(ROC)(Z-Transforms - adasda PDF

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Module 19 Region of Convergence (ROC) (Z-Transforms) Objective : To understand the meaning of ROC in Z transforms and the need to consider it. Introduction : As we are aware that the Z- transform of a discrete signal x(n) is given by 𝑋𝑧 =

∞

𝑥 𝑛 𝑧 −𝑛

𝑛 =−∞

The Z-transform has two parts which are, the expression and Region of Convergence respectively. Whether the Z-transform X(z) of a signal x(n) exists or not depends on the complex variable „z‟ as well as the signal itself. All complex values of „z=rejω‟ for which the summation in the definition converges form a region of convergence (ROC) in the z-plane. A circle with r=1 is called unit circle and the complex variable in z-plane is represented as shown below.

Description : The concept of ROC can be understood easily by finding z transform of two functions given below: a) 𝑥 𝑛 = 𝑎𝑛 𝑢(𝑛) 𝑋𝑧 =

∞

𝑎 𝑢(𝑛)𝑧 𝑛

𝑛 =−∞

−𝑛

∞

= 𝑎 𝑧

𝑛 −𝑛

𝑛 =0

= 𝑎𝑧

∞

−1 𝑛

𝑛 =0

< ∞. Consequently, the For convergence of X(z), we require that ∞ 𝑎𝑧−1 𝑛 𝑛 =0 region of convergence is that range of values of z for which 𝑎𝑧 −1 < 1, or equivalently, 𝑧 > 𝑎 and is shown in figure below

Then 𝑋 𝑧 =

∞ −1 𝑛 𝑎𝑧 𝑛 =0

=

b) 𝑥 𝑛 = −𝑎 𝑛 𝑢(−𝑛 − 1) 𝑋𝑧 =

∞

−𝑎 𝑢(−𝑛 − 1)𝑧 𝑛

𝑛 =−∞

1

1−𝑎𝑧 −1

=−

−𝑛

∞

= 1 − 𝑎−1 𝑧𝑛

=1−

𝑛 =0

𝑧

= 𝑧−𝑎 −1

𝑎 𝑧 =−𝑎 𝑛 −𝑛

𝑛 =−∞

𝑧

∞

−1 𝑛

𝑛 =1

1 1 𝑧 = = −1 −1 𝑧−𝑎 1 − 𝑎 𝑧 1 − 𝑎𝑧

This result converges only when 𝑎−1 𝑧 < 1, or equivalently, |z| < |a|. The ROC is shown below

Need to consider region of convergence while determining the z-transform If we consider the signals anu(n) and -anu(-n-1), we note that although the signals are 𝑧 differing, their z Transforms are identical which is 𝑧−𝑎 . Thus we conclude that to distinguish zTransforms uniquely their ROC's must be specified.

Zeros and Poles of the Laplace Transform LikeLaplace transforms as studied earlier in Module-15, the z-transforms in the above examples are rational, i.e., they can be written as a ratio of polynomials of variable „z‟in the general form 𝑋𝑧 =  

𝑁(𝑧)

𝐷𝑧

=

𝑀 𝑏 𝑘 𝑧 −𝑘 𝑘=0 𝑁 𝑎 𝑘 𝑧 −𝑘 𝑘=0

=

𝑀 𝑧−𝑧 𝑘=1 𝑧𝑘 𝑁 𝑧−𝑧𝑝𝑘 𝑘=1

.

N(z) is the numerator polynomial of order M withzzk ,(k=1,2,…,M) roots D(z) is the denominator polynomial of order N with zpk (k=1,2,…,N) roots

Roots of numerator polynomial are called zeros and the roots of denominator polynomial are called poles. Poles in z-plane are indicated with „x‟ and zeros with‟o‟ similar to s-plane. The representation of X(z) through its poles and zeros in the z-plane is referred to as the pole-zero plot of X(z). In general, we assume the order of the numerator polynomial is always lower than that of the denominator polynomial, i.e., Mro, so that r1-n decays quickly than ro-n for increasing n as illustrated in the figure below.

Consequently, x(n)r1-n is absolutely summable. For right sided sequences in general 𝑋 𝑧 = positive or negative.

