HUZAIRI MOHAMED TS. SHARITA ABD GHONI CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS: A Problem-Solving Approach SULTAN ABDUL HALIM MU'ADZAM SHAH KEMENTERIAN PENDIDIKAN TINGGI JABATAN PENDIDIKAN POLITEKNIK DAN KOLEJ KOMUNITI
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS : A PROBLEM-SOLVING APPROACH
HUZAIRI BIN MOHAMED SHARITA BINTI ABD GHONI e-ISBN 978-967-0055-22-0 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the Publisher. Polytechnic Sultan Abdul Halim Mu’adzam Shah, Bandar Darulaman 06000, Jitra, Kedah. 16 March 2023
PREFACE Assalamu'alaikum and peace be upon you Alhamdulillah, we have finally been able to publish our valuable masterpiece, Circuit Analysis with Laplace Transforms(L.T): A Problem-solving approach for Polytechnic Students. This eBook is targeted to be used as a reference and mainly focuses on Polytechnic students in Malaysia. This eBook was created to cover the syllabus for the course of Circuit Analysis and Engineering Mathematic 3. The contents of this eBook about Laplace Transform. The advantage of this eBook is that there are additional materials in the form of reference notes that contain basics of mathematical and electrical knowledge related to this course. Our main intention for writing this eBook is to help and assist polytechnic students to achieve greater knowledge and to develop the students’ confidence. Furthermore, step-by-step working examples are provided for better understanding of this topic. This eBook may also be utilized as a source of information and guidance for advance education. Lastly, we would like to thank everyone in Electrical Engineering Department, Polytechnic Sultan Abdul Halim Mu'adzam Shah (POLIMAS) who has provided us with the necessary assistance and encouragement. Any comments and suggestions from students or other users are welcomed to make sure this book can be improved upon newer editions. Huzairi Bin Mohamed - Laplace Transform and Inverse L.T Sharita Binti Abd. Ghoni - Analyze RLC circuits using L.T
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 1 Introduction: Laplace Transform • Integration Method • Converting using Laplace Table 1. Explain Laplace Transform and Inverse Laplace Transform • Inverse Laplace Transform of a Function by using table • Inverse Laplace Transform of some standard functions • Inverse Laplace Transforms by completing the square • Inverse Laplace Transforms using partial fractions 2. Calculate Theorem and property of Laplace Transform and Inverse Laplace Transform • Series Circuit & Parallel Circuit • Analyze the circuit using any network analysis theorem 3. Analyze RLC circuits using Laplace Transform
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 2 1.0 – LAPLACE TRANSFORM 1.1 - Laplace Transform for simple functions using integral definition The Laplace Transform is an integral transformation of a function of () from the time domains to complex frequency domain (), giving by: () = ℒ[()] = ∫ () − ∞ 0 () = Complex frequency domain or Laplace Transform,() = Complex variable () = Time Domains − = Complex Factor a. Unit Step function () = () = ℒ[()] = ∫ () − ∞ 0 Sub; () = () = ℒ[()] = ∫ − ∞ 0 () = ∫ − ∞ 0 Integration Method STEP 1
