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Nguyễn Công Phương PHYSIOLOGICAL SIGNAL PROCESSING The z – Transform Contents I Introduction II Introduction to Electrophysiology III Signals and Systems IV Fourier Analysis V Signal Sampling and Reconstruction VI The z – Transform VII.Discrete Filters VIII.Random Signals IX Time-Frequency Representation of Physiological Signals X Physiological Signal Processing s i tes.google.com/site/ncpdhbkhn The z – Transform The z – Transform The Inverse z – Transform Properties of the z – Transform s i tes.google.com/site/ncpdhbkhn The z – Transform (1) x[ n] = ∞ x[k ]δ [n − k ] ֏ y[ n] = k =−∞ x[ n] = z , for all n z = Re( z ) + j Im( z ) n ֏ y[n] = ∞ ∞ k =−∞ k =−∞ x[k ]h[n − k ] = h[k ]x[n − k ] ∞ h[k ]z k =−∞ n −k ∞ = h[k ] z − k z n , k =−∞ H ( z) = ∞ h[k ]z for all n −k k =−∞ → y[ n ] = H ( z ) z n , x[ n ] = ck zkn , k for all n ֏ y[ n ] = ck H ( z k ) zkn , for all n for all n k s i tes.google.com/site/ncpdhbkhn The z – Transform (2) X ( z) = ∞ Im( z) z = e jω z – plane x[ n ]z − n ω n =−∞ • ROC (region of convergence): the set of values of z for which X(z) converges • Zeros: the values of z for which X(z) = • Poles: the values of z for which X(z) is infinite s i tes.google.com/site/ncpdhbkhn Re( z ) Unit circle Im( z) z – plane z = re jω r sin ω r ω Re( z ) r cos ω The z – Transform (3) Ex Given x1[n] = {1 5}, x2 [n] = {1 5} ↑ ↑ Determine their z – transforms? X ( z) = ∞ x[ n ] z − n n =−∞ −4 X ( z ) = x1[0]z + x1[1] z −1 + x1[2]z −2 + x1[3]z −3 + x1[4] z = + z − + 3z −2 + z −3 + z −4 ROC: entire z – plane except z = X ( z ) = x2[ −2]z − ( −2 ) + x2 [−1] z − ( −1) + x2[0]z + x2 [1]z −1 + x2[ 2]z −2 = 1z + z + 3z + z −1 + 5z − = z2 + z + + z −1 + z −2 ROC: entire z – plane except z = & z = ∞ s i tes.google.com/site/ncpdhbkhn The z – Transform (4) Ex x1[n] = δ [n], x2 [n] = δ [n − k ], x3[n ] = δ [n + k ], k > Determine their z – transforms? X (z) = ∞ x[n]z − n n =−∞ X ( z ) = + x1[− 1]z − ( −1) + x1[0]z + x1[1]z −1 + x1[2]z −2 + = + z − ( − 1) + 1z + z −1 + z −2 + = ROC: entire z – plane s i tes.google.com/site/ncpdhbkhn The z – Transform (5) Ex 1, 0, Find the z – transform of the square – pulse sequence x[ n] = X (z) = ∞ x[n]z −n 0≤n≤M otherwise M = 1z −n n= n =−∞ − AN +1 + A + A + A + + A = , if A < 1− A Im N − z − ( M +1) → X ( z) = − z −1 z −1 Re 1 ROC: |z| > s i tes.google.com/site/ncpdhbkhn ROC The z – Transform (6) Ex Find the z – transform of the sequence x[n] = anu[n]? X (z) = ∞ n =−∞ ∞ ∞ n= n =0 x[n]z − n = a n z − n = (az − )n + A + A2 + A3 + = , if A < 1− A → X ( z) = az −1 < → z > a