SIGNALS AND SYSTEMS 15. Z TRANSFORM SOLUTION OF BACKWARD DE S FROM DEQ S WITH INITIAL CONDITIONS
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1 73 SIGNALS AND SYSTEMS 5. Z TRANSFRM SLUTIN F BACKWARD DE S FRM DEQ S WITH INITIAL CNDITINS Intrductin In the previus handut, The Z Transfrm Slutin f Difference Equatins, the difference equatins were btained frm differential equatins where the derivatives were given by apprximatins. Thse apprximatins were based upn the frward difference. This develpment will cmplete that effrt with the backward difference apprximatin. Anther handut, Laplace t Z, develped techniques that allw direct substitutin fr the Laplace expressin t btain the Z expressin fr bth the frward and backward apprximatins. In that handut the backwards difference fr the Z transfrm was develped with the exceptin f n initial cnditins. Thus the backwards difference Z transfrm slutin including initial cnditins is piece that is missing. This handut will start with the analysis develped in Laplace t Z and include initial cnditins. Then the illustrative Examples 3, 4 and 5 that were wrked in the handut, The Z Transfrm Slutin f Difference Equatins, will be rewrked as the backward difference. Cncept Develpment Frm the handut Laplace t Z we btain Equatins 3 and 4, repeated here as Equatin and : ( u t ( In that handut, Equatin was cnverted int discrete time and the backward derivatives plugged in. ( n ( n ( n ( n ( n ( ( n u n t t ( n ( n ( t ( n( t t t u( n ( We start in here with Equatin and take the Z transfrm including initial cnditins. ( ( ( ( ( Z n (3a Z n (3b Equatins 3a and 3b are btained frm Frmulas and frm Table f the Z transfrm handut. Equatins 3a and 3b are applied t Equatin and after simplificatin we btain Equatin 4.
2 74 t t t ( (t ( t t ( t t (4 FRCED RESPNSE ( t t t ( t t t t t INITIAL CNDITIN RESPNSE Where: (, ( The slutin fr the Frced Respnse term in Equatin 4 was dne in the handut Laplace t Z. The slutin fr the Initial Cnditin term is als straight frward as sn as the initial cnditins f ( and ( can be determined. We assume that the cntinuus initial cnditins f ( t 0 and ( t 0 are given. We assume that: ( ( t 0, ( ( t 0 t ( t 0 (5 Example 3: (frm the handut, The Z Transfrm Slutin f Difference Equatins nd rder real distinct rts, ( t ( t t ( t t ( ( (6 Given the fllwing characteristic plynmial that factrs int distinct real rts: ( s 3( s (7 The DEQ that this characteristic plynmial culd cme frm is given by: 5 6 6u( t (8 Bundary cnditins are given by: ( 0, (0 (9 5, 6
3 75 Use the same time step: small relative t, t. 0 and using Equatin 5 we btain the bundary cnditins: ( ( t 0, ( (.0.96 Since the rts are real and distinct, the characteristic plynmial given by Equatin 6 will factr. The MATLAB syntax is: % rts f Z plynmial w=6^.5; =5/(*w; dt=.0; d=; ddd=(+**w*dt+w^*dt^^-; dd=-(**w*dt+*c; r=rts([d, dd, ddd] % r = [ ] Equatin 4 is prepared fr a PFE: t g ( t t (, (. 96 gg ( t t, r ( , r ( ,, ggg t g gg( ( ggg ( r(( r( ( r(( r( (0 TERM A TERM B TERM TERM A B C A ( r( ( r( ( AA BB B ( r( ( r( ( The cmplete MATLAB cde t perfrm PFE s, cmbine term A & term B and plt results : % Example 3 nd rder ver Damped Backward Z Slutin with ICs N=400; w=6^.5; t=5/(*w; dt=.0;
4 76 d=; ddd=(+*t*w*dt+w^*dt^^-; dd=-(*t*w*dt+*ddd; % rts f Z plynmial r=rts([d, dd, ddd] % r = [ ] g=w^*dt^*ddd; gg=ddd; ggg=(*t*w*dt+; % Term A PFE =; A=g*^/((-r(*(-r(; =r(; B=g*^/((-*(-r(; =r(; C=g*^/((-*(-r(; =; =.96; % Term B PFE =r(; AA=gg*(*(*ggg--/(-r(; =r(; BB=gg*(*(*ggg--/(-r(; % Cmbine Terms A and B B=B+AA; C=C+BB; % Get ut and plt fr i=:n t(i=dt*(i-; tt=t(i; (i=-*exp(-tt*3+*exp(-tt*; % S slutin (i=(a+b*r(^(i-+c*r(^(i-; end % Z Plt plt(t, title('nd rder ver Damped Backward Z Slutin with ICs' xlabel('time, sec, dt=.0sec' ylabel('(t' % Laplace Plt plt(t, title('nd rder ver Damped S Slutin with ICs' xlabel('time, sec, dt=.0sec' ylabel('(t'
5 77 The Z backward difference plt is given by Figure and the Laplace slutin plt is given by Figure. Figure Z slutin backward difference plt f Example 3
