CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 47 From table, Time domain, () Frequency domain, () 2 + 2 2 + 2 Since, () = ℒ −1 {()} = ℒ −1 { 6 − 3 2 + 5 } = ℒ −1 { 6 2 + 5 } − ℒ −1 { 3 2 + 5 } = 6ℒ−1 { 2 + 5 } − 3ℒ−1 { 1 2 + 5 } = 2 + 2 = 2 + 2 = √5 2 = √5 2 = 6ℒ−1 { 2 + (√5 2 ) 2 } − 3 √5 2 ℒ −1 { √5 2 2 + (√5 2 ) 2 } Inverse Transform Frequency domain, () () use Laplace Transform Table () = 6 cos √5 2 − 3 √5 2 sin √5 2 Separate the constant and the solve the function. STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 48 c. First Shift Properties Example 33 Determine the time domain function, f (t) for the following Laplace function, () = 3 2 + 4 + 13 From table, Time domain, () Frequency domain, () 2 + 2 − ( + ) 2 + 2 Since, () = ℒ −1 {()} = ℒ −1 { 3 2 + 4 + 13} Factorized the equation = ℒ −1 { 3 ( + 2) 2 + 9 } = ℒ −1 { 3 ( + 2) 2 + 3 2 } Separate the constant and the solve the function. Find the exact similar denominator for function in the Laplace Table STEP 1: STEP 2: The first shift theorem can be written in inverse form as: ℒ −1 {( − )} = () = ℒ −1 {()}
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 49 Example 34 Determine the time domain function, f (t) for the following Laplace function, () = 5 2 − 6 + 7 From table, Time domain, () Frequency domain, () 2 − 2 − ( + ) 2 − 2 Since, Inverse Transform Frequency domain, () () use Laplace Transform Table ( + ) 2 + 2 = − () = − () = ℒ −1 {()} = ℒ −1 { 5 2 − 6 + 7 } Factorized the equation = ℒ −1 { 5 ( − 3) 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 50 Example 35 Determine the time domain function, f (t) for the following Laplace function, () = 5 (2 − 3) 4 From table, Time domain, () Frequency domain, () − ! ( + ) +1 = ℒ −1 { 5 ( − 3) 2 − (√2 2 ) 2 } = 5 √2 2 ℒ −1 { √2 2 ( − 3) 2 − (√2 2 ) 2 } Inverse Transform Frequency domain, () () use Laplace Transform Table ( + ) 2 − 2 = − = 5 √2 2 ℒ −1 { √2 2 ( − 3) 2 − (√2 2 ) 2 } () = 5 √2 2 3(√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 51 Since, () = ℒ −1 {()} = ℒ −1 { 5 (2 − 3) 4 } = 5 ℒ −1 { 1 (2 − 3) 4 } Factorized the equation = 5ℒ −1 { 1 (2 ( − 3 2 )) 4 } = 5ℒ−1 { 1 2 4 ( − 3 2 ) 4 } = 5 2 4 ℒ −1 { 1 ( − 3 2 ) 4 } Inverse Transform Frequency domain, () () use Laplace Transform Table ! ( + ) +1 = − = 3 2 = 4 = ( 5 (2 4)(3!) ) ℒ −1 { 3! ( − 3 2 ) 3+1 } = ( 5 96) 3 2 () = ( 5 96) 3 2
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 52 d. First Order & Second Order Derivatives First Order & Second Order Derivatives Theorem Inverse Laplace using Partial Fraction. Example 36 Express the function, ′ () + () = into frequency domain, Y(s), if (0) = 3 ′ () + () = if () = 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(()) = 12 − 3 Sub value of: (0) = 3 (() − 3) + 2(()) = 12 − 3 ()( + 2) − 3 = 12 − 3 ()( + 2) = 12 − 3 + 3 ()( + 2) = 12 + 3 − 9 − 3 STEP 1 STEP 2
