(a) Explanation Concept: 1. A piecewise defined continuous function of the form f ( t ) = { g ( t ) , 0 ≤ t < a h ( t ) , t ≥ a can be expressed as f ( t ) = g ( t ) − g ( t ) U ( t − a ) + h ( t ) U ( t − a ) . If F ( s ) = L { f ( t ) } and a > 0 , then L { f ( t − a ) U ( t − a ) } = e − a s F ( s ) L { U ( t − a ) } = e − a s s 2. The Laplace transformation of a linear differential equation with constant coefficient can be expressed as, a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] 3. The formula for the Laplace transform is as follows, L { 1 } = 1 s L { sin k t } = k s 2 + k 2 L { cos k t } = s s 2 + k 2 4. The formula for the inverse Laplace transform is as follows, L − 1 { 1 s − a } = e a t 5. If F ( s ) = L { f ( t ) } and a > 0 , then L − 1 { F ( s − a ) } = e a t f ( t ) 6. The linearity property of the Laplace transformation is as below. If a , b be any constant and f , g be any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } Calculation: The expression for the linear differential L R series circuit is as follows: L d i d t + R i = E ( t ) Substitute the value of R and L in the above expression. d i d t + 10 i = E ( t ) The step function for the given graph is as follows: E ( t ) = { sin t , 0 ≤ t < 3 π 2 0 , t ≥ 3 π 2 Use the definition 1 of Laplace transforms to write the unit step function for the function E ( t ) . E ( t ) = sin t − sin t U ( t − 3 π 2 ) E ( t ) = sin t − cos ( t − 3 π 2 ) U ( t − 3 π 2 ) Use the definition 2 of Laplace transforms to write the Laplace transform of the above linear differential L R series circuit and substitute E ( t ) = sin t − cos ( t − 3 π 2 ) U ( t − 3 π 2 ) , then equation becomes s L { i } − i ( 0 ) + 10 L { i } = L { sin t − cos ( t − 3 π 2 ) U ( t − 3 π 2 ) } … … . ( 1 ) Substitute the initial given value i ( 0 ) = 0 in equation (1) and solve. ( s + 10 ) L { i } = L { sin t } − L { cos ( t − 3 π 2 ) U ( t − 3 π 2 ) } ( s + 10 ) L { i } = 1 s 2 + 1 − e − 3 π s 2 s s 2 + 1 L { i } = 1 ( s 2 + 1 ) ( s + 10 ) − e − 3 π s 2 s ( s 2 + 1 ) ( s + 10 ) From the above expression, the result obtained is as follows: L { i } = 1 ( s 2 + 1 ) ( s + 10 ) − e − 3 π s 2 s ( s 2 + 1 ) ( s + 10 ) … … ( 2 ) The term 1 ( s 2 + 1 ) ( s + 10 ) in equation (2) is simplify as follows using partial fraction, 1 ( s + 10 ) ( s 2 + 1 ) = a 1 s + a 0 s 2 + 1 + a 2 s + 10 … … ( 3 ) Multiply with ( s + 10 ) ( s 2 + 1 ) on both side of above expression. 1 ⋅ ( s + 10 ) ( s 2 + 1 ) ( s + 10 ) ( s 2 + 1 ) = ( a 1 s + a 0 ) ( s + 10 ) ( s 2 + 1 ) s 2 + 1 + a 2 ( s + 10 ) ( s 2 + 1 ) s + 10 1 = ( a 1 s + a 0 ) ( s + 10 ) + a 2 ( s 2 + 1 ) From the above expression, the result obtained is as follows: 1 = ( a 1 s + a 0 ) ( s + 10 ) + a 2 ( s 2 + 1 ) … … ( 4 ) Substitute s = − 10 in equation (4). 1 = ( a 1 ( − 10 ) + a 0 ) ( ( − 10 ) + 10 ) + a 2 ( ( − 10 ) 2 + 1 ) 1 = 101 a 2 a 2 = 1 101 Substitute the value of a 2 = 1 101 in equation (4) and solve. 1 = ( a 1 s + a 0 ) ( s + 10 ) + 1 101 ( s 2 + 1 ) 1 = a 1 s 2 + 10 a 1 s + a 0 s + 10 a 0 + 1 101 s 2 + 1 101 1 = s 2 ( a 1 + 1 101 ) + s ( a 0 + 10 a 1 ) + ( 10 a 0 + 1 101 ) Compare the similar coefficients of on both sides of the above expression. 10 a 0 + 1 101 = 1 10 a 1 + a 0 = 0 a 1 + 1 101 = 0 Solve the above system of equation. a 0 = 10 101 a 1 = − 1 101 Substitute the value of a 0 , a 1 and a 2 in equation (3) (b) To determine To draw: The graph for the current i ( t ) for 0 ≤ t ≤ 6 , and then find the i max and i min from the graph.

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.3, Problem 75E

(a)

To determine

The expression for the current i(t) from the linear differential equation of LR series circuit using Laplace transform.

(b)

To determine

To draw: The graph for the current i(t) for 0≤t≤6, and then find the imax and imin from the graph.

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Chapter 4.3, Problem 74E

Chapter 4.3, Problem 76E

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The expression for the current i ( t ) from the linear differential equation of L R series circuit using Laplace transform.

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The expression for the current i(t) from the linear differential equation of LR series circuit using Laplace transform.

To draw: The graph for the current i(t) for 0≤t≤6, and then find the imax and imin from the graph.

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Chapter 4 Solutions