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Lecture 04: Discrete Frequency Domain Analysis (Z-Transform)

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Lecture 04: Discrete Frequency Domain Analysis (Z-Transform) Lecture 04: Discrete Frequency Domain Analysis (z-transform) John Chiverton School of Information Technology Mae Fah Luang University 1st Semester 2009/ 2552 Outline Overview Lecture Contents Introduction z-Transform Time Delays in z-Transform Representations The Inverse z-Transform Stability BIBO Stability Lecture Contents Overview Lecture Contents Introduction z-Transform Time Delays in z-Transform Representations The Inverse z-Transform Stability BIBO Stability Outline Overview Lecture Contents Introduction z-Transform Time Delays in z-Transform Representations The Inverse z-Transform Stability BIBO Stability Introduction to the z-transform The z-transform I Transforms a digital signal to a frequency respresentation I Used for stability analysis using the poles and zeros I Also used for frequency analysis Outline Overview Lecture Contents Introduction z-Transform Time Delays in z-Transform Representations The Inverse z-Transform Stability BIBO Stability z-Transform Two z-transforms are common in digital signal processing. The one-sided or unilateral z-transform: 1 X X(z) = x[n]z−n n=0 and the bilateral z-transform: 1 X X(z) = x[n]z−n: n=−∞ The one-sided transform is considered here only. z-Transform Example Q. Find the z-Transform of a step function: 1 for n ≥ 0 x[n] = 0 elsewhere A. 1 1 X X X(z) = x[n]z−n = z−n n=0 n=0 =(z0 + z−1 + z−2 + z−3 + z−4 + :::) „ 1 1 1 1 « = 1 + + + + + ::: z z2 z3 z4 This is a geometric series of the form: 1 X a s = ark = 1 − r k=0 where a = 1 and r = z−1 so that 1 z X(z) = = : 1 − z−1 z − 1 z-Transform Example Q. Find the z-Transform of a square pulse: 0:2 for 0 ≤ n < 5 x[n] = 0 elsewhere A. 1 4 X X X(z) = x[n]z−n = 0:2 z−n n=0 n=0 =0:2(z0 + z−1 + z−2 + z−3 + z−4) „ 1 1 1 1 « =0:2 1 + + + + z z2 z3 z4 This is a geometric series of the form: n−1 X 1 − rn s = ark = a 1 − r k=0 where a = 0:2, n = 5 and r = z−1 so that 1 − z−5 z5 − 1 z5 − 1 X(z) = 0:2 = 0:2 = 0:2 : 1 − z−1 z5 − z4 z4(z − 1) Time delays in z-Transform Representations Each z−1 in a z-Transform can be considered as a single time delay. Consider the time-shifted or delayed unit impulse: x[n] = δ[n − τ] then the z-Transform is given by: 1 X X(z) = δ[n − τ]zn = z−τ n=0 indicating a delay of τ samples. The z-Transform and Time Delay Example Q. Find the signal corresponding to the z-Transform: z X(z) = : z − 0:5 1 P k a A. Remember the geometric series formula: s = ar = 1−r : Need to find the form k=0 of X(z) to easily find r and a... Dividing the numerator and denominator by z gives 1 X(z) = : 1 − 0:5z−1 So that r = 0:5z−1 and a = 1 then 1 X k X(z) = `0:5z−1´ = 1 + 0:5z−1 + (0:5z−1)2 + (0:5z−1)3 + (0:5z−1)4 + ::: k=0 =1 + 0:5z−1 + 0:25z−2 + 0:125z−1 + 0:0625z−4 + ::: Remembering that each z−1 is a delay of 1 time instance, the signal x[n] is then given by the coefficients for each time instance, i.e. x[0] = 1, x[1] = 0:5, x[2] = 0:25, x[3] = 0:125, x[4] = 0:0625 etc. The z-Transform and Time Delay Example 2 Q. Find the signal corresponding to the z-Transform: z2 − 0:2 X(z) = z(z − 0:2) n−1 P k 1−rn A. Remember the geometric series formula: s = ar = a 1−r . Need to find the k=0 form of X(z) to easily find r, a and n... Dividing through by z2 gives 1 − 0:2z−2 X(z) = : 1 − 0:2z−1 So that r = 0:2z−1, a = 1 and n = 2 resulting in: n−1 1 X X X(z) = ark = (0:2z−1)k = 1 + 0:2z−1: k=0 k=0 Therefore the original signal, x[n] is given by x[0] = 1 and x[1] = 0:2. Outline Overview Lecture Contents Introduction z-Transform Time Delays in z-Transform Representations The Inverse z-Transform Stability BIBO Stability The Inverse z-Transform The inverse z-Transform Z−1(X(z)) is given by 1 Z