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 53 () = 3 + 3 ( − 3)( + 2) () = ℒ −1 (()) = ℒ −1 ( 3 + 3 ( − 3)( + 2) ) Now using Partial Fraction to get last Inverse Laplace Equation 3 + 3 ( − 3)( + 2) = ( − 3) + ( + 2) 3 + 3 = ( + 2) + ( − 3) , = 3 3 + 3(3) = (3 + 2) + (3 − 3) 12 = (5) = 12 5 , = −2 3 + 3(−2) = (−2 + 2) + (−2 − 3) −3 = (−5) = 3 5 3 + 3 ( − 3)( + 2) = 12 5 ⁄ ( − 3) + 3 5 ⁄ ( + 2) Refer Laplace Table to get Function in Time Domain y(t) () = ℒ −1 (()) = ( 12 5 ⁄ ) 3 + ( 3 5 ⁄ ) −2 Example 37 Express the function, ′ () + () = () into frequency domain, Y(s), if. (0) = −2 STEP 3 ANSWER STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 54 ′ () + () = if () = − 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 ′ () + 2() = 10 2(() − (0)) + 2(()) = 10 Sub value of: (0) = −2 2(() − (−2)) + 2(()) = 10 (2() + 4) + 2(()) = 10 ()(2 + 2) + 4 = 10 ()(2 + 2) = 10 − 4 ()(2 + 2) = 10 − 4 () = 10 − 4 (2 + 2) () = 10 − 4 (2 + 2) () = ℒ −1 (()) = ℒ −1 ( 10 − 4 (2 + 2) ) Now using Partial Fraction to get last Inverse Laplace Equation 10 − 4 (2 + 2) = + (2 + 2) 10 − 4 = (2 + 2) + , = 0 10 − 4(0) = (2(0) + 2) + (0) 10 = (2) = 5 STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 55 , = −1 10 − 4(−1) = (2(−1) + 2) + (−1) 15 = (−1) = −15 3 + 3 ( − 3)( + 2) = 5 − 15 (2 + 2) Refer Laplace Table to get Function in Time Domain y(t) () = ℒ −1 (()) = 5 + ( 15 2 ⁄ ) − Example 38 Express the function, ′ () + () = () into frequency domain, Y(s), if. (0) = −2 ′ () + () = if () = − 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 ′ () + 2() = 10 2(() − (0)) + 2(()) = 10 Sub value of: (0) = −2 2(() − (−2)) + 2(()) = 10 (2() + 4) + 2(()) = 10 STEP 1 STEP 2 ANSWER STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 56 ()(2 + 2) + 4 = 10 ()(2 + 2) = 10 − 4 ()(2 + 2) = 10 − 4 () = 10 − 4 (2 + 2) () = 10 − 4 (2 + 2) () = ℒ −1 (()) = ℒ −1 ( 10 − 4 (2 + 2) ) Now using Partial Fraction to get last Inverse Laplace Equation 10 − 4 (2 + 2) = + (2 + 2) 10 − 4 = (2 + 2) + , = 0 10 − 4(0) = (2(0) + 2) + (0) 10 = (2) = 5 , = −1 10 − 4(−1) = (2(−1) + 2) + (−1) 15 = (−1) = −15 3 + 3 ( − 3)( + 2) = 5 − 15 (2 + 2) Refer Laplace Table to get Function in Time Domain y(t) () = ℒ −1 (()) = 5 + ( 15 2 ⁄ ) − Example 39 Express the function, ′′ + ′ () + () = into frequency domain, Y(s), if. () = if ′ () = ′′ + ′ () + () = if () = if ′ () = STEP 3 ANSWER STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 57 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)) + (()) = Sub value of: () = if ′ () = ( 2() − (0) − ′ (0)) + (() − (0)) + (()) = ( 2() − (1) − (0)) + (() − 1) + (()) = ( 2() − ) + (3() − 3) + (()) = ()( 2 + 3 + 2) − − 3 = ()( 2 + 3 + 2) = + () = + ( 2 + 3 + 2) () = + ( + )( + ) () = ℒ −1 (()) = ℒ −1 ( + ( + )( + ) ) Now using Partial Fraction to get last Inverse Laplace Equation + ( + )( + ) = ( + ) + ( + ) + = ( + ) + ( + ) , = −1 − + = (− + ) + (− + ) STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 58 = () = , = −2 − + = (− + ) + (− + ) = (−) = − Refer Laplace Table to get Function in Time Domain y(t) () = ℒ −1 ( + ( + )( + ) ) = ℒ −1 ( ( + ) + ( + ) ) = ℒ −1 ( ( + ) + ( + ) ) = ℒ −1 ( −1 ( + ) + 2 ( + ) ) () = − −2 + 2 − Example 40 Express the function, ′′ + ′ () = () into frequency domain, Y(s), if. () = , ′ () = ′′ + ′ () = () if () = , ′ () = 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 STEP 4 ANSWER STEP 1
