Time-Domain Analysis and z Transform

CHAPTER 2 Time-Domain Analysis and z Transform 2.1 A LINEAR, TIME-INVARIANT SYSTEM The purpose of analysis of a discrete-time system is to find the output in either the time or frequency domain of the system due to a discrete-time input signal. In Chapter 1, we defined the discrete-time signal as a function of the integer variable n, which represents discrete time, space, or some other physical variable. Given any integer value in −∞ <n<∞, we can find the value of the signal according to some well-defined relationship. This can be described as a mapping of the set of integers to a set of values of the discrete-time signal. Description of this relationship varied according to the different ways of modeling the signal. In this chapter, we define the discrete-time system as a mapping of the set of discrete- time signals considered as the input to the system, to another set of discrete-time signals identified as the output of the system. This mapping can also be defined by an analytic expression, formula, algorithm, or rule, in the sense that if we are given an input to the system, we can find the output signal. The mapping can therefore be described by several models for the system. The mapping or the input–output relationship may be linear, nonlinear, time-invariant, or time- varying. The system defined by this relationship is said to be linear if it satisfies the following conditions. Assume that the output is y(n) due to an input x(n) according to this rela- tionship. If an input Kx(n) produces an output Ky(n), the system satisfies the condition of homogeneity, where K is any arbitrary constant. Let K 1 y 1 (n) and K 2 y 2 (n) be the outputs due to the inputs K 1 x 1 (n) and K 2 x 2 (n), respectively, where K 1 and K 2 are arbitrary constants. If the output is K 1 y 1 (n) + K 2 y 2 (n) when the input is K 1 y 1 (n) + K 2 y 2 (n), then the system satisfies the superposition property. A system that satisfies both homogeneity and superposition is defined as a linear system. If the output is y(n − M) when the input is delayed by M sam- ples, that is, when the input is x(n − M), the system is said to be time-invariant or shift-invariant. If the output is determined by the weighted sum of only the previous values of the output and the weighted sum of the current and previous Introduction to Digital Signal Processing and Filter Design, by B. A. Shenoi Copyright © 2006 John Wiley & Sons, Inc. 32 A LINEAR, TIME-INVARIANT SYSTEM 33 values of the input, then the system is defined as a causal system. This means that the output does not depend on the future values of the input. We will discuss these concepts again in more detail in later sections of this chapter. In this book, we consider only discrete-time systems that are linear and time-invariant (LTI) systems. Another way of defining a system in general is that it is an interconnection of components or subsystems, where we know the input–output relationship of these components, and that it is the way they are interconnected that determines the input–output relationship of the whole system. The model for the DT system can therefore be described by a circuit diagram showing the interconnection of its components, which are the delay elements, multipliers, and adders, which are introduced below. In the following sections we will use both of these definitions to model discrete-time systems. Then, in the remainder of this chapter, we will discuss several ways of analyzing the discrete-time systems in the time domain, and in Chapter 3 we will discuss frequency-domain analysis. 