∞ 𝑥 𝑛 𝑧 −𝑛 𝑛 =𝑁1

, where N1 is finite and may be

If N1 is negative, then the summation above includes terms with positive powers of z, which become unbounded as |z|→∞. Consequently, for right sided sequences in general, ROC will not include infinity. However, for causal sequences, i.e., sequences that are zero for n 1 {causal but not stable} Fig (b) indicates ROC for left sided sequence if b > 1 {non causal and unstable} Fig (c) indicates ROC for right sided sequence if 0

; 𝑅𝑂𝐶: 𝑧 > −6

𝑛

1

1 2

1−

2

3

{shown in figure a} and

{Shown in figure b} 1 − 2 𝑧 −1 3

1 1

1

𝑧 −1

=

1−

1 3

𝑧 −1 1 −

1

2

𝑧 −1

=

𝑧𝑧−

𝑧−

1

3

3 2 1

𝑧−

2

For convergence of X(z), both sums must converge, which requires that the ROC should be an intersection of 𝑧 >

1 3

and 𝑧 >

1 2

. i.e., 𝑧 >

1 2

{shown in figure c}

The pole zero plot and ROC are shown in the figure below

Problem 2:Draw the pole-zero plot and graph the frequency response for the system whose

transfer function is𝐻 𝑧 = Solution:

𝑧 2 −0.96𝑧 +0.9028 . 𝑧 2 −1.56𝑧+0.8109

Is the system both causal and stable?

The transfer function can be factored into 𝐻𝑧 =

𝑧 2 − 0.96𝑧 + 0.9028 𝑧2

− 1.56𝑧 + 0.8109 The Pole-zero diagram is shown below

𝑧 − 0.48 + 𝑗0.82(𝑧 − 0.48 − 𝑗0.82) = 𝑧 − 0.78 + 𝑗0. 45(𝑧 − 0.78 − 𝑗0.45)

The magnitude and phase frequency responses of the system are obtained by substituting z=ejωand varying the frequency variable ω over a range of 2𝜋. The frequency response plots are illustrated in Figure below

The system is both causal and stable because all the poles are inside the unit circle Problem 3:For the following algebraic expression for the z-transform of a signal, determine the number of zeros in the finite z-plane and the number of zeros at infinity 𝑧 −1 (1 − 2 𝑧 −1 ) 1

(1 − 3 𝑧 −1 )(1 − 𝑧 −1 ) 4 1

1

Solution: The given z-transform may be written as 𝑋(𝑧) =

𝑧−

1

𝑧− 3

1 2

𝑧−

1 4

Clearly, X(z) has a zero at z=1/2. Since the order of the denominator polynomial exceeds the order of the numerator polynomial by 1, X(z) has a zero at infinity. Therefore, X(z) has one zero in the finite z-plane and one zero at infinity. Problem 4:Suppose that the algebraic expression for the z-transform of x(n) is 𝑋𝑧 =

1 − 4 𝑧 −1 1

1+

1 4

𝑧 −2 1 +

5

4

𝑧 −1 − 8 𝑧 −2 3

How many different regions of convergence could correspond to X(z) Solution: We may find different signals using given z-transform by choosing different regions of convergence. The poles of the z-transform are 𝑧𝑜 =

1 𝑗, 2

1 𝑧1 = − 𝑗 , 2

1 𝑧2 = − , 2

𝑧3 =

3 4

Based on these pole locations, we may choose from the following regions of convergence: (i) (ii) (iii)

0 1 and the function within the summation grows towards infinity with

increasing n. Also, the summation does not converge. But if 𝑟 > , then the summation 2 converges 1

Problem 6:Consider the following system function for stable LTI systems. Without utilizing the inverse z-transform, determine in each case whether or not the corresponding system is causal

1 − 𝑧 −1 + 1 𝑧 −2 4

𝑋𝑧 =

𝑧 −1 1 −

3

1

2

𝑧 −1 1 − 2

1

3

𝑧 −1

Solution: For a system to be both causal and stable, the corresponding z-transform must not have any poles outside the unit circle. The given z-transform has a pole at infinity. Therefore, it is not causal Problem 7: Suppose we are given the following five facts about a particular LTI systems „S‟ with impulse response h(n) and z-transform H(z) 1. 2. 3. 4. 5.

H(n) is real H(n) is right sided lim𝑧→∞𝐻 𝑧 = 1 𝐻 𝑧 has two zeros 𝐻 𝑧 has one of its poles at a nonreal location on the circle defined by |z|=3/4.

Answer the following two questions: (a) Is „S‟ causal? (b) Is „S‟ stable? Solution: (a) Since lim𝑧→∞𝐻 𝑧 = 1, H(z) has not poles at infinity. Furthermore, since h(n) is given to be right sided, h(n) must be causal. (b) Since h(n) is causal, the numerator and denominator polynomials of H(z) have the same order. Since H(z) is given to have two zeros, we may conclude that it also has two poles. Since h(n) is real, the poles must occur in conjugate pairs. Also, it is given that one of the poles lies on the circle defined by |z|=3/4. Therefore, the other pole also lies on the same circle. Clearly, the ROC for H(z) will be of the form |z|>3/4, and will include the unit circle. Therefore, we may conclude that the system is stable Problem 8: Determine the location of poles and zeros of X(z) for the signal 𝑥𝑛 = 𝑎

