Signals and Systems Lecture 8: Z Transform
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1 Signals and Systems Lecture 8: Z Transform Farzaneh Abdollahi Department of Electrical Engineering Amirkabir University of Technology Winter 2012 Farzaneh Abdollahi Signal and Systems Lecture 8 1/29
2 Introduction Relation Between LT and ZT ROC Properties The Inverse of ZT ZT Properties Analyzing LTI Systems with ZT Geometric Evaluation LTI Systems Description Unilateral ZT Farzaneh Abdollahi Signal and Systems Lecture 8 2/29
3 Z transform (ZT) is extension of DTFT Like CTFT and DTFT, ZT and LT have similarities and differences. We had defined x[n] = z n as a basic function for DT LTI systems,s.t. z n H(z)z n In Fourier transform z = e jω, in other words, z = 1 In Z transform z = re jω By ZT we can analyze wider range of systems comparing to Fourier Transform The bilateral ZT is defined: X (z) = x[n]z n X (re jω ) = x[n](re jω ) n = = F{x[n]r n } {x[n]r n }e jωn arzaneh Abdollahi Signal and Systems Lecture 8 3/29
4 Region of Convergence (ROC) Note that: X (z) exists only for a specific region of z which is called Region of Convergence (ROC) ROC: is the z = re jω by which x[n]r n converges: ROC : {z = re jω s.t. n= x[n]r n < } Roc does not depend on ω Roc is absolute summability condition of x[n]r n If r = 1, i,e, z = e jω X (z) = F{x[n]} ROC is shown in z-plane arzaneh Abdollahi Signal and Systems Lecture 8 4/29
5 Example Consider x[n] = a n u[n] X (z) = n= an u[n]z n = n=0 (az 1 ) n If z > a X (z) is bounded X (z) = z z a, ROC : z > a arzaneh Abdollahi Signal and Systems Lecture 8 5/29
6 Example Consider x[n] = a n u[ n 1] X (z) = n= an u[ n 1]z n 1 n=0 (a 1 z) n If a 1 z < 1 z < a, X (z) is bounded X (z) = z z a, ROC : z < a arzaneh Abdollahi Signal and Systems Lecture 8 6/29
7 In the recent two examples two different signals had similar ZT but with different Roc To obtain unique x[n] both X (z) and ROC are required If X (z) = N(z) D(z) Roots of N(z) zeros of X(z); They make X(z) equal to zero. Roots of D(z) poles of X(z); They make X(z) to be unbounded. arzaneh Abdollahi Signal and Systems Lecture 8 7/29
8 Relation Between LT and ZT In LT: x(t) L X (s) = x(t)est dt = L{x(t)} Now define t = nt : X (s) = lim T 0 n= x(nt )(est ) n.t = lim T 0 T n= x[n](est ) n In ZT: x[n] Z X (z) = n= x[n]z n = Z{x[n]} by taking z = e st ZT is obtained from LT. jω axis in s-plane is changed to unite circle in z-plane arzaneh Abdollahi Signal and Systems Lecture 8 8/29
9 z-plane z < 1 (insider the unit circle) special case: z = 0 z > 1 (outsider the unit circle) special case: z = z = cte (a circle) s-plane Re{s} < 0 (LHP) special case: Re{s} = Re{s} > 0 (RHP) special case: Re{s} = Re{s} = cte (a vertical line) arzaneh Abdollahi Signal and Systems Lecture 8 9/29
