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

Does the number of zeros affect ∠

ELT2035 Signals & Systems Lesson 13: z-transform exercises Hoang Gia Hung Faculty of Electronics and Telecommunications University of Engineering and Technology, VNU Hanoi Exercise 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 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 Exercise 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 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 > Similarly, if 𝑥[𝑛] is left-sided, the ROC is given by 𝑧 < b If the Fourier transform of x[n] converges, the ROC must include the unit circle 𝑧 = Since the ROC is a connected region and bounded by poles, the ROC must be < 𝑧 < Similarly, if the Fourier transform of x[n] does not converge, there are possibilities: i ii 𝑧 < 3 < 𝑧 < iii 𝑧 > 2 Exercise Find x[n] from X(z) below using partial fraction expansion, where x[n] is known to be causal + 2𝑧 −1 𝑋 𝑧 = + 3𝑧 −1 + 𝑧 −2 SOLUTION We have: 𝑋 𝑧 = = = Hence, 𝑥 𝑛 = 𝑛 −2 𝑢 3+2𝑧 −1 2+𝑧 −1 1+𝑧 −1 1 + 2+𝑧 −1 1+𝑧 −1 −1 1+2𝑧 + 1+𝑧 −1 𝑛 + (−1)𝑛 𝑢[𝑛] since x[n] is causal Exercise 𝑤 − σ∞ 𝑖=1 𝑖 Using the power-series expansion log − 𝑤 = determine the inverse of the following z-transforms a 𝑋 𝑧 = log − 2𝑧 , 𝑧 < 2 b 𝑋 𝑧 = log − 𝑧 −1 , 𝑧 > SOLUTION: a Applying the power-series expansion to 𝑋 𝑧 yields log − 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 < 𝑖 , 𝑤 < 1, Exercise solution b Applying the power-series expansion to 𝑋 𝑧 yields log − 𝑧 −1 = − σ∞ 𝑖=1 = Hence, 𝑥 𝑛 = ቐ− 𝑛 𝑛 0, , 𝑛 > 𝑛≤0 −1 𝑖 𝑧 − σ∞ 𝑖=1 𝑖 𝑖 𝑖 , −1 𝑧 <1 𝑧 −𝑖 , 𝑧 > Exercise Consider the pole-zero plot of 𝐻(𝑧) given in the below figure, where 𝐻 a Sketch 𝐻 𝑒 𝑗Ω 𝑎 = as the number of zeros at z = increases from to 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 = Exercise 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+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 Exercise 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 = is the perpendicular bisector of (0,a) and is depicted below Exercise Use z-transforms to compute the zero-input response of the system 𝑦 𝑛 − 2𝑦 𝑛 − = 3𝑥 𝑛 + 4𝑥[𝑛 − 1] with initial condition 𝑦 −1 = SOLUTION: Zero-input response is the solution of the equation 𝑦 𝑛 − 2𝑦 𝑛 − = Taking (unilateral) z-transform on both sides of the equation yields 𝑌 𝑧 − 𝑧 −1 𝑌 𝑧 + 𝑦 −1 = ⟺ − 2𝑧 −1 𝑌 𝑧 = 1 ⟺ 𝑌 𝑧 = − 2𝑧 −1 Therefore, 𝑦 𝑛 = 𝒵 −1 𝑌(𝑧) = 2𝑛 𝑢[𝑛] Exercise A system is described by 𝑦 𝑛 − 𝑦 𝑛 − + 𝑦 𝑛 − = 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 Hence 𝐻 𝑧 = 1+2𝑧 −1 1 − 𝑧 −1 + 𝑧 −2 = 𝑋(𝑧) + 2𝑧 −1 𝑌 𝑧 1−4𝑧 −1 +8𝑧 −2 = 𝑧(𝑧+2) 1 𝑧−2 𝑧−4 The system has two zeros at and -2, two poles at ½ and ¼ Since both poles are inside the unit circle, the system is BIBO stable Exercise 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) − 𝑧 −2 It’s obvious that 𝐻 𝑧 = − 𝑧 −2 , and therefore 𝐻 𝑒 𝑗Ω = − 𝑒 −𝑗2Ω On the other hand, we have 𝑥 𝑛 = 𝑒 𝑗𝑛 𝜋 +𝑒 −𝑗𝑛 𝜋 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 𝜋 𝑦 𝑛 = 𝐻 𝑒 𝑗Ω0 𝑒 𝑗𝑛Ω0 + 𝐻 𝑒 −𝑗Ω0 𝑒 −𝑗𝑛Ω0 , with Ω0 = Hence, 𝑦 𝑛 = 𝜋 1−𝑗 𝑒 −𝑗𝑛 𝜋 1−𝑒 𝜋 −𝑗 𝜋 𝑒 𝑗𝑛 𝜋 + 1−𝑒 𝜋 𝑗2 𝜋 𝑒 −𝑗𝑛 = ൤ 𝜋 1+𝑗 𝑒 𝜋 𝑗𝑛 𝜋 + ൨ = cos 𝑛 − sin 𝑛 , or equivalently, cos 𝑛 + Exercise 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 + 𝑧 −2 = 𝑋(𝑧) − 𝑧 −2 Taking inverse z-transform gives: 𝑦 𝑛 + 𝑦 𝑛 − = 𝑥 𝑛 − 𝑥[𝑛 − 2]