𝑛

0 ≤ 𝑛 ≤ 𝑁 − 1, 𝑎 > 0 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Solution: 𝑁−1

𝑋(𝑧) = 𝑎 𝑧 = 𝑎𝑧 𝑛 −𝑛

𝑛 =0

𝑁−1

−1 𝑛

=

𝑛 =0

1 − 𝑎𝑧 −1 𝑁

1 − 𝑎𝑧−1

=

1 𝑧𝑁 − 𝑎𝑁 𝑧 𝑁−1 𝑧 − 𝑎

Since x(n) is of finite length, it follows from property 3 that the ROC included the entire zplane except possible the origin and /or infinity. Since x(n) is zero for n |b| (ii) |a| < |b| Problem 3:Let 𝑥 𝑛 = −1𝑛 𝑢 𝑛 + 𝛼 𝑛 𝑢(𝑛 − 𝑛𝑜 ) Determine the constraints on the complex number 𝛼 and the integer no, given that the ROC of X(z) is 1 < |z| < 2 Problem 4:Consider the signal𝑥 𝑛 =

𝑛

1 3

cos

𝑛𝜋

0

4

,

𝑛≤0

𝑛>0

Determine the poles and ROC for X(z) Problem 5: For the following algebraic expression for the z-transform of a signal, determine the number of zeros in the finite z-plane and the number of zeros at infinity 𝑧 −2 (1 − 𝑧−1 )

(1 − 4 𝑧 −1 )(1 + 𝑧 −1 ) 4 1

1

Problem 6: Let x(n) be an absolutely summable signal with rational z-transform X(z). If X(z) is known to have a pole at z=1/2, could x(n) be a) b) c) d)

A finite duration signal? A left sided signal? A right sided signal? A two sided signal?

Problem 7: Determine the constraint on r=|z| for1 the following sum to converge ∞

−𝑛+1

1

𝑛 =1

2

𝑧𝑛

Problem 8: Let x(n) be asignal whose rational z-transform X(z) contains a pole at z=1/2. Given that 𝑥1 𝑛 =

𝑛

𝑛

𝑥(𝑛) is absolutely summable and 𝑥2 𝑛 = 𝑥(𝑛) is not 4 8 absolutely summable, determine whether x(n) is left sided, right sided or two sided. 1

1

Problem 9: Consider the following system function for stable LTI systems. Without utilizing the inverse z-transform, determine in each case whether the corresponding system is causal. 𝑋𝑧 =

𝑧+

4 3

𝑧+1

− 𝑧 −2 − 3 𝑧 −3 2 1

2

Problem 10: If 𝑥[𝑛] = (1/3)𝑛 − (1/2)𝑛 𝑢 [𝑛], then find the region of convergence (ROC) of itsz -transform in the z -plane. Simulation: Pole-Zero Plot:  

A Pole-Zero plot displays the “poles” and “zeros” of the rational transform byplacing an „x‟ at each pole location and an „o‟ at each zero location in thecomplex z-plane. Poles and zeros can be found out by using tf2zpfunction and pole-zero plot is obtained using zplane function in MATLAB. For Example:

% Pole – Zero Map in Z- Domain clc; clear all;

close all; %z-domain LTI system %y(n)=(3/8)y(n-1)+(2/3)y(n-2)+x(n)+(1/4)x(n-1) a= input('Enter the Numerator coefficients'); b = input('Enter the Denominator coefficients'); [z1,p1,k]=tf2zp(a,b); disp('pole locations are');p1 disp('zero locations are');z1 figure, zplane(a,b); INPUT: Enter the Numerator coefficients[1,1/4] Enter the Denominator coefficients[1 -3/8,-2/3] OUTPUT: pole locations are p1 = 1.0252 -0.6502 zero locations are z1 = -0.2500

Practice: Locate poles and zeros in z-plane using MATLAB for the following function

References:

𝑧 3 − 0.8𝑧 2 𝐻(𝑧) = z3 − 1.1 z2 − 0.22 z + 0.06

[1] Alan V.Oppenheim, Alan S.Willsky and S.Hamind Nawab, “Signals & Systems”, Second edition, Pearson Education, 8th Indian Reprint, 2005. [2] M.J.Roberts, “Signals and Systems, Analysis using Transform methods and MATLAB”, Second edition,McGraw-Hill Education,2011 [3] John R Buck, Michael M Daniel and Andrew C.Singer, “Computer explorations in Signals and Systems using MATLAB”,Prentice Hall Signal Processing Series [4] P Ramakrishna rao, “Signals and Systems”, Tata McGraw-Hill, 2008 [5] Tarun Kumar Rawat, “Signals and Systems”, Oxford University Press,2011...