8. z-domain Analysis of Discrete-Time Signals and Systems

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1 8. z-domain Analysis of Discrete-Time Signals and Systems 8.. Definition of z-transform ( ) 8.2. Properties of z-transform (0.5) 8.3. System Function (0.7) 8.4. Classification of a Linear Time-Invariant Discrete-Time System by its System Function (0.7) 8.5. Linear Constant-Coefficient Difference Equations (0.7) 8.6. Implementation of Discrete-Time Systems (0.8)

2 8.. Definition of z-transform The z-transform of a sequence x(n) is defined as X(z) n n x(n)z, (8.) where z is a complex variable. x(n) can be recovered from X(z) by x(n) n X(z)z dz. (8.2) 2j ( z r) (8.2) is called the inverse z-transform Derivation of Inverse z-transform Letting z=re j in (8.), we obtain j n jn x(n)r e. n X re (8.3)

3 That is, X(re j ) is the Fourier transform of x(n)r n. Thus, x(n)r n j X re jn e d. (8.4) 2 2 Multiplying both the sides of (8.4) by r n, we obtain and thus (8.2). j j n x (n) X re re d (8.5) Relation of z-transform to Fourier Transform The z-transform of x(n) on the circle z =r is equal to the Fourier transform of x(n)r n, i.e., X(re j )=FT[x(n)r n ]. (8.6) Especially, the z-transform of x(n) on the circle z = is equal to the Fourier transform of x(n), i.e.,

4 8..3. Zeros and Poles of z-transform X(e j )=FT[x(n)]. (8.7) Assume that X(z) is rational, i.e., X(z)=P (z)/p 2 (z), where P (z) and P 2 (z) are two polynomials. The roots of P (z)=0 are called the zeros of X(z), and the roots of P 2 (z)=0 are called the poles of X(z). Zeros and poles are indicated with and in the complex plane, respectively. The algebraic expression of X(z) can be specified by its zeros and poles except for a scale factor Region of Convergence of z-transform The open region where X(z) converges is referred to as the region of convergence (ROC) of X(z). In other discussions, the ROC may mean the region where X(z) converges. It should be noted that both the algebraic expression and the ROC of X(z) are required to specify

5 x(n) uniquely. Assume that x(n) has the z-transform X(z). For different types of x(n), the ROC of X(z) has different types. () If x(n) is of finite duration, then the ROC of X(z) is the entire plane, i.e., zc. Especially, z=0 needs to be excluded from the ROC if x(n) has nonzero values for n>0. (2) If x(n) is right-sided, then the ROC of X(z) is the exterior of a circle centered about the origin, i.e., z >r 0. (3) If x(n) is left-sided, then the ROC of X(z) is the interior of a circle centered about the origin, i.e., z <r 0. Especially, z=0 needs to be excluded from the ROC if x(n) has nonzero values for n>0. (4) If x(n) is two-sided, then the ROC of X(z) is a ring centered about the origin, i.e., r < z <r 2. In addition, if X(z) is rational and has poles, then the ROC of X(z)

6 is bounded by the poles. Example. Find the z-transforms of the following signals, including the algebraic expressions and the ROCs. () x(n)=(n). (2) x(n)=(n). (3) x(n) n a 0,, 0 (4) x(n)=a n u(n). (5) x(n)=a n u(n). (6) x(n) 3 (7) x(n)=a n, a <. n n N. otherwise sin nu(n). 4

7 8.2. Properties of z-transform Linearity If x (n)x (z) and x 2 (n)x 2 (z), then a x (n)+a 2 x 2 (n)a X (z)+a 2 X 2 (z), (8.8) where a and a 2 are two arbitrary constants Differentiation If x(n)x(z), then Shifting If x(n)x(z), then nx(n) dx(z) z. (8.9) dz x(nm)x(z)z m, (8.0)

8 where m is an arbitrary integer. Example. Find the inverse z-transforms of the following signals: () X(z)=/(az ) m, z > a. (2) X(z)=/(az ) m, z < a Scaling If x(n)x(z), then y(n) x(n / m), n a multiple of 0, n a multiple of where m is a nonzero integer. m m Y(z) m X z, (8.) Letting m= in (8.), we can obtain the time-reversal property of the z-transform, i.e., x(n)x(z ). (8.2)

9 From (8.2), the following conclusions can be drawn: () x(n) even X(z)=X(z ). (2) x(n) odd X(z)=X(z ). If x(n)x(z), then a n x(n) z X. (8.3) a Letting a=exp(j 0 ) in (8.3), we can obtain the rotation property of the z-transform, i.e., Conjugation If x(n)x(z), then x(n)exp(j 0 n)x[zexp(j 0 )]. (8.4) x * (n)x * (z * ). (8.5)

