VI. Z Transform and DT System Analysis

Transcription

1 Summer 2008 Signals & Systems S.F. Hsieh VI. Z Transform and DT System Analysis Introduction Why Z transform? a DT counterpart of the Laplace transform in CT. Generalization of DT Fourier transform: z n (re jω ) n Transfer function of LTI system: Laplace transform converts a differential equation y (t) + a y (t) + a 0 y(t) bx(t) to get its transfer function H(s). Z transform converts a difference equation: y[n] α 2 y[n 2] + α y[n ] + βx[n] to get H(z). System stability analysis: poles & zeros 2 Z-transforms. Definition(2-sided Z-transform): X(z) x[n] z n [Ex] If x[n] 3δ[n + ] 2δ[n] + 5δ[n ] + 4δ[n 2] δ[n 3], then its Z-transform X(z) 3z 2 + 5z + 4z 2 z 3 is simply the polynomial(in z ) with coefficients being x[n]. Likewise, if X(z) 3z 2+5z +4z 2 z 3, then we know(by extracting its coefficients before z n ) its inverse Z-transform is x[n] 3δ[n + ] 2δ[n] + 5δ[n ] + 4δ[n 2] δ[n 3]. Note that a less useful Unilateral(one-sided) Z-transform is defined as: X(z) x[n] z n n0 If x[n] is a causal sequence, then two-sided and one-sided Z-transforms are identical. 2. z re jω is a complex variable (recall s σ + jω in the Laplace transform), X(z) x[n]r n e jωn DTFT{r n x[n]} If r, i.e., z e jω, then Z-transform becomes the DTFT (discrete-time Fourier transform): X DTFT (e jω ) x[n]e jωn X Z trans (z) ze jω We may also consider X(z) as the projection of x[n] onto the damped complex exponential r n e jωn, or consider x[n] as a linear combination of many damped complex exponentials r n e jωn. VI -

2 They are some DT sequences whose DTFT do not converge, such as x[n] 2 n u[n]. By imposing a exponential weighting function r n, its Z-transform does exist: X(z) n0 (2z ) n 2z so long as 2z < or r > 2. As a result, Z-transform allows us to perform transform analysis of unstable systems. 3. Inverse Z-transform: x[n] 2πj À X(z)z n dz [Pf] Because X(z) zre jω X(re jω ) DTFT{r n x[n]}, r n x[n] Inverse DTFT {X(re jω )} X(re jω )e jωn dω 2π 2π x[n] X(re jω )(re jω ) n dω 2π 2π because z re jω and dz rje jω dω jzdω Á X(z)z n dz 2πj z Just like Inverse Laplace transform, you need to study the Complex Variables course before solving inverse Z-transform s contour integral. 4. Region of convergence(roc): the values of z in the complex domain for which X(z) converges. [Ex] x[n] α n u[n], then X(z) n0 αn z n αz if αz <. Thus ROC : z > α. Note: For two-sided Z-transform, ROC must be specified to assure unique inverse Z-transform. [Ex] x [n] α n u[n] X (z) x 2 [n] α n u[ n ] X 2 (z) x [n]z n x 2 [n]z n Though X (z) X 2 (z), but ROC ROC 2, and x [n] x 2 [n]. 5. Z-transform Examples: (a) δ[n]. (b) δ[n k] z k. (c) u[n] z, if z >. (d) α n u[n] n0 αn z n n0 (αz ) n αz, if z > α. αz, ROC : z > α αz, ROC 2 : z < α VI - 2

