Q2 Topics for MC PDF

Z-Transform The Direct 𝔃-Transform The 𝔃-transform of a discrete-time signal 𝑥 power series is defined as the ∞ 𝑋 = ෍ 𝑥 𝑛=−∞ 𝑧−𝑛 where 𝑧 is a complex variable. The relation above is sometimes called transform. because it transforms the time- domain signal 𝑥 into its complex-plane representation 𝑋. T...

Z-Transform The Direct 𝔃-Transform The 𝔃-transform of a discrete-time signal 𝑥 power series is defined as the ∞ 𝑋 = ෍ 𝑥 𝑛=−∞ 𝑧−𝑛 where 𝑧 is a complex variable. The relation above is sometimes called transform. because it transforms the time- domain signal 𝑥 into its complex-plane representation 𝑋. The inverse procedure [i.e., obtaining 𝑥 inverse 𝔃-transform. the direct 𝔃- from 𝑋 ] is called the The Direct 𝔃-Transform For convenience, the 𝔃-transform of a signal 𝑥is denoted by 𝑋 ≡𝑍 whereas the relationship between 𝑥 𝑧 and 𝑋 is indicated by 𝑥 𝑋 Since the 𝑧-transform is an infinite power series, it exists only for those values of 𝑧 for which this series converges. The region of convergence (ROC) of 𝑋 𝑋 is the set of all values of 𝑧 for which attains a finite value. Thus, any time we cite a 𝔃-transform we should also indicate its ROC. Radius of Convergence in Complex Plane Let us express the complex variable z in polar form as 𝑧 = 𝑟𝑒𝑗𝜃 Where 𝑟 = |𝑧| and 𝜃∡𝑧. Then 𝑋 ∞ can be expressed as, ∞ = ෍ 𝑧−𝑛 = ෍ 𝑥 𝑋 𝑛=−∞ 𝑛=−∞ −𝑛 In the ROC of 𝑋 𝑋 ,𝑋 = ෍ 𝑛=−∞ < ∞. So 𝑥 ∞ −𝑛 = 𝑛=−∞ ෍ 𝑥 𝑥 𝑟−𝑛 Hence 𝑋 is finite if the sequence 𝑥 𝑟−𝑛 is absolutely summable. Radius of Convergence in Complex Plane ∞ 𝑋 =෍𝑥 𝑛=1 ∞ 𝑟𝑛 +෍𝑥 𝑛=0 𝑟−𝑛 If 𝑋 converges in some region of the complex plane, both summations in the equation above must be finite in that region. If the first sum converges, there must exist values of r small enough such that the product sequence 𝑥 absolutely summable. 𝑟𝑛 , 1 𝑟1, there is no common region of convergence for the two sums and hence X(z) does not exist. 𝔃-Transform Issue The previous examples raises an important issue regarding the uniqueness of the z-transform. Again, from the previous examples, the causal signal 𝛼𝑛𝑢 and its anti-causal signal −𝛼𝑛𝑢 have identical closed-form expressions for the z-transform, that is, 𝑍 =𝑍 1 = 1 − 𝛼𝑧−1 This implies that a closed-form expression for the z-transform does not uniquely specify the signal in the time domain. 𝔃-Transform Issue The ambiguity can be resolved only if in addition to the closed-form expression, the ROC is specified. In summary, a discrete-time signal 𝑥 is uniquely determined by its z-transform 𝑋 and the region of convergence of 𝑋. Moving forward, the term “z-transform” is used to refer to both the closed- form expression and the corresponding ROC. The previous examples also illustrates the point that the ROC of a causal signal is the exterior of a circle of some radius 𝑟2 while the ROC of an anti-causal signal is the interior of a circle of some radius 𝑟1. Example In determining the convergence of 𝑋 , we consider two different cases: Case 1: 𝑏 < 𝛼 In this case the two ROC above do not overlap. Consequently, we cannot find values of z for which both power series converge simultaneously. Clearly, in this case, X(z) does not exist. Example In determining convergence of the 𝑋 , we consider two different cases: Case 2: 𝑏 > 𝛼 In this case there is a ring in the z-plane where both power series converge (overlapping area). simultaneously, as shown in the figure. Example So if Case 2 is true, that is, 𝑏 > 𝛼 , then 𝑋 = 1 1 − 𝛼𝑧−1 − 1 1 − 𝑏𝑧−1 𝑋 = 𝑏−𝛼 𝛼 + 𝑏 − 𝑧 − 𝛼𝑏𝑧−1 ROC: 𝛼 < 𝑧 < 𝑏 This example shows that if there is an ROC for an infinite-duration two-sided signal, it is a ring (annular region) in the z-plane. Poles and Zeros The zeros of a 𝑧-transform 𝑋 are the values of 𝑧 for which 𝑋 = 0. The poles of a 𝑧-transform are the values of 𝑧 for which 𝑋 = ∞.. Zeros - The value(s) for z where