Laplace and Z Transforms - MIT

1 : Signals and SystemsLaplace and Z TransformsOctober 1, 2009 Mid-term Examination #1 Wednesday, October 7, 7:30-9:30pm, Walker recitations on the day of the :DT Signals and SystemsLectures 1 5 Homeworks 1 4 Homework 4 will include practice problems for mid-term , it will not collected or graded. Solutions will be book: 1 page of notes (812 11inches; front and back).Designed as 1-hour exam; two hours to sessions during open o ce ict? Contact before Friday, October 2, TimeMany continuous-time systems can be represented with di : leaky tankr0(t)r1(t)h1(t)Di erential equation representation: r1(t) =r0(t) r1(t)Last time we considered two methods to solve di erential equations: solving homogeneous and particular equations singularity matchingSolving Di erential Equations with Laplace TransformThe Laplace transform provides a particularly powerful method ofsolving di erential equations it Transforms a di erential equationinto an algebraic (whereLrepresents the Laplace transform ):di erentialalgebraicalgebraicdi erential equation solve di erentialequationalgebraicequationalgebra icanswersolution todi erential equationL solveL 1 Laplace transform .

2 De nitionLaplace transform maps a function of timetto a function (s) = x(t)e stdtThere are two important variants:Unilateral ( )X(s) = 0x(t)e stdtBilateral ( )X(s) = x(t)e stdtBoth share important properties will discuss di erences TransformsExample: Find the Laplace transform ofx1(t):0tx1(t)x1(t) ={Ae tift 00otherwiseX1(s) = x1(t)e stdt= 0Ae te stdt=Ae (s+ )t (s+ ) 0=As+ provided Re{s+ }>0which implies that Re{s}> . s-planeROCAs+ ;Re{s}> Check Yourself0tx2(t)x2(t) ={e t e 2tift 00otherwiseWhich of the following is the Laplace transform ofx2(t)? (s) =1(s+1)(s+2);Re{s}> (s) =1(s+1)(s+2);Re{s}> (s) =s(s+1)(s+2);Re{s}> (s) =s(s+1)(s+2);Re{s}> 25. none of the aboveCheck YourselfX2(s) = 0(e t e 2t)e stdt= 0e te stdt 0e 2te stdt=1s+ 1 1s+ 2=(s+ 2) (s+ 1)(s+ 1)(s+ 2)=1(s+ 1)(s+ 2)These equations converge if Re{s+ 1}>0and Re{s+ 2}>0, thusRe{s}> 1.}}

3 1 2s-planeROC1(s+ 1)(s+ 2);Re{s}> 1 Check Yourself0tx2(t)x2(t) ={e t e 2tift 00otherwiseWhich of the following is the Laplace transform ofx2(t)? (s) =1(s+1)(s+2);Re{s}> (s) =1(s+1)(s+2);Re{s}> (s) =s(s+1)(s+2);Re{s}> (s) =s(s+1)(s+2);Re{s}> 25. none of the aboveRegions of ConvergenceLeft-sided signals have left-sided Laplace Transforms (bilateral only).Example:tx3(t) 1x3(t) ={ e tift 00otherwiseX3(s) = x3(t)e stdt= 0 e te stdt= e (s+1)t (s+ 1) 0 =1s+ 1provided Re{s+ 1}<0which implies that Re{s}< 1. 1s-planeROC1s+ 1;Re{s}< 1 Left- and Right-Sided SignalsWe can concisely express left- and right-sided signals by multiplica-tion with step (t)x1(t) ={e tift 00otherwise e tu(t)tx3(t) 1x3(t) ={ e tift 00otherwise e tu( t)Left- and Right-Sided ROCsLaplace Transforms of left- and right-sided exponentials have thesame form (except ); with left- and right-sided ROCs, tu(t)time functionLaplace transform 1s-planeROC1s+ 1t e tu( t) 1 1s-planeROC1s+ 1 Check YourselfFind the Laplace transform ofx4(t).}}}}

