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

<span class='text_page_counter'>(1)</span>ELT2035 Signals & Systems. Lesson 13: z-transform exercises. Hoang Gia Hung Faculty of Electronics and Telecommunications University of Engineering and Technology, VNU Hanoi.

<span class='text_page_counter'>(2)</span> Exercise 1 𝑧. Given 𝒵 𝑢[𝑛] = . By applying appropriate property of z-transform, 𝑧−1 determine the following signals a. 𝑋 𝑧 = 𝒵 𝑛𝑢[𝑛] b. 𝑌 𝑧 = 𝒵 𝑛𝑎𝑛 𝑢[𝑛] SOLUTION a. Applying the z-derivative property to 𝒵 𝑢[𝑛] , we have 𝒵 𝑛𝑢 𝑛. 𝑑. =. −𝑧 𝑑𝑧 𝒵 𝑢 𝑛 𝑑. = −𝑧 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 𝑛𝑎𝑛 𝑢 𝑛. = =. 𝑧 𝑧 𝑎 𝑋 = 𝑧 𝑎 𝑎−1 𝑎𝑧 𝑧−𝑎 2. 2.

<span class='text_page_counter'>(3)</span> 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..

<span class='text_page_counter'>(4)</span> 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 2 region and bounded by poles, the ROC must be < 𝑧 < 2. 3 Similarly, if the Fourier transform of x[n] does not converge, there are 3 possibilities:. i. ii.. 𝑧 < 1 3. 1 3. < 𝑧 <. iii. 𝑧 > 2. 2 3.

<span class='text_page_counter'>(5)</span> Exercise 3 Find x[n] from X(z) below using partial fraction expansion, where x[n] is known to be causal. 3 + 2𝑧 −1 𝑋 𝑧 = 2 + 3𝑧 −1 + 𝑧 −2 SOLUTION We have: 𝑋 𝑧. = = =. Hence, 𝑥 𝑛 =. 1 2. 1 𝑛 −2 𝑢. 3+2𝑧 −1 2+𝑧 −1 1+𝑧 −1 1 1 + 2+𝑧 −1 1+𝑧 −1 1 2 1 −1 1+2𝑧. 1. + 1+𝑧 −1. 𝑛 + (−1)𝑛 𝑢[𝑛] since x[n] is causal..

<span class='text_page_counter'>(6)</span> Exercise 4 𝑤 − σ∞ 𝑖=1 𝑖. Using the power-series expansion log 1 − 𝑤 = determine the inverse of the following z-transforms. 1 2. a. 𝑋 𝑧 = log 1 − 2𝑧 , 𝑧 < . 1 2. 1 2. b. 𝑋 𝑧 = log 1 − 𝑧 −1 , 𝑧 > .. SOLUTION: a. Applying the power-series expansion to 𝑋 𝑧 yields log 1 − 2𝑧. = =. Hence, 𝑥 𝑛 =. 2−𝑛 − , ቐ 𝑛. 0,. 𝑛<0 . 𝑛≥0. 2𝑧 𝑖 ∞ − σ𝑖=1 , 2𝑧 𝑖 2𝑖 𝑖 ∞ − σ𝑖=1 𝑧 , 𝑧 𝑖. <1 <. 1 2. .. 𝑖. , 𝑤 < 1,.

<span class='text_page_counter'>(7)</span> Exercise 4 solution b. Applying the power-series expansion to 𝑋 𝑧 yields 1 2. log 1 − 𝑧 −1. = − σ∞ 𝑖=1 =. Hence, 𝑥 𝑛 = ቐ−. 1 𝑛 2. 𝑛. 0,. , 𝑛 > 0. 𝑛≤0. 1 −1 𝑖 𝑧 2. − σ∞ 𝑖=1. 𝑖 1 𝑖 2. 𝑖. ,. 1 −1 𝑧 2. <1 .. 𝑧 −𝑖 , 𝑧 >. 1 2.

<span class='text_page_counter'>(8)</span> Exercise 5 Consider the pole-zero plot of 𝐻(𝑧) given in the below figure, where 𝐻 a. Sketch 𝐻 𝑒 𝑗Ω. 𝑎 2. = 1.. as the number of zeros at z = 0 increases from 1 to 5.. b. Does the number of zeros affect ∠𝐻 𝑒 𝑗Ω ? If so, specifically in what way? c. Find the region of the z plane where 𝐻 𝑧. = 1..