3 3 Properties of Z-transform. Linear: If x [n] X (z) and x 2 [n] X 2 (z), then α x [n] + α 2 x 2 [n] α X (z) + α 2 X 2 (z). [Ex] We know e ±jω 0n u[n] e ±jω 0z, Therefore, the Z-transform of cos ω 0 n u[n] 2 [ejω 0n + e jω 0n ] u[n] is [ ] 2 e jω 0 z + e jω 0 z 2. Time-delay: If x[n] X(z), then x[n m] z m X(z). 2 (e jω 0 + e jω 0 )z 2 (e jω 0 + e jω 0)z + z 2 cos ω 0 z 2cos ω 0 z + z 2 [Pf] Z{x[n m]} x[n m]z n x[n m]z (n m) m z m x[k]z k k z m X(z) For this reason, the delay operator is often represented by a z operator: x[n] Unit Delay y[n] x[n ] is equivalent to (take Z-transform) X(z) z z X(z) As for the -sided Z-transform, if x[n]u[n] X(z), then x[n m]u[n] z m X(z)+z m m k x[ k]z+k. (Proof is omitted.) 3. Convolution: [Pf] x [n] x 2 [n] X (z) X 2 (z) Z{x [n] x 2 [n]} Z{ l x [l]x 2 [n l]} n z n l x [l] x 2 [n l] l l l x [l] x 2 [n l]z n n x [l] z l x 2 [n l]z (n l) n x [l] z l X 2 (z) X (z)x 2 (z) VI - 3

4 [Ex] Recall the polynomial method. Express x[n] δ[n + ] + 2δ[n] + 3δ[n ] and h[n] δ[n] + 2δ[n ] + δ[n 2] as X(z) x[n]z n x[ 2]z 2 + x[ ]z + x[0]z 0 + x[]z + x[2]z 2 + z + 2 z z H(z) h[n]z n h[ 2]z 2 + h[ ]z + h[0]z 0 + h[]z + h[2]z 2 + z z + z 2 Y (z) y[n]z n y[ 2]z 2 + y[ ]z + y[0]z 0 + y[]z + y[2]z 2 + z + 4 z z + 8 z z 3 If y[n] x[n] h[n], then Y (z) X(z) H(z), and y[n] can be found from the coefficients of the polynomial Y (z) X(z) H(z). Thus y[n],4,8,8,3 for n,0,,2,3, respectively. 4. Modulation(Frequency-shift): α n x[n] X(z/α) [Pf] n x[n]αn z n n x[n](α z) n X(z) z:α z X(z/α). [Ex] We know cos ω 0 n u[n] cos ω 0z 2 cos ω 0 z +z 2, thus α n cos ω 0 n u[n] αcos ω 0 z 2α cos ω 0 z +α 2 z 2. This property is useful in designing a BPF from a LPF prototype. (by choosing α as e ±jω 0 ) [Ex] (a) Consider a -point averager with impulse response h [n] u[n ] u[n + ], then its Z-transform is H (z) + z + z z 0 and its frequency response is H (e jω ) H (z) ze jω (b) Now, let h 2 [n] [e jω 0 ] n h [n], then H 2 (z) H (e jω 0 z) ω sin 2 sin ω e j5ω 2 and the frequency response is obtained by (freq-)shifting H (e jω ) to ω 0 : H 2 (e jω ) H (e jω 0 e jω ) H (e j(ω ω 0) ) which is a BPF with midband frequency located at ω 0. (c) If ω 0 π, then h 3 [n] e jπn h [n] ( ) n h [n], and H 3 (z) z + z 2 z z 0 and frequency-shift H (ω) to π: H 3 (e jω ) H (e j(ω π) ) which is a HPF.(note that π is the highest frequency allowed for a digital signal.) VI - 4

5 5. Differentiation: nx[n] z d dz X(z) [Pf] d dz X(z) nx[n] z n z (nx[n])z n. 6. Accumulation(Integration): n x[l] X(z) z l0 [Pf] n l0 x[l] x[n] u[n] X(z)Z{u[n]}. Notice that for a causal sequence x[n], we have x[n] u[n] l0 x[l]u[n l] n l0 x[l]. For continuous-time system, we say that the impulse response of an integrator(y(t) t x(τ)dτ, an LTI system) is a step function u(t), because y(t) u(t) when x(t) δ(t). 7. Conjugation: x [n] X (z ) 8. Time reversal: x[ n] X(/z). 9. Initial value: x[0] lim z X(z). because lim z [x[0] + x[]z + x[2]z 2 + ] x[0]. 0. Final value: x[ ] lim z (z ) X(z), if poles of (z )X(z) are inside the unit circle. [Pf] x[n + ] z{x(z) x[0]} lim (z )X(z) zx[0] lim N n0 N n0 x[n + ] z n (x[n + ] x[n])z n ( as z ) lim (x[n + ] x[0]) N VI - 5

