Signal and System (Chapter 3. Continuous-Time Systems)

Transcription

1 Signal and Sysem (Chaper 3. Coninuous-Time Sysems) Prof. Kwang-Chun Ho Tel: Fax: Dep. Elecronics and Informaion Eng. 1

2 Nodes, Branches, Loops A nework wih b branches, n nodes, and l independen loops(meshes) will saisfy he fundamenal heorem of nework opology: b l n 1 NODE: poin where wo, or more, elemens are joined (e.g., node 1,,3,4,5). LOOP: A closed pah ha never goes wice over a node. MESH: A special ype of loop, which does no conain any oher loops wihin i. BRANCH: Componen conneced beween wo nodes (e.g., componen ). R, R, R, R, R How many branches, nodes and loops are here? Branches: Nodes : Meshes : b 5 n 3 l 3 Dep. Elecronics and Informaion Eng.

3 Sysem Equaions In elecrical sysems, a cenral law obeyed he physical properies is Kirchhoff s law: Algebraic sum of volages around any closed pah equals zero: v () 0 ( A form of differenial equaion) Algebraic sum of he currens enering a node is equal o he sum of he currens leaving he node : i () 0 ( A form of differenial equaion) A node is a juncion poin! 3 Dep. Elecronics and Informaion Eng. 3

4 Sysem Equaions Applicable examples: Circui consised of passive devices as follows: v () Ri (), i () v () R di() 1 v () L, i () vd () d L 1 dv( ) v () id (), i () C C d ( Inegro-differenial relaionships beween currens and volages ) 4 Dep. Elecronics and Informaion Eng. 4

5 Sysem Equaions Example 3.1: Wrie he differenial equaion governing circui behavior Soluion: Applying Kirchhoff s volage law, we have 1 di( ) vg () i() d L i() R C d dvg () di () di () i () L R d d d C Differeniaing ( We have a differenial equaion! ) 5 Dep. Elecronics and Informaion Eng. 5

6 Sysem Equaions If he capacior is iniially uncharged, hen deermine v () for 0 g di() vr i() R 1 e, vl L 61 e d 0 1 vc i() d 6 e dv (0) C e C v () v v v 6[V] g R L C 6 Dep. Elecronics and Informaion Eng. 6

7 Sysem Equaions Also, we can find he soluion by solving direcly differenial equaions. How can do i? Firs, consider general form of differenial equaion M N d y() dy() d x() dx() b1 b0y() an a1 a0x() M N d d d d where x() is inpu signal (or forcing funcion), and y() is he resuling oupu Homogeneous linear differenial equaions: if x()=0 7 Dep. Elecronics and Informaion Eng. 7

8 Sysem Equaions Example 3.: Find he soluion o equaion dy() sy() 0 d Assume he soluion o be of he form Le s check he soluion is rue! Thus, in general, he soluion consiss of exponenial si erms of form Ce Tha is, for i M s leing y () Ce, we have h s Since e canno be zero, i reduces o sc e y () Ce h Subsiuing, we have d y() dy() b1 b0y() 0 M d d s M Ce s b s b M s bs 1 b0 0 s s e 0 s ( Characerisic Equaion ) 8 Dep. Elecronics and Informaion Eng. 8

9 Sysem Equaions The characerisic equaion is zero a only is roos. Thus, he facorizaion gives s s0ss1 ssm 1 0 s 0 s1 sm 1 We noe ha Ce are all possible 0, Ce 1,, CM 1e soluions Then, he homogeneous soluion is given by he superposiion principle Example 3.3: where can be obained by he iniial condiions Find he soluion for y() 5 y() 4 y() 0 given he iniial condiions y(0), y(0) 1 Soluion: M y () C e Ce C e h C 0 ~ C M 1 s 0 s1 s M1 9 Dep. Elecronics and Informaion Eng. 9

10 Sysem Equaions The oupu of a sysem is called is response The characerisic equaion is s s s s The roos are Then, he homogeneous soluion becomes y () Ce C e h 1 s 1; s 4 Applying he iniial condiions o y(), we have y(0) C1C, Solving 7 1 C1, C y(0) C 4C 1 simulaneously Thus, he naural response (complemenary soluion) is defined as yn() e e 3 3 MaLab scrip: y=dsolve('dy+5*dy+4*y=0','y(0)=','dy(0)=-1',''); 4 10 Dep. Elecronics and Informaion Eng. 10

