Slide Tín hiệu và Hệ thống – Lesson 13 z-transform exercises – Hoàng Gia Hưng – UET

Does the number of zeros affect ∠

ELT2035 Signals & Systems

Hoang Gia Hung Faculty of Electronics and Telecommunications University of Engineering and Technology, VNU Hanoi

Lesson 13: z-transform exercises

Exercise 1

Given 𝒵 𝑢[𝑛] = 𝑧

𝑧−1 By applying appropriate property of z-transform,

determine the following signals

a 𝑋 𝑧 = 𝒵 𝑛𝑢[𝑛]

b 𝑌 𝑧 = 𝒵 𝑛𝑎𝑛𝑢[𝑛]

SOLUTION

a Applying the z-derivative property to 𝒵 𝑢[𝑛] , we have

𝑑𝑧𝒵 𝑢 𝑛

𝑑𝑧

𝑧 𝑧−1 = 𝑧

𝑧−1 2

b Applying the z-scaling (multiplication by an exponential

sequence) property to 𝒵 𝑛𝑢[𝑛] obtained in part a, we have:

𝑧 𝑎 𝑧

𝑎 − 1

2

𝑧 − 𝑎 2

Exercise 2

Shown in below figure is the pole-zero plot for the z-transform 𝑋(𝑧) of

a sequence 𝑥[𝑛]

Determine what can be inferred about the associated ROC from each

of the following statements

a x[n] is right-sided/left-sided

b The Fourier transform of x[n] converges/does not converge

Exercise 2 solution

a If 𝑥[𝑛] is right-sided, the ROC is given by 𝑧 > 𝛼 Since the ROC cannot include poles, for this case the ROC is given by

𝑧 > 2 Similarly, if 𝑥[𝑛] is left-sided, the ROC is given by 𝑧 <

1

3.

b If the Fourier transform of x[n] converges, the ROC must

include the unit circle 𝑧 = 1 Since the ROC is a connected region and bounded by poles, the ROC must be 2

3 < 𝑧 < 2 Similarly, if the Fourier transform of x[n] does not converge, there are 3 possibilities:

i 𝑧 < 1

3

ii. 1

3 < 𝑧 < 2

3

iii 𝑧 > 2

Exercise 3

Find x[n] from X(z) below using partial fraction expansion, where x[n] is known to be causal

−1

2 + 3𝑧−1 + 𝑧−2

SOLUTION

We have:

2+𝑧−1 1+𝑧−1

2+𝑧−1 + 1

1+𝑧−1

=

1 2

1+1

2 𝑧−1 + 1

1+𝑧−1

Hence, 𝑥 𝑛 = 1

2 −1

2 𝑛

𝑢 𝑛 + (−1)𝑛𝑢[𝑛] since x[n] is causal

Exercise 4

Using the power-series expansion log 1 − 𝑤 = − σ𝑖=1∞ 𝑤𝑖

𝑖 , 𝑤 < 1, determine the inverse of the following z-transforms

a 𝑋 𝑧 = log 1 − 2𝑧 , 𝑧 < 1

2.

b 𝑋 𝑧 = log 1 − 1

2𝑧−1 , 𝑧 > 1

2.

SOLUTION :

a Applying the power-series expansion to 𝑋 𝑧 yields

log 1 − 2𝑧 = − σ𝑖=1∞ 2𝑧 𝑖

𝑖 , 2𝑧 < 1

= − σ𝑖=1∞ 2𝑖

𝑖 𝑧𝑖 , 𝑧 < 1

2

.

Hence, 𝑥 𝑛 = ቐ −

2−𝑛

𝑛 , 𝑛 < 0

0, 𝑛 ≥ 0 .

Exercise 4 solution

b Applying the power-series expansion to 𝑋 𝑧 yields

log 1 − 1

2𝑧−1 = − σ𝑖=1∞

1

2 𝑧−1 𝑖

𝑖 , 1

2𝑧−1 < 1

= − σ𝑖=1∞

1 2 𝑖

𝑖 𝑧−𝑖 , 𝑧 > 1

2

.

Hence, 𝑥 𝑛 = ቐ−

1 2 𝑛

𝑛 , 𝑛 > 0

.

Exercise 5

Consider the pole-zero plot of 𝐻(𝑧) given in the below figure, where 𝐻 𝑎

2 = 1

a Sketch 𝐻 𝑒𝑗Ω as the number of zeros at z = 0 increases from 1 to 5

b Does the number of zeros affect ∠𝐻 𝑒𝑗Ω ? If so, specifically in what way?

c Find the region of the z plane where 𝐻 𝑧 = 1

Exercise 5 solution

a For the number of zeros is one, we have 𝐻 𝑧 = 𝑧

𝑧−𝑎, thus 𝐻 𝑒𝑗Ω =

cos Ω+𝑗 sin Ω

cos Ω−𝑎 +𝑗 sin Ω Hence, 𝐻 𝑒𝑗Ω = 𝐻 𝑒𝑗Ω 𝐻∗ 𝑒𝑗Ω = 1

1+𝑎2−2𝑎 cos Ω For the number of zeros is two, we have 𝐻 𝑧 = 𝑧2

𝑧−𝑎, thus 𝐻 𝑒𝑗Ω =

cos 2Ω+𝑗 sin 2Ω

cos Ω−𝑎 +𝑗 sin Ω Therefore, we see that the magnitude of 𝐻 𝑒𝑗Ω does not change as the number of zeros increases as depicted in the below figure.

