HOMEWORK_9

# HOMEWORK_9 - SCHOOL OF ENGINEERING ‐MECHATRONICS GRADUATE...

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Unformatted text preview: SCHOOL OF ENGINEERING ‐MECHATRONICS GRADUATE PROGRAM HOME WORK #9 MTR 500 ADVANCED ENGINEERING MATHEMATICS 1. What is the z‐transform F(z) of the triangular pulse in the figure if the sampling period has the following values: (a) Δt = 5 s (b) Δt = 10 s 0, 0.2 2 0.2 6 0 0, 0.2 ∆ 0.2 ∆ 0 5 : 10 20 30 10 10 20 30 2 6 10 20 ∆ 2 4 6 0· 2 6 3 2 2 4 0.2 · 5 · 2 5 6 2 20 30 ∆ ∆ ∆ 30 10 20 30 ∆ For ∆ at ∆ at ∆ at ∆ 0.2 · 5 · 6 0· 2 2 6 6 For ∆ at ∆ at ∆ at ∆ 10 : 10 20 30 1 2 3 0· 0.2 · 10 · 2 2·2 2 6 2·3 6 2 0.2 · 10 · 6 0· 2 2 2 2. Suppose that 2 2 F(z) = 1 − 0.2 z −1 (1 + 0.6 z −1 )(1 − 0.3z −1 )(1 − z −1 ) (a) Calculate the corresponding time‐domain response f*(t). 1 0.2 1 0.6 1 1 0 .3 1 1 0.7143 3 21 1 1 0.6 1 0.3 1 1 ∆ 0.3 0.7143 0.6 3 21 (b) As a check, use the final value theorem to determine the steady‐state value of f*(t). 1 0.2 lim 1 lim ∆ 1 0.6 1 0.3 1 1 0.2 lim 1 0.3 1 0.6 1 0.2 0.8 0.7143 1 0.6 1 0.3 1.6 0.7 3. Determine the inverse transform of z ( z + 1) ( z − 1)( z 2 − z + 1) By the following methods: (a) Partial fraction expansion. 1 1 1 2 2 1 1 1 1 2 1 1 From table: ∆ 2 2 2 cos 3 2 2 2 1 1 1 0.5 1 cos ∆ 2 cos ∆ cos ∆ (b) Long division. 1 1 1 3 1 2 2 √ 2 3 3 2 2 6 4 4 6 5 8 3 3 3 3 8 5 6 4 4 6 2 2 3 3 2 ∆ ∆ 3 ∆ ∆ 3 4 7∆ ∆ 3 2∆ 4 0 ∆ 3∆ 3 ∆ 4∆ ∆ 5∆ 4 3 0 2 1 2 1 2 2 1 4 3 1 ∆t 2∆t 3∆t 4∆t 5∆t 6∆t 7∆t 8∆t 9∆t 10∆t 11∆t 12∆t 4. Calculate the z‐transform of the rectangular pulse shown in the drawing. Assume that the sampling period is Δt = 2 min. The pulse is f = 3 for 2 ≤ t < 6. For ∆ 2 : at ∆ 2 1 the sampled value is 3 at ∆ 4 2 the sampled value is 3 3 the sampled value is 0 (f = 3 for 2 ≤ t < 6) at ∆ 6 4 the sampled value is 0 at ∆ 8 The sampled values: 3 1 ∆t 2∆t 3∆t 4∆t 0· 3 3 3 3 0· 5. The pulse transfer function of a process is given by 5( z + 0.6) Y ( z) =2 X ( z ) z − z + 0.41 (a) Calculate the response y ( nΔt ) to a unit step change in x using the partial fraction method. 5 0.6 5 0.6 0.41 1 0.41 5 0.6 1 0.41 1 1 1 5 5 0.6 0.6 0.41 1 0.41 1 5 0.6 0.41 1 9.7561 5.9451 9.7561 5.9451 19.5122 1 0.5 0.4 0.5 0.4 9.7561 5.9451 9.7561 5.9451 19.5122 1 0.5 0.4 0.5 0.4 . 11.4248 . 11.4248 19.5122 1 0.5 0.4 0.5 0.4 From table: ∆ 19.5122 22.8496 0.6403 cos 0.6747 2.5943 (b) Check your answer in part (a) by using long division. 5 0.6 1 0.41 1 1 5 0.6 0.41 1 3 5 0.41 0.41 3 5 2 1.41 0.41 Long division: 2 1.41 5 0.41 √5 5 13 3 10 13 13 21.62 7.05 2.05 7.05 2.05 26 18.33 5.33 5.33 18.95 16.28 18.95 37.9 26.7195 21.62 21.3895 43.24 21.62 21.8505 18.95 7.7695 7.7695 30.4842 21.62 ∆ 8.8642 ∆ ∆ 5 ∆ 13 ∆ 2∆ 18.95 ∆ 3∆ ∆ n 0 0 1 5 2 13 3 18.95 4 21.62 Using part (a) 2.5943 ∆ 19.5122 22.8496 0.6403 cos 0.6747 ∆ n 0 0 1 5 2 13 3 18.95 4 21.62 So, it’s proven. (c) What is the steady‐state value of y? 5 0.6 1 lim ∆ lim 1 0.41 1 1 5 0.6 lim 0.41 1 5 1.6 5 1 0.6 19.51 0.41 1 1 0.41 4∆ ...
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