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6.003: Signals and SystemsZ TransformSeptember 22, 20111

Concept Map: Discrete-Time SystemsMultiple representations of DT systems.Delay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .Difference EquationSystem Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]2

Concept Map: Discrete-Time SystemsRelations among representations.Delay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .Difference EquationSystem Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]3

Concept Map: Discrete-Time SystemsTwo interpretations of “Delay.”Delay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .Difference EquationSystem Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]4

Concept Map: Discrete-Time SystemsRelation between System Functional and System Function.Delay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .Difference EquationR z1System Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]5

Check YourselfWhat is relation of System Functional to Unit-Sample ResponseDelay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .Difference EquationSystem Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]6

Check YourselfExpand functional in a series:Y1 H(R) X1 R R21 R 2R2 3R3 5R4 8R5 · · ·1 R R2 11 R R2R R2R R2 R32R2 R32R2 2R3 2R43R3 2R43R3 3R4 3R5···H(R) 1 1 R 2R2 3R3 5R4 8R5 13R6 · · ·1 R R27

Check YourselfCoeﬃcients of series representation of H(R)1 1 R 2R2 3R3 5R4 8R5 13R6 · · ·H(R) 1 R R2are the successive samples in the unit-sample response!h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .If a system is composed of (only) adders, delays, and gains, thenH(R) h[0] h[1]R h[2]R2 h[3]R3 h[4]R4 · · · h[n]RnnWe can write the system function in terms of unit-sample response!8

Check YourselfWhat is relation of System Functional to Unit-Sample Response?Delay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayH(R) Ph[n]RnUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .Difference EquationSystem Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]9

Check YourselfWhat is relation of System Function to Unit-Sample Response?Delay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayH(R) Ph[n]RnUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .Difference EquationSystem Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]10

Check YourselfStart with the series expansion of system functional:H(R) Xh[n]RnnSubstitute R H(z) X1:zh[n]z nn11

Check YourselfWhat is relation of System Function to Unit-Sample Response?Delay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayH(R) Ph[n]RnUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .H(z) Difference EquationPh[n]z nSystem Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]12

Check YourselfStart with the series expansion of system functional:H(R) Xh[n]RnnSubstitute R H(z) X1:zh[n]z nnToday: thinking about a system as a mathematical function H(z)rather than as an operator.13

Z TransformWe call the relation between H(z) and h[n] the Z transform.H(z) X h[n]z nnZ transform maps a function of discrete time n to a function of z.Although motivated by system functions, we can deﬁne a Z trans form for any signal.X(z) X x[n]z nn Notice that we include n 0 as well as n 0 bilateral Z transform(there is also a unilateral Z transform with similar but not identicalproperties).14

Simple Z transformsFind the Z transform of the unit-sample signal.δ[n]nx[n] δ[n]X(z) Xx[n]z n x[0]z 0 1n 15

Simple Z transformsFind the Z transform of a delayed unit-sample signal.x[n]nx[n] δ[n 1]X(z) Xx[n]z n x[1]z 1 z 1n 16

Check YourselfWhat is the Z transform of the following signal. nx[n] 78 u[n]n 4 3 2 1 0 1 2 3 41.11 78 z2.11 78 z 13.z1 78 z174.z 11 78 z 15. none

Check YourselfWhat is the Z transform of the following signal. nx[n] 78 u[n]n 4 3 2 1 0 1 2 3 4X(z) Xn X 7 n nz u[n] 8n 0187 n n1z 81 78 z 1

Check YourselfWhat is the Z transform of the following signal.2 nx[n] 78 u[n]n 4 3 2 1 0 1 2 3 41.11 78 z2.11 78 z 13.z1 78 z194.z 11 78 z 15. none

Z Transform PairsThe signal x[n], which is a function of time n, maps to a Z transformX(z), which is a function of z. n7x[n] u[n]8 X(z) 11 78 z 1For what values of z does X(z) make sense?The Z transform is only deﬁned for values of z for which the deﬁningsum converges. n n XX 771 nX(z) z u[n] z n 881 78 z 1n n 0Therefore77 1z 1, i.e., z .8820

Regions of ConvergenceThe Z transform X(z) is a function of z deﬁned for all z inside aRegion of Convergence (ROC).x[n] n7u[n]8ROC: z X(z) 11 78217 z 18; z 78

Z Transform MathematicsBased on properties of the Z transform.Linearity:ifx1 [n] X1 (z)for z in ROC1andx2 [n] X2 (z)for z in ROC2then x1 [n] x2 [n] X1 (z) X2 (z) for z in (ROC1 ROC2 ).Let y[n] x1 [n] x2 [n] then X Y (z) y[n]z n n X(x1 [n] x2 [n])z nn Xx1 [n]z nn Xx2 [n]z nn X1 (z) X2 (z)22