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 59 ( 2() − (0) − ′ (0)) + 4(() − (0)) = 1 Sub value of: () = , ′ () = ( 2() − (0) − 5) + 4(() − 0) = 1 ( 2() − 5) + 4(()) = 1 ()( 2 + 4) − 5 = 1 () = 6 ( 2 + 4) () = ℒ −1 (()) = ℒ −1 ( 6 ( 2 + 4) ) Refer Laplace Table to get Function in Time Domain y(t) () = ℒ −1 ( ( + ) ) = 3ℒ−1 ( ( + ) ) = 3ℒ−1 ( ( + ) ) () = 3 sin 2 Example 41 Express the function, ′′ − ′ () + () = into frequency domain, Y(s), if. () = − if ′ () = STEP 2 STEP 3 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 60 ′′ − 10 ′ () + 9() = 5 if (0) = −1 if ′ (0) = 2 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)) − 10(() − (0)) + 9(()) = 5 2 Sub value of: () = − if ′ () = ( 2() − (−1) − 2) − 10(() − (−1)) + 9(()) = 5 2 2() + s − 2 − 10() − 10 + 9() = 5 2 ()( 2 − 10 + 9) + s − 12 = 5 2 ()( 2 − 10 + 9) = 5 2 + 12 − () = 5 2( 2 − 10 + 9) + 12 − ( 2 − 10 + 9) () = 5 + 12 2 − 3 2( 2 − 10 + 9) Combine 2 term’s () = 5 + 12 2 − 3 2( − 1)( − 9) () = ℒ −1 (()) = ℒ −1 ( 5 + 12 2 − 3 2( − 1)( − 9) ) Now using Partial Fraction to get last Inverse Laplace Equation 5 + 12 2 − 3 2( − 1)( − 9) = + 2 + ( − 1) + ( − 9) 5 + 12 2 − 3 = ( − 1)( − 9) + ( − 1)( − 9) + 2 ( − 9) + 2 ( − 1) 5 + 12 2 − 3 = ( 3 − 10 2 + 9) + ( 2 − 10 + 9) + ( 3 − 9 2 ) + ( 3 − 2 ) STEP 1 STEP 2 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 61 5 + 12 2 − 3 = 3 ( + + ) + 2 (−10 + − 9 − ) + (9 − 10) + (9) USING Comparison Coefficient 3 −1 = + + 2 12 −10 + − 9 − 1 0 9 − 10 0 5 9 = 50 81 = 5 9 = 31 81 = −2 Refer Laplace Table to get Function in Time Domain y(t) () = ℒ −1 ( 5 + 12 2 − 3 2( − 1)( − 9) ) = ℒ −1 ( + 2 + ( − 1) + ( − 9) ) = ℒ −1 ( + 2 + ( − 1) + ( − 9) ) = ℒ −1 ( 50 81 + 5 9 2 + 31 81 ( − 1) − 2 ( − 9) ) () = 50 81 + 5 9 + 31 81 − 2 9 e. completing the square Method. Example 42 Find the Inverse Laplace for the function given, F(s) = 3 s 2 + 4s + 5 STEP 4 ANSWER A complex Laplace function that cannot be factorised with exact value, so completing square is the final solution.
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 62 F(s) = 3 s 2 + 4s + 5 Build a conjugate with the middle number with “s” that is divided by 2 with power of 2 = 3 2 + 4 + ( 4 2 ) 2 − ( 4 2 ) 2 + 5 = 3 2 + 4 + (2) 2 − (2) 2 + 5 Factorize the 1st three characters = 3 ( + 2) 2 − 4 + 5 Simplify the equation so that the equation looks alike the function available in the table = 3 ( + 2) 2 + 1 2 = 3 ( 1 ( + 2) 2 + 1 2 ) = 3 −2 sin Example 43 Find the Inverse Laplace for the function given, F(s) = 7 s 2 + 12s + 45 F(s) = 7 s 2 + 12s + 45 Build a conjugate with the middle number with “s” that is divided by 2 with power of 2 = 7 2 + 12 + ( 12 2 ) 2 − ( 12 2 ) 2 + 45 = 7 2 + 12 + (6) 2 − (6) 2 + 45 Factorize the 1st three characters STEP 2 STEP 1 STEP 3 STEP 2 STEP 1 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 63 = 7 ( + 6) 2 − 36 + 45 = 7 ( + 6) 2 + 9 Simplify the equation so that the equation looks alike the function available in the table = ℒ −1 { 7 ( + 6) 2 + 3 2 } = 7ℒ −1 { 1 ( + 6) 2 + 3 2 } = 7 3 ℒ −1 { 3 ( + 6) 2 + 3 2 } = 7 3 −6 sin 3 Example 44 Find the Inverse Laplace for the function given, F(s) = 6 s 2 − 8s + 5 F(s) = 6 s 2 − 8s + 5 Build a conjugate with the middle number with “s” that is divided by 2 with power of 2 = 6 2 − 8 + ( 8 2 ) 2 − ( 8 2 ) 2 + 5 = 6 2 − 8 + (4) 2 − (4) 2 + 5 Factorize the 1st three characters = 6 2 − 8 + (4) 2 − (4) 2 + 5 = 6 ( − 4) 2 − 16 + 5 = 6 ( − 4) 2 − 11 Simplify the equation so that the equation looks alike the function available in the table = ℒ −1 { 6 ( − 4) 2 − 11} STEP 3 ANSWER STEP 2 STEP 1 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 64 = ℒ −1 { 6 ( − 4) 2 − 11} = 6ℒ −1 { 1 ( − 4) 2 − 11} = 6ℒ −1 { 1 ( − 4) 2 − 11} = 6 √11 2 ℒ −1 { √11 2 ( − 4) 2 − (√11 2 ) 2 } = 6 √11 2 −4 sinh √11 2 Example 45 Find the Inverse Laplace for the function given, F(s) = 10s 5s2 + 20s + 30 F(s) = 10s 5s2 + 20s + 30 Build a conjugate with the middle number with “s” that is divided by 2 with power of 2 F(s) = ( 10s 5s2 + 20s + 30) ( 1 5 ⁄ 1 5 ⁄ ) = 2s s 2 + 4s + 6 = 2s s 2 + 4s + ( 4 2 ) 2 − ( 4 2 ) 2 + 6 = 2s s 2 + 4s + (2) 2 − (2) 2 + 6 Factorize the 1st three characters = 2 ( + 2) 2 − 4 + 6 = 2 ( + 2) 2 + 2 Simplify the equation so that the equation looks alike the function available in the table = ℒ −1 { 2 ( + 2) 2 + 2 } = 2ℒ −1 { ( + 2) 2 + 2 } = 2ℒ −1 { ( + 2) 2 + 2 } = 2ℒ −1 { + 2 − 2 ( + 2) 2 + 2 } = 2ℒ −1 { + 2 ( + 2) 2 + 2 − 2 ( + 2) 2 + 2 } STEP 2 STEP 1 STEP 3 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 65 = 2ℒ −1 { + 2 ( + 2) 2 + 2 } − 2ℒ −1 { 2 ( + 2) 2 + 2 } = 2ℒ −1 { + 2 ( + 2) 2 + (√2 2 ) 2 } − 4 √2 2 ℒ −1 { √2 2 ( + 2) 2 + (√2 2 ) 2 } = 2 −2 sin √2 2 − 4 √2 2 −2 cos √2 2 Example 46 Find the Inverse Laplace for the function given, F(s) = 8 + 4s 4s2 − 8s − 12 F(s) = 8 + 4s 4s2 − 8s − 12 Build a conjugate with the middle number with “s” that is divided by 2 with power of 2 F(s) = ( 8 + 4s 4s2 − 8s − 12) ( 1 4 ⁄ 1 4 ⁄ ) = 2 + s s 2 − 2s − 3 = 2 + s s 2 − 2s + ( −2 2 ) 2 − ( −2 2 ) 2 − 3 = 2 + s s 2 − 2s + (−1) 2 − (−1) 2 − 3 Factorize the 1st three characters = 2 + ( − 1) 2 − 1 − 3 = 2 + ( − 1) 2 − 4 Simplify the equation so that the equation looks alike the function available in the table = ℒ −1 { 2 + ( − 1) 2 − 4 } = ℒ −1 { 2 + ( − 1) 2 − 4 } = ℒ −1 { 2 + + 1 − 1 ( − 1) 2 − 4 } = ℒ −1 { − 1 + 3 ( − 1) 2 − 4 } = ℒ −1 { − 1 + 3 ( − 1) 2 − 4 } = ℒ −1 { − 1 ( − 1) 2 − 4 + 3 ( − 1) 2 − 4 } = ℒ −1 { − 1 ( − 1) 2 − 4 + 3 ( − 1) 2 − 4 } ANSWER STEP 2 STEP 1 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 66 = ℒ −1 { − 1 ( − 1) 2 − 4 } + 3ℒ −1 { 1 ( − 1) 2 − 4 } = ℒ −1 { − 1 ( − 1) 2 − 2 2 } + 3 2 ℒ −1 { 2 ( − 1) 2 − 2 2 } = − sinh 2 + 3 2 − cosh 2 Exercise: Determine the Inverse Laplace for the function given, 1. F(s) = 5 s 2 + 2s + 10 2. F(s) = 10 s 2 − 2s + 5 3. F(s) = s s 2 + 4s + 5 ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 67 f. Partial Fractions. P(x) and Q(x) are polynomials, and the degree of P(x) is smaller than the degree of Q(x). Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. That is important to remember. () () So, once we’ve determined that partial fractions can be done, we factor the denominator as completely as possible. Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition. Term in partial fraction Factor in denominator LINEAR = () ( − )( − ) = ( − ) + ( − ) REPITATION = () ( − ) = − + ( − ) 2 + ( − ) 3 + ⋯ + ( − ) QUADRATIC = () ( 2 + + ) = + ( 2 + + ) COMBINATION = () ( 2 + + ) = + ( 2 + + ) + + ( 2 + + ) 2 +. . + + ( 2 + + ) The denominator of the function must be factorised first in order to be solved.