10 From (8.5), the following conclusions can be drawn: () Im[x(n)]=0 X(z)=X * (z * ). (2) Re[x(n)]=0 X(z)=X * (z * ) Convolution If x (n)x (z) and x 2 (n)x 2 (z), then x (n)x 2 (n)x (z)x 2 (z). (8.6) Example. Let x (n)=a n u(n), x 2 (n)=b n u(n) and ab. Find x (n)x 2 (n) Initial-Value Theorem and Final-Value Theorem Assume x(n)=0 for n<0. If x(n)x(z), then Proof. X(z) can be expressed as x(0) lim X(z). (8.7) z

11 X(z) n x(n)z n x(0) n x(n)z n. (8.8) Letting z, we obtain x( ) lim (z )X(z). z Proof. (z)x(z) can be expressed as (z )X(z) zx(z) X(z) n [x(n ) x(n)]z lim X(z) x(0). z Assume x(n)=0 for n<0. If x(n)x(z), then Letting z, we obtain n n x(0)z x(n )z n0 n n x(n)z [x(n ) x(n)]z n n. (8.9) (8.20) (8.2)

12 lim[(z )X(z)] z lim x(0) N N n0 x(0) n0 [x(n ) x(n)] [x(n ) x(n)] lim x(n ) N x( ). (8.22) Example. Find the inverse z-transform of X(z)=ln(az ), z > a Inverse of Rational z-transforms Let us consider how to find x(n) if X(z) is a rational function. This can be carried out by the partial-fraction expansion. () By the long division or inspection, expand X(z) into X(z)= P 0 (z )+ P (z )/P 2 (z ), (8.23) where P 0 (z ), P (z ) and P 2 (z ) are polynomials of z, and the order of P (z ) is lower than that of P 2 (z ). (2) Expand P (z )/P 2 (z ) into partial fractions of z. Assume that

13 z=a is an Mth-order root of P 2 (z )=0. It then corresponds to M terms in the expansion, i.e., M P z A m. P z az (8.24) m 2 m Here A m is determined by the comparison method or the elimination method (multiplying (8.24) by (az ) M, taking the derivative of order Mm with respect to z, and substituting a for z to obtain A m ). (3) Determine x(n) according to the basic z-transform pairs and the properties of the z-transform. The basic z-transform pairs are C C m nm m nm a a n n u(n) (n), (8.25) u( n ) az m az, z a, m, z a. (8.26) (8.27)

14 When m=, (8.26) and (8.27) become, respectively, u(n), z a, (8.28) az n a n a u( n ), z a. (8.29) az Example. Find the inverse z-transform of X(z) 8.3. System Function z 6 z A linear time-invariant discrete-time system can be characterized by the system function. It is defined as the z-transform of the impulse response. 3 z. (8.30)

15 The input-output relation of a linear time-invariant discrete-time system can be expressed by the system function, i.e., Y(z)=X(z)H(z), (8.3) where X(z) and Y(z) are the z-transforms of the input and the output, respectively, and H(z) is the system function. The frequency response H() is actually the system function H(z) on the circle z =, i.e., H()=H(e j ). (8.32) 8.4. Classification of a Linear Time-Invariant Discrete-Time System by its System Function Memoryless Systems versus Systems with Memory Assume that H(z) is the system function of a linear time-invariant discrete-time system. The system is memoryless if and only if H(z) is a constant.

16 Causal Systems versus Noncausal Systems Let H(z) be the system function of a linear time-invariant discretetime system. Then, the system is causal if and only if H(z) converges as z. Example. Determine the causality of the following systems: () H(z)=. 3 2 z 2z z ( 2) H(z). (3) H(z) 2 z z z 2z Stable Systems versus Unstable Systems Let H(z) be the system function of a linear time-invariant discretetime system. Then, the system is stable if and only if the ROC of H(z) includes the unit circle centered about the origin..

17 Example. Determine the stability of the following systems: () H(z) z 2z 2 (2) H(z).Causal. az (3) H(z) (2r cos)z. r 2 z 2. Causal Invertible Systems versus Noninvertible Systems Assume that two linear time-invariant discrete-time systems A and B have the system functions G(z) and H(z), respectively. A and B are mutually inverse if and only if G(z)H(z)=. (8.33) (8.33) can be used to construct the inverse of a given system.