s i tes.google.com/site/ncpdhbkhn z = − az − z − a Zero: z = Pole: z = a ROC: |z| > a The z – Transform (7) Ex Find the z – transform of the sequence x[n] = anu[n]? X (z) = z = − az −1 z − a Zero: z = 0; pole: z = a; ROC: |z| > a 0< a1 a =1 … … … … … … n n Im a n Im Re Im Re 0 ROC ROC s i tes.google.com/site/ncpdhbkhn a Re ROC 10 The z – Transform (8) Ex n≥0 0, Find the z – transform of the sequence x[ n] = −a u[ −n − 1] = n −a , n X (z) = ∞ x[n]z n =−∞ −n −1 =−a z n −n n=−∞ −1 = − (az ) −1 n n a s i tes.google.com/site/ncpdhbkhn 11 The z – Transform (9) Ex n a , Find the z – transform of the sequence x[ n] = −bn , X (z) = ∞ n =−∞ x[n]z −n −1 =−b z n =−∞ n −n −1 n a → a n z − n = z−a n =0 −1 n≥0 n −n z z → X ( z) = + z−b z−a Zero: z = Pole: z = a, b ROC: a < |z| < b s i tes.google.com/site/ncpdhbkhn 12 The z – Transform (10) a … … … … n … n n … b Im a Im Re Re ROC ROC: |z| > a Im 0 b ROC ROC: |z| < b s i tes.google.com/site/ncpdhbkhn a Re b ROC ROC: a < |z| < b 13 The z – Transform (11) Ex Find the z – transform of the sequence x[n] = rn (cos ω0n)u[n ], r > 0, ≤ ω0 ≤ 2π X (z) = ∞ ∞ n −n x[n]z − n = r (cos ω0n) z n =−∞ n =0 jθ − jθ e = cos θ + j sin θ → cos θ = e + e 2 jθ ∞ ∞ jω0 − n → X ( z ) = (re z ) + (re − jω0 z −1 )n n= n =0 e jθ = cos θ + j sin θ = cos θ + sin2 θ = re jω0 z −1 < & re − jω0 z −1 < → rz −1 < → z > r → X ( z) = 1 1 + , ROC : z > r jω0 − − jω0 −1 − re z − re z s i tes.google.com/site/ncpdhbkhn 14 The z – Transform (12) Ex Find the z – transform of the sequence x[n] = rn (cos ω0n)u[n ], r > 0, ≤ ω0 ≤ 2π 1 1 X (z) = + jω0 − − re z − re − jω0 z −1 (1 − re − jω0 z −1 ) + (1 − re jω0 z −1 ) − rz −1(e − jω0 + e jω0 ) = = jω0 − − jω0 −1 2(1 − re z )(1 − re z ) 2[1 − 2(r cos ω0 ) z −1 + r z −2 ] e jθ = cos θ + j sin θ → e jθ + e − jθ = cos θ Im p1 r Re z1 z2 − r (cos ω0 ) z −1 → X ( z) = − 2( r cos ω0 ) z −1 + r z −2 z ( z − r cos ω0 ) = ( z − re jω0 )( z − re− jω0 ) Zero: z1 = 0; z2 = rcosω0 p2 ROC Pole : p1 = re jω0 ; p2 = re − jω0 s i tes.google.com/site/ncpdhbkhn ROC: |z| > r 15 The z – Transform (13) • ROC – There is no pole inside a ROC – The ROC is a connected region – For finite duration sequences, the ROC is the entire z – plane, sometimes except for z = and z=∞ • The z – transform – We need both X(z) and its ROC – X(z) is not defined outside the ROC s i tes.google.com/site/ncpdhbkhn 16 The z – Transform (14) s i tes.google.com/site/ncpdhbkhn 17 The z – Transform The z – Transform The Inverse z – Transform Properties of the z – Transform s i tes.google.com/site/ncpdhbkhn 18 The Inverse z – Transform (1) x[n] = 2π j n −1 