6 4 Inverse Z-transform. Power-series expansion: (a) X(z) e α/z + α z + α2 2! z 2 +, thus x[n] αn n! u[n]. (b) X(z) log(+az ), with z > a, then X(z) ( ) n+ a n z n n n and x[n] ( ) n+ an n u[n]. 2. Long division. If ROC is exterior of circle, then x[n] is right-sided, express X(z) as ratio of polynomials in z and carry out long division(arrange the polynomials in order of,z,z 2,...: X(z) 2 + 2z + z 2 + z, z > X(z) z 2 z 3 + z 4 +. If ROC is interior of circle, then x[n] is left-sided, express X(z) as ratio of polynomials in z X(z) 2 + z + 2z + z + z, z < X(z) + z z2 + z Partial fraction expansion/table look-up X(z) M k0 b kz k N k0 a kz k (a) if M < N and all poles are first-order: X(z) k A k d k z, where d k s are the nonzero poles of X(z), and then A k ( d k z )X(z) zdk, x[n] A k (d k ) n u[n]. k [Ex] (Quadratic form, complex-conjugate poles) 2 + 2z X(z) z + z 2 2( + z ) ( e jπ/3 z )( e jπ/3 z ) partial fraction expansion ( ) ( ) 2+2e jπ/3 2+2e jπ/3 e j2π/3 e jπ/3 z + e j2π/3 e jπ/3 z ( ) + ( ) 2 e jπ/3 e jπ/3 z + 2 e jπ/3 e jπ/3 z x[n] 2e jπ/3 e j(π/3)n u[n] + ( ) 2R{ } 4cos[2π(n )/6]u[n] VI - 6

7 (b) if M N and all first-order poles, M N X(z) B r z r A k + d k z, r0 k where B r s are obtained by long division of the numerator by the denominator. Then M N x[n] B r δ[n r] + A k (d k ) n u[n] [Ex] r0 k X(z) 4 + 3z 2 z z z > 2z 2, + z + + z z x[n] δ[n] + δ[n ] + ( ) n u[n] + 2( 2 )n u[n] (c) if M N and X(z) has a pole of order s at z d i : M N X(z) B r z r A k s + d k z + C m ( d i z ) m, r0 where the coefficients C m are C m (s m)!( d i ) s m k,k i (d) Several poles of multiple orders(omitted). m { d s m [ ( di dw s m w) s X(w ) ] } wd i Note that in carrying out partial fraction expansion for H(z), a technique is commonly used to expand X(z)/z first. Examples are: X(z) z i k i z b i, then X(z) k i i b i, and x[n] z i k ib n i. [Ex] X(z) z(2z2 z+2) (z )(z 2) 3 X(z) z X(z) x[n] Quadratic factor: X(z)/z 2z2 z + 2 (z )(z 2) 3 α z + α 2 (z 2) 3 + α 3 (z 2) 2 + α 4 (z 2) 3 z + 2 (z 2) 3 + (z 2) (z 2) 3 2z 2 + z ( 2z ) 3 + z ( 2z ) ( 2z ) table look up [ n(n ) 3 2 (2) n n 8 2 (2)n + 3(2) ]u[n] n {3 + 4 } [n2 + n 2] (2) n u[n] 2(3z + 7) (z )(z 2 6z + 25) 2 z + Az + B z 2 6z + 25 A 2,B 6 x[n] { (5) n cos[0.927n 2.246]} u[n] VI - 7