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 68 Example 47 Find the Inverse Laplace for the function given, F(s) = 3 + 5 2s2 − 5 − 3 F(s) = 3 + 5 2s2 − 5 − 3 Factorize the denominator of the function, if done skip F(s) = 3 + 5 (2 + 1)( − 3) Assuming the function equal to another function have the same denominator 3 + 5 (2 + 1)( − 3) = 2 + 1 + − 3 Equalizing both side function’s denominator 3 + 5 = ( − 3) + (2 + 1) For finding the unknown variable, equalizing each denominator fraction with “0” 3 + 5 = ( − 3) + (2 + 1) Equalize the denominator of each fraction = 0 ; = − 1 2 3 (− 1 2 ) + 5 = ((− 1 2 ) − 3) + (2 (− 1 2 ) + 1) 7 2 = 7 2 1 = ; = 3 3(3) + 5 = (3 − 3) + (2(3) + 1) 14 = 7 2 = STEP 2 STEP 1 STEP 3
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 69 Substitute the unknown variable value and solve like standard function. Substitute the value of ‘A’ and ‘B’ 3 + 5 (2 + 1)( − 3) = 1 2 + 1 + 2 − 3 = ℒ −1 { 3 + 5 (2 + 1)( − 3) } = ℒ −1 { 1 2 + 1 + 2 − 3 } = ℒ −1 { 1 2 + 1 + 2 − 3 } = ℒ −1 { 1 2 + 1 + 2 − 3 } = ℒ −1 { 1 2 + 1 + 2 − 3 } = 1 2 ℒ −1 { 1 + 1 2 } + 2ℒ −1 { 1 − 3 } = 1 2 − 1 2 + 2 3 Example 48 Find the Inverse Laplace for the function given, F(s) = 3 + 2 ( − 1)( 2 + 1) F(s) = 3 + 2 ( − 1)( 2 + 1) Check the cases numerator is a polynomial of lower degree than the denominator proper fraction. The function denominator is Linear and Quadratic = 3 + 2 ( − 1)( 2 + 1) = ( − 1) + + ( 2 + 1) Equalizing both side function’s denominator 3 + 2 = ( 2 + 1) + ( + )( − 1) Expand the equation and solving by using comparison coefficient. 3 + 2 = ( 2 + 1) + ( + )( − 1) 3 + 2 = 2 + + 2 − + − 3 + 2 = 2 ( + ) + (− + ) + ( − ) ANSWER STEP 4 STEP 2 STEP 3 STEP 1
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 70 2 0 = + 1 3 = − + 0 2 = − Using Matrix to solved equation. [ 1 1 0 0 −1 1 1 0 −1 ][ ] = [ 0 3 2 ] Using Calculator find A, B and C = 2 1 2 = −2 1 2 = 1 2 Substitute the unknown variable value and solve like standard function. Substitute the value of ‘A’ ‘B’ and ‘C’ = 3 + 2 ( − 1)( 2 + 1) = 2 1 2 ( − 1) + −2 1 2 + 1 2 ( 2 + 1) = ℒ −1 { 3 + 2 ( − 1)( 2 + 1) } = ℒ −1 { 2 1 2 ( − 1) + −2 1 2 + 1 2 ( 2 + 1) } = ℒ −1 { 2 1 2 ( − 1) + −2 1 2 + 1 2 ( 2 + 1) } = ℒ −1 { 2 1 2 ( − 1) + −2 1 2 ( 2 + 1) + 1 2 ( 2 + 1) } = ℒ −1 = 2 1 2 ℒ −1 { 1 − 1 } − 2 1 2 ℒ −1 { 2 + 1 } + 1 2 ℒ −1 1 ( 2 + 1) = 2 1 2 −− 2 1 2 cos + 1 2 ANSWER sin STEP 5 STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 71 Example 49 Find the Inverse Laplace for the function given, F(s) = 11 2 + 14 + 5 ( + 1) 2(2 + 1) F(s) = 11 2 + 14 + 5 ( + 1) 2(2 + 1) Check the cases numerator is a polynomial of lower degree than the denominator proper fraction. The function denominator is Linear and Quadratic = 11 2 + 14 + 5 ( + 1) 2(2 + 1) = ( + 1) + ( + 1) 2 + (2 + 1) Equalizing both side function’s denominator 11 2 + 14 + 5 = ( + 1)(2 + 1) + (2 + 1) + ( + 1) 2 11 2 + 14 + 5 = ( + 1)(2 + 1) + (2 + 1) + ( + 1)( + 1) 11 2 + 14 + 5 = ( + 1)(2 + 1) + (2 + 1) + ( + 1)( + 1) 11 2 + 14 + 5 = (2 2 + 3 + 1) + (2 + 1) + ( 2 + 2 + 1) 11 2 + 14 + 5 = 2 (2 + ) + (3 + 2 + 2) + ( + + ) Expand the equation and solving by using comparison coefficient. 2 11 = 2 + 1 14 = 3 + 2 + 2 0 5 = + + Using Matrix to solved equation. [ 2 0 1 3 2 2 1 1 1 ][ ] = [ 11 14 5 ] Using Calculator find A, B and C STEP 2 STEP 3 STEP 1 STEP 4