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 3 Integrate Laplace Transform Equation = ∫ − ∞ 0 = [ − (−) ] 0 ∞ = [ − − ] 0 ∞ Sub = ∞ = () = [ − − ] 0 ∞ = [ −(∞) − − −(0) − ] () = [ −(∞) − − −(0) − ] = [(0 − 0) − (0 − 1 − )] = () = b. Ramp function () = () = ℒ[()] = ∫ () − ∞ 0 Sub; () = () = ℒ[()] = ∫ − ∞ 0 () = ∫ − ∞ 0 STEP 2 STEP 3 ANSWER STEP 1
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 4 Solve by using integration by part formula ∫ ∞ 0 = [uv]0 ∞ − ∫ ∞ 0 = = − = 1 = = − − ;, , to integration by part formula () = ∫ − ∞ 0 = [ − − ] 0 ∞ − ∫ − − ∞ 0 Integrated for: ∫ − − ∞ 0 = [ − 2 ] 0 ∞ () = ∫ − ∞ 0 = [ − − ] 0 ∞ − [ − 2 ] 0 ∞ Sub = ∞ = () = ∫ − ∞ 0 = [(0 − 0) − (0 − ( 1 2 ))] = 2 () = ∫ − ∞ 0 = 2 STEP 2 STEP 3 STEP 4 ANSWER G
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 5 c. Exponential function () = () = ℒ[()] = ∫ () − ∞ 0 Sub; () = () = ∫ . − ∞ 0 () = ∫ . − ∞ 0 Index laws: . − = − () = ∫ − ∞ 0 () = ∫ −(−) ∞ 0 Integrated for: () = ∫ −(−) ∞ 0 = [ −(−) −( − ) ] 0 ∞ Sub = ∞ = () = [( −(−)∞ −( − ) ) − ( −(−)0 −( − ) )] () = [0 − ( 1 −( − ) )] = 1 ( − ) () = 1 ( − ) STEP 1 STEP 2 STEP 3 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 6 d. Sine function () = () = ℒ[()] = ∫ () − ∞ 0 Sub; () = () = ∫ . − ∞ 0 () = ∫ . − ∞ 0 ′ = = ( − −) Sub: ′ in formula F(s) () = ℒ[()] = ∫ 1 2 ( − −). − ∞ 0 () = 2 ∫ ( −− −−) ∞ 0 Integrated for F(s) () = 2 ∫ ( −(−)− −(+) ) ∞ 0 () = 2 ([ −(−) −( − ) − −(+) −( + ) ] 0 ∞ ) Sub = ∞ = () = 2 ([ −(−)∞ −( − ) − −(+)∞ −( + ) ] − [ −(−)0 −( − ) − −(+)0 −( + ) ]) () = 2 ([0 − 0] − [ 1 −( − ) − 1 −( + ) ]) () = 2 ([ 1 ( − ) − 1 ( + ) ]) = 2 ([ ( + ) − ( − ) ( − )( + ) ]) STEP 1 STEP 2 STEP 3 STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 7 () = 2 ([ 2 ( − )( + ) ]) = 2 ([ 2 2 + − − 22 ]) () = 2 ([ 2 2 − (−1)2 ]) = 2 ([ 2 2 + 2 ]) = ( 2 + 2 ) () = 2 + 2 e. Cosine function () = () = ℒ[()] = ∫ () − ∞ 0 Sub; () = () = ℒ[()] = ∫ . − ∞ 0 () = ℒ[()] = ∫ . − ∞ 0 ′ = = ( + −) ; () () = ℒ[()] = 2 ∫ ( + −). − ∞ 0 () = 2 ∫ ( − + −−) ∞ 0 () = 2 ∫ ( −(−) + −(+) ) ∞ 0 Integrated for ANSWER STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 8 () = 2 ([ −(−) −( − ) + −(+) −( + ) ] 0 ∞ ) Sub = ∞ = () = 2 ([ −(−)∞ −( − ) + −(+)∞ −( + ) ] − [ −(−)0 −( − ) + −(+)0 −( + ) ]) () = 2 ([0 + 0] − [ 1 −( − ) + 1 −( + ) ]) () = 2 ([ 1 ( − ) + 1 ( + ) ]) = 2 ([ ( + ) + ( − ) ( − )( + ) ]) () = 2 ([ + + − ( − )( + ) ]) = 2 ([ 2 2 + − − 22 ]) () = 2 ([ 2 2 − (−1)2 ]) = 2 ([ 2 2 + 2 ]) = ( 2 + 2 ) () = 2 + 2 Example 1 Calculate Laplace Transform for: () = 3 Solution () = ℒ[()] = ∫ () − ∞ 0 () = ∫ () − ∞ 0 () = () = ∫ 3 − ∞ 0 = 3 ∫ − ∞ 0 STEP 4 ANSWER STEP 1
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 9 Example 2 Calculate Laplace Transform for: () = −2 Solution Integrated for () = 3 [ − − ] 0 ∞ Sub = ∞ = () = 3 [ −(∞) − − −(0) − ] () = 3 [0 − 1 − ] () = 3 [ 1 ] () = 3 () = −2 () = ℒ[()] = ∫ () − ∞ 0 () = − () = ∫ −2 . − ∞ 0 Index laws: − . − = −− () = ∫ −2− ∞ 0 Integrated () STEP 2 STEP 3 ANSWER STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 10 Example 3 Calculate Laplace Transform for: () = 4 − 3 8 Solution: () = ∫ −(2+) ∞ 0 () = [ −(2+) −( + 2) ] 0 ∞ Sub = ∞ = () = ( −(2+)∞ −( + 2) ) − ( −(2+)0 −( + 2) ) () = 0 − ( 1 −( + 2) ) = 1 ( + 2) () = 1 ( +) () = 4 − 3 8 () = ℒ[()] = ∫ () − ∞ 0 () = 4 − 3 8 () = ∫ 4 − 3 8 . − ∞ 0 () = 4 ∫ − 3 8 . − ∞ 0 STEP 4 ANSWER STEP 1