Exercise 5 solution (cont.)

b For one zero at z = 0, we have 𝐻 𝑧 = 𝑧

𝑧−𝑎, thus 𝐻 𝑒𝑗Ω = 𝑒𝑗Ω

𝑒𝑗Ω−𝑎 The phase of 𝐻 𝑒𝑗Ω , hence is the subtraction of the phase of the

denominator from Ω (the phase of the numerator)

For two zeros, the phase of 𝐻 𝑒𝑗Ω is the subtraction of the

phase of the denominator from 2Ω

Hence, the phase changes by a linear factor with the number of

zeros

c The region of the z-plane that makes

𝐻 𝑧 = 1 is the perpendicular bisector of

(0,a) and is depicted below

Exercise 6

Use z-transforms to compute the zero-input response of the system 𝑦 𝑛 −

2𝑦 𝑛 − 1 = 3𝑥 𝑛 + 4𝑥[𝑛 − 1] with initial condition 𝑦 −1 = 1

2

SOLUTION:

Zero-input response is the solution of the equation 𝑦 𝑛 − 2𝑦 𝑛 − 1 = 0

Taking (unilateral) z-transform on both sides of the equation yields

𝑌 𝑧 − 2 𝑧−1𝑌 𝑧 + 𝑦 −1 = 0

1 − 2𝑧−1 Therefore, 𝑦 𝑛 = 𝒵−1 𝑌(𝑧) = 2𝑛𝑢[𝑛]

Exercise 7

A system is described by 𝑦 𝑛 − 3

4𝑦 𝑛 − 1 + 1

8𝑦 𝑛 − 2 = 𝑥 𝑛 + 2𝑥[𝑛 − 1] Determine its poles, zeros and whether or not it is BIBO stable

SOLUTION:

Taking z-transform on both sides of the equation gives

𝑌 𝑧 1 − 3

4𝑧−1 + 1

8𝑧−2 = 𝑋(𝑧) 1 + 2𝑧−1 Hence 𝐻 𝑧 = 1+2𝑧−1

1−34𝑧−1+18𝑧−2 = 𝑧(𝑧+2)

𝑧−1

2 𝑧−1

4

The system has two zeros at 0 and -2, two poles at ½ and ¼

Since both poles are inside the unit circle, the system is BIBO stable

Exercise 8

Compute the response of the LTI system 𝑦[𝑛] = 𝑥[𝑛] − 𝑥[𝑛 − 2] to input

𝑥 𝑛 = cos 𝜋𝑛 4 Τ

SOLUTION:

Taking z-transform on both sides of the equation gives

𝑌 𝑧 = 𝑋(𝑧) 1 − 𝑧−2 It’s obvious that 𝐻 𝑧 = 1 − 𝑧−2, and therefore 𝐻 𝑒𝑗Ω = 1 − 𝑒−𝑗2Ω

On the other hand, we have 𝑥 𝑛 = 1

2 𝑒𝑗𝑛𝜋4 + 𝑒−𝑗𝑛𝜋4 Since 𝑒𝑗𝑛Ω0 is the

𝑒𝑗𝑛Ω0 is also a complex sinusoid scaled by a (complex) constant), thus the output of the system in this case is

𝑦 𝑛 = 1

2 𝐻 𝑒𝑗Ω0 𝑒𝑗𝑛Ω0 + 𝐻 𝑒−𝑗Ω0 𝑒−𝑗𝑛Ω0 , with Ω0 = 𝜋

4 Hence, 𝑦 𝑛 = 1

2 1 − 𝑒−𝑗

𝜋

2 𝑒𝑗𝑛

𝜋

4 + 1 − 𝑒𝑗

𝜋

2 𝑒−𝑗𝑛

𝜋

4 = 1

2൤ ൨

1 + 𝑗 𝑒𝑗𝑛

𝜋

4 +

1 − 𝑗 𝑒−𝑗𝑛𝜋4 = cos 𝑛𝜋

4 − sin 𝑛𝜋

4 , or equivalently, 2 cos 𝑛𝜋

4 + 𝜋

4

Exercise 9

An LTI system has 𝐻 𝑒𝑗Ω = 𝑗 tan Ω Compute the difference equation that characterizes this system

SOLUTION:

We have 𝐻 𝑒𝑗Ω = 𝑗 sin Ω

cos Ω = 𝑒𝑗Ω−𝑒−𝑗Ω

𝑒𝑗Ω+𝑒−𝑗Ω Substituting z for 𝑒𝑗Ω gives the transfer function 𝐻 𝑧 = 𝑧−𝑧−1

𝑧+𝑧−1 = 1−𝑧−2

1+𝑧−2 Therefore: 𝑌 𝑧 1 + 𝑧−2 = 𝑋(𝑧) 1 − 𝑧−2

Taking inverse z-transform gives: 𝑦 𝑛 + 𝑦 𝑛 − 2 = 𝑥 𝑛 − 𝑥[𝑛 − 2]