11 Sysem Equaions Nex, examine when he characerisic equaion has k muliple roos like s s ss Then, he soluion is of he form k 1 y () C e C e C e Ce h Example 3.4: Solve he equaiony() 8 y() 16 y() 0 Soluion: s 0 s 0 s 0 s1 0,0 0,1 0, k1 1 The characerisic equaion is s s s Since i has a double roo, he soluion mus ake he form of y () Ce C e h Anoher Case: when he characerisic equaion has complex conjugae roos ss ss 0, s, s j Dep. Elecronics and Informaion Eng. 11

12 Sysem Equaions If he characerisic equaion has complex muliple roos, which form is he soluion? Then, he soluion is of he form Applying Euler s ideniy, i becomes () y C C e cos j C C e sin Seing: where M, N can deermine from he iniial condiions Example 3.5: Find he soluion for Soluion: M 0 N0 y () M e cosn e sin h The characerisic equaion is s s5 s1 j s1 j 0 y () Ce Ce j j 0 1 y() y() 5 y() 0 Since 1,, he soluion mus ake he form y () M e cosn e sin h Dep. Elecronics and Informaion Eng. 1

13 Sysem Equaions Non-homogeneous differenial equaions: if Since he forcing funcion x() is no zero, here exiss a paricular soluion so ha we can ge he forced response ( y () y () y () wih zero iniial condiions): Example 3.6: Find he paricular soluion for Soluion: f h p y() 3 y() 3e Paricular soluion mus have he same form as he forced funcion! 4 Thus, he soluion mus be y () Ke p 4 x () 0 13 Dep. Elecronics and Informaion Eng. 13

14 Sysem Equaions Subsiuing his ino differenial equaion, unknown consan K is deermined 4Ke 3Ke 3e Consequenly, he paricular soluion is y () 3e p 4 K 3 Forced response? Then, he complee response of nonhomogeneous equaions consiss of he naural and forced responses like y() y () y () n f Example 3.7: Find he soluion for Soluion: y() 3 y() y() 4e 14 Dep. Elecronics and Informaion Eng. 14

15 Sysem Equaions Characerisic equaion is s 3s s1 s 0 so ha he homogeneous soluion is y () Ce C e h 1 If iniial condiions are y(0) 1, y(0) 1, we have 1 1 Subsiuing his equaion and solving gives C C 1, C C 1 C1 1, C 0 Solving Thus, he naural response becomes yn() e Nex paricular soluion has a form of y () Ke Then, he nonhomogeneous soluion becomes ynh() yh() yp () Ce 1 Ce e 3 p K 3 15 Dep. Elecronics and Informaion Eng. 15

16 Sysem Equaions Applying zero iniial condiions ( y(0) 0, y(0) 0 ) o deermine C gives 1, C C1 C 0, C1C 0 Solving C1, C 3 Thus, he complee response becomes 4 y () yn() yf () e e e e e e e 3 3 Simple way o ge he complee response: The nonhomogeneous soluion is ynh() yh() yp () Ce C e e Dep. Elecronics and Informaion Eng. 16

17 Sysem Equaions We obain he complee response of he sysem by solving for C, C so ha he iniial condiions are saisfied 1 This implies y(0) 1, y(0) 1 C1 C 1, C1C 1 Solving C1 1, C 3 Thus, he complee response is 4 y() e e e 3 3 Be exacly equal o he response we already obained!! MaLab scrip: y=dsolve('dy+3*dy+*y=4*exp()','y(0)=1','dy(0)=-1',''); 17 Dep. Elecronics and Informaion Eng. 17

18 Sysem Equaions Example 3.8: Find he curren i() in he circui shown for zero inpu signal, i.e. vg () 0 The iniial condiions are i(0) A, and i(0) A/s Soluion: The loop equaion is i() 4 i() 40() i 0 The characerisic equaion is s 4s40 s j6 s j6 0 Thus, he soluion becomes Applying o he iniial condiions gives = 1/40 F = 1 H 4 () i Me 0 cos6ne 0 sin6 M, N 1039 / Dep. Elecronics and Informaion Eng. 18