10 ROC Properties ROC of X (z) is a ring in z-plane centered at origin ROC does not contain any pole If x[n] is of finite duration, then the ROC is the entire z-plane, except possibly z = 0 and/or z = X (z) = N 2 n=n 1 x[n]z n If N 1 < 0 x[n] has nonzero terms for n < 0, when z positive power of z will be unbounded If N2 > 0 x[n] has nonzero terms for n > 0, when z 0 negative power of z will be unbounded If N1 0 only negative powers of z exist z = ROC If N 2 0 only positive powers of z exist z = 0 ROC Example: ZT LT δ[n] 1 1 ROC: all z δ(t) 1 ROC: all s δ[n 1] z 1 ROC: z 0 δ(t T ) e st ROC: Re{s} δ[n + 1] z ROC: z δ(t + T ) e st ROC: Re{s} arzaneh Abdollahi Signal and Systems Lecture 8 10/29
11 ROC Properties If x[n] is a right-sided sequence and if the circle z = r 0 is in the ROC, then all finite values of z for which z > r 0 will also be in the ROC. If x[n] is a left-sided sequence and if the circle z = r 0 is in the ROC, then all finite values of z for which z < r 0 will also be in the ROC. If x[n] is a two-sided sequence and if the circle z = r 0 is in the ROC, then ROC is a ring containing z = r 0. If X (z) is rational The ROC is bounded between poles or extends to infinity, no poles of X (s) are contained in ROC If x[n] is right sided, then ROC is in the out of the outermost pole If x[n] is causal and right sided then z = ROC If x[n] is left sided, then ROC is in the inside of the innermost pole If x[n] is anticausal and left sided then z = 0 ROC If ROC includes z = 1 axis then x[n] has FT arzaneh Abdollahi Signal and Systems Lecture 8 11/29
12 The Inverse of Z Transform (ZT) By considering r fixed, inverse of ZT can be obtained from inverse of FT: x[n]r n = 1 2π 2π X (rejω }{{})e jωn dω z x[n] = 1 2π 2π X (rejω )r n e (jω)n dω assuming r is fixed dz = jzdω x[n] = 1 2πj X (z)z n 1 dz Methods to obtain Inverse ZT: 1. If X (s) is rational, we can use expanding the rational algebraic into a linear combination of lower order terms and then one may use X (z) = A i x[n] = A 1 a i z 1 i a i u[n] if ROC is out of pole z = a i X (z) = A i x[n] = A 1 a i z 1 i a i u[ n 1] if ROC is inside of z = a i Do not forget to consider ROC in obtaining inverse of ZT! arzaneh Abdollahi Signal and Systems Lecture 8 12/29
13 Methods to obtain Inverse ZT: 2. If X is nonrational, use Power series expansion of X(z), then apply δ[n + n 0 ] z n 0 Example: X (z) = 5z 2 z + 3z 3 x[n] = 5δ[n + 2] δ[n + 1] + 3δ[n 3] 3. If X is rational, power series can be obtained by long division Example: X (z) = 1 1 az 1, z > a 1 + az 1 az 1 az 1 + a 2 z az 1 1+az 1 +(az 1 ) x[n] = a n u[n] arzaneh Abdollahi Signal and Systems Lecture 8 13/29
14 Methods to obtain Inverse ZT: Example: X (z) = 1 1 az 1, z < a X (z) = a 1 z( 1 1 a 1 z ) 1 + a 1 z a 1 z a 1 z + a 2 z a 1 z 1+a 1 z+(a 1 z) x[n] = a n u[ n 1] arzaneh Abdollahi Signal and Systems Lecture 8 14/29
15 ZT Properties Linearity: ax 1 [n] + bx 2 [n] ax 1 (z) + bx 2 (z) ROC contains: R 1 R2 If R1 R2 = it means that ZT does not exit By zeros and poles cancelation ROC can be lager than R1 R2 Time Shifting:x[n n 0 ] z n 0 X (z) with ROC=R (maybe 0 or is added/omited) If n0 > 0 z n0 may provide poles at origin 0 ROC If n0 < 0 z n0 may eliminate from ROC Time Reversal: x[ n] X ( 1 z ) with ROC= 1 R Scaling in Z domain: z n 0 x[n] X ( z z 0 ) with ROC = z 0 R If X (z) has zero/poles at z = a X ( z z 0 ) has zeros/poles at z = z 0 a dx (z) Differentiation in the z-domain: nx[n] z dz with ROC = R Convolution: x 1 [n] x 2 [n] X 1 (z)x 2 (z) with ROC containing R 1 R 2 arzaneh Abdollahi Signal and Systems Lecture 8 15/29