8 5 Transfer Functions: H(z). H(z) is defined as the ratio of Z-transforms of the output and the input: H(z) Y (z)/x(z) 2. The system function H(z) is defined as the Z-transform of the impulse response: h[n] H(z). (a) (Pf) because Y (z) H(z)X(z) and y[n] h[n] x[n]. (b) (Pf2) Eigen interpretation(why the system function H(z) is defined as the Z-transform of h[n]): If the input is a complex exponential sequence x[n] z n r n e jωn, y[n] h[n] x[n] h[k]x[n k] k k { k h[k]z (n k) h[k]z k } H(z) z n //x[n] z n We find that z n r n e jωn is an eigenfunction for the LTI system, and H(z) corresponds to its eigenvalue. [Ex] Suppose an LTI system has H(z) + z + 3z 2. Find its output y[n] if x[n] 2 n + (3e j π 2 ) n. (Hint: eigenvectors) y[n] H(z) z2 2 n + H(z) z3e j π 2 (3ej π 4 ) n 9 4 2n The frequency response H(e jω ) is the transfer function H(z) evaluated at z e jω H(e jω ) H(z) ze jω For a system equation: a 0 y[n] + a y[n ] + + a N y[n N] b 0 x[n] + b x[n ] + + b M x[n M] Take Z-transform on the difference equation (note that y[n l] z l Y (z)), (a 0 + a z + + a N z N )Y (z) (b 0 + b z + + b M z M )X(z), then the system function is H(z) Y (z) X(z) M r0 b rz r N k0 a kz k (b 0 M i ) ( c iz ) a N 0 j ( d jz ), where c i s and d j s are zeros and poles, respectively, of H(z). If then H(z) M N r0 B r z r + k A k d k z, VI - 8

9 (a) Impulse response is h[n] M N r0 (b) Magnitude response: B r δ[n r] + A k d n k u[n]. k 20log 0 H(e jω ) 20log 0 b 0 a 0 + (c) Phase response: H(e jω ) [ b0 a 0 M 20log 0 c k e jω k ] M + [ c k e jω ] k [ d k e jω ]. 4. Transfer function H(z) fully characterizes an LTI system, others are: k System equation: constant-coefficient difference equation 20log 0 d k e jω (db). System block diagram: made up with adders, delays, and constant gain elements. Impulse response h[n] to relate i/o as y[n] x[n] h[n] k DTFT Impulse Resp. h[n] Z{ } Z { } Freq. Resp. H(e jω ) Transfer Func H(z) z e jω Diff. Eqn y[n]...x[n] Block Diagram D, gain, Note (a) the real or actual domain is the n domain(time domain) where the signals are generated and processed, and where the implementation of filters takes place. (b) The frequency domain ω has physical significance when analyzing sound, but is seldom used for implementation(fast Fourier transform + multiplications can replace convolution) (c) The z domain exists primarily for its convenience in mathematical analysis and synthesis. 5. Ex [Ref: G.E. Carlson, Signal and Linear System Analysis, Example 4.22](Continued from above, except for zero initial conditions) VI - 9

10 (a) Block diagram(in time-domain or Z-domain): w[n] x[n] 0.25 Delay v[n] 0.4 Delay y[n] note that In general, the state(dummy)-variables w[n] s are placed before or after the delay elements. All loops in a block diagram must have delays. (b) Difference eqn(in time-domain): w[n] 0.25x[n] + 0.4v[n ] v[n] w[n ] + 0.5x[n] + 0.3v[n ] y[n] 0.2v[n ] + 0.x[n] (c) Take Z-transform first(instead of directly finding y[n] in terms of x[n]) W(z) 0.25X(z) + 0.4z V (z) V (z) z W(z) + 0.5X(z) + 0.3z V (z) Y (z) 0.2z V (z) + 0.X(z) which can also be obtained directly from the block diagram in Z-domain. (d) Transfer function: H(z) Y (z) X(z) z + 0.0z 2 0.3z 0.4z 2 Poles are 0.8 and 0.5, both are inside unit-circle. This is a stable causal LTI system. (e) Difference eqn relating y[n] and x[n]: From above Y (z) [ 0.3z 0.4z 2 ] X(z) [ z + 0.0z 2 ] Take Inverse Z-transforms on both sides, y[n] 0.3y[n ] 0.4y[n 2] 0.x[n] x[n ] + 0.0x[n 2] (f) Impulse response h[n] is the inverse Z-transform of H(z): A h[n] Z { 0.8z + A z 0.025} 0.025δ[n] + [A (0.8) n + A 2 ( 0.5) n ] u[n]. 6. If the system function H(z) is known, then we can obtain its difference equation relating y[n] and x[n] as follows: H(z) Y (z) X(z) 4z z + z 2 VI - 0