6 78 Figure Laplace slutin plt f Example 3 Example 4: nd rder real repeated rts, ( t ( ( (3 t t ( t t bserve that when, Equatin 55 becmes a perfect square: ( t ( t t ( t t ( t (3a Given the fllwing characteristic plynmial that factrs int repeated real rts: ( s (4 The DEQ that this characteristic plynmial culd cme frm is given by: 4 4 4u( t (5 Bundary cnditins are given by:
7 79 ( 0, (0 (6 The backwards bundary cnditins are the same as the last example: ( ( t 0, ( (.0.96 Fr this case:,, t.005 Equatin 3a results in the repeated rts f: (, r t t Equatin 4 is prepared fr a PFE: t ( (( t t ( ( r ( r TERM A TERM B k k3 k4 k ( ( r ( r TERM A TERM B (7 (8 k r k t, k3 ( t,, k (8a 4 TERM A B C A (9 ( r ( r TERM AA BB B (0 ( r ( r n AA B n ( n k A ( C BB r nr ( r The cmplete MATLAB cde t perfrm PFE s, cmbine term A & term B and plt results: % Example 4 Critical damped
8 80 N=600; w=; t=; dt=.005; d=; ddd=(+*t*w*dt+w^*dt^^-; dd=-(*t*w*dt+*ddd; % rts f Z plynmial rts([d, dd, ddd] r = (+w*dt^- =; =-*dt; k=r^; k=w^*dt^; k3=**(+w*dt-; k4=; % PFE A=k*(-r^-; B=k*(r-^-; C=-k*(r-^- AA=r*k3-k4; BB=k3; % Get ut and plt fr i=:n t(i=dt*(i-; tt=t(i; (i=+*tt*exp(-tt*; % S slutin (i=k*(a+(c+bb*r^(i-+(aa+b/r*(i-*r^(i-; end % Z Plt plt(t, title('nd rder Critical Damped Backward Z Slutin with ICs' xlabel('time, sec, dt=.005sec' ylabel('(t' % Laplace Plt plt(t, title('nd rder Critical Damped S Slutin with ICs' xlabel('time, sec, dt=.005sec' ylabel('(t' The backward Z is Figure 3 and the Laplace is Figure 4
9 Figure 3 backward Z slutin fr Example 4 8
10 8 Figure 4 Laplace slutin fr Example 4 The Z slutin is very clse t the Laplace slutin. As the time step gets small the Z slutin gets clser t the Laplace slutin. Example 5: nd rder real cmplex rts, ( t ( j ( j ( ( t t ( t t Given the fllwing characteristic plynmial that factrs int cmplex cnjugate rts: ( s j( s j (3 The DEQ that this characteristic plynmial culd cme frm is given by: 4 8 8u( t (4
11 83, 8 Bundary cnditins are given by: ( 0, (0 (5 Use the same time step: small relative t, t. 005 and using Equatin 5 we btain the bundary cnditins: ( ( t 0, ( t Since the rts are cmplex, the characteristic plynmial given by Equatin 6 will be left as a quadratic fr PFE. The MATLAB syntax is t cnfirm this is: % rts f Z plynmial w=6^.5; =5/(*w; dt=.0; d=; ddd=(+**w*dt+w^*dt^^-; dd=-(**w*dt+*c; r=rts([d, dd, ddd] % r = [ ] Equatin 4 is prepared fr a PFE: t g ( t t (, ( t k ( t t AA k k, BB k, k ( t g ( t ( t t ( t t TERM A AA BB ( t ( t t ( t t TERM B (6
12 84 TERM TERM C A B A (7 ( t ( t t ( t t AA BB B (8 ( t ( t t ( t t Upn perfrming the PFE n term A the quadratic prtin f term A will be cmbined with term B. The cmplete MATLAB cde t perfrm PFE s, cmbine the quadratic part f term A with term B and plt results : % Example 5 Backwards Difference % <, N=600; w=*^.5; =^-.5; dt=.005; =; =-*dt; dd=(+**w*dt+w^*dt^^-; d=-(+**w*dt*dd; g=w^*dt^*dd; % PFE C=g*(+d+dd^-; A=g-C; B=C*dd; % IC terms k=(+**w*dt+w^*dt^^-; k=(+**w*dt*-; AA=k*k; BB=-k*; % Cmbine frced plus IC A=A+AA; B=B+BB; % Cmpute inverse Z cnstants alph=dd^.5; a=d/; p=((a^*dd+b^-*a*a*b/(dd-a^^.5; % the minus frm A and B is here beta=acs(-a/alph; theta=atan((a*a-b/(a*(dd-a^^.5;
13 85 % Make a time recrd t plt % (i=(c+p*alph^(i-*cs(beta*(i-+theta This is the respnse fr i=:n t(i=dt*(i-; tt=t(i; vv(i=+exp(-*tt*sin(*tt; % the Laplace slutin (i=(c+p*alph^(i-*cs(beta*(i-+theta; end % The Z plt plt(t, title('nd rder Under Damped Backward Z Slutin ' xlabel('time, sec, dt=.005sec' ylabel('(t' % The Laplace plt plt(t,vv title('nd rder Under Damped S Slutin ' xlabel('time, sec, dt=.005sec' ylabel('(t' Figure 5 gives the backwards Z plt f the slutin and Figure 6 gives the Laplace plt f the slutin. Figure 5 backward Z slutin fr Example 5
14 86 Figure 6 Laplace slutin fr Example 5 As with the ther examples Figure 5 the backward Z f Example 5 is very clse t the Laplace slutin given by Figure 6. As with all the examples as the time step gets small the Z slutin appraches the Laplace slutin.
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