C X ( z ) z dz b0 + b1 z −1 + b2 z −2 + bN −1 z − ( N −1) X ( z) = + a1 z −1 + a2 z −2 + a N z − N s i tes.google.com/site/ncpdhbkhn 19 The Inverse z – Transform (2) Ex 1 + z −1 X ( z) = (1 − z −1 )(1 − 0.2 z −1) + z −1 K1 K2 = + (1 − z −1)(1 − 0.2 z −1) − z− 1 − 0.2 z −1 → + z −1 = K1(1 − 0.2 z −1 ) + K2 (1 − z −1 ) z = → + = K1 (1 − 0.2 × 1) + K (1 − 1) → K1 = 2.5 z = 0.2 → + = K1 (1 − 0.2 × 5) + K (1 − 5) → K2 = − 1.5 → X ( z) = s i tes.google.com/site/ncpdhbkhn 1.5 − − z −1 − 0.2 z −1 20 The Inverse z – Transform (3) Ex 1 + z −1 5 X ( z) = = − (1 − z −1 )(1 − 0.2 z −1 ) − z −1 − 0.2 z −1 2.5 1 − z −1 → 5u[ n ] a u[n ] → , ROC : z > a −1 − az → −1.5 → −1 5( 0.2 )n u[ n] If z > 1 − 0.2 z − n → x[n ] = 5u[n ] − 5(0.2 )n u[n] − z − → −2.5u[ −n − 1] −a u[− n − 1] → , ROC : z < a −1 − az → −1 → 1.5(0 2) n u[ − n − 1] If z < − 0.2 z −1 n → x[ n ] = −2 5u[ − n − 1] + 1.5( 0.2 )n u[ −n − 1] 2.5 1 − z −1 → −2.5u[ −n − 1] If < z < → −1.5 → −1 5( 0.2 )n u[ n] 1 − 0.2 z − → x[ n ] = −2 5u[ − n − 1] − 1.5( 0.2 )n u[ n] s i tes.google.com/site/ncpdhbkhn 21 The Inverse z – Transform (4) Ex + z −1 X ( z) = − z −1 + 2.5z −2 − z −1 + z −2 = → p1,2 = ± j1.5 = 1.58e ± j1 25 + z −1 K1 K2 = + − z −1 + 2.5z −2 − p1z −1 − p2 z −1 → + z −1 = K1(1 − p2 z− 1) + K (1 − p1z −1 ) z = p1 → + / p1 = K1(1 − p2 / p1 ) + K2 (1 − 1) → K1 = − j 0.67 = 0.83e − j 93 z = p2 → + / p = K (1 − 1) + K (1 − p1 / p ) → K = + j 67 = 83e j 0.93 → x[n] = 0.83e − j 0.93 (1.58e j1.25 )n u[n] + 0.83e j 0.93 (1.58e − j1.25 )n u[n ] = 0.83(1 58 )n ( e j (1 25 n −0 93) + e − j (1.25 n−0 93 ) ) e j (1.25n −0.93) + e− j (1.25n− 0.93) = cos(1.25n − 0.93) → x[ n ] = 1.67(1 58) n cos(1.25n − 93)u [n ] = 67 (1.58) n cos(1 25 n − 53 13 o )u[ n ] s i tes.google.com/site/ncpdhbkhn 22 The z – Transform The z – Transform The Inverse z – Transform Properties of the z – Transform s i tes.google.com/site/ncpdhbkhn 23 Properties of the z – Transform s i tes.google.com/site/ncpdhbkhn 24 ... + = , if A < 1− A z Zero: z = = − a − 1z z − a Pole: z = a a −1 z < → z < a ROC: |z| < a → X ( z ) = −a − 1z x[n] = a nu[n] → X ( z) = z z−a Zero: z = 0; pole: z = a; ROC: |z| > a s i tes.google.com/site/ncpdhbkhn... = X ( z ) = x2[ −2 ]z − ( −2 ) + x2 [−1] z − ( −1) + x2[0 ]z + x2 [1 ]z −1 + x2[ 2 ]z −2 = 1z + z + 3z + z −1 + 5z − = z2 + z + + z −1 + z −2 ROC: entire z – plane except z = & z = ∞ s i tes.google.com/site/ncpdhbkhn... their z – transforms? X ( z) = ∞ x[ n ] z − n n =−∞ −4 X ( z ) = x1[0 ]z + x1[1] z −1 + x1[2 ]z −2 + x1[3 ]z −3 + x1[4] z = + z − + 3z −2 + z −3 + z −4 ROC: entire z – plane except z = X ( z ) =