11 multiply Y (z) and X(z) with the denominator and numerator polynomials, respectively: Y (z) [ 3 2 z + z 2 ] X(z) [ 4z 2 ] Inverse Z-transform on both sides (note that z Y (z) y[n ]) y[n] 3 y[n ] + y[n 2] + x[n] 4x[n 2] 2 7. If an LTI system is stable & causal, then all poles of H(z) must lie inside the unit circle. [Pf] From the inverse Z-transform of partial fraction expansion, a pole at α corresponds to a causal exponential time response α n u[n], hence α < assures stability. H(z) k h[n] k A k α k z, A k α n k u[n] which is causal & stable if α k <, k 8. Inverse systems: Let H i (z) be the inverse of H(z), i.e., H(z)H i (z). (a) If x[n] is the input to H(z) and y[n] x[n] h[n] is its output, obtained from convolution. The purpose of H i (z) is to process y[n] so that x[n] can be recovered: y[n] h i [n] x[n] h[n] h i [n] x[n]. Thus, H(z) acts as a filter of convolution and its inverse filter H i (z) undoes convolution, which is called deconvolution. (e.g., modulation/demodulation, encoder/decoder, preemphasis/deemphasis, etc.) (b) An LTI system H(z) is stable and causal and also has a stable causal inverse if and only if both the poles and zeros of H(z) are inside the unit circle. (Minimum-phase systems) (c) (Ex) Given a channel: h[n] δ[n] + 2δ[n ] H(z) + 2z, we aim to find its inverse g[n] for equalization. i. G (z) 2z + 4z 2 8z 3 + g +2z [n] ( 2) n u[n] is a causal but unstable system. ii. G 2 (z) +2z 2 z 4 z2 + 8 z3 6 z4 + g 2 [n] ( 2) n u[ n ] is a stable but anticausal system. iii. By incorporating a delay of 4 samples and truncation on g 2 [n], a finite-length causal stable equalizer G 3 (z) z 4 z z 3, we find H(z) G 3 (z) 6 +z 4 h[n] g 3 [n] 6δ[n]+δ[n 4], which is very close to an ideal delayed impulse function. iv. We may use the least-squares criterion to design the finite-length equalizer g 4 [n], by expressing the convolution of h[n] and g 4 [n] as g 4 [0] g 4 [] g 4 [2] g 4 [3] From MATLAB, the least-squares solution of Hg d is given by H\d, which is g 4 [n] [ ]. The convolution of h g 4 is [ ], while h g 3 [ ]. The 2-norm distances h g δ[n 4] are and for g 4 and g 3, respectively. VI -

12 6 Pole/Zero plot of H(z). For a rational transfer function H(z) Q(z) P(z) (z q )(z q 2 ) (z q m ) (z p )(z p 2 ) (z p n ) Poles: p,p 2,...,p n, because H(z) zpi H(p i ). Zeros: q,q 2,...,q m, because H(z) zqj H(q j ) The no. of poles equals the no. of zeros, counting all poles/zeros at infinity z and the origin z 0. 2 [Ex] H(z) 2z 0.5z z 0.5 has one pole at 0.5 and one zero at 0. [Ex] H(z) 3 z 0.5 has one pole at 0.5 and one zero at. 3. Poles and Time responses: According to the location of poles of H(z), we can have a rough picture about the behavior of its impulse response h[n]: (a) Single-order pole: p α is real A αz Aαn u[n] (b) Single-order Complex-conjugate poles: p αe jω 0 and p αe jω 0 E + Fz + c z + c 2 z 2 E + Fz ( αe jω 0 z )( αe jω 0 z ) [E (α)n cos ω 0 n + D(α) n sin ω 0 n] u[n] (c) The poles magnitude α determines the signal s exponentially growing/decaying rate in time. Poles close to unit circle z, corresponds to a slow growing( α >, outside unit circle) or decaying( α <, inside unit circle) time sequence. (d) The angles of the complex poles, ω 0, determines the oscillation frequency of the time sequence. (e) For a causal LTI system, all poles must be within unit circle, poles <, to assure BIBO stability. 4. Geometrical evaluation of frequency response H(ω) from the pole/zero plot: H freq resp (e jω ) H transfer func (z) ze jω poles will push up the frequency response zeros will pull down the frequency response Consider a rational system function: M r H(z) K (z z r) N k (z p k) Its frequency response is: M H(e jω r ) K (ejω z r ) M N k (ejω p k ) K ej K r ejω z r e j (ejω z r) N k ejω p k e j (ejω p k ) H(ejω ) e j H(ejω), with VI - 2