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 11 Example 4 Calculate Laplace Transform for: () = 5 () = 5 () = ℒ[()] = ∫ () − ∞ 0 Index laws: − 3 8 . − = − 3 8 − () = 4 ∫ − 3 8 − ∞ 0 () = 4 ∫ −( 3 8 +) ∞ 0 Integrate for () = [ −( 3 8 +) − ( + 3 8 ) ] 0 ∞ Sub = ∞ = () = ( −( 3 8 +)∞ − ( + 3 8 ) ) − ( −( 3 8 +)0 − ( + 3 8 ) ) () = 0 − ( 1 − ( + 3 8 ) ) = 1 ( + 3 8 ) STEP 2 STEP 3 STEP 4 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 12 () = 5 () = ℒ[()] = 5 ∫ − ∞ 0 () = 5 ∫ − ∞ 0 Solve by using integration by part formula 5 ∫ ∞ 0 = 5 [[uv]0 ∞ − ∫ ∞ 0 ] Define and find value of = = − = 5 = = − − Sub , , () = 5 ∫ − ∞ 0 = [5 [ − − ] 0 ∞ − ∫ − − ∞ 0 ] () = 5 ∫ − ∞ 0 = [5 [ − − ] 0 ∞ − [ − 2 ] 0 ∞ ] Sub = ∞ = () = 5 ∫ − ∞ 0 = 5 [(0 − 0) − (0 − ( 1 2 ))] = 5 2 STEP 1 STEP 2 STEP 3 STEP 4 STEP 5
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 13 () = 5 ∫ − ∞ 0 = 5 2 Example 5 Calculate Laplace Transform for: () = 3 2 () = 3 2 () = ℒ[()] = ∫ () − ∞ 0 () = 3 2 () = ℒ[()] = 3 2 ∫ − ∞ 0 () = 3 2 ∫ − ∞ 0 Solve by using integration by part formula 3 2 ∫ ∞ 0 = 3 2 [[uv]0 ∞ − ∫ ∞ 0 ] Define and find value of = = − = 1 = = − − Sub , , () = 3 2 ∫ − ∞ 0 = [ 3 2 [ − − ] 0 ∞ − ∫ − − ∞ 0 ] Sub = ∞ = () = 3 2 ∫ − ∞ 0 = [ 3 2 [ − − ] 0 ∞ − [ − 2 ] 0 ∞ ] ANSWER STEP 1 STEP 2 STEP 3 STEP 4 STEP 5
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 14 () = 3 2 ∫ − ∞ 0 = 3 2 [(0 − 0) − (0 − ( 1 2 ))] = 3 2 2 () = 3 2 ∫ − ∞ 0 = 3 2 2 Example 6 Calculate Laplace Transform for: () = 2 sin 4 () = 2 sin 4 () = ℒ[()] = ∫ () − ∞ 0 () = () = ℒ[()] = 2 ∫ 4. − ∞ 0 ′ = = ( − −) ; Sub: Euler rule in formula F(s) () = ℒ[()] = 2 2 ∫ ( 4− −4 ). − ∞ 0 Index laws: ( 4− −4 ). − = 4−− −4− () = 2 2 ∫ ( 4−− −4−) ∞ 0 () = 2 2 ∫ ( −(−4)− −(+4) ) ∞ 0 Integrated for () = 2 2 ([ −(−4) −( − 4) − −(+4) −( + 4) ] 0 ∞ ) Sub = ∞ = ANSWER STEP 1 STEP 2 STEP 3 STEP 4 STEP 5
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 15 () = 2 2 ([ −(−4)∞ −( − 4) − −(+4)∞ −( + 4) ] − [ −(−4)0 −( − 4) − −(+4)0 −( + 4) ]) () = 2 2 ([0 − 0] − [ 1 −( − 4) − 1 −( + 4) ]) () = 2 2 ([ 1 ( − 4) − 1 ( + 4) ]) = 2 2 ([ ( + 4) − ( − 4) ( − )( + ) ]) () = 1 ([ 4 + 4 ( − 4)( + 4) ]) = 1 ([ 8 2 + 4 − 4 − 24 2 ]) () = 1 ([ 8 2 − (−1)4 2 ]) = 1 ([ 8 2 + 4 2 ]) = 8 2 + 16 () = (2) [ 4 2 + 16] Example 7 Calculate Laplace Transform for: () = 5 sin 2 () = 5 sin 2 () = ℒ[()] = ∫ () − ∞ 0 () = () = ℒ[()] = ∫ 5 sin 2. − ∞ 0 () = ℒ[()] = 5 ∫ 2. − ∞ 0 ′ = = ( − −) ; Sub: Euler rule in formula F(s) () = ℒ[()] = 5 2 ∫ ( − −). − ∞ 0 : ( − −). − = −− −− ANSWER STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 16 () = 5 2 ∫ ( 2−− −2−) ∞ 0 () = 5 2 ∫ ( −(−2)− −(+2) ) ∞ 0 Integrated for () = 5 2 ([ −(−2) −( − 2) − −(+2) −( + 2) ] 0 ∞ ) Sub = ∞ = () = 5 2 ([ −(−2)∞ −( − 2) − −(+2)∞ −( + 2) ] − [ −(−2)0 −( − 2) − −(+2)0 −( + 2) ]) () = 5 2 ([0 − 0] − [ 1 −( − 2) − 1 −( + 2) ]) () = 5 2 ([ 1 ( − 2) − 1 ( + 2) ]) = 5 2 ([ ( + 2) − ( − 2) ( − 2)( + 2) ]) () = 5 2 ([ 2 + 2 ( − 2)( + 2) ]) = 5 2 ([ 4 2 + 2 − 2 − 22 2 ]) () = 5 2 ([ 4 2 − (−1)2 2 ]) = 5 2 ([ 4 2 + 2 2 ]) () = (5) [ 2 2 + (2) 2 ] () = 10 2 + 4 STEP 4 STEP 5 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 17 Example 8 Calculate Laplace Transform for: () = 3 2 3 () = 3 2 3 () = ℒ[()] = ∫ () − ∞ 0 () = 3 2 3 () = ℒ[()] = ∫ 3 2 3 . − ∞ 0 () = ℒ[()] = 3 ∫ 2 3 . − ∞ 0 ′ = = ( 2 3 + − 2 3 ) ; : () () = ℒ[()] = 3 2 ∫ ( 2 3 + − 2 3 ) . − ∞ 0 : ( + − ) . − = −(− ) + −(+ ) () = 3 2 ∫ ( 2 3 + − 2 3 ) . − ∞ 0 () = 3 2 ∫ ( −(− ) + −(+ ) ) ∞ 0 STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 