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 72 = 4 = −2 = 3 Substitute the unknown variable value and solve like standard function. Substitute the value of ‘A’ ‘B’ and ‘C’ = 11 2 + 14 + 5 ( + 1) 2(2 + 1) = 4 ( + 1) − 2 ( + 1) 2 + 3 (2 + 1) = ℒ −1 { 11 2 + 14 + 5 ( + 1) 2(2 + 1) } = ℒ −1 { 4 ( + 1) − 2 ( + 1) 2 + 3 (2 + 1) } = ℒ −1 { 4 ( + 1) − 2 ( + 1) 2 + 3 (2 + 1) } = ℒ −1 { 4 ( + 1) } − ℒ −1 { 2 ( + 1) 2 } + ℒ −1 { 3 (2 + 1) } = 4ℒ−1 { 1 ( + 1) } − 2ℒ −1 { 1 ( + 1) 2 } + 3 2 ℒ −1 { 1 ( + 1 2 ) } = 4 −− 2 − + 3 2 − 1 2 Exercise: Find the Inverse Laplace for the function given, 1 3 − ( − 1)( + 1) ( + 4)( − 2)( − 1) 2 6 + ( + 1)( 2 − 5 + 11) ( − 1)( − 2)( + 1) 3 8 − ( + 2)(4 + 10) ( + 1)( + 2) 2 4 2 − 5 + 7 ( + 2) 3 5 8 + 3 ( 2 + 1)( 2 + 4) ANSWER STEP 5
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 73 2.2 Analyze RLC circuits using Laplace Transform The differentiation and integration properties may also be used to transform circuits involving capacitive and inductive elements so that the circuit may be solved directly in term of Laplace Transforms. This is accomplished by replacing resistive, capacitive, and inductive elements with their Laplace Transform Equivalents. Resistor Capacitor Inductor Time Domain Time Domain Time Domain Kirchhoff’s voltage law with initial current Transfomation the equation, we write () = () () = 1 ∫ () 0 () = () () = () () = () () = 1 ∫ () 0 + 0 Frequency Domain Frequency Domain Frequency Domain () = () () = 1 () + 0 0 = Initial voltage () = [() − 0 ] () = () − 0 () = () () = [() − 0 ] () = () − 0 () = 1 () + 1 0 Table: Laplace Transform time domain to frequency domain for resistor, inductor, and capacitor. 0 = Initial current
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 74 a. RC Circuit Apply Laplace Transform based on figure 2.31(a), find () by assuming zero initial condition Figure: 2.31(a) Construct the equation in f(t) from the circuit () = () + 1 ∫ () 0 ℒ{ ()} = ℒ {() + 1 ∫ () 0 } Convert the function using derivative and integrals rules. () = () + () 1 () = () + () 1 () = () ( + 1 ) () = () ( + 1 ) () = () ( + 1 ) Construct a transfer function and solve according to Inverse Laplace Method STEP 3: STEP 2: STEP 1: I(t) ()
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 75 Find the laplace (FREQUENCY DOMAIN) expressionn for in the RC circuit below figure 2.3.1(b). The input is sine function, () = figure 2.31(b) Convert circuit to the frequency domain. () = + () () = ( 1 1 + ) () = ( 1 1 + ) () = ( 1 + ) () = ( 1 + ) Construct a transfer function and solve according to Inverse Laplace Method STEP 2: STEP 1: () = = () = I(t) () ()