18 8.5. Linear Constant-Coefficient Difference Equations The zero-state response of a linear constant-coefficient difference equation can be found using the z-transform. Example. A causal discrete-time system is given by y(n)0.5y(n)=x(n), (8.34) where x(n)=0.25 n u(n) and y()=2. Find the zero-input response, the zero-state response and the complete response. First, using the method in section 5.5, we can obtain the zero-input response y zi (n)=0.5 n. (8.35) Then, let us consider the zero-state response. The zero-state response satisfies (8.34), i.e., y zs (n)0.5y zs (n)=x(n). (8.36)

19 Taking the z-transform of (8.36), we obtain From (8.37), we obtain (z) 0.5z Yzs(z). (8.37) 0.25z Yzs 2 (z). (8.38) 0.5z 0.25z Yzs The ROC of 2/(0.5z ) is determined by the system. The system is causal, and thus the ROC of 2/(0.5z ) is z >0.5. The ROC of /(0.25z ) is determined by the input. The input is right-sided, and thus the ROC of /(0.25z ) is z >0.25. Hence, the inverse z- transform of Y zs (z) is y zs (n)=(2 0.5 n 0.25 n )u(n). (8.39) The complete response equals the sum of the zero-input response and the zero-state response, i.e.,

20 y(n)=0.5 n +(2 0.5 n 0.25 n )u(n). (8.40) Example. A causal discrete-time system is given by y(n)0.5y(n)=x(n). (8.4) Find the impulse response. The zero-state response satisfies (8.4), i.e., y zs (n)0.5y zs (n)=x(n). (8.42) Taking the z-transform of (8.42), we obtain Yzs(z). (8.43) X(z) 0.5z Thus, (z). (8.44) 0.5z H

21 Since the system is causal, the ROC of H(z) is z >0.5. Thus, h(n)=0.5 n u(n). (8.45) Example. Consider a linear time-invariant discrete-time system. If the input is then the output is If the input is then the output is x (n)=(/6) n u(n), (8.46) y (n)=[a(/2) n +0(/3) n ]u(n). (8.47) x 2 (n)=() n, (8.48) y 2 (n)=(7/4)() n. (8.49) Use a difference equation to characterize the system.

22 X 2 (z) X (z) + Adder X (z)+x 2 (z) X(z) a ax(z) Coefficient Multiplier X(z) z Delayer z X(z) Figure 8.. Elements of a Block Diagram.

23 8.6. Implementation of Discrete-Time Systems Let us discuss the implementation of a causal linear time-invariant discrete-time system characterized by a rational system function. We only consider the implementation using adders, coefficient multipliers and delayers. In the implementation, various structures can be chosen. These structures are different in accuracy, speed, cost, and others. A structure can be described by a block diagram. A block diagram consists of three elements, that is, the adder, the coefficient multiplier and the delayer. Figure 8. shows these elements. Example. Suppose that a causal linear time-invariant discrete-time system is characterized by (7 / 4)z (/ 2)z H(z) 2 (/ 4)z (/ 8)z 2. (8.50)

24 X(z) + + Y(z) z z 7/4 + + /4 z z /2 /8 From (8.50), we obtain Figure 8.2. Direct Form I. Y(z)=(/4)z Y(z)+(/8)z 2 Y(z)+ X(z)(7/4)z X(z)(/2)z 2 X(z). (8.5) The direct-form I structure of the system is based on (8.5). Draw the block diagram.

25 The block diagram is shown in figure 8.2. Example. From (8.50), we obtain (8.52) is also written as where Thus, Y(z) 7 z 4 z 4 z 2 z X(z). (8.52) Y(z)=[(7/4)z (/2)z 2 ]W(z), (8.53) W(z) 4 z z 8 2 X(z). (8.54)

26 W(z)=(/4)z W(z)+(/8)z 2 W(z)+X(z), (8.55) Y(z)=W(z)(7/4)z W(z)(/2)z 2 W(z). (8.56) The direct-form II structure of the system is obtained from (8.55) and (8.56). Draw the block diagram. The block diagram is shown in figure 8.3. X(z) + W(z) + Y(z) z + /4 7/4 + z /8 /2 Figure 8.3. Direct Form II.

27 Example. (8.50) is also written as. z 4 2z z 2 z 4 H(z) (8.57) A cascade-form structure of the system is obtained from (8.57). The system is expressed as the cascade of 2 subsystems. Each subsystem is implemented in direct form I or II. Draw the block diagram. Example. (8.50) is also written as. z 4 4/ 3 z 2 5/ 3 4 H(z) (8.58) A parallel-form structure of the system is obtained from (8.58). The system is expressed as the parallel of 3 subsystems. Each subsystem

28 is implemented in direct form I or II. Draw the block diagram.

5. Time-Domain Analysis of Discrete-Time Signals and Systems

5. Time-Domain Analysis of Discrete-Time Signals and Systems 5.1. Impulse Sequence (1.4.1) 5.2. Convolution Sum (2.1) 5.3. Discrete-Time Impulse Response (2.1) 5.4. Classification of a Linear Time-Invariant