2.1.1 Models of the Discrete-Time System First let us consider a discrete-time system as an interconnection of only three basic components: the delay elements, multipliers, and adders. The input–output relationships for these components and their symbols are shown in Figure 2.1. The fourth component is the modulator, which multiplies two or more signals and hence performs a nonlinear operation. A simple discrete-time system is shown in Figure 2.2, where input signal x(n) ={x(0), x(1), x(2), x(3)} is shown to the left of v 0 (n) = x(n). The signal Delay Element Multiplier Adder Modulator m 6 (n) X 1 (n) Y 1 (n) = X 1 (n − 1) Y 2 (n) = KX 2 (n) Y 3 (n) = X 3 (n) + X 4 (n) Y 5 (n) = X 5 (n)m 6 (n) X 2 (n) X 3 (n) X 4 (n) X 5 (n) Σ Σ K z −1 Figure 2.1 The basic components used in a discrete-time system. 34 TIME-DOMAIN ANALYSIS AND z TRANSFORM x(0) x(1) x(2) x(3) x(0) x(1) x(2) x(3) 2 1 034 21034567 n n X(n) y(n) V 0 (n) b(0) b(1) b(2) b(3) V 1 (n) V 2 (n) V 3 (n) z −1 z −1 z −1 Σ x(0) x(1) x(2) x(3) 210 3 4 56 7 n 210 x(0) x(1) x(2) x(3) 34567 n Figure 2.2 Operations in a typical discrete-time system. v 1 (n) shown on the left is the signal x(n) delayed by T seconds or one sam- ple, so, v 1 (n) = x(n − 1). Similarly, v(2) and v(3) are the signals obtained from x(n) when it is delayed by 2T and 3T seconds: v 2 (n) = x(n − 2) and v 3 (n) = x(n − 3). When we say that the signal x(n) is delayed by T,2T ,or3T seconds, we mean that the samples of the sequence are present T,2T ,or3T seconds later, as shown by the plots of the signals to the left of v 1 (n), v 2 (n), and v 3 (n). But at any given time t = nT ,thesamplesinv 1 (n), v 2 (n),andv 3 (n) are the samples of the input signal that occur T,2T ,and3T seconds previous to t = nT . For example, at t = 3T , the value of the sample in x(n) is x(3),andthe values present in v 1 (n), v 2 (n) and v 3 (n) are x(2), x(1),andx(0), respectively. A good understanding of the operation of the discrete-time system as illustrated above is essential in analyzing, testing, and debugging the operation of the sys- tem when available software is used for the design, simulation, and hardware implementation of the system. It is easily seen that the output signal in Figure 2.2 is y(n) = b(0)v(0) + b(1)v(1) + b(2)v(2) + b(3)v(3) = b(0)x(n) + b(1)x(n − 1) + b(2)x(n − 2) + b(3)x(n − 3) A LINEAR, TIME-INVARIANT SYSTEM 35 X(z) ΣΣ Σ 0.8 0.5 0.3 −0.2 −0.4 0.6 −0.1 y 1 (z) y 3 (z) y 2 (z) z −1 z −1 z −2 y 1 (z) z −1 y 2 (z) z −1 z −1 Figure 2.3 Schematic circuit for a discrete-time system. where b(0), b(1), b(2), b(3) are the gain constants of the multipliers. It is also easy to see from the last expression that the output signal is the weighted sum of the current value and the previous three values of the input signal. So this gives us an input–output relationship for the system shown in Figure 2.2. Now we consider another example of a discrete-time system, shown in Figure 2.3. Note that a fundamental rule is to express the output of the adders and generate as many equations as the number of adders found in this circuit diagram for the discrete-time system. (This step is similar to writing the node equations for an analog electric circuit.) Denoting the outputs of the three adders as y 1 (n), y 2 (n),andy 3 (n),weget y 1 (n) = 0.3y 1 (n − 1) − 0.2y 1 (n − 2) − 0.1x(n − 1) y 2 (n) = y 1 (n) + 0.5y 1 (n − 1) − 0.4y 2 (n − 1) y 3 (n) = y 2 (n) + 0.6y 2 (n − 1) + 0.8y 1 (n) (2.1) These three