16 ZT Properties Conjugation: x [n] X (z ) with ROC = R If x[n] is real X (z) = X (z ) If X (z) has zeros/poles at z 0 it should have zeros/poles at z0 as well Initial Value Theorem: If x[n] = 0 for n < 0 then x[0] = lim z X (z) lim z X (z) = lim z n=0 x[n]z n = x[0] + lim z n=1 x[n]z n = x[0] For a casual x[n], if x[0] is bounded it means # of zeros are less than or equal to # of poles (This is true for CT as well) Final Value Theorem: If x[n] = 0 for n < 0 and x[n] is bounded when n then x( ) = lim z 1 (1 z 1 )X (z) arzaneh Abdollahi Signal and Systems Lecture 8 16/29
17 Analyzing LTI Systems with ZT ZT of impulse response is H(z) which is named transfer function or system function. Transfer fcn can represent many properties of the system: Casuality: h[n] = 0 for n < 0 It is right sided ROC of a causal LTI system is out of a circle in z-plane, it includes Note that the converse is not always correct For a system with rational transfer fcn, causality is equivalent to ROC being outside of the outermost pole (degree of nominator should not be greater than degree of denominator) Stability: h[n] should be absolute summable its FT converges An LTI system is stable iff its ROC includes unit circle A causal system with rational H(z) is stable iff all the poles of H(z) are inside the unit circle DC gain in DT is H(1) and in CT is H(0) arzaneh Abdollahi Signal and Systems Lecture 8 17/29
18 Geometric Evaluation of FT by Zero/Poles Plot Consider X 1 (z) = z a X1 : length of X 1 X 1 : angel of X 1 Now consider X 2 (s) = 1 z a = 1 X 1 (s) X2 = 1 X 1(z) X2 = X 1 arzaneh Abdollahi Signal and Systems Lecture 8 18/29
19 For higher order fcns: X (z) = M R i=1 (z β i ) P j=1 (z α j ) R X (z) = M i=1 z β i P j=1 z α j X (z) = M + R i=1 (z β i) R j=1 (z α j) Example:H(z) = 1, z > a, a < 1+az 1 1, a real h(t) = a n u[n] H(e jω ) = v1 v 2, H(e jω ) = 1 v 2 at ω = 0, H(e jω ) = 1 is max 1 a 0 < ω < π: ω H(e jω ) at ω = π, H(e jω ) = 1 1+a H(e jω ) = v 1 v 2 = ω v 2 at ω = 0, H(e jω ) = 0 0 < ω < π, ω H(e jω ) at ω = π, H(e jω ) = 0 Farzaneh Abdollahi Signal and Systems Lecture 8 19/29
20 First Order Systems a in DT first order systems plays a similar role of time constant τ in CT a H(e jω ) at ω = 0 Impulse response decays more rapidly Step response settles more quickly In case of having multiple poles, the poles closer to the origin, decay more rapidly in the impulse response arzaneh Abdollahi Signal and Systems Lecture 8 20/29
21 arzaneh Abdollahi Signal and Systems Lecture 8 21/29