<span class='text_page_counter'>(9)</span> Exercise 5 solution 𝑧. a. For the number of zeros is one, we have 𝐻 𝑧 = 𝑧−𝑎, thus 𝐻 𝑒 𝑗Ω = cos Ω+𝑗 sin Ω . cos Ω−𝑎 +𝑗 sin Ω. Hence, 𝐻 𝑒 𝑗Ω. =. 𝐻 𝑒 𝑗Ω 𝐻∗ 𝑒 𝑗Ω =. the number of zeros is two, we have 𝐻 𝑧 = cos 2Ω+𝑗 sin 2Ω . cos Ω−𝑎 +𝑗 sin Ω. 𝑧2 , 𝑧−𝑎. 1 1+𝑎2 −2𝑎 cos Ω. . For. thus 𝐻 𝑒 𝑗Ω =. Therefore, we see that the magnitude of 𝐻 𝑒 𝑗Ω. does not change as the number of zeros increases as depicted in the below figure..

<span class='text_page_counter'>(10)</span> Exercise 5 solution (cont.) 𝑧. 𝑒 𝑗Ω. b. For one zero at z = 0, we have 𝐻 𝑧 = 𝑧−𝑎, thus 𝐻 𝑒 𝑗Ω = 𝑒 𝑗Ω −𝑎. The phase of 𝐻 𝑒 𝑗Ω , hence is the subtraction of the phase of the denominator from Ω (the phase of the numerator). For two zeros, the phase of 𝐻 𝑒 𝑗Ω 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..

<span class='text_page_counter'>(11)</span> Exercise 6 Use z-transforms to compute the zero-input response of the system 𝑦 𝑛 − 1 2𝑦 𝑛 − 1 = 3𝑥 𝑛 + 4𝑥[𝑛 − 1] with initial condition 𝑦 −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 −1 𝑌 𝑧 + 𝑦 −1 = 0 ⟺ 1 − 2𝑧 −1 𝑌 𝑧 = 1 1 ⟺ 𝑌 𝑧 = 1 − 2𝑧 −1 Therefore, 𝑦 𝑛 = 𝒵 −1 𝑌(𝑧) = 2𝑛 𝑢[𝑛]..

<span class='text_page_counter'>(12)</span> Exercise 7 3. 1. A system is described by 𝑦 𝑛 − 4 𝑦 𝑛 − 1 + 8 𝑦 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 3. Hence 𝐻 𝑧 =. 1+2𝑧 −1 3. 1. 1 − 4 𝑧 −1 + 8 𝑧 −2 = 𝑋(𝑧) 1 + 2𝑧 −1 .. 𝑌 𝑧 1. 1−4𝑧 −1 +8𝑧 −2. =. 𝑧(𝑧+2) 1. 1. 𝑧−2 𝑧−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..

<span class='text_page_counter'>(13)</span> Exercise 8 Compute the response of the LTI system 𝑦[𝑛] = 𝑥[𝑛] − 𝑥[𝑛 − 2] to input 𝑥 𝑛 = cos 𝜋𝑛Τ4 . 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 𝑒 𝑗Ω = 1 − 𝑒 −𝑗2Ω . On the other hand, we have 𝑥 𝑛 =. 1 2. 𝑒. 𝑗𝑛. 𝜋 4. +𝑒. −𝑗𝑛. 𝜋 4. . Since 𝑒 𝑗𝑛Ω0 is the. eigenfunction of any DT LTI system (i.e. the output of a DT LTI system to 𝑒 𝑗𝑛Ω0 is also a complex sinusoid scaled by a (complex) constant), thus the output of the system in this case is 1. 𝜋. 𝑦 𝑛 = 2 𝐻 𝑒 𝑗Ω0 𝑒 𝑗𝑛Ω0 + 𝐻 𝑒 −𝑗Ω0 𝑒 −𝑗𝑛Ω0 , with Ω0 = 4 . Hence, 𝑦 𝑛 = 𝜋. 1−𝑗 𝑒. −𝑗𝑛 4. 1 2. 𝜋. 1−𝑒 𝜋. −𝑗 2. 𝜋. 𝑒. 𝑗𝑛 4. 𝜋. + 1−𝑒 𝜋. 𝑗2. 𝜋. 𝑒. −𝑗𝑛 4. =. 1 ൤ 2. 𝜋. 1+𝑗 𝑒 𝜋. 𝑗𝑛 4. 𝜋. +. ൨ = cos 𝑛 4 − sin 𝑛 4 , or equivalently, 2 cos 𝑛 4 + 4 ..

<span class='text_page_counter'>(14)</span> Exercise 9 An LTI system has 𝐻 𝑒 𝑗Ω = 𝑗 tan Ω. Compute the difference equation that characterizes this system. SOLUTION: We have 𝐻 𝑒. 𝑗Ω. Substituting z for. =. sin Ω 𝑗 cos Ω. 𝑒 𝑗Ω. =. 𝑒 𝑗Ω −𝑒 −𝑗Ω . 𝑒 𝑗Ω +𝑒 −𝑗Ω. 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]..

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