equations give us a mathematical model derived from the model shown in Figure 2.3 that is schematic in nature. We can also derive (draw the circuit realization) the model shown in Figure 2.3 from the model given in Equations (2.1). We will soon describe a method to obtain a single input–output relationship between the input x(n) and the output y(n) = y 3 (n), after eliminat- ing the internal variables y 1 (n) and y 2 (n); that relationship constitutes the third model for the system. The general form of such an input–output relationship is y(n) =− N  k=1 a(k)y(n − k) + M  k=0 b(k)x(n − k) (2.2) 36 TIME-DOMAIN ANALYSIS AND z TRANSFORM or in another equivalent form N  k=0 a(k)y(n − k) = M  k=0 b(k)x(n − k); a(0) = 1 (2.3) Equation (2.2) shows that the output y(n) is determined by the weighted sum of the previous N values of the output and the weighted sum of the current and previous M + 1 values of the input. Very often the coefficient a(0) as shown in (2.3) is normalized to unity. Soon we will introduce the z transform to represent the discrete-time signals in the set of equations above, thereby generating more models for the system, and from these models in the z domain, we will derive the transfer function H(z −1 ) and the unit sample response or the unit impulse response h(n) of the system. From any one of these models in the z domain, we can derive the other models in the z domain and also the preceding models given in the time domain. It is very important to know how to obtain any one model from any other given model so that the proper tools can be used efficiently, for analysis of the discrete-time system. In this chapter we will elaborate on the different models of a discrete- time system and then discuss many tools or techniques for finding the response of discrete-time systems when they are excited by different kinds of input signals. 2.1.2 Recursive Algorithm Let us consider an example of Equation (2.2) as y(n) = y(n − 1) − 0.25y(n − 2) + x(n), where the input sequence x(n) = δ(n), and the two initial conditions are y(−1) = 1.0andy(−2) = 0.4. We compute y(0), y(1), y(2), . in a recursive manner as follows: y(0) = y(−1) − 0.25y(−2) + x(0).Sincex(n) = δ(n), we substitute x(0) = 1 and get y(0) = 1.0 − 0.25(0.4) + 1 = 1.9. Next y(1) = y(0) − 0.25y(−1) + x(1).We know y(0) = 1.9 from the step shown above, and also that x(1) = 0. So we get y(1) = 1.9 − 0.25(1.0) + 0 = 1.65. Next, for n = 2, when we compute y(2) = y(1) − 0.25y(0) + x(2). Substituting the known values from above, we get y(2) = 1.65 − 0.25(1.9) + 0 = 1.175. Next, when n = 3, we obtain y(3) = y(2) − 0.25y(1) + x(3) = 1.175 − 0.25(1.65) + 0 = 0.760 We can continue to calculate the values of the output y(n) for n = 4, 5, 6, 7, This is known as the recursive algorithm, which we use to calculate the output when we are given an equation of the form (2.2); it can be used when there is any other input. For a system modeled by an equation of the form (2.2), the output is infinite in length in general. As a special case, when the input is the unit impulse function δ(n), and the initial conditions are assumed to be zero, the A LINEAR, TIME-INVARIANT SYSTEM 37 resulting output is called the unit impulse response h(n) (or more appropriately the unit sample response) and is infinite in length. Consider a system in which the multiplier constants a(k) = 0fork = 1, 2, 3, .,N. Then Equation (2.2) reduces to the form y(n) = M  k=0 b(k)x(n − k) (2.4) = b(0)x(n) + b(1)x(n − 1) + b(2)x(n − 2) +···+b(M)x(n − M) Let us find the unit