19 Sysem Equaions Finally, we have 4 e 1039 i () e cos6 e sin e cos6 A 3 Damping erm Oscillaing erm clear all; =0:0.01:; y=*exp(-*).*cos(6*)+1039/300*exp(-*).*sin(6*); plo(,y,'b') hold on; plo(,4*exp(-*),'r'); hold off; xlabel(' (second)'); ylabel('i() (Amps)') 19 Dep. Elecronics and Informaion Eng. 19

20 Exponenial Signal General form of exponenial signal: If s is real, he signal decays (s<0) or grows (s>0) exponenially If s=0, he signal has consan ampliude (D-C value) If s j, he signal can express as sinusoidal form in erms of Euler s ideniy Example 3.9: Find he paricular soluion for equaion y y e 3 () 3 () cos( 50) 0 Dep. Elecronics and Informaion Eng. 0 s Ae Key word: Replace cosine funcion o exponenial form!!

21 Exponenial Signal Soluion: Since firs we need o solve he equaion * 3 j50 ( 3 j) j50 ( 3 j) e cos(50 ) e e e e, Thus, a soluion of he form () ( 3 j ) y1 Ke is expeced Subsiuing his soluion ino he differenial equaion gives ( 3 j) ( 3 j) j50 ( 3 j) ( 3 j) Ke 3Ke e e e e j K e, K 0.5e j e j40 ( 3 j) y () 0.5e e 1 y() 3 y() e e j50 ( 3 j) j50 j50 j50 j40 j90 1 Dep. Elecronics and Informaion Eng. 1

22 Exponenial Signal Consequenly, he paricular soluion becomes y () 0.5e e 0.5 e e p e 3 * j40 ( 3 j) j40 ( 3 j) cos(40 ) Dep. Elecronics and Informaion Eng.

23 Transfer Funcions Saring wih he general differenial equaion M N d y() dy() d x() dx() b1 b0y() an a1 a0x() M N d d d d s we may propose he inpu signal x () Xe and he s oupu signal y() Ye Assuming zero iniial condiions, and subsiuing and y() ino he equaion above, i becomes s Ye bsye bye a s Xe asxe a Xe M s s s N s s s 1 0 N 1 0 s Canceling common e erms and arranging gives H() s N an s as 1 a0 M 1 0 Y X s bs b x () 3 Dep. Elecronics and Informaion Eng. 3

24 Transfer Funcions, which is called Transfer Funcion presening by complex frequency s because i describes how he inpu is ransferred o he oupu in a domain The descripion is said o be in s-domain, or frequency domain, or phasor domain Example 3.10: Deermine he ransfer funcion for y() 3 y() y() 3 x() x() Y 3s1 H() s X s 3s 3s 1/3 s1 s Soluion: Pole-Zero Diagram 4 Dep. Elecronics and Informaion Eng. 4

25 Transfer Funcions Now, le s sudy how o find he soluion of equaion from he poin of view of ransfer funcion Example 3.11: Deermine he forced response for y() y() 4 y() x() x() when x () cos Soluion: Y s1 Transfer funcion is H() s X s s 4 For he paricular soluion, he phasor ransformaion of inpu signal gives s X 0, s j1 x () ReXe 5 Dep. Elecronics and Informaion Eng. 5

26 Transfer Funcions 1 j Thus, Y and 3 j For naural response, he characerisic equaion is s s s j 4 0, 1 3 Since 1, 3, he soluion becomes y () M e cos 3N e sin 3 h 0 0 Evenually, he nonhomogeneous soluion is y ( ) y ( ) y ( ) 1.403cos( 9.74 ) M e cos 3N e sin 3 nh h p s y ( ) Re Ye 1.403cos( 9.74 ) p Applying iniial condiions gives he naural response! 0 0 Applying he zero iniial condiions o ge yf () gives M01.403cos(9.74 ), N0 cos(9.74 ) sin(9.74 ) 3 6 Dep. Elecronics and Informaion Eng. 6