18 Integrated for () = 3 2 ([ −(− 2 3 ) − ( − 2 3 ) + −(+ 2 3 ) − ( + 2 3 ) ] 0 ∞ ) Sub = ∞ = () = 3 2 ([ −(− 2 3 )∞ − ( − 2 3 ) + −(+ 2 3 )∞ − ( + 2 3 ) ] − [ −(− 2 3 )0 − ( − 2 3 ) + −(+ 2 3 )0 − ( + 2 3 ) ]) () = 3 2 ([0 + 0] − [ 1 − ( − 2 3 ) + 1 − ( + 2 3 ) ]) () = 3 2 ([ 1 ( − 2 3 ) + 1 ( + 2 3 ) ]) = 2 ([ ( + 2 3 ) + ( − 2 3 ) ( − 2 3 ) ( + 2 3 ) ]) () = 3 2 ([ + 2 3 + − 2 3 ( − 2 3 ) ( + 2 3 ) ]) = 3 2 ([ 2 2 + 2 3 − 2 3 − 2 ( 2 3 ) 2 ]) () = 3 2 ([ 2 2 − (−1) ( 2 3 ) 2 ]) = 3 ([ 2 + ( 2 3 ) 2 ]) = 3 2 + ( 2 3 ) 2 () = 3 2 + 4 9 = 27 9 2 + 4 STEP 4 STEP 5 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 19 Exercise: Calculate Laplace Transform for: () 1. () = 20 2. () = 100 3. () = 5 4. () = 13 7 5. () = 2 10 6. () = 3 √4 7. () = 10 2 8. () = 2 √ 3 10 9. () = 10 −4 10. () = 7 9 2
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 20 1.2 Laplace Transform Table Laplace Table Time domain, () Frequency domain, () () () − + ! + − ( + ) − ! ( + ) + + + − ( + ) + − + ( + ) + Converting Using Laplace Table
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 21 Laplace Table Time domain, () Frequency domain, () ( − ) ! ( − ) + ( − ) + + ( − ) + − − − ( + ) − − ( + ) − () () ′ () () − () ′′() () − () − ′ ()
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 22 Example 9 Determine the value of ℒ{()} = ℒ{7} by using Laplace transform Table. From table, So, Example 10 Determine the value of ℒ{()} = ℒ{4 −2 } by using Laplace transform Table. Time domain, () Frequency domain, () () 1 () = ℒ{()} = ℒ{7} () = 7ℒ{1} From L.T table () = = = 7 ℒ { 1 } = 7 ( 1 ) () = 7 Look into table and find the similar equation between the function asked and the function in the table. STEP 1: Change the figure of the table according to function asked. Look into table and find the similar equation between the function asked and the function in the table. STEP 1: STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 23 From table, So, Example 11 Determine the value of ℒ{()} = ℒ{5 sin 3} by using Laplace transform Table. Time domain, () Frequency domain, () − 1 + () = ℒ{()} = ℒ{4 −2 } () = 4ℒ{ −2 } From L.T Table − = + () = 4 ℒ{ −2 } = −2 : () = 4 ( 1 + 2 ) () = 4 + 2 Change the figure of the table according to function asked. STEP 2: Look into table and find the similar equation between the function asked and the function in the table. STEP 1:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 24 From table, So, Example 12 Determine the value of ℒ{()} = ℒ {3 cos 4 7 } by using Laplace transform Table. Time domain, () Frequency domain, () 2 + 2 () = ℒ{()} = ℒ{5 sin 3} = 5ℒ{sin 3} From L.T Table = 2 + 2 () = ℒ{()} = 5ℒ { 2 + 2 } = 3 : () () = 5 ( 3 2 + 3 2 ) () = 15 2 + 9 Change the figure of the table according to function asked. STEP 2: Look into table and find the similar equation between the function asked and the function in the table. STEP 1:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 25 From table, So, Example 13 Determine the value of ℒ{()} = ℒ{ 7 } by using Laplace transform Table. Time domain, () Frequency domain, () 2 + 2 () = ℒ{()} = ℒ {3 cos 4 7 } = 3 ℒ {cos 4 7 } From L.T Table = { 2 + 2 } = 3 = 4 7 : () () = 3 ( 2 + ( 4 7 ) 2) () = 3s 2 + 16 49 Change the figure of the table according to function asked. STEP 2: Look into table and find the similar equation between the function asked and the function in the table. STEP 1:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 26 From table, So, Example 14 Determine the value of ℒ{()} = ℒ{5 −3 + 2 + 4 4 − 10} by using Laplace transform Table. Time domain, () Frequency domain, () ! +1 () = ℒ{()} = ℒ{ 7 } From L.T Table ! +1 = 7 : = = ( 7! 