impulse response of this system, using the recursive algo- rithm, as before: y(0) = b(0)(1) + 0 + 0 + 0 +···=b(0) y(1) = b(0)x(1) + b(1)x(0) + b(2)x(−1) + 0 + 0 +···=b(1) y(2) = b(0)x(2) + b(1)x(1) + b(2)x(0) + 0 + 0 + 0 +···=b(2) Continuing this procedure recursively, we would get y(3) = b(3) y(4) = b(4) · · · y(M) = b(M) This example leads to the following two observations: (1) the samples of the unit impulse response are the same as the coefficients b(n), and (2) therefore the unit impulse response h(n) of the system is finite in length. So we have shown without proof but by way of example that the unit impulse response of the system modeled by an equation of the form (2.2) is infinite in length, and hence such a system is known as an infinite impulse response (IIR) filter, whereas the system modeled by an equation of the form (2.4), which has an unit impulse response that is finite in length, is known as the finite impulse response (FIR) filter. We will have a lot more to say about these two types of filters later in the book. Equation (2.3) is the ordinary, linear, time-invariant, difference equation of N th order, which, if necessary, can be rewritten in the recursive difference equation form (2.2). The equation can be solved in the time domain, by the following four methods: 1. The recursive algorithm as explained above 2. The convolution sum, to get the zero state response, as explained in the next section 38 TIME-DOMAIN ANALYSIS AND z TRANSFORM 3. The classical method of solving a difference equation 4. The analytical solution using the z transform. We should point out that methods 1–3 require that the DT system be modeled by a single-input, single-output equation. If we are given a large number of difference equations describing the DT system, then methods 1–3 are not suitable for finding the output response in the time domain. Method 4, using the z transform, is the only powerful and general method to solve such a problem, and hence it will be treated in greater detail and illustrated by several examples in this chapter. Given a model in the z-transform domain, we will show how to derive the recursive algorithm and the unit impulse response h(n) so that the convolution sum can be applied. So the z-transform method is used most often for time-domain analysis, and the frequency-domain analysis is closely related to this method, as will be discussed in the next chapter. 2.1.3 Convolution Sum In the discussion above, we have assumed that the unit impulse response of a discrete-time system when it is excited by a unit impulse function δ(n),exists (or is known), and we denote it as h(n). Instead of using the recursive algo- rithm to find the response due to any input, let us represent the input sig- nal x(n) not by its values in a sequence {x(0), x(1), x(2), x(3), .} but as the values of impulse function at the corresponding instants of time. In other words, we consider the sequence of impulse functions x(0)δ(n), x(1)δ(n − 1), x(2)δ(n − 2), . as the input—and not the sequence of values {x(0), x(1), x(2), x(3), .