Problem 2. (Problem 2.5-2 from textbook)2.5-2. Find the z-transform, in closed form, of the number sequence generated by sampling the timefunction e(t) every T seconds beginning at i = 0. The function e(t) is specified by its Laplacetransform,2(1-e-5:E(s) =s(s + 2)T = 1s

Answered: 2.5-2. Find the z-transform, in closed…

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.16P

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Problem 2. (Problem 2.5-2 from textbook)
2.5-2. Find the z-transform, in closed form, of the number sequence generated by sampling the time
function e(t) every T seconds beginning at i = 0. The function e(t) is specified by its Laplace
transform,
2(1-e-5:
E(s) =
s(s + 2)
T = 1s

Expert Solution

Step 1

The given function is specified by its Laplace transform given as:

$E (s) = \frac{2 (1 - e^{- 5 s})}{s (s + 2)}$

The above function can be represented in the time domain using the delay property as shown below:

$E (s) = (1 - e^{- 5 s}) \times \frac{2}{s (s + 2)} \frac{2}{s (s + 2)} = F (s) E (s) = (1 - e^{- 5 s}) F (s) f (t) = L^{- 1} \{F (s)\} f (t) = L^{- 1} \{\frac{2}{s (s + 2)}\} f (t) = L^{- 1} \{\frac{1}{s} - \frac{1}{(s + 2)}\} f (t) = L^{- 1} \{\frac{1}{s}\} - L^{- 1} \{\frac{1}{(s + 2)}\} f (t) = u (t) - e^{- 2 t} u (t) f (t) = (1 - e^{- 2 t}) u (t) e (t) = L^{- 1} \{E (s)\} e (t) = L^{- 1} \{(1 - e^{- 5 s}) F (s)\} e (t) = L^{- 1} \{F (s)\} - L^{- 1} \{(e^{- 5 s}) F (s)\} L^{- 1} \{F (s)\} = (1 - e^{- 2 t}) u (t) L^{- 1} \{(e^{- 5 s}) F (s)\} = f (t - 5) u (t - 5) L^{- 1} \{(e^{- 5 s}) F (s)\} = (1 - e^{- 2 (t - 5)}) u (t - 5) e (t) = (1 - e^{- 2 t}) u (t) - (1 - e^{- 2 (t - 5)}) u (t - 5)$

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2.5-2. Find the z-transform, in closed form, of the number sequence generated by sampling the time function e(t) every T seconds beginning at i = 0. The function e(t) is specified by its Laplace transform, 2(1-e-5: E(s) = s(5+2) T = 1s