27 Transfer Funcions Example 3.1: Find he oupu volage y() of he sysem, if he inpu volage is he periodic signal x ( ) 4coscos All he iniial condiions are zero! Soluion: Applying KVL o he circui yields R R y() y() x() L L The ransfer funcion is / sy R Y R X H () s R L L L s R/ L 7 Dep. Elecronics and Informaion Eng. 7

28 Transfer Funcions Using he phasor ransformaion for x() gives Thus, and s 1 s 1 x ( ) Re Xe Re Xe X 4, X, s j, where and Y Y sj, R/ L1 sj, R/ L1 j R/ L X, s R/ L j1 R/ L 1 X 63 s R/ L j1 5 y1() cos 45, y() cos 63 5 Then, he paricular soluion becomes yp () y1() y() cos 45 cos Dep. Elecronics and Informaion Eng. 8 s

29 Transfer Funcions From he characerisic equaion, he homogeneous soluion is y () Ce h Applying he zero iniial condiion gives C=0, so ha he naural response vanishes! Evenually, he nonhomogeneous soluion is expressed as ynh() Ce cos45 cos63 5 Applying zero iniial condiions, he oupu volage (complee response) is 8 y () e cos45 cos Dep. Elecronics and Informaion Eng. 9

30 Transfer Funcions Sable sysem: One of he mos imporan conceps in he sudy of sysems is he noaion of sabiliy BIBO (bounded-inpu bounded-oupu) sabiliy: Signal x() is said o be bounded if is magniude does no grow wihou bound, i.e., x () B, for all A sysem is BIBO sable if, for any bounded inpu x(), he response y() is also bounded. Tha is x () B1 implies y () B 30 Dep. Elecronics and Informaion Eng. 30

31 Transfer Funcions Example 3.13: Deermine which of hese sysems is sable: (a) y() exp x(), (b) y() x( ) d Soluion: ( BIBO sable! ) (a) For x () B, y () exp x () exp x () exp B (b) Consider he uni sep funcion u() as inpu x(). Then, he oupu y() is equal o u () ( Unbounded oupu! ) y() u( ) d r() r () 1 1 (Uni sep funcion) 1 (Ramp funcion) 31 Dep. Elecronics and Informaion Eng. 31

32 Transfer Funcions Sable or unsable concep based on ransfer funcion having poles in s-plane If all poles have n 0 (LHP), he naural response dies ou wih ime so ha he forced response dominaes he sysem performance. Such sysems are BIBO sable! H() s ( Will be discussed 1 in deails a Laplace s a () a h e u () ransform! ) ( s-domain! ) ( ime-domain! ) 3 Dep. Elecronics and Informaion Eng. 3

33 Transfer Funcions If any one pole has n 0 and a conjugae pair of poles, he naural response oscillaes a sinusoidal form. The sysem is marginally sable! H() s b b h () sinb s b s jbs jb If any one pole has n 0 (RHP), he naural response grows exponenially wih ime. Such sysems are unsable! H() s 1 s a h () e a u () 33 Dep. Elecronics and Informaion Eng. 33

34 Transfer Funcions s j n n n 34 Dep. Elecronics and Informaion Eng. 34

35 Sabiliy Tess Direc es of unsable sysems is boh expensive and dangerous Insead, we can use compuer simulaion ools o evaluae sabiliy In conrol heory: use roo locus diagram, which draws he movemen of poles as gain varies Mahemaically, a sabiliy es requires Rouh- Hurwiz Crierion! Classificaion of ess Rouh: England in 1877, Hurwiz: Germany in 1895 Tes 1: Be sufficien for polynomials of 1 s - or nd - order characerisic equaions 35 Dep. Elecronics and Informaion Eng. 35

36 Sabiliy Tess For as, if here is a negaive erm in 1 s - or nd as 1 a0 - degree polynomials, a roo in RHP has he form ( s r) Tes : Sep and impulse funcions are frequenly used es signals in circuis 0, 0 Sep funcion: u () 1, 0 Impulse (or Dela) funcion: () d1, () 0 for 0 When hey urn on, hey acivae a circui s response Sep produces a D-C (because s=0) forced response On he conrary, impulse dumps energy ino circui a =0 and hen les circui s naural response ake over 36 Dep. Elecronics and Informaion Eng. 36