Delay PropertyIf x[n] X(z) for z in ROC then x[n 1] z 1 X(z) for z in ROC.We have already seen an example of this property.δ[n] 1δ[n 1] z 1More generally, XX(z) x[n]z nn Let y[n] x[n 1] then X nY (z) y[n]z x[n 1]z nn n Substitute m n 1 X Y (z) x[m]z m 1 z 1 X(z)m 23

Rational PolynomialsA system that can be described by a linear diﬀerence equation withconstant coeﬃcients can also be described by a Z transform that isa ratio of polynomials in z.b0 y[n] b1 y[n 1] b2 y[n 2] · · · a0 x[n] a1 x[n 1] a2 x[n 2] · · ·Taking the Z transform of both sides, and applying the delay propertyb0 Y (z) b1 z 1 Y (z) b2 z 2 Y (z) · · · a0 X(z) a1 z 1 X(z) a2 z 2 X(z) · · ·H(z) Y (z)a0 a1 z 1 a2 z 2 · · · X(z)b0 b1 z 1 b2 z 2 · · · a0 z k a1 z k 1 a2 z k 2 · · ·b0 z k b1 z k 1 b2 z k 2 · · ·24

Rational PolynomialsApplying the fundamental theorem of algebra and the factor theo rem, we can express the polynomials as a product of factors.H(z) a0 z k a1 z k 1 a2 z k 2 · · ·b0 z k b1 z k 1 b2 z k 2 · · · (z z0 ) (z z1 ) · · · (z zk )(z p0 ) (z p1 ) · · · (z pk )where the roots are called poles and zeros.25

Rational PolynomialsRegions of convergence for Z transform are delimited by circles inthe Z-plane. The edges of the circles are at the poles.Example: x[n] αn u[n]X(z) Xαn u[n]z n n Xαn z nn 01; αz 1 11 αz 1z; z α z αROCx[n] αn u[n]zz αn 4 3 2 1 0 1 2 3 426z -planeα

Check YourselfWhat DT signal has the following Z transform?z -planez7; z 78z 8ROC7827

Check YourselfRecall that we already know a function whose Z transform is theouter region.ROC 7 nx[n] 8u[n]zz 78n 4 3 2 1 0 1 2 3 4What changes if the region changes?The original sumX(z) n X7n 08z ndoes not converge if z 87 .28z -plane78

Check YourselfThe functional form is still the same,H(z) Y (z)z. X(z)z 78Therefore, the diﬀerence equation for this system is the same,7y[n 1] y[n] x[n 1] .8Convergence inside z 78 corresponds to a left-sided (non-causal)response. Solve by iterating backwards in time:8y[n] (y[n 1] x[n 1])729

Check YourselfSolve by iterating backwards in time:8y[n] (y[n 1] x[n 1])7Start “at rest”:nx[n]y[n] 000010 10 87 20 3···n0 n8y[n] ;72 873 87··· n 87 n7n 0 u[ 1 n]830

Check YourselfPlot ny[n] 87 u[ 1 n]z -planen 4 3 2 101zz 87231ROC78

Check YourselfWhat DT signal has the following Z transform?z -planez7; z 78z 8ROC78 ny[n] 87 u[ 1 n]n 4 3 2 132012

Check YourselfTwo signals and two regions of convergence.ROC 7 nx[n] 8u[n]zz 78z -plane78n 4 3 2 1 0 1 2 3 4 ny[n] 87 u[ 1 n]z -planen 4 3 2 101zz 87233ROC78

Check YourselfFind the inverse transform of 3zX(z) 22z 5z 2given that the ROC includes the unit circle.34

Check YourselfFind the inverse transform of 3zX(z) 22z 5z 2given that the ROC includes the unit circle.Expand with partial fractions: 3z12 X(z) 2 2z 1 z 22z 5z 2Not a standard form!35

Check YourselfStandard forms:ROC 7 nx[n] 8u[n]zz 78z -plane78n 4 3 2 1 0 1 2 3 4 ny[n] 87 u[ 1 n]z -planen 4 3 2 101zz 87236ROC78

Check YourselfFind the inverse transform of 3zX(z) 22z 5z 2given that the ROC includes the unit circle.Expand with partial fractions: 3z12 X(z) 2 2z 1 z 22z 5z 2Not a standard form!Expand it diﬀerently: as a standard form: 3z2zzzzX(z) 2 12z 1 z 2z 22z 5z 2z 2Standard form: a pole at 12 and a pole at 2.37

Check YourselfRatio of polynomials in z: 3zzzX(z) 2 1z 22z 5z 2z 2– a pole at 12 and a pole at 2.z -planeROC122Region of convergence is “outside” pole at 12 but “inside” pole at 2. n1x[n] u[n] 2n u[ 1 n]238