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 76 b. RL Circuit Apply Laplace Transform based on figure 2.3.2(a), find () by assuming zero initial condition figure 2.3.2(a) Construct the equation in f(t) from the circuit () = () + () ℒ{ ()} = ℒ {() + () } Convert the function using derivative and integrals rules. () = () + () () = ()( + ) () = () ( 1 + ) Construct a transfer function and solve according to Inverse Laplace Method I(t) STEP 3: STEP 2: STEP 1: R L Vi(t)
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 77 Find the laplace (FREQUENCY DOMAIN) expressionn for in the RL circuit below figure 2.3.2(b). The input is sine function, () = sin , = 1 figure 2.3.2(b) Convert circuit to the frequency domain. () = + () () = ( + ) ( 2 + 2 ) , = 1 () = ( + ) ( 2 + 1 ) () = ( ( + )( 2 + 1) ) STEP 1: R L Vi(t) () R Ls () () = ( 1 2 + 2 ) ()
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 78 Construct a transfer function and solve according to Inverse Laplace Method c. RLC Circuit Apply Laplace Transform based on figure 2.3.3(a), find () by assuming zero initial condition. figure 2.3.3(a) Construct the equation in f(t) from the circuit () = () + 1 ∫ () 0 + () ℒ{ ()} = ℒ {() + 1 ∫ () 0 + () } Convert the function using derivative and integrals rules. () = () + () 1 + () () = () ( + + 1 ) () = () ( + 2 + 1 ) Construct a transfer function and solve according to Inverse Laplace Method STEP 2: I(t) STEP 3: STEP 1: STEP 2: Vi(t) C + R L
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 79 Apply Laplace Transform based on figure 2.3.3(b), find () by assuming zero initial condition. The input is sine function, () = cos , = 1 figure 2.3.3(b) Convert circuit to the frequency domain. Construct the equation in f(t) from the circuit () = + + () () = ( 1 + + 1 )( 2 + 2 ) , = 1 () = ( 1 1 + + 2) ( 2 + 1 ) () = ( 1 1 + + 2) ( 2 + 1 ) () = ( 1 2 + + 1 )( 2 + 1 ) Construct a transfer function and solve according to Inverse Laplace Method I(t) STEP 3: STEP 2: Vi(t) C + R L STEP 1: 1 () = ( 2 + 2 ) + ()
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 80 Example 50 Apply Laplace Transform based on figure 2.50, find i(t) by assuming zero initial condition. Figure 2.50 () = 5() + 1 () ℒ{()} = ℒ {5() + () } 1 s = 5I(s) + sI(s) 1 s = (s + 5)I(s) I(s) = 1 s(s + 5) I(s) = A s + B s + 5 Convert the function using derivative and integrals rules. Construct the equation in f(t) from the circuit. STEP 1: STEP 2: Construct a transfer function and solve according to Inverse Laplace Method STEP 3: 5Ω 1H u(t) V
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 81 ∴ A = sI(s)|s=0 = 1 s(s + 5) (s) = 1 s + 5 = 1 0 + 5 = 1 5 ∴ B = (s + 5)I(s)|s=−5 = 1 s(s + 5) (s + 5) = 1 s = 1 −5 = − 1 5 I(s) = A s + B s + 5 = 1 5s − 1 5(s + 5) ℒ −1 {()} = ℒ −1 { 1 5 − 1 5( + 5) } () = 1 5 () − 1 5 −5 Look into table and invert the Laplace into time domain. STEP 4:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 82 Example 51 By applying Laplace Transform based on Figure 2.51, determine i(t) (Assume zero initial condition) Figure 2.51 10() = 5() + () + 1 0.25 ∫ () ℒ{10()} = ℒ {5() + () + 4 ∫ ()} 10 s = 5I(s) + [sI(s) − I(0)] + 4 ( 1 s I(s) ) 10 s = 5I(s) + sI(s) + 4 s I(s) 10 s = I(s)(5 + s + 4 s ) 10 s = I(s) (5 + s + 5 s ) 10 s = I(s)( 5s + s 2 + 4 s ) 10 = I(s)(s 2 + 5s + 4) 10V 0.25F + 5Ω 1H Construct the equation in f(t) from the circuit. STEP 1: Convert the function using derivative and integrals rules. STEP 2:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 83 I(s) = 1 (s + 1)(s + 4) I(s) = A s + 1 + B s + 4 ∴ A = (s + 1)I(s)|s=−1 = 1 (s+1)(s+4) (s + 1) = 1 s + 4 = 1 −1 + 4 = 1 3 ∴ B = (s + 4)I(s)|s=−4 = 1 (s + 1)(s + 4) (s + 4) = 1 + 1 = 1 −4 + 1 = − 1 3 I(s) = A s + 1 + B s + 4 = 1 3(s + 1) − 1 3(s + 4) Construct a transfer function and solve according to Inverse Laplace Method STEP 3: Look into table and invert the Laplace into time domain. STEP 4:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 84 ℒ −1 {()} = ℒ −1 { 1 3( + 1) − 1 3( + 4) } () = 1 3 −− 1 3 −4 Example 52 Apply Laplace Transform based on figure xxx, find () by assuming zero initial condition. The input is sine function, () = 10 () Vi(t)=10 R L C 10/3 5H 0.1F Convert circuit to the frequency domain. Vi(t)=10 R L C 10/3 5s 10/s Construct the equation in () from the circuit using nodal analysis. () − 10 ⁄ 10⁄3 + () 5 + () 10⁄ = 0 () 10⁄3 + () 5 + () 10⁄ = 10 ⁄ 10⁄3 () ( 3 10 + 1 5 + 10) = 3 () (3 + 2 + ) = 30 STEP 2: STEP 1:
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 85 () ( 3 + 2 + 2 ) = 30 () = 30 ( 3 + 2 + 2 ) () = ( 30 3 + 2 + 2 ) = 30 2 + 3 + 2 () = 30 2 + 3 + 2 = 30 ( + 2)( + 1) Construct a transfer function and solve according to Inverse Laplace Method () = 30 ( + 2)( + 1) = + 2 + + 1 30 ( + 2)( + 1) = + 2 + + 1 30 = ( + 1) + ( + 2) 30 = + + + 2 30 = ( + ) + ( + 2) By using comparison Coefficient 1 0 = ( + ) 0 30 = ( + 2) ℎ ∶ ( 1 1 1 2 ) ( ) = ( 0 30) = −30, = 30 () = 30 ( + 2)( + 1) = −30 + 2 + 30 + 1 ℒ −1 {()} = ℒ −1 { −30 + 2 + 30 + 1 } () = −30 −2 + 30 − STEP 3: ANSWER
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 86 Exercise: Q1. Use Laplace Transform based on figure Ex.1, Find the VL (t) by assuming zero initial condition if () = 4 −3 Figure Ex 1 Q2. By using Laplace Transform based on figure Ex.2, calculate the value of V0(t) when t=0, q0=0 and i0=0 Figure Ex 2 2Ω 1H Vs 1H 14u(t)V 1/50F + 2Ω
CIRCUIT ANALYSIS WITH LAPLACE TRANSFORMS A PROBLEM-SOLVING APPROACH Electrical Engineering Department POLIMAS 87 Q3. Determine the Inverse Laplace Transform f(t) for the following function by using Partial Fraction. i. F(s) = s (s + 4)(s + 3) ii. F(s) = 2 s(s + 1) 2 iii. F(s) = 8( s + 2 ) (s + 1)(s + 3)(s + 5) iv. F(s) = 5s2+14s+2 (s − 2)(s + 3) 2 v. F(s) = s 2 (s − 1) 4 vi. F(s) = 1 (s + 2)(s + 5)(s +11) vii. F(s) = 1 s 2(s+3) viii. F(s) = 2s2 − 16 s 3 − 16s Q4. By using Laplace transform, produce the time domain y(t) for the given function. yʹʹ(t) + 4yʹ(t) + 3y(t) = 0; given y (0) = 3 and yʹ (0) = 1 Q5. By using Laplace Transform, solve for fʹʹ(t) + 2fʹ(t) – 3f(t) = 0; if given f (0) =1 dan f ʹ (0) = 1 Q6. Produce Laplace Transform y(t) of first derivatives for function. yʹ(t) – 2y(t) = 4; given y (0) = 1 Q7. Solve – 4x = 8 ; given x (0) = 2 when t = 0 Q8. By using Laplace Transform, produce the time domain b(t) for the given function. b''(t) – 3b'(t) + 2b(t) = 4e3t ; given b(0) = 0 and b'(0) = 5