37 Sabiliy Tess Thus, he acquired responses of circui can es if i is sable or unsable Tes 3 (Rouh-Hurwiz crierion) : Firs sep: selec he coefficiens for wo-rows from he given characerisic equaion as as as as as as as M M1 M M3 M4 M5 M Row 1 a a a a a a a a a a Row Nex sep: complee he R-H array of numbers. For M 6, his is 37 Dep. Elecronics and Informaion Eng. 37

38 Sabiliy Tess s a a a a s a1 a3 a5 4 s b1 b3 b s c1 c3 0 0 s d1 d s e s f Dep. Elecronics and Informaion Eng. 38

39 Sabiliy Tess The R-H heorem saes The number of changes in he sign of he firs column (as we scan from op o boom) is equal o he number of roos wih posiive real pars (in RHP) Be unsable! s s s s s column Thus, if and only if here are no changes of sign in he 1 s column of he array, he sysem is sable Example 3.14: 3 consider s s s400, which is known o have wo roos wih posiive real pars There are wo changes of sign ( o 18 and 18 o 40) as required Be unsable! Why? 39 Dep. Elecronics and Informaion Eng. 39

40 Sabiliy Tess Example 3.15: Deermine he condiions on K for sabiliy if a sysem 3 has characerisic equaion s s Ks80 Soluion: The array is s s s s If K 8 0, here are no changes of sign so ha here are no roos in RHP Thus, be sable for K 4! K 8 (K 8) / Check he sign of 1 s column! 40 Dep. Elecronics and Informaion Eng. 40

41 Sabiliy Tess Example 3.16: Deermine he condiions on K for sabiliy if a sysem 3 has characerisic equaion s s sk 0 Soluion: The array is s s s s K (4 K) / 0 K 0 If we check he sign of 1 s column, we see ha he sysem is sable for and 4 K 0 K 0 0 K 4 Thus, i is sable only as long as! 41 Dep. Elecronics and Informaion Eng. 41

42 LTI Sysems Linear Time-Invarian (LTI) Sysems The sysems saisfy he lineariy and ime-invarian properies, which lead o simplificaions of mahemaical analysis and greaer insigh and undersanding of sysem behavior Lineariy: Linear sysem have he propery ha if x () y () and x () y (), hen x () x() x() y() y () y () Consider he square-law sysem of x y() () Linear sysem? 4 Dep. Elecronics and Informaion Eng. 4

43 LTI Sysems x () 1 x () y () 1 w () y () x () 1 x () y() x () w () will equal y() when he sysem is linear! Time Invariance: A sysem whose oupu does no depend explicily on ime. 43 Dep. Elecronics and Informaion Eng. 43

44 LTI Sysems A sysem is said o be ime-invarian if, when an inpu is delayed (shifed) by 0, he oupu is delayed by he same amoun x ( ) y( ) 0 0 Consider wo sysems of y() x() and y() 10 x() Sysem A: Sar wih a delay of he inpu xd () x( 0) w () xd () x ( ) y( 0) ( 0) x( 0) w () y( ) Then, 0 Now delay he oupu by 0 Clearly, herefore 0 he sysem is no ime-invarian. (I is ime-varying sysem!) x () 0 x ( ) 0 w () y( 0) y() w () will equal when he sysem is ime-invarian! 0 y( ) 0 44 Dep. Elecronics and Informaion Eng. 44

45 Homework Assignmen #3 [Problem 1] For he following condiions, find he forced response of he differenial equaion d y dy 4 y x ( ) d d x () e (a), (b) x () 3e [Problem ] Find he oupu phasor given ha x () 3e sin and he sysem is described by d y dy dx 3 y x d d d Sae he phasor ransformaion used, and ideniy X and s 45 Dep. Elecronics and Informaion Eng. 45

46 Homework Assignmen #3 [Problem 3] Deermine he ransfer funcions of he following sysems and Skech he pole-zero diagrams (a) d 3 y 4 d y 16 dy 8y d x 3x, (b) d y 3 dy d x 9x 3 d d d d d d d [Problem 4] Deermine he naural response of he sysem. Classify each as sable, unsable, or marginally sable Y s s Y ss (a) (b) 3 X s 4s8 X s s 3s 46 Dep. Elecronics and Informaion Eng. 46