P(z)=0. Poles - The value(s) for z where Q(z)=0. Inversion of the 𝔃-Transform There are three methods that are often used for the evaluation of the inverse 𝔃-transform in practice: Direct evaluation of inverse 𝔃-Transform equation, by contour integration. Expansion into a series of terms, in the variables 𝑧, and 𝑧−1 (Power Series Expansion) Partial-fraction expansion and table lookup. The Inverse 𝔃-Transform by Partial- Fraction Expansion In the table lookup method, we attempt to express the function 𝑋 linear combination as a where 𝑋1 , … , 𝑋𝑘 are expressions with inverse transforms a table 𝑥1 , … , 𝑥𝑘 available in decomposition is possible, then 𝑥 , the inverse z-transform of 𝑋 , can of 𝑧 -transform pairs. If such a easily be found using the linearity property as The Inverse 𝔃-Transform by Partial- Fraction Expansion This approach is particularly useful if 𝑋 𝑧 is a rational function. Without loss of generality, we assume that 𝑎0 = 1, The expression above is called proper if 𝑎𝑁 ≠ 0 and 𝑀 < 𝑁. This is equivalent to saying that the number of finite zeros is less than the number of finite poles. The Inverse 𝔃-Transform by Partial- Fraction Expansion In general, any improper rational function (𝑀 ≥ 𝑁) can be expressed as where the degree of 𝐵1 is less than 𝐴. The Inverse 𝔃-Transform by Partial- Fraction Expansion If we already have our rational expression, Next thing to do is to eliminate negative powers of 𝑧 by multiplying both the numerator and denominator of 𝑧𝑁. This results in, The Inverse 𝔃-Transform by Partial- Fraction Expansion To make sure that the transform is proper, we factor out 𝑧 on the numerator to achieve the form, We then express the denominator in factored form to get the partial fraction decomposition of the transform. There are two cases: Distinct poles Repeating poles Distinct Poles Suppose that the poles 𝑝1, 𝑝2, … , 𝑝𝑛 are all different (distinct). Then we seek an expansion of the form, Now the problem is finding the value of the unknown coefficients 𝐴1, 𝐴2, … , 𝐴𝑁. Repeating Poles Suppose that the factored form of the denominator of the rational expression is in the form 𝑚, which means that there is 𝑚 multiplicity of poles. Then we seek an expansion of the form, Inversion of Transform with Real and Distinct Poles For transforms in the form, 𝑋 = 𝑧 𝑧 − 𝑝1 + 𝐴2 𝐴1 𝑧 − 𝑝2 + ⋯ + 𝐴𝑘 𝑧 − 𝑝𝑘 Where 𝑝1, 𝑝2, … , 𝑝𝑘 are real and distinct. Manipulate into the form, 𝑋 = 𝐴1 1 1 − 𝑝1 𝑧−1 + 𝐴2 1 1 − 𝑝2 𝑧−1 + ⋯ + 𝐴𝑘 1 1 − 𝑝𝑘 𝑧−1 Inversion of Transform with Real and Distinct Poles Take the inversion on both sides 𝑍−1 = 𝐴1 𝑋 1 1 − 𝑝1𝑧−1 + 𝐴2 1 𝑧 − 𝑝2𝑧−1 + ⋯ + 𝐴𝑘 From the table, 𝑍−1 𝑝𝑘 𝑛𝑢 =ቊ 𝑛𝑢 𝑖𝑓 𝑅𝑂𝐶: 𝑧 > 𝑝𝑘 𝑖𝑓 𝑅𝑂𝐶: 𝑧 < 𝑝𝑘 Inversion of Transform with Real and Distinct Poles Therefore, 𝑥 = 𝐴1𝑝𝑛 + 𝐴2𝑝2 + ⋯ + 𝐴𝑘𝑝𝑛 𝑢 1 2 𝑘 If the signal 𝑥 is causal, the ROC is 𝑧 > 𝑝𝑚𝑎𝑥 , where 𝑝𝑚𝑎𝑥 = max. Thus, a causal signal, having a 𝒛-transform that contains real and distinct poles, is a linear combination of real exponential signals. Inversion of Transform with Complex Poles For transforms in the form, 𝑋 = 𝑧 𝑧 − 𝑝𝑘 𝐴𝑘 ∗ 𝑘 𝑧 − 𝑝∗ Where 𝑝𝑘, 𝑝∗ are complex. Then its inverse transform is 𝑘 𝑥 = 𝐴 𝑝𝑛 + 𝐴∗ 𝑛 𝑢 𝑘 𝑘 𝑘 We have to express 𝐴𝑖 and 𝑝𝑖 in complex exponential form 𝐴𝑘 = 𝐴𝑘 𝑒𝑗𝛼𝑘 𝑝𝑘 = 𝑟𝑘𝑒𝑗𝛽𝑘 Where 𝛼𝑘 and 𝛽𝑘 are phases of 𝐴𝑘 and 𝑝𝑘. Inversion of Transform with Complex Poles Substitute in the inverse transform, 𝑥 = 𝐴𝑘 𝑛 𝑒𝑗 𝛽𝑘𝑛+𝛼𝑘 + 𝑒−𝑗 𝛽𝑘𝑛+𝛼𝑘 𝑢 Which is also equal to 𝑥 Therefore, = 2 𝐴𝑘 𝑟𝑛 cos 𝑢 𝑍−1 ∗ 𝑘 𝑧 − 𝑝∗ = 2 𝐴𝑘 𝑟𝑛 cos 𝑢 if the ROC is 𝑧 > 𝑝𝑘 = 𝑟𝑘 Inversion of Transform with Double Poles For transforms in the form, Manipulate into the form, 𝑋 = 𝑧 2 𝐴𝑘 𝑧−1 𝑋 = 𝐴𝑘 2 𝐴 𝑋 = 𝑘 𝑝𝑘 𝑝𝑘𝑧 2 Inversion of Transform with Double Poles Recall that, 𝑍−1 =ቊ 𝑛 𝑖𝑓 𝑅𝑂𝐶: 𝑧 −𝑛 Therefore, = 𝐴𝑘 𝑛 𝑝𝑘 𝑛𝑢 = 𝐴𝑘𝑛 𝑛−1𝑢 𝑝𝑘 < 𝑝𝑘 𝑢 𝑖𝑓 𝑅𝑂𝐶: 𝑧 𝑍−1 > If the signal 𝑥 is causal or the ROC is 𝑧 > 𝑝𝑘