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EE2S11 Signals and Systems, Part 2 Ch.10 the Z-Transform

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EE2S11 Signals and Systems, Part 2 Ch.10 the Z-Transform EE2S11 Signals and Systems, part 2 Ch.10 The z-transform Contents definition of the z-transform region of convergence convolution property, stability inverse z-transform Skip sections 10.5.3, 10.6, 10.7 1 12a z-transform January 7, 2020 The z-transform Sampled sequence Suppose that we have a sampled signal: xs (t) = x[n]δ(t − nTs ) , x[n] := x(nTs ) X The Laplace transform L{xs (t)} is −sTs n −n Xs (s) = x[n]L{δ(t − nTs )} = x[n]e = x[n]z , X X X where z := esTs . jΩTs jω For s = jΩ we obtain z = e = e , with ω = ΩTs . More generally: s = σ + jΩ becomes z = eσTs ejΩTs = rejω. 2 12a z-transform January 7, 2020 The z-transform Note that the mapping s → z = esTs is not one-to-one. For a given z = ejω we can take −π ≤ ω ≤ π, this corresponds to π π − ≤ Ω ≤ : the fundamental interval. Ts Ts Complex numbers s = jΩ with Ω outside this interval are mapped onto the same z. Left half-plane is mapped to the inside of the unit circle. s-vlak z-vlak (aliasing) π/Ts ω Ω π 0 −π −π/Ts (aliasing) 3 12a z-transform January 7, 2020 The z-transform From now on, we will work with z and apply this transform to time series, even if there is no connection to continuous-time signals. The z-transform The (two-sided) z-transform of a time series x[n] is defined as ∞ X(z) = Z(x[n]) := x[n]z −n , z ∈ ROC n=X−∞ We also need to indicate the region of convergence (ROC). For example: x =[· · · , 0, 1, 2, 3 , 4, 5, 0, · · · ] ⇒ X(z) = z 2 + 2z 1 +3+4z −1 + 5z −2 ROC: z ∈ C| \{0, ∞} ∞ 1 z x[n] = anu[n] ⇒ X(z) = anz −n = = 1 − az −1 z − a Xn=0 ROC: |z| >a 4 12a z-transform January 7, 2020 The z-transform The z-transform A few properties: ax[n] + by[n] ⇔ aX(z) + bY (z) x[n − k] ⇔ z −k X(z) n z jω0 a x[n] ⇔ X( a ) often a = e (modulation) x[−n] ⇔ X(z −1) x[n] = δ[n] ⇔ X(z) = 1 ROC: z ∈ C| 1 x[n] = u[n] ⇔ X(z) = 1−z −1 ROC: |z| > 1 See Chaparro p.589-590 (2nd ed. p.653) for tables and more properties. 5 12a z-transform January 7, 2020 The z-transform Region of convergence The region of convergence (ROC) of the z-transform of a signal x[n] contains those values of z for which the summation converges. With z = rejω we find ROC: |X(z)| = | x[n]z −n| ≤ |x[n]| r −n < ∞ X X The ROC is the area where |X(z)| < ∞, this depends on r but not on ω. Hence, the ROC is limited by circles. X(z) and the ROC together uniquely determine x[n]. Poles pk are the locations whereX(pk ) → ∞: these are never in the ROC. Zeros zk are the locations where X(zk ) = 0. 6 12a z-transform January 7, 2020 The z-transform Example Determine the poles and zeros of z + 2 X(z)=1+2z −1 = z Answer: 1 pole at z = 0; 1 zero at z = −2. Same for 1 + 2z −1 z(z + 2) X(z) = = 1 + z −2 z 2 + 1 Answer: poles at z = ±j; 1 zero at z = −2, 1 zero at z = 0. Theory says that for rational functions, the number of poles equals the number of zeros (also taking into account those at z = 0 and z = ∞). If X(z) is a rational function with real-valued coefficients, then the complex poles ∗ and zeros appear in conjugated pairs: if pk is a complex pole, then so is pk . 7 12a z-transform January 7, 2020 The z-transform ROC for a finite sequence If x[n] = 0 outside an interval −∞ < N0 ≤ n ≤ N1 < ∞, i.e. −N0 −N1 X(z) = x[N0]z + · · · + x[N1]z then the sum has a finite number of terms, and the ROC is all of C,| except perpaps at z = 0 or |z| = ∞: −1 0 , if N1 ≥ 0 , e.g.: X(z) = z + 1 + z ∞ , if N0 ≤ 0 8 12a z-transform January 7, 2020 The z-transform ROC of an infinite sequence Split the sequence x[n] into the sum of a causal and an anti-causal term, and use the linearity of the z-transform. The causal part Xc (z) has ROC: |z| > R1, the largest radius of the poles. The anti-causal part Xac (z) has ROC: |z| < R2, the smallest radius of the poles. Hence, the ROC of X(z) is the intersection: R1 < |z| < R2. All poles are outside the ROC. r r 1 1 R2 R1 1 R2 R1 causal anti-causal not causal 9 12a z-transform January 7, 2020 The z-transform Example Causal signal n ∞ n 1 1 −1 1 z 1 x1[n] = u[n] ⇔ X1(z) = z = = ROC: |z| > 2 2 1 − 1 z −1 z − 1 2 Xn=0 2 2 1 1/2 1 n 10 12a z-transform January 7, 2020 The z-transform Example Anti-causal signal n −1 n ∞ 1 1 −n m −1 z x2[n] = − u[−n − 1] ⇔ X2(z) = − z = − (2z) + 1 = + 1 = 2 2 1 − 2z z − 1 n=X−∞ mX=0 2 1 ROC: |z| < 2 −2 n 1 The same X(z) corresponds to different x[n] depending on the ROC. The z-transform of x1[n]+x2[n] does not exist, because the intersection of the ROCs is empty. 11 12a z-transform January 7, 2020 The z-transform Example Compute the z-transform of the two-sided signal: 1 |n| x[n] = . 2 For the causal part we find: n ∞ n 1 1 −n 1 z 1 xc [n] = u[n] ⇔ Xc (z) = z = = , ROC: |z| > 2 2 1 − 1 z −1 z − 1 2 Xn=0 2 2 For the anti-causal part: −n ∞ n 1 1 n 1 xac [n] = u[−n] ⇔ Xac (z) = z = , ROC: |z| < 2 2 2 1 − 1 z Xn=0 2 12 12a z-transform January 7, 2020 The z-transform For x[n] we find 1 z 1 z z −1 2 z X(z) = 1 + 1 − 1 = 1 − = 1 z − 2 1 − 2 z z − 2 z − 2 (z − 2 )(z − 2) 1 with ROC: 2 < |z| < 2. 1 1 2 n 13 12a z-transform January 7, 2020 The z-transform Region of convergence With X(z), formally we should always also specify the region of convergence (ROC), indicating where the function is valid. Different sequences x[n] can have the same X(z), but with different ROCs. Later, it will turn out that we are mostly interested in an ROC which contains the unit circle, because the Fourier Transform is defined here, and in that case we can consider the frequency spectrum of a signal x[n] or system h[n]. Convergence on the unit circle also corresponds to BIBO stability. Thus, without further specifications, we will always assume that the ROC contains the unit circle. 14 12a z-transform January 7, 2020 The z-transform Exponential signals and corresponding poles 1 z x[n] = anu[n] ⇔ X(z) = = , ROC: |z| >a 1 − az −1 z − a Pole at z = a, zero at z = 0. a =0.5 a =1 a =2 1 1 1 2 1 1 1200 a=0.5 a=1 a=2 0.8 0.8 1000 800 0.6 0.6 600 0.4 0.4 400 0.2 0.2 200 0 0 0 −5 0 5 10 −5 0 5 10 −5 0 5 10 15 12a z-transform January 7, 2020 The z-transform a = −0.5 a = −1 a = −2 1 1−2 1 1 1 a=−1 1500 a=−0.5 0.8 a=−2 0.5 0.6 1000 0.4 500 0 0.2 0 0 −0.5 −0.2 −500 −0.4 −1 −0.6 −5 0 5 10 −1000 −5 0 5 10 −5 0 5 10 16 12a z-transform January 7, 2020 The z-transform Harmonic (exponentially damped) signals and corresponding poles ejθ e−jθ x[n] = r n cos(ω n + θ)u[n] = r nejω0n + r ne−jω0n u[n]=[γαn + γ∗(α∗)n] u[n] 0 2 2 jθ jω0 e with α = re and γ = 2 both complex. γz γ∗z z(z cos(θ) − r cos(ω − θ)) X(z) = + = · · · = 0 , ROC: |z| > |a| z − α z − α∗ (z − rejω0 )(z − re−jω0 ) This is a second-order rational function with real-valued coefficients. Poles at z = rejω0 and z = re−jω0 . Special case: r = 1, now x[n] is an undamped (causal) sinusoid, with its two poles on the unit circle. r cos(ω − θ) Zeros at z = 0 and z = 0 . cos(θ) 17 12a z-transform January 7, 2020 The z-transform r = 1, ω0 = 1, θ = 0 r = 1, ω0 = 0 (one pole and zero cancel each other) r = 0.5, ω0 = 1 r =1, ω0 =1 r =1, ω0 =0 r =0.5, ω0 =1 θ =0 θ =0 θ =0 1 1 1 1 1 1.2 ω =1 ω =0 0 0 ω =1 0 r=1 0.8 r=1 1 0.5 r=0.5 0.8 0.6 0.6 0 0.4 0.4 −0.5 0.2 0.2 0 −1 0 −5 0 5 10 −5 0 5 10 −0.2 −5 0 5 10 18 12a z-transform January 7, 2020 The z-transform Double poles For a causal x[n]: ∞ X(z) = x[n]z −n Xn=0 ∞ ∞ dX(z) dz −n = x[n] = −z −1 nx[n]z −n dz dz Xn=0 Xn=0 Hence dX(z) nx[n]u[n] ⇔ −z dz Taking a derivative often leads to double poles.
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