13 Magnitude response: H(e jω ) K ejω z e jω z 2 e jω z M e jω p e jω p 2 e jω p N, Phase response: H(e jω ) K + M r (ejω z r ) N k (ejω p k ), 5. Examples: (a) h[n] a n u[n] H(z) z 0 az z a, then H(ω) H(z) ze jω z zero z pole I(z) ze jω zero vector vector from unit-circle to zero pole vector vector from unit-circle to pole z e jω pole-vector zero-vector R(z) unit circle (b) H (z) z is a HPF, H +0.9z 2 (z) +0.8z a LPF, H ( 0.9z ) 2 3 (z) a BPF. +0.8z 2 (c) Multiple poles and zeros: i. Magnitude response is raised(suppressed) by poles(zeros): ii. Higher-order IIR filters can realize frequency responses with flatter passband and stopbands(less ripples) and sharper transition bands. 0.8 a[0.6, *j, *j]; b[, j,j]; zplane(poly(b),poly(a)); Imaginary Part Real Part VI - 3

14 4 magnitude response 2 0 amplitude a[0.6, *j, *j]; b[, j,j]; w pi:0.0:pi; hfreqz(poly(b),poly(a),w); plot(w/pi,abs(h)) title( magnitude response ); xlabel( w/pi );ylabel( amplitude ); w/pi 7 Z sided -Transform and Difference Eqns with Initial Conditions [Ex] Find the output of an LTI system: y[n] 5y[n ] + 6y[n 2] 3x[n ] + 5x[n 2] with initial conditions: y[ ] /6,y[ 2] 37/36 and a causal input sequence x[n] 2 n u[n]. [Sol] Take Z-transform on both sides, and note that y[n] u[n] y[n ] u[n] y[n 2] u[n] Y (z) z Y (z) + y[ ] z 2 Y (z) + z y[ ] + y[ 2] Thus, ( 5z + 6z 2) Y (z) 5y[ ] + 6z y[ ] + 6y[ 2] 3z X(z) + 5z 2 X(z) Y (z) 5y[ ] 6z y[ ] 6y[ 2] 3z + 5z 2 } 5z {{ + 6z 2 + } 5z X(z) + 6z 2 }{{} Y ZIR (z):zero-input response Y ZSR (z):zero-state response Plug in the initial conditions y[ ],y[ 2] and the input X(z) (/2)z, Y (z) Y (z) z y[n] 3 z 5z + 6z 2 + 3z + 5z 2 5z + 6z 2 0.5z 26/5 z 0.5 7/3 z 2 + 8/5 z 3 [ 26 5 ( 2 ) n 7 3 (2)n (3)n ] u[n] VI - 4

15 8 Transient and Steady-State Responses Suppose H(z) b 0 a z.. Complex sinusoidal input(eigenfunction): If x[n] e jω 0n, then y[n] H(e jω 0 )e jω 0n 2. Causal(truncated) complex sinusoidal input: If x[n] e jω 0n u[n], X(z) e jω 0 z Y (z) H(z)X(z) partial fraction expansion b 0 a y[n] a e jω b 0 an 0 u[n] + a e jω ejω 0n u[n] 0 transient response + H(e jω 0 ) x[n] as long as the pole falls into the unit circle a <. b 0 a b 0 a e jω 0 a z + a e jω 0 e jω 0 z 3. For any N th order IIR system with a causal sinusoidal input x[n] e jω 0n u[n], its output can be written as y[n] A k p n k u[n] + H(ejω 0 ) e jω0n u[n] k p k < n H(e jω 0 ) e jω 0n VI - 5