7+1 ) () = 5040 8 Change the figure of the table according to function asked. STEP 2: Change the figure of the table according to function asked. Look into table and find the similar equation between the function asked and the function in the table. STEP 1: STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 27 So, Example 15 Determine the value of ℒ{()} = ℒ{2 −5 + 3 sin 2 + cos 10 − 2 } by using Laplace transform Table. So, () = ℒ{()} = ℒ{5 −3 + 2 + 4 4 − 10} ℒ{()} = ℒ{5 −3 } + ℒ{ 2 } + ℒ{4 4 } − ℒ{10} = 5ℒ{ −3 } + ℒ{ 2 } + 4ℒ{ 4 } − 10ℒ{1} ℒ{()} = 5 ( 1 + ) + ( 1 − ) + 4 ( ! +1 ) + 10 ( 1 ) = ( 5 s + 3 ) + ( 1 s − 2 ) + 4 ( 24 s 5 ) + ( 10 s ) () = 5 + 3 + 1 − 2 + 96 5 + 10 () = ℒ{()} = ℒ{2 −5 + 3 sin 2 + cos 10 − 2 } = ℒ{ −5 } + ℒ{3 sin 2} + ℒ{10} − ℒ{ 2 } = 2ℒ{ −5 } + 3ℒ{sin 2} + ℒ{10} − ℒ{ 2 } Change the figure of the table according to function asked. Look into table and find the similar equation between the function asked and the function in the table. STEP 1: STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 28 1.3 Laplace Transform Properties (First Shift Theorem) Example 2.16 By using First Shift Theorem, express () = −3 2 into frequency domain. = 2 ( 1 + ) + 3 ( 2 + 2 ) + ( 2 + 2 ) + ( ! +1 ) () = ( 2 + 5 ) + ( 3 2 + 2 2 ) + ( 2 + 102 ) + ( 2! 2+1 ) () = 2 + 5 + 3 2 + 4 + 2 + 100 + 2 3 () = ℒ{()} = ℒ{ −3 2} : Break the function into 2 (two) part () = 2 −3 1 = 2 2 = −3 : For function consisting of exponent (e), take note the value of exponential power ℎ1 () = sin 2 ℎ2 () = −3 Convert the other function, ℎ1 () and ℎ2 () like the normal function use Laplace Transform Table 1() = 2 2 + 2 2 2() = 1 + 3 If function, g() = −()then the Laplace transform of the function is,() = ( + ) First Shift Theorem STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 29 Example 17 By using First Shift Theorem, get the equation of F(s) if () = 2 3 + − . Denominator 2 = + 3 Substitute, the denominator of function, 2 into “s” in function, 1() () = 2 ( + 3) 2 + 2 2 () = ℒ{()} = ℒ{ 2 3 + − } : Break the function into 4 (four) part () = 3 2 + − 1. 3 2. 2 3. 4. − : For function consisting of exponent (e), take note the value of exponential power ℎ1 () = cos 3 ℎ2 () = 2 1 () = 2 () = − Convert the other function, ℎ1 () and ℎ2 () like the normal function, use Laplace Transform Table 1 () = 2 + 3 2 2 () = 1 − 2 1 () = 1 2 2 () = 1 + 1 Denominator 2 = − 2 2 = + 1 Substitute, the denominator of function, 2 into “s” in function, 1() and denominator of function, 2into “s” in function, 1() STEP 4 STEP 1 STEP 2 STEP 3 STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 30 Example 18 By using First Shift Theorem, get the equation of F(s) if () = 4 3 + 2 −3 . () = ( − 2) 2 + 3 2 + 1 ( + 1) 2 () = ℒ{()} = ℒ{ 4 3 + 2 −3 } : Break the function into 4 (four) part () = 4 3 + 2 −3 1. sin 4 2. 3 3. 2 4. −3 : For function consisting of exponent (e), take note the value of exponential power ℎ1 () = sin 4 ℎ2 () = 3 1 () = 2 2 () = −3 Convert the other function, ℎ1 () and ℎ2 () like the normal function. 