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Digital Signal Processing (DSP) Inverse Z-Transform
- 1. Inverse Z-Transform Digital SignalProcessing (DSP)
- 2. Content 2 ❑ Difference Equationvs Differential Equati ❑ Z-Transform ▪ Examples ▪ Properties ❑ Inverse Z-Transform ▪ Long Division ▪ Partial Fraction
- 3. Difference Equation vsDifferential Equation 3 ❑ A difference equation expresses the change in some variable as a result of a finite change in the other variable. ❑ A differential equation expresses the change in some variable as a result of an infinitesimal change in the other variable.
- 4. Z-Transform 4 ❑ Difference equationscan be solved using z-transforms which provide a convenient approach for solving LTI equations. ❑ The z-transform is an important tool in the analysis and design of discrete-time systems. ❑ It simplifies the solution of discrete-time problems by converting LTI difference equations to algebraic equations and convolution to multiplication. ❑ Thus, it plays a role similar to that served by Laplace transforms in continuous-time problems.
- 5. Laplace Transform andZ-Transform ❑ Given the sampled impulse train of a signal 𝑥∗ 𝑡 𝑥∗ 𝑡 = 𝑥 0 𝛿 𝑡 + 𝑥 𝑇 𝛿 𝑡 − 𝑇 + ⋯ + 𝑥 𝑘𝑇 𝛿(𝑡 − 𝑘𝑇) + ⋯ ∞ = ∑ 𝑥 𝑘𝑇 𝛿 𝑡 − 𝑘𝑇 𝑘=0 12
- 6. Laplace Transform andZ-Transform 𝑋∗ 𝑆 = 𝑥 0 + 𝑥 𝑇 𝑒−𝑠𝑇 + 𝑥 2𝑇 𝑒−𝑠2𝑇 + ⋯ + 𝑥 𝑘𝑇 𝑒−𝑠𝑘𝑇 + ⋯ ∞ ∞ 𝑋∗ 𝑆 = ∑ 𝑥 𝑘𝑇 𝑒−𝑠𝑘𝑇 = ∑ 𝑥 𝑘𝑇 (𝑒𝑠𝑇)−𝑘 → (1) 𝑘=0 𝑘=0 ❑ The z-transform is ∞ 𝑋 𝑧 = ∑ 𝑥 𝑘𝑇 𝑧−𝑘 𝑘=0 6 → (2) ❑ Comparing (1) and (2) yields 𝑧 = 𝑒𝑠𝑇 where 𝑇 is the sample period
- 7. A Note ❑ Ingeneral, given a transfer function in s-domain you cannot just replace 𝑠 by s = 𝑙𝑛𝑧 (from 𝑧 = 𝑒𝑠𝑇) to get its z-domain transfer function. 7 𝑇 ❑ The reason is that 𝑧 = 𝑒𝑠𝑇 is true with the sampled signal.
- 8. Identities Used Repeatedly 1 1− 𝑎 , 𝑎 < 1 ∞ ∑ 𝑎𝑘 = 𝑘=0 𝑛 ∑ 𝑎𝑘 = 𝑘=0 1 − 𝑎𝑛+1 1 − 𝑎 , 𝑎 = 1 ❑ Special Cases: ∞ ∑ 𝑘𝑎𝑘 = 𝑘=0 1 (1 − 𝑎)2 , 𝑎 < 1 8