x[n] = Z−1(X(z)) = X(z)zn−1dz 2πj I The inverse z-Transform is not usually computed directly. I Instead the z-Transform is split into parts using partial fractions I And then the inverse z-Transform of the parts are found using a table of z-Transform pairs. Table of (unilateral) z-Transform Pairs Lynn and Fuerst give the following table of z-Transform pairs: Signal x[n] z-Transform X(z) δ[n] 1 1 for n ≥ 0 u[n] = z 0 elsewhere z−1 n for n ≥ 0 r[n] = z 0 elsewhere (z−1)2 n z a u[n] z−a n z(1−a) (1 − a )u[n] (z−a)(z−1) z(z−cos(Ω0) cos(nΩ0)u[n] 2 z −2z cos(Ω0)+1 z sin(Ω0) sin(nΩ0)u[n] 2 z −2z cos(Ω0)+1 n az sin(Ω0) a sin(nΩ0)u[n] 2 2 z −2az cos(Ω0)+a The z-Transform representation has therefore to be separated using partial fractions into parts, each of the form of the expressions on the right hand side of the above table. Method of Partial Fractions Example Q. Decompose the following function into partial fractions: 1 : (z + 3)(z − 2) A. Let 1 A B = + : (z + 3)(z − 2) z + 3 z − 2 Then A(z − 2) + B(z + 3) = 1: So that Az − 2A + Bz + 3B = 1 z(A + B) − 2A + 3B = 1 Therefore z(A + B) = 0 ) A = −B and −2A + 3B = 1 so that −2A − 3A = 1 1 1 giving A = − 5 and B = 5 . Check: A B − 1 1 1 (5) 1 + = 5 + 5 = 5 = z + 3 z − 2 z + 3 z − 2 (z + 3)(z + 2) (z + 3)(z + 2) Method of Partial Fractions: Cover up method Example Q. Decompose the following function into partial fractions: z : (z + 3)(z − 2) z A B A. Let (z+3)(z−2) = z+3 + z−2 : To find A ) z + 3 = 0 ) z = −3. ˛ z ˛ −3 3 A = ˛ = = : (z + 3)(z − 2) ˛z=−3 −3 − 2 5 To find B ) z − 2 = 0 ) z = 2. ˛ z ˛ 2 2 B = ˛ = = : (z + 3)(z − 2) ˛z=2 2 + 3 5 Hence z 3 2 = 5 + 5 : (z + 3)(z − 2) z + 3 z − 2 Check: 3 2 3 (z − 2) + 2 (z + 3) 3 z − 6 + 2 z + 6 z 5 + 5 = 5 5 = 5 5 5 5 = : z + 3 z − 2 (z + 3)(z − 2) (z + 3)(z − 2) (z + 3)(z − 2) Inverse z-Transform Example Q. Find the inverse z-Transform of: 1 1 „ 1 1 « X(z) = = − : (1) (z + 3)(z − 2) 5 z − 2 z + 3 A. Re-writing (1) to z−1 „ z z « X(z) = − : (2) 5 z − 2 z + 3 Enables us to find inverse z-Transforms for the two terms inside the brackets: „ z « Z−1 = 2nu[n] z − 2 and „ z « Z−1 − = −((−3)n)u[n]: z + 3 The two terms are multiplied by z−1 which is equivalent to a time delay hence the final signal is given by: 1 “ ” x[n] = Z−1(X(z)) = 2(n−1)u[n − 1] − ((−3)(n−1))u[n − 1] : 5 Inverse z-Transform Example Q. Find the inverse z-Transform of: z X(z) = : (z + 3)(z − 2) 3 2 z 5 5 A. From earlier the partial fraction expansion is given by: (z+3)(z−2) = z+3 + z−2 : (i) However it is more convenient if we divide both sides by z first. Hence (ii) We saw earlier: „ « X(z) 1 −1 z n = : Z = 2 u[n] z (z + 3)(z − 2) z − 2 and The Right Hand Side (RHS) has partial fractions (see earlier slide): „ z « Z−1 − = −((−3)n)u[n] z + 3 X(z) − 1 1 = 5 + 5 : z z + 3 z − 2 so that Multiplying both sides by z then gives: x[n] =Z−1(X(z)) “ ” 1 „ −z z « 1 (n) (n) X(z) = + : = 2 u[n] − ((−3) )u[n] : 5 z + 3 z − 2 5 Inverse z-Transform: Algebraic Long Division The numerator and the denominator of the z-Transform can be divided using algebraic long division to find coefficients that correspond to the original signal. Example Q. Given H(z) = z = z , determine the coefficients. (z−1)(z+2) z2+z−2 A. Via algebraic or polynomial long So the coefficients of the original signal division: are given by: x[0] = 0, x[1] = 1, x[2] = −1, x[3] = 3, z−1 − z−2 + 3z−3 − 5z−4::: x[4] = −5, etc. 2 This can be checked by performing the z + z − 2dz inverse z-Transform on H(z). z + 1 − 2z−1 Expansion with partial fractions gives: 1 “ z z ” −1 H(z) = − − 1 + 2z 3 z−1 z+2 −1 −2 Inverse z-Transform: x[n] = −1 − z + 2z −1 1 n Z (H(z)) = 3 (u[n] − (−2) u[n]) 3z−1 − 2z−2 1 Then x[0] = 3 (1 − 1) = 0, −1 −2 −3 1 3z + 3z − 5z x[1] = 3 (1 + 2) = 1, x[2] = 1 (1 − 4) = −1, − 5z−2 + 5z−3 3 1 x[3] = 3 (1 + 8) = 3, ::: 1 x[4] = 3 (1 − 15) = −5, etc.
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