1 () = sin 4 = 4 2 + 4 2 2 () = 3 = 1 − 3 1 () = 2 = 1 3 2 () = −3 = 1 + 3 Denominator 2 = − 3 2 = + 3 Substitute, the denominator of function, 2 into “s” in function, 1() and denominator of function, 2into “s” in function, 1() () = 4 ( − 3) 2 + 16 + 1 ( + 3) 3 STEP 1 STEP 2 STEP 3 STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 31 Example 19 By using First Shift Theorem, get the equation of F(s) if () = 4 3 3 2 + 4 − 5 4 () = ℒ{()} = ℒ { 4 3 3 2 + 4 − 5 4 } : Break the function into 4 (four) part () = 4 3 3 2 + 4 − 5 4 1. 4 3 2. 3 2 3. 4 4. − 5 4 : For function consisting of exponent (e), take note the value of exponential power ℎ1 () = 4 3 ℎ2 () = 3 2 1 () = 4 2 () = − 5 4 Convert the other function, ℎ1 () and ℎ2 () like the normal function. 1 () = 4 3 = 2 + ( 4 3 ) 2 2 () = 3 2 = 1 − 3 2 2 () = 2 2 − 3 1 () = 4 = 1 5 2 () = − 5 4 = 1 + 5 4 2 () = 4 4 + 5 Denominator 2 = − 3 2 2 = + 5 4 STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 32 Exercise Determine the value of ℒ{()}by using Laplace transform Table. 1. () = 23 + 3 5 2. () = 120 3. () = 7 4. () = 6 5. () = 5 3 6. () = 4 2 + 10 7. () = 2 10 8. () = −2 3 7 9. () = 3 ℎ √4 10. () = 10 2 11. () = 2 3 √ 8 3 12. () = 5 −4 + 3 −7 + 2 3 13. () = 7 9 2 + 3 5 7 + 9 7 −2 + 5 2 −3 Substitute, the denominator of function, 2 into “s” in function, 1() and denominator of function, 2into “s” in function, 1() () = ( − 3 2 ) 2 + ( 4 3 ) 2 + 1 ( + 5 4 ) 5 STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 33 1.4 Derivative Laplace Transform. Example 20 Express the function, ′′() + ′ () − ()into frequency domain, F(s), if (0) = 1 and ′ (0) = 0 ′′() + ′ () − () if (0) = 1 and ′ (0) = 0 Derivative Rules Function in Time Domain Function in Frequency Domain () = () () = ′ () = () − (0) 2() 2 = ′′() = 2() − (0) − ′ (0) Sub Function in Frequency Domain to Time Domain in equation ( 2() − (0) − ′ (0)) + (() − (0)) − (()) = Derivative Rules Function in Time Domain Function in Frequency Domain () () () = ′ () () − (0) 2() 2 = ′′() 2() − (0) − ′ (0) Derivative Function STEP 1
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 34 Sub value of: () = and ′ () = ( 2() − (1) − (0)) + (() − (1)) − (()) = 2() − 2(1) + 4() − 4(1) − () = ()( 2 + 4 − 3) − 2 − 4 = 0 FACTORIZED () = 2 + 4 ( 2 + 4 − 3) Example 21 Express the function, () − ′ () + ′′()into frequency domain, G(s), if (0) = 2 and ′ (0) = 1 () − ′ () + ′′() if (0) = 2 and ′ (0) = 1 Derivative Rules Function in Time Domain Function in Frequency Domain () = () () = ′ () = () − (0) 2() 2 = ′′() = 2() − (0) − ′ (0) Sub Function in Frequency Domain to Time Domain in equation (()) − (() − (0)) + ( 2() − (0) − ′ (0)) = 0 Sub value of: (0) = 2 and ′ (0) = 1 (()) − (() − 2) + ( 2() − 2 − 1) = 0 4() − 2() + 4 + 4 2() − 8s − 4 = 0 4() − 2() + 4 2() − 8s = 0 STEP 2 ANSWER STEP 1 STEP 2
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 35 ()(4 2 − 2 + 4) − 8 = 0 FACTORIZED () = 8 (4 2 − 2 + 4) Example 22 Express the function, () + ′ () + ′′() into frequency domain, Y(s), if. (0) = 1 and ′ (0) = −1 () + ′ () + ′′() if () = and ′ () = − Derivative Rules Function in Time Domain Function in Frequency Domain () = () () = ′ () = () − (0) 2() 2 = ′′() = 2() − (0) − ′ (0) Sub Function in Frequency Domain to Time Domain in equation () + (() − (0)) + ( 2() − (0) − ′ (0)) = Sub value of: (0) = 1 and ′ (0) = −1 () + (() − 1) + ( 2() − 1 − (−1)) = () + 