- 9. Signal, 𝐱(𝒌𝑻) Z-Transform,𝐗(𝐳) Z-Transform, 𝐗(𝐳) ROC 1 ✿(𝒌𝑻) 1 1 all Z 2 ✿(𝒌𝑻 − 𝒒) 𝐳−𝒒 𝐳−𝒒 𝐳 ≠ 𝟎 3 𝐮(𝒌𝑻) 𝟏 𝟏 − 𝒛−𝟏 𝐳 𝐳 − 𝟏 𝐳 > 𝟏 4 𝐚𝒌 𝐮(𝐤𝐓) 𝟏 𝟏 − 𝒂𝒛−𝟏 𝐳 𝐳 − 𝐚 𝐳 > 𝐚 5 𝒌 𝐮(𝐤𝐓) 𝟏 (𝟏 − 𝒛−𝟏)𝟐 𝐳 (𝐳 − 𝟏)𝟐 𝐳 > 𝟏 6 𝒌 𝐚𝒌 𝐮(𝐤𝐓) 𝐚𝒛−𝟏 (𝟏 − 𝐚𝒛−𝟏)𝟐 𝐚𝐳 (𝐳 − 𝐚)𝟐 𝐳 > 𝐚 7 𝐜𝐨𝐬(𝜔𝟎𝐤𝐓) 𝟏 − 𝒛−𝟏𝐜𝐨𝐬(𝜔𝟎) 𝟏 − 𝟐𝒛−𝟏 𝐜𝐨𝐬 𝜔𝟎 + 𝒛−𝟐 𝐳𝟐 − 𝐳𝐜𝐨𝐬(𝜔𝟎) 𝐳𝟐 − 𝟐𝐳 𝐜𝐨𝐬 𝜔𝟎 + 𝟏 𝐳 > 𝟏 8 𝐬𝐢𝐧(𝜔𝟎𝐤𝐓) 𝒛−𝟏𝐬𝐢𝐧(𝜔𝟎) 𝟏 − 𝟐𝒛−𝟏 𝐜𝐨𝐬 𝜔𝟎 + 𝒛−𝟐 𝐳 𝐬𝐢𝐧(𝜔𝟎) 𝐳𝟐 − 𝟐𝐳 𝐜𝐨𝐬 𝜔𝟎 + 𝟏 𝐳 > 𝟏 9 𝐫𝒌 𝐜𝐨𝐬(𝜔𝟎𝐤𝐓) 𝟏 − 𝐫 𝒛−𝟏𝐜𝐨𝐬(𝜔𝟎) 𝟏 − 𝟐𝐫 𝒛−𝟏 𝐜𝐨𝐬 𝜔𝟎 + 𝐫𝟐 𝒛−𝟐 𝐳𝟐 − 𝒓 𝐳 𝐜𝐨𝐬(𝜔𝟎) 𝐳𝟐 − 𝟐𝒓 𝐳 𝐜𝐨𝐬 𝜔𝟎 + 𝐫𝟐 𝐳 > 𝐫 10 𝐫𝒌 𝐬𝐢𝐧(𝜔𝟎𝐤𝐓) 𝒓 𝒛−𝟏𝐬𝐢𝐧(𝜔𝟎) 𝟏 − 𝟐𝐫 𝒛−𝟏 𝐜𝐨𝐬 𝜔𝟎 + 𝐫𝟐 𝒛−𝟐 𝒓 𝐳 𝐬𝐢𝐧(𝜔𝟎) 𝐳𝟐 − 𝟐𝒓 𝐳 𝐜𝐨𝐬 𝜔𝟎 + 𝐫𝟐 𝐳 > 𝐫 9
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- 11. The Inverse Z-Transform ❑Formal inverse z-transform is based on a Cauchy integral ❑ Less formal ways sufficient most of the time 1) Direct or Long Division Method 2) Partial fraction expansion and Look-up Table 3) Inversion Integral Method (Residue-theorem) x(kT)= Z −1 X (z)= 1 X (z)zk−1 dz 2j C 11
- 12. Inverse Z-Transform: PowerSeries Expansion 12 ❑ Using Long Division to expand X(z) as a series X (z)= x(0)+ x(T )z−1 + x(2T)z−2 +... = x(kT)z−k k=0 ❑ Write the inverse transform as the sequence x(kT)={x(0), x(T), x(2T),...}
- 13. Inverse Z-Transform: PowerSeries Expansion ❑ Solution: −1 35 z −1)(z − 0.2) 10z + 5 ❑ Example: Find x(n) for n = 0, 1, 2, 3, 4, when X(z) is given by: X (z)= ( −2 1−1.2z−1 + 0.2z 10z−1 +5z−2 X (z)= ▪ First, rewrite X(z) as a ratio of polynomial in 𝑧 , as follows: ▪ Dividing the numerator by the denominator, we have: 10z−1 +17z−2 +18.4z−3 +18.68z−4 1−1.2z−1 + 0.2z−2 10z−1 + 5z−2 17z−2 − 2z−3 17z−2 − 20.4z−3 + 3.4z−4 18.4z−3 −3.4z−4 18.4z−3 − 22.08z−4 + 3.68z−5 18.68z−4 −3.68z−5 18.68z−4 − 22.416z−5 + 3.736z−6 10z−1 −12z−2 + 2z−3 Therefore, • x(0) = 0, • x(1) = 10, • x(2) = 17, • x(3) = 18.4 • x(4) = 18.68
- 16. Home work :3.10 , 3.11, 3.13 (a, b c, d)