4 0tx4(t)x4(t) =e |t| (s) =21 s2; <Re{s}< (s) =21 s2; 1<Re{s}< (s) =21+s2; <Re{s}< (s) =21+s2; 1<Re{s}<15. none of the aboveCheck YourselfX4(s) = e |t|e stdt= 0 e(1 s)tdt+ 0e (1+s)tdt=e(1 s)t(1 s) 0 +e (1+s)t (1 +s) 0=11 s Re{s}<1+11 +s Re{s}> 1=1 +s+ 1 s(1 s)(1 +s)=21 s2; 1<Re{s}<1 The ROC is the intersection of Re{s}<1and Re{s}> YourselfThe Laplace transform of a signal that is both-sided a vertical (t)x4(t) =e |t| 11s-planeROCX4(s) =21 s2 1<Re{s}<1 Check YourselfFind the Laplace transform ofx4(t).20tx4(t)x4(t) =e |t| (s) =21 s2; <Re{s}< (s) =21 s2; 1<Re{s}< (s) =21+s2; <Re{s}< (s) =21+s2; 1<Re{s}<15. none of the aboveSolving Di erential Equations with Laplace TransformsSolve the following di erential equation: y(t) +y(t) = (t)Take the Laplace transform of this { y(t) +y(t)}=L{ (t)}The Laplace transform of a sum is the sum of the Laplace Transforms (prove this as an exercise).

5 L{ y(t)}+L{y(t)}=L{ (t)}What s the Laplace transform of a derivative? Laplace transform of a derivativeAssume thatX(s)is the Laplace transform ofx(t):X(s) = x(t)e stdtFind the Laplace transform ofy(t) = x(t).Y(s) = y(t)e stdt= x(t)e stdt=x(t)e st x(t)( se st)dtThe rst term must be zero sinceX(s)converged. ThusY(s) =s x(t)e stdt=sX(s)Solving Di erential Equations with Laplace TransformsBack to the previous problem:L{ y(t)}+L{y(t)}=L{ (t)}LetY(s)represent the Laplace transform ofy(t).ThensY(s)is the Laplace transform of y(t).sY(s) +Y(s) =L{ (t)}What s the Laplace transform of the impulse function? Laplace transform of the impulse functionLetx(t) = (t).X(s) = (t)e stdt= (t)e st t=0dt= (t) 1dt= 1 Sifting property: (t)siftsout the value of the integrand att= Di erential Equations with Laplace TransformsBack to the previous problem:sY(s) +Y(s) =L{ (t)}= 1 This is a simple algebraic expression.

6 Solve forY(s):Y(s) =1s+ 1We ve seen this Laplace transform (t) =e tu(t)(why noty(t) = e tu( t) ?)Notice that we solved the di erential equation y(t) +y(t) = (t)with-out computing homogeneous and particular solutions and withoutsingularity Di erential Equations with Laplace TransformsSummary of with di erential equation: y(t) +y(t) = (t)Take the Laplace transform of this equation:sY(s) +Y(s) = 1 Solve forY(s):Y(s) =1s+ 1 Take inverse Laplace transform (by recognizing form of transform ):y(t) =e tu(t)Solving Di erential Equations with Laplace TransformsRecognizing the form ..Is there a more systematic way to take an inverse Laplace transform ?Yes .. and ,x(t) =12 j +j j X(s)estdsbut this integral is not generally easy to equation can be useful to prove will nd better ways ( , partial fractions) to compute inversetransforms for common Di erential Equations with Laplace TransformsExample 2: y(t) + 3 y(t) + 2y(t) = (t) Laplace transform :s2Y(s) + 3sY(s) + 2Y(s) = 1 Solve:Y(s) =1(s+ 1)(s+ 2)=1s+ 1 1s+ 2 inverse Laplace transform .

7 Y(t) =(e t e 2t)u(t)These forward and inverse Laplace Transforms are easy if di erential equation is linear with constant coe cients, and the input signal is an impulse of Laplace TransformsThe use of Laplace Transforms to solve di erential equations de-pends on several important (t)X(s)ROCL inearityax1(t) +bx2(t)aX1(s) +bX2(s) (R1 R2)Delay byTx(t T)X(s)e sTRMultiply byttx(t) dX(s)dsRMultiply bye tx(t)e tX(s+ )shiftRby Di erentiate intdx(t)dtsX(s)Integrate int t x( )d X(s)s (R {Re{s}>0})Convolve int x1( )x2(t )d X1(s)X2(s) (R1 R2)Z TransformThe Z transform is the DT analog of the Laplace Z transform converts a di erence (recurrence) equation into asimple algebraic (whereZrepresents the Z transform ):di erentialalgebraicalgebraicdi erential equation solve di erenceequationalgebraicequationalgebraic answersolution todi erence equationZ solveZ 1 Z transform : De nitionZ transform maps a function of discrete timento a function (z) = x(n)z nThere are two important variants:UnilateralX(z) = n=0x[n]z nBilateralX(z) = n= x[n]z nDi erences are analogous to those for the Laplace TransformsExample: Find the Z transform ofx1[n]: 4 3 2 1 0 1 2 34nx1[n]x1[n] ={(78)nifn 00otherwise (78)nu[n]X1(z) = n= (78)nz nu[n] = n=0(78)nz n=11 78z 1=zz 78provided|z|>78.}