2() − 2 + 2 2() − 2 + 2 = () + 2() + 2 2() − 2 + 2 − 2 = ()(2 2 + 2 + 1) − 2 = 0 FACTORIZED () = 2 (2 2 + 2 + 1) ANSWER STEP 1 STEP 2 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 36 Example 23 Express the function, 5() − ′ () + ′′() = − − into frequency domain, Y(s), if (0) = 2 and ′ (0) = 12 () − ′ () + ′′() = − − if () = and ′ () = Derivative Rules Function in Time Domain Function in Frequency Domain () = () () = ′ () = () − (0) 2() 2 = ′′() = 2() − (0) − ′ (0) Sub Function in Frequency Domain to Time Domain in equation (()) − (() − (0)) + ( 2() − (0) − ′ (0)) = −8 + 2 Sub value of: (0) = 2 and ′ (0) = 12 (()) − (() − 2) + ( 2() − 2 − 12) = −8 + 2 (5()) − (2() − 4) + ( 2() − 2 − 12) = −8 + 2 (5()) − 2() + 4 + 2() − 2 − 12 = −8 + 2 ()( 2 − 2 + 5) − 2 − 12 = −8 + 2 FACTORIZED () = 2 + 4 2 − 2 + 5 STEP 1 STEP 2 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 37 Exercise: Express the Frequency Domain function from the function given. 1. () + 2 ′ () + 3 ′′(), if (0) = 2 and ′ (0) = 1 2. 2() − ′ () + 4 ′′(), if zero initial condition 3. () + ′′() = cos 2, if (0) = 0 and ′ (0) = 1 4. y′′(t) + y′(t) − 2y(t) = 4, (0) = 2 ′(0) = 1 2.0 Theorem and property of Inverse Laplace Transform Inverse Laplace Transform of a Function by using table. Inverse Laplace transform is an essential operation for all applications of Laplace transformation. This transformation is used for the retrieval of the original function. ℒ −1() = () The inverses of functions can be obtained by inspection. We discuss the LT inverse of functions, which are not so obvious there are various methods of determining the inverse Laplace transform of a function. a. Standard Function A function that exactly almost in similar state to the function available in Laplace Table Example 24 Determine the time domain function, f (t) for the following Laplace function, () = 3 + 5 Find the exact similar denominator for function in the Laplace Table STEP 1:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 38 From table, Time domain, () Frequency domain, () − 1 + Since, Example 25 Determine the time domain function, f (t) for the following Laplace function, () = 3 2 + 9 () = ℒ −1 {()} = ℒ −1 { 3 + 5 } = 3ℒ−1 { 1 + 5 } = 3ℒ−1 { 1 − (−5) } = −5 Inverse Transform Frequency domain, () () use Laplace Transform Table () = 3 −5 Separate the constant and the solve the function. STEP 2: Find the exact similar denominator for function in the Laplace Table STEP 1:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 39 From table, Time domain, () Frequency domain, () 2 + 2 Since, Example 26 Determine the time domain function, f (t) for the following Laplace function, () = 8 2 + 7 () = ℒ −1 {()} = ℒ −1 { 3 2 + 9 } = ℒ −1 { 3 2 + 3 2 } = 3 Inverse Transform Frequency domain, () () use Laplace Transform Table () = 3 Separate the constant and the solve the function. STEP 2: Find the exact similar denominator for function in the Laplace Table STEP 1:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 40 From table, Time domain, () Frequency domain, () 2 + 2 Since, Example 27 Determine the time domain function, f (t) for the following Laplace function, () = 7 5 () = ℒ −1 {()} = ℒ −1 { 8 2 + 7 } = 8ℒ−1 { 2 + (√7 2 ) 2 } = √7 2 Inverse Transform Frequency domain, () () use Laplace Transform Table () = 8 √7 2 Separate the constant and the solve the function. STEP 2: Find the exact similar denominator for function in the Laplace Table STEP 1:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 41 From table, Time domain, () Frequency domain, () ! +1 Since, Example 2.28 Determine the time domain function, f (t) for the following Laplace function, () = 4 7 + 5 4 + 3 3 8 () = ℒ −1 {()} = ℒ −1 { 7 5 } = 7ℒ−1 { 1 5 } = 7ℒ−1 { ! +1 } = 4 = 7 4! ( 4! 