Check YourselfPlot. n1u[n] 2n u[ 1 n]x[n] 2x[n]n39

Check YourselfAlternatively, stick with non-standard form: 3z12 X(z) 2 2z 1 z 22z 5z 2Make it look more standard:1zzX(z) z 1 2z 11z 22z 2Now 1 nu[n] 2R { 2n u[ 1 n]}2 n 111 u[n 1] 2 2n 1 u[ n]22 1 n u[n 1] { 2n u[ n]}2x[n]1x[n] R2n40

Check YourselfAlternative 3: expand as polynomials in z 1 : 3z 3z 1X(z) 2 2z 5z 22 5z 1 2z 22111 1 11 2z 12 z 1 1 2z 11 2zNow n1u[n] 2n u[ 1 n]x[n] 2x[n]n41

Check YourselfFind the inverse transform of 3zX(z) 22z 5z 2given that the ROC includes the unit circle.x[n]n42

Solving Diﬀerence Equations with Z TransformsStart with diﬀerence equation:1y[n] y[n 1] δ[n]2Take the Z transform of this equation:1Y (z) z 1 Y (z) 12Solve for Y (z):1Y (z) 1 12 z 1Take the inverse Z transform (by recognizing the form of the trans form): n1y[n] u[n]243

Inverse Z transformThe inverse Z transform is deﬁned by an integral that is not partic ularly easy to solve.Formally,1X(z)z n 1 dz2πj Cwhere C represents a closed contour that circles the origin by runningin a counterclockwise direction through the region of convergence.This integral is not generally easy to compute.x[n] This equation can be useful to prove theorems.There are better ways (e.g., partial fractions) to compute inversetransforms for the kinds of systems that we frequently encounter.44

Properties of Z TransformsThe use of Z Transforms to solve diﬀerential equations depends onseveral important properties.Propertyx[n]Linearityax1 [n] bx2 [n]Multiply by nz 1 X(z)RdX(z)dzR znx[n] XROCaX1 (z) bX2 (z) (R1 R2 )x[n 1]DelayConvolve in nX(z)x1 [m]x2 [n m]m 45X1 (z)X2 (z) (R1 R2 )

Check Yourself Find the inverse transform of Y (z) 46 2z;z 1 z 1.

Check Yourself Find the inverse transform of Y (z) 2z;z 1 z 1.y[n] corresponds to unit-sample response of the right-sided system 2 2 2Yz11 Xz 11 R1 z 1 1 R R2 R3 · · · 1 R R2 R3 · · ·1R2RR3R2 R3R3 R4R4 R5R5 R6··· ··· X Y(n 1)Rn 1 2R 3R2 4R3 · · · X1RR2R3···1RR2R3···RR2R3R4···n 0y[n] h[n] (n 1)u[n]47··················

Check YourselfTable lookup method. 2zY (z) z 1z z 1y[n] ?u[n]48

Properties of Z TransformsThe use of Z Transforms to solve diﬀerential equations depends onseveral important properties.Propertyx[n]Linearityax1 [n] bx2 [n]Multiply by nz 1 X(z)RdX(z)dzR znx[n] XROCaX1 (z) bX2 (z) (R1 R2 )x[n 1]DelayConvolve in nX(z)x1 [m]x2 [n m]m 49X1 (z)X2 (z) (R1 R2 )

Check YourselfTable lookup method. 2zY (z) z 1zz 1 21dz z zz 1dz z 1 2zdz z zz 1dz z 1 50 y[n] ? u[n] nu[n] (n 1)u[n 1] (n 1)u[n]

Concept Map: Discrete-Time SystemsRelations among representations.Delay RBlock DiagramX DelaySystem FunctionalY1Y H(R) X1 R R2DelayUnit-Sample Responseindex shifth[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .Z transformDifference EquationSystem Functionz2Y (z) 2H(z) X(z)z z 1y[n] x[n] y[n 1] y[n 2]51

MIT OpenCourseWarehttp://ocw.mit.edu6.003 Signals and SystemsFall 2011For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

Block Diagram System Functional Di erence Equation System Function Unit-Sample Response Delay Delay. strong X Y /strong . strong Y X /strong H (R ) 1 1 RR. 2. strong y /strong [ strong n /strong ] strong x /strong [ strong n /strong ] strong y /strong [ strong n /strong 1] strong y /strong [ strong n /strong 2] H (z) /p div class "b_factrow b_twofr" div class "b_vlist2col" ul li div strong File Size: /strong 796KB /div /li /ul ul li div strong Page Count: /strong 52 /div /li /ul /div /div /div

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