}. The difference between the values of the samples as a sequence of numbers and the sequence of impulse functions described above should be clearly understood. The first operation is simple sampling operation, whereas the second is known as impulse sampling, which is a mathematical way to repre- sent the same data, and we represent the second sequence in a compact form: x(n) =  ∞ k=0 x(k)δ(n − k). The mathematical way of representing impulse sam- pling is a powerful tool that is used to analyze the performance of discrete-time systems, and the values of the impulse functions at the output are obtained by analytical methods. These values are identified as the numerical values of the output signal. Since h(n) is the response due to the input δ(n),wehavex(0)h(n) as the response due to x(0)δ(n) because we have assumed that the system is linear. Assuming that the system is time-invariant as well as linear, we get the output due to an input x(1)δ(n − 1) to be x(1)h(n − 1). In general, the output due to an input x(k)δ(n − k) is given by x(k)h(n − k). Adding the responses due to all the impulses in x(n) =  ∞ k=0 x(k)δ(n − k), we get the total output as the sum y(n) = ∞  k=0 x(k)h(n − k) (2.5) A LINEAR, TIME-INVARIANT SYSTEM 39 This is known as the convolution sum, denoted by a compact notation y(n) = x(n) ∗ h(n). The summation formula can be used to find the response due to any input signal. So if we know the unit impulse response h(n) of the system, we can find the output y(n) due to any input x(n)—therefore it is another model for the discrete-time system. In contrast to the recursive algorithm, however, note that the convolution sum cannot be used to find the response due to given initial conditions. When and if the input signal is defined for −∞ <n<∞ or −M ≤ n<∞, obviously the lower index of summation is changed to −∞.In this case the convolution sum formula takes the general form y(n) = ∞  k=−∞ x(k)h(n − k) (2.6) For example, even though we know that h(n) = 0for−∞ <n<0, if the input sequence x(n) is defined for −M<n<∞,thenwehavetousetheformula y(n) =  ∞ k=−∞ x(k)h(n − k).Ifx(n) = 0for−∞ <n<0, then we have to use the formula y(n) =  ∞ k=0 x(k)h(n − k). To understand the procedure for implementing the summation formula, we choose a graphical method in the following example. Remember that the recur- sive algorithm cannot be used if the DT system is described by more than one difference equation, and the convolution sum requires that we have the unit pulse response of the system. We will find that these limitations are not present when we use the z-transform method for analyzing the DT system performance in the time domain. Example 2.1 Given an h(n) and x(n), we change the independent variable from n to k and plot h(k) and x(k) as shown in Figure 2.4a,b. Note that the input sequence is defined for −2 ≤ k ≤ 5 but h(k) is a causal sequence defined for 0 ≤ k ≤ 4. Next we do a time reversal and plot h(−k) in Figure 2.4c. When n ≥ 0, we obtain h(n − k) by delaying (or shifting to the right) h(−k) by n samples; when n<0, the sequence h(−k) is advanced (or shifted to the left). For every value of n, we have h(n − k) and x(k) and we multiply the samples of h(n − k) and x(k) at each value of k and add the products. For our example, we show the summation of the product when n =−2in Figure 2.4d, and show the summation of the product when n = 3inFigure2.4e. The output y(−2) has only one nonzero product = x(−2)h(0). But the output sample y(3) is equal to x(0)h(3) + x(1)h(2) + x(2)h(1) + x(3)h(0). But note that when n>9, and n<−2, the sequences h(n − k) and x(k) do not have overlapping samples, and therefore y(n) = 0forn>9andn<−2. Example 2.2 As another example, let us assume that the input sequence x(n) and also the unit impulse response h(n) are given for 0 ≤ n<∞. Then output y(n) given 40 TIME-DOMAIN ANALYSIS AND z TRANSFORM −2 −6 −5−4 −3−2 −1 −1 0 0 −4 −3 −2 −1 0 1 1 2 2 −10 1 2 3 012345 109876 3 4 5 0 1 2 3 4 k k k k k k X(k) h(k) h(−2 − k) h(− k) h(3 − k) h(10 − k) (a)(b) (d) (c) (e) (f ) Figure 2.4 Convolution sum explained. by (2.5) can be computed for each value of n as shown below: y(0) = x(0)h(0) y(1) = x(0)h(1) + x(1)h(0) y(2) = x(0)h(2) + x(1)h(1) + x(2)h(0) y(3) = x(0)h(3) + x(1)h(2) + x(2)h(1) + x(3)h(0) y(4) = x(0)h(4) + x(1)h(3) + x(2)h(2) + x(3)h(1 ) + x(4)h(0) · · · y(n) = x(0)h(n)+x(1)h(n − 1)+x(2)h(n − 2)+x(3)h(n − 3)+· · ·+x(n)h(0) · (2.7) · · z TRANSFORM THEORY 41 It is interesting to note the following pattern. In the expressions for each value of the output y(n) above, we have x(0), x(1), x(2) . and h(n), h(n − 1), h(n − 2) . multiplied term by term in order and the products are added, while the indices of the two samples in each product always add to n. Convolution is a fundamental operation carried out by digital signal processors in hardware and in the processing of digital signals by software. The design of digital signal processors and the software to implement the convolution sum have been developed to provide us with very efficient and powerful tools. We will discuss this subject again in Section 2.5, after we learn the theory and application of z transforms. 2.2 z TRANSFORM THEORY 2.2.1 Definition In many textbooks, the z transform of a sequence x(n) is simply defined as Z[x(n)] = X(z) = ∞  n=−∞ x(n)z −n (2.8) and the inverse z transform defined as Z −1 [X(z)] = x(n) = 1 2πj  C X(z)z n−1 dz (2.9) Equation (2.8) represents the (double-sided or) bilateral z transform of a sequence x(n) defined for −∞ <n<∞. The inverse z transform given in (2.9) is obtained by an integration in the complex z plane, and this integration in the z plane is beyond the scope of this book. We prefer to consider signals that are of interest in digital signal processing and hence consider a sequence obtained by sampling a continuous-time signal x(t) with a constant sampling period T (where T is the sampling period), and generate a sequence of numbers x(nT ). Remember that according to the sifting theorem, we have x(t)δ(t) = x(0)δ(t). We use this result to carry out a proce- dure called impulse sampling by multiplying x(t) with an impulse train p(t) =  ∞ n=0 δ(t − nT ). Consequently we consider a sequence of delayed impulse func- tions weighted by the strength equal to the numerical values of the signal instead of a sequence of numbers. By doing so, we express the discrete sequence as a function of the continuous variable t, which allows us to treat signal processing mathematically. The product is denoted as x ∗ (t) = ∞  n=0 x(t)δ(t − nT ) = ∞  n=0 x(nT )δ(t − nT ) (2.10) [...]... z( 2z2 − 1 1z + 12) (z − 1) (z − 2)3 (2.44) ( 2z2 − 1 1z + 12) k C0 C1 C2 G (z) = + = + + 3 3 2 z (z − 1) (z − 2) (z − 2) (z − 2) (z − 2) (z − 1) k= C0 = C1 = C2 = ( 2z2 − 1 1z + 12) (z − 2)3 z= 1 ( 2z2 − 1 1z + 12) (z − 1) z= 2 = −3 = −2 ( 2z2 − 1 1z + 12) (z − 1) d dz z= 2 ( 2z2 − 1 1z + 12) (z − 1) 1 d2 2 dz2 2z2 − 4z − 1 (z − 1)2 = = z= 2 1 d 2 dz = −1 z= 2 2z2 − 4z − 1 (z − 1)2 =3 z= 2 Therefore we have G (z) = − 3z. .. we obtain Y0i (z) = [0.28 8z2 − 0.0 2z] z[ 0.28 8z − 0.02] [0.288 − 0.0 2z 1 ] = 2 = −1 + 0.0 2z 2 ] [1 − 0. 3z z − 0. 3z + 0.02 (z − 0.1) (z − 0.2) and Y0s (z) = = z [1 − 0. 1z 1 ] X (z) [1 − 0. 1z 1 ] = [1 − 0. 3z 1 + 0.0 2z 2 ] z + 0.2 [1 − 0. 3z 1 + 0.0 2z 2 ] z[ z2 − 0. 1z] z2 (z − 0.1) = (z + 0.2) (z2 − 0. 3z + 0.02) (z + 0.2) (z − 0.1) (z − 0.2) We notice that there is a pole and a zero at z = 0.1 in the second term... 9 X (z) δ(n) δ(n − m) n2 a n 10 11 45 1 z m z z−1 az z 1 z z−a az (z − a)2 z( z + 1) (z − 1)3 n(n − 1) n−2 a 2! n(n − 1)(n − 2) · · · · · · · (n − m + 2) n−m+1 a (m − 1)! 