8 4 3 2 1 0 1 2 34nx1[n] =(78)nu[n]78z-planeROCzz 78 Check YourselfLetX2(z)represent the Z transform ofx2[n]: 4 3 2 1 0 1 2nx2[n] = (78)nu[ 1 n]What is the relation betweenX2(z)andX1(z)? (z) = X1(z) (z) =X1(z) (z) =1X1(z)4. none of the aboveCheck YourselfLetX2(z)represent the Z transform ofx2[n].What is the relation betweenX2(z)andX1(z)? 4 3 2 1 0 1 2nx2[n] = (78)nu[ 1 n]X2(z) = n= (78)nu[ 1 n]z n= 1 n= (78)nz nLetl= 1 n:X2(z) = l=0(78) 1 lz1+l= 8z7 l=0(8z7)l= 8z711 8z7=zz 78provided 8z7 <1or|z|< Transforms : Check Yourself 4 3 2 1 0 1 2nx2[n] = (78)nu[ 1 n] (87)1 (87)2 (87)3 (87)478z-planeROCzz 78 4 3 2 1 0 1 2 34nx1[n] =(78)nu[n]78z-planeROCzz 78 Check YourselfLetX2(z)represent the Z transform ofx2[n]: 4 3 2 1 0 1 2nx2[n] = (78)nu[ 1 n]What is the relation betweenX2(z)andX1(z)?

9 (z) = X1(z) (z) =X1(z) (z) =1X1(z)4. none of the aboveSolving Di erence Equations with Z TransformsSolve the following di erence equation:y[n] 12y[n 1] = [n]Take the Z transform of this {y[n] 12y[n 1]}=Z{ [n]}The Z transform of a sum is the sum of the Z Transforms (provethis as an exercise).Z{y[n]} Z{12y[n 1]}=Z{ [n]}What s the Z transform of a delay?Z transform of a delayAssume thatX(z)is the Z transform ofx[n]:X(z) = n= x[n]z nFind the Z transform ofy[n] =x[n 1].Y(z) = n= y[n]z n= n= x[n 1]z nLetl=n 1:Y(z)= l= x[l]z 1 l=z 1 l= x[l]z l=z 1X(z)Solving Di erence Equations with Z TransformsBack to the previous problem:Z{y[n]} Z{12y[n 1]}=Z{ [n]}LetY(z)represent the Z transform ofy[n]. Thenz 1Y(z)is the Ztransform ofy[n 1].Y(z) 12z 1Y(z) =Z{ [n]}What s the Z transform of the unit sample function?

10 Z transform of the unit sample functionLetx[n] = [n]. Substituting in the de nition of the Z transform :X(z) = n= [n]z n=z 0= 1 Solving Di erence Equations with Z TransformsBack to the previous problem:Y(z) 12z 1Y(z) =Z{ [n]}= 1 This is a simple algebraic expression. Solve forY(z):Y(z) =11 12z 1By analogy with the Z transform ofx1[n]:y[n] =(12)nu[n]This method for solving di erence equations is analogous to usingthe Laplace transform to solve a di erential Di erence Equations with Z TransformsSummary of with di erence equation:y[n] 12y[n 1] = [n]Take the Z transform of this equation:Y(z) 12z 1Y(z) = 1 Solve forY(z):Y(z) =11 12z 1 Take the inverse Z transform (by recognizing the form of the trans-form):y[n] =(12)nu[n] inverse Z transformThe inverse Z transform has de ned by an integral that is not par-ticularly easy to ,x[n] =12 j CX(z)zn 1dswereCrepresents a closed contour that circles the origin by runningin a counterclockwise direction through the region of integral is not generally easy to equation can be useful to prove will nd better ways ( , partial fractions) to compute inversetransforms for the kinds of systems that we most frequently of Z TransformsThe use of Z Transforms to solve di erential equations depends onseveral important [n]X(z)ROCL inearityax1[n] +bx2[n]aX1(z) +bX2(z) (R1 R2)Delayx[n 1]z 1X(z)RMultiply bynnx[n] zdX(z)dzRConvolve inn m= x1[m]x2[n m]X1(z)X2(z) (R1 R2)Concept Map.