22 Second Order Systems Consider a second order system with poles at z 1 = re jθ, z 2 = re jθ H(z) = z 2 (z z 1 )(z z 2 ) = 1 1 2rcosθz 1 +r 2 z 2, 0 < r < 1, 0 < θ < π h[n] = r n sin(n+1)θ sinθ u[n] H = 2 v 1 v 2 v 3 Starting from ω = 0 to ω = π: At first v 2 is decreasing v2 is min at ω = θ max H Then v2 is increasing If r is closer to unit, then H is greater at poles and change of H is sharper If r is closer to the origin, impulse repose decays more rapidly and step response settles more quickly. arzaneh Abdollahi Signal and Systems Lecture 8 22/29
23 Bode Plot of H(jω) arzaneh Abdollahi Signal and Systems Lecture 8 23/29
24 LTI Systems Description N k=0 a ky[n k] = M k=0 b kx[n k] N k=0 a kz k Y (z) = M k=0 b kz k X (z) H(z) = Y (z) X (z) = M k=0 b kz k N k=0 a kz k ROC depends on placement of poles boundary conditions (right sided, left sided, two sided,...) arzaneh Abdollahi Signal and Systems Lecture 8 24/29
25 Feedback Interconnection of two LTI systems: Y (z) = Y1 (z) = X 2 (z) X 1 (z) = X (z) Y 2 (z) = X (z) H 2 (z)y (z) Y (z) = H 1 (z)x 1 (z) = H 1 (z)[x (z) H 2 (z)y (z)] Y (z) X (z) = H(z) = H 1(z) 1+H 2(z)H 1(z) ROC: is determined based on roots of 1 + H2 (z)h 1 (z) arzaneh Abdollahi Signal and Systems Lecture 8 25/29
26 Block Diagram Representation for Casual LTI Systems We can represent a transfer fcn by different methods: Example: H(z) = 1 2z z 1 = ( z 1 )(1 2z 1 ) arzaneh Abdollahi Signal and Systems Lecture 8 26/29
27 Unilateral ZT It is used to describe casual systems with nonzero initial conditions: X (z) = 0 x[n]z n = UZ{x[n]} If x[n] = 0 for n < 0 then X (z) = X (z) Unilateral ZT of x[n] = Bilateral ZT of x[n]u[n] If h[n] is impulse response of a casual LTI system then H(z) = H(z) ROC is not necessary to be recognized for unilateral ZT since it is always outside of a circle For rational X (z), ROC is outside of the outmost pole arzaneh Abdollahi Signal and Systems Lecture 8 27/29
28 Similar Properties of Unilateral and Bilateral ZT Convolution: Note that for unilateral ZT, If both x 1 [n] and x 2 [n] are zero for t < 0, then X (z) = X 1 (z)x 2 (z) Time Scaling Time Expansion Initial and Finite Theorems: they are indeed defined for causal signals Differentiating in z domain: The main difference between UL and ZT is in time differentiation: UZ{x[n 1]} = 0 x[n 1]z n = x[ 1] + n=1 x[n 1]z n = x[ 1] + m=0 x[m]z m 1 UZ{x[n 1]} = x[ 1] + z 1 X (z) UZ{x[n 2]} = x[ 2] + z 1 x[ 1] + z 2 X (z) UZ{x[n + 1]} = n=0 x[n + 1]z n = m=1 x[m]z m+1 ±x[0]z UZ{x[n + 1]} = zx (z) zx[0] UZ{x[n + 2]} = z 2 X (z) z 2 x[0] zx[1] Follow the same rule for higher orders arzaneh Abdollahi Signal and Systems Lecture 8 28/29
29 Example Consider y[n] + 2y[n 1] = x[n], where y[ 1] = β, x[n] = αu[n] Take UZ: Y(z) + 2[β + z 1 Y(z)] = X (s) Y(z) = 2β 1 + 2z 1 + X (z) 1 + 2z 1 }{{} ZSR } {{ } ZIR Zero State Response (ZSR): is a response in absence of initial values H(z) = Y(z) X (z) Transfer fcn is ZSR ZSR: Y 1(z) = α (1+2z 1 )(1 z 1 ) Zero Input Response (ZIR): is a response in absence of input (x[n] = 0) ZIR: Y 2 (z) = 2β 1+2z 1 y 2 [n] = 2β( 2) n u[n] y[n] = y 1 [n] + y 2 [n] arzaneh Abdollahi Signal and Systems Lecture 8 29/29
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