4+1 ) Inverse Transform Frequency domain, () () use Laplace Transform Table () = 7 24 4 Find the exact similar denominator for function in the Laplace Table STEP 1: Separate the constant and the solve the function. STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 42 From table, Time domain, () Frequency domain, () ! +1 Since, () = ℒ −1 {()} = ℒ −1 { 4 7 + 5 4 + 3 3 8 } = ℒ −1 ({ 4 7 8 } + { 5 4 8 } + { 3 3 8 }) = ℒ −1 ({ 4 } + { 5 4 } + { 3 5 }) = ℒ −1 { 4 } + ℒ −1 { 5 4 } + ℒ −1 { 3 5 } = 4 + ( 5 3! ) 3! 3+1 + ( 3 4! ) 4! 4+1 = 4 + ( 5 3! ) 3 + ( 3 4! ) 4 Inverse Transform Frequency domain, () () use Laplace Transform Table () = = 4 + 5 3 3! + 3 4 4! Separate the constant and the solve the function. STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 43 b. Linearity Property Example 29 Determine the time domain function, f (t) for the following Laplace function, () = 17 + 2 + 1 3 − 5 From table, Time domain, () Frequency domain, () − 1 + Since, () = ℒ −1 {()} = ℒ −1 { 17 + 2 + 1 3 − 5 } = 17ℒ−1 { 1 + 2 } + ℒ −1 { 1 3 − 5 } = 17 ( 1 − (−2) ) + 1 3 ( 1 − 5 3 ) Inverse Transform Frequency domain, () () use Laplace Transform Table ℒ −1 {1() + 2()} = 1ℒ −1 {()} + 2ℒ −1 {()} Separate the constant and the solve the function. Find the exact similar denominator for function in the Laplace Table STEP 1: STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 44 Example 30 Determine the time domain function, f (t) for the following Laplace function, () = 6 2 + 25 + 3 2 + 49 From table, Time domain, () Frequency domain, () 2 + 2 2 + 2 Since, () = 17 −2 + 1 3 5 3 () = ℒ −1 {()} = ℒ −1 { 6 2 + 25 + 3 2 + 49} = 6ℒ−1 { 2 + 25} + 3ℒ−1 { 1 2 + 49} = 2 + 2 = 2 + 2 Separate the constant and the solve the function. Find the exact similar denominator for function in the Laplace Table STEP 1: STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 45 Example 31 Determine the time domain function, f (t) for the following Laplace function, () = 2 4 + 3 2 + 3 From table, Time domain, () Frequency domain, () 2 + 2 ! +1 = 5 = 7 () = ℒ −1 {()} = 6ℒ−1 { 2 + 5 2 } + 3ℒ−1 { 1 2 + 7 2 } () = ℒ −1 {()} = 6ℒ−1 { 2 + 5 2 } + 3 7 ℒ −1 { 7 2 + 7 2 } Inverse Transform Frequency domain, () () use Laplace Transform Table () = 6 cos 5 + 3 7 sin 7 Separate the constant and the solve the function. Find the exact similar denominator for function in the Laplace Table STEP 1: STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 46 Since, Example 32 Determine the time domain function, f (t) for the following Laplace function, () = 6 − 3 2 + 5 () = ℒ −1 {()} = ℒ −1 { 2 4 + 3 2 + 3 } = ℒ −1 { 2 4 } + ℒ −1 { 3 2 + 3 } () = ℒ −1 {()} = ℒ −1 { 2 4 } + ℒ −1 { 3 2 + 3 } () = ℒ −1 {()} = 2ℒ−1 { 1 4 } + 3ℒ−1 { 1 2 + 3 } = 2 3! ℒ −1 { 3! 3+1 } + 3 √3 2 ℒ −1 { √3 2 2 + (√3 2 ) 2 } = 2 + 2 = ! +1 = √3 2 = 3 Inverse Transform Frequency domain, () () use Laplace Transform Table () = 2 3! 3 + 3 √3 2 sin √3 2 Find the exact similar denominator for function in the Laplace Table STEP 1: