu. Note that y i negative for all value a of a indicated by equation (ii). Hence only the lower part of the circle give the polar plot. For the given problem 40 and a hence the 40 centre i at,0 i.e. (0, 0) and radiu i, a 40 0 4 0 (0,0) 0 0 Fig.. (b) Polar plot of G() a for40anda
Principle of Control Sytem - 6 Polar and Nyquit Plot.4 Step to Solve Problem by Nyquit Criterion Q. Solution : The open loop T.F. i ( 0) G()H() G()H() j ( 0)( j 0) ( j) There i no pole on right hand ide hence for the tability NP0.TheNyquit plot hould not encircle critical point j0. The Nyquit path i, j Section-I Section I Section II j M0 90º j0 Section IV j0 M 70º O Rotation 70º ( 90º) j0 80º Clockwie Section III - plane Section-II j j0 70º Fig.. j0 70º Rotation 70º ( 70º) 540 Anticlockwie Section-III Section-III i mirror image of ection-i To find interection of Nyquit plot with negative real axi, rationaliing G(j)H(j). G(j)H(j) j ( 0)( j 0)( j ) ( j)( j) j 00 0 j 0 j ( 00 ) 6 6 For interection with negative real axi, 00 0 00 pc 0 4 6 R
Principle of Control Sytem - 7 Polar and Nyquit Plot Interection point Q 0 0 0 6 4 Point Q 5 Hence the Nyquit plot i : 70º I Q j0 O N0 900º Fig..4 i) For tability, Q > > 5 > 5 for tability. ii) For 7, Q 7 5.4 G.M. 0 log OQ G.M. 0 log.9 db 4. Q. Solution : Hj G j j8j j Step : No pole in r.h.. of -plane, P 0 Step : For tability, NP0
Principle of Control Sytem - 8 Polar and Nyquit Plot Step : Nyquit path i, j I IV j0 j0 III j II Fig..5 (a) Step 4 : Section I : Section - II : j 90 90 0 º º 70º j0 00 º º 70º 70º 0 90º 70º (90º) 80º Clockwie j0i.e. 0 70º 70º (70º) 540º j0i.e. 0 70º Anticlockwie Section - III i mirror image of ection - I, about real axi. Section - IV not required. Step 5 : Calculate point Q which i interection of Nyquit plot with negative real axi. j8j GjHj j... j j 6 0 j 0 j 6 j 6 0... for making imaginary part zero. pc 4 Point Q 0 6 0.65 Step 6 : The Nyquit plot i hown in the Fig..5 (b)