12 r n ej θn 13 r n cos(θn) 14 r n sin(θn) 15 e−αn cos(θn) Proof : X (z) = we get z( z − e−α cos(θ )) z2 − (2e−α cos(θ ) )z + e−2α ∞ n=0 dX (z) = dz z z (z2 + 4z + 1) (z − 1)4 az (z + a) (z − a)3 z (z − a)3 z (z − a)m z z − rej θ z( z − r... = z 1 Y (z) + y(−1) and Z[ x(n − 1)] = z 1 X (z) + x(−1) where X (z) = z/ (z − 0.2) and x(−1) = 0, since x(n) is zero for −∞ < n < 0 Substituting these results, we get Y (z) − 0.5 z 1 Y (z) + y(−1) = 5 z 1 X (z) + x(−1) Y (z) − 0.5 z 1 Y (z) + y(−1) = 5z 1 X (z) 48 TIME-DOMAIN ANALYSIS AND z TRANSFORM Y (z) 1 − 0. 5z 1 = 0.5y(−1) + 5z 1 X (z) (2.34) 0.5y(−1) 5z 1 + X (z) (1 − 0. 5z 1 ) (1 − 0. 5z 1 ) 0.5y(−1 )z. .. z: G (z) = C0 z C1 z Cr−1 z + + ··· + r r−1 (z − z0 ) (z − z0 ) (z − z0 ) k2 z km z k1 z + + ··· + + (z − z1 ) (z − z2 ) (z − zm ) Then we find the inverse z transform of each term to get g(n), using the z transform pairs given in Table 2.1 To illustrate this method, we consider the 56 TIME-DOMAIN ANALYSIS AND z TRANSFORM function G (z) , which has a simple pole at z = 1 and a triple pole at z = 2: G (z) ... and Y0s (z) reduces to z2 /[ (z + 0.2) (z − 0.2)] We divide Y0i (z) by z, expand it into its normal partial fraction form [0.28 8z − 0.02] 0.376 0.088 Y0i (z) = = − z (z − 0.1) (z − 0.2) (z − 0.2) (z − 0.1) and multiply by z to get Y0i (z) = 0.08 8z 0.37 6z − (z − 0.2) (z − 0.1) Similarly, we expand Y0s (z) /z = z/ [ (z + 0.2) (z − 0.2)] in the form −0.5/ (z + 0.2) + 0.5/ (z − 0.2) and get Y0s (z) = −0. 5z 0. 5z. .. G (z) N (z) = r (z − z ) (z − z ) (z − z ) · · · (z − z ) z (z − z0 ) 1 2 3 m Its normal partial fraction is in the form G (z) C0 C1 Cr−1 = + + ··· + z (z − z0 )r (z − z0 )r−1 (z − z0 ) k2 km k1 + + ··· + + (z − z1 ) (z − z2 ) (z − zm ) The residues k1 , k2 , , km for the simple poles at z1 , z2 , are obtained by the normal method of multiplying G (z) /z by (z − zi ), i = 1, 2, 3, , m and evaluating... the product at z = zi The residue C0 is also found by the same method: C0 = (z − z0 )r G (z) z z =z0 The coefficient C1 is found from G (z) d (z − z0 )r dz z z =z0 and the coefficient C2 is found from 1 d2 G (z) (z − z0 )r 2 dz2 z z =z0 The general formula for finding the coefficients Cj , j = 1, 2, 3, , (r − 1) is Cj = 1 dj j ! dzj (z − z0 )r G (z) z (2.43) z= z0 After obtaining the residues and the coefficients,... residues R1 and R2 using the normal procedure and get R1 = R2 = Y0s (z) (z − 0.5) z Y0s (z) (z − 0.2) z = z= 0.5 = z= 0.2 5 (z − 0.2) z= 0.5 5 (z − 0.5) z= 0.2 = 16.666 = −16.666 Therefore Y0s (z) 5 16.666 16.666 = = − z (z − 0.5) (z − 0.2) z − 0.5 z − 0.2 z TRANSFORM THEORY 49 Multiplying both sides by z, we get Y0s (z) = 16.66 6z 16.66 6z − z − 0.5 z − 0.2 (2.38) Now we obtain the inverse z transform y0s... = 0 and expressing them in a matrix equation as (1 + 0. 2z 1 + 0. 4z 2 ) − 2z 1 0 (1 + 0. 1z 1 ) Y1 (z) Y2 (z) = z 1 X (z) 0 from Cramer’s rule, we find the z transform of the output y2 (n): Y2 (z) = (1 + 0. 2z 1 + 0. 4z 2 ) − 2z 1 (1 + 0. 2z 1 + 0. 4z 2 ) − 2z 1 z 1 X (z) 0 0 (1 + 0. 1z 1 ) SOLVING DIFFERENCE EQUATIONS USING THE CLASSICAL METHOD = 2z 2 X (z) 1 + 0. 3z 1 + 0.4 2z 2 + 0.0 4z 3 = 63 2z X (z) z3 + 0. 3z2 . 0 X (z) 1 δ(n) 1 2 δ(n − m) z −m 3 u(n) z z − 1 4 au(n) az z − 1 5 a n z z − a 6 na n az ( z − a ) 2 7 n 2 z( z + 1) (z − 1) 3 8 n 3 z( z 2 + 4z + 1) (z −. X (z) Y (z) = 0.5y(−1 )z (z − 0.5) + 5 (z − 0.5) X (z) Substituting y(−1) = 2andX (z) = z/ (z − 0.2) in this last expression, we get Y (z) = z (z − 0.5) + 5z (z