Çözüldü Ters Laplace Dönüşümü (3 Soru)

3. a)
ℒ^(-1) [ (s + 2) / (s^2 - 3s + 4) ] = ℒ^(-1) [ (s + 2) / (s^2 - 3s + 9 / 2 - 9 / 2 + 4) ] =
ℒ^(-1) { (s + 2) / [ (s - 3 / 2)^2 + 7 / 4 ] } = ℒ^(-1) { (s + 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } =
ℒ^(-1) { (s + 2 - 3 / 2 + 3 / 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } =
ℒ^(-1) { (s - 3 / 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } + ℒ^(-1) { (7 / 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } =
ℒ^(-1) { (s - 3 / 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } + (7 / 2)·ℒ^(-1) { (√7 / 2)·(2 / √7) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } =
ℒ^(-1) { (s - 3 / 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } + (7 / 2)·(2 / √7)·ℒ^(-1) { (√7 / 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } =
ℒ^(-1) { (s - 3 / 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } + (√7)·ℒ^(-1) { (√7 / 2) / [ (s - 3 / 2)^2 + (√7 / 2)^2 ] } =
[ e^(3t / 2) ]·cos[ √7(t) / 2 ] + (√7)·[ e^(3t / 2) ]·sin[ √7(t) / 2 ] =
[ e^(3t / 2) ]·{ cos[ √7(t) / 2 ] + (√7)·sin[ √7(t) / 2 ] }

Check by WolframAlpha:
https://www.wolframalpha.com/input/?i=inverse laplace transform [ (s + 2) / (s^2 - 3s + 4) ]
(1st result given under the "Alternate forms" section)
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3. b)
Converting 1 / [ (s^2 + 1)(s^2 + 4s + 8) ] into partial rational fractions as (A, B, C, D ∈ R);
1 / [ (s^2 + 1)(s^2 + 4s + 8) ] = (A·S + B) / (s^2 + 1) + (C·s + D) / (s^2 + 4s + 8)
1 = (A + C)(s^3) + (4A + B + d)(s^2) + (8A + 4B + C)s + D
A + C = 0
4A + B + D = 0
8A + 4B + C = 0
D = 1
Solving the first three equations using D = 1, other coefficients are found as A = -4 / 9, B = 7 / 9, C = 4 / 9.
1 / [ (s^2 + 1)(s^2 + 4s + 8) ] = (-4s / 9 + 7 / 9) / (s^2 + 1) + (4s / 9 + 1) / (s^2 + 4s + 8) =
(-1 / 65)·[ (4s - 7) / (s^2 + 1) - (4s + 9) / (s^2 + 4s + 8) ]
Then, ℒ^(-1) { 1 / [ (s^2 + 1)(s^2 + 4s + 8) ] } =
(-1 / 65)·ℒ^(-1) { (4s - 7) / (s^2 + 1) } - [ (4s + 9) / (s^2 + 4s + 8) ] } =
(-1 / 65)·ℒ^(-1) [ 4s / (s^2 + 1) ] + (1 / 65)·ℒ^(-1) [ 7 / (s^2 + 1) ] - (4 / 65)·ℒ^(-1) { (s - 9 / 4) / (s^2 + 4s + 4 + 4) ] } =
(-4 / 65)·cos(t) + (7 / 65)·sin(t) - (4 / 65)·ℒ^(-1) (s + 2 - 2 + 9 / 4) / { [ s - (-2) ]^2 + 4 } =
(-4 / 65)·cos(t) + (7 / 65)·sin(t) - (4 / 65)·ℒ^(-1) [ s - (-2) ] / { [ s - (-2) ]^2 + 4 } - (4 / 65)·ℒ^(-1) (1 / 4) / { [ s - (-2) ]^2 + 4 } =
(-4 / 65)·cos(t) + (7 / 65)·sin(t) - (4 / 65)·[ e^(-2t) ]·cos(2t) - (4 / 65)·ℒ^(-1) (1 / 4)·2·(1 / 2) / { [ s - (-2) ]^2 + 4 } =
(-4 / 65)·cos(t) + (7 / 65)·sin(t) - (4 / 65)·[ e^(-2t) ]·cos(2t) - (1 / 65)·(1 / 2)·ℒ^(-1) 2 / { [ s - (-2) ]^2 + 4 } =
(-4 / 65)·cos(t) + (7 / 65)·sin(t) - (4 / 65)·[ e^(-2t) ]·cos(2t) - (1 / 130)·[ e^(-2t) ]·sin(2t) =
(1 / 130)·[ e^(-2t) ]·{ 14·[ e^(2t) ]·sin(t) + sin(2t) - 8·[ e^(2t) ]·cos(t) + 8·cos(2t) }

Check by WolframAlpha:
https://www.wolframalpha.com/input/?i=inverse laplace transform (1 / ((s^2+1)(s^2+4s+8)))
(Last result given under the "Alternate form assuming t is real:" section)
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3. c
(2s - 13) / [ s(s^2 - 4s + 13) ] ≡ A / s + (B·s + C) / (s^2 - 4s + 13)
2s - 13 = A(s^2 - 4s + 13) + s·(B·s + C) and doing as shown in the solution 3.b;
(2s - 13) / [ s(s^2 - 4s + 13) ] = -1 / s + (s - 2) / (s^2 - 4s + 13) =
-1 / s + (s - 2) / (s^2 - 4s + 4 + 9) =
-1 / s + (s - 2) / [ (s - 2)^2 + 9 ] and taking inverse Laplace transform;
ℒ^(-1) { (2s - 13) / [ s(s^2 - 4s + 13) ] } =
ℒ^(-1) (-1 / s) + ℒ^(-1) { (s - 2) / [ (s - 2)^2 + 9 ] } =
-1 + [ e^(2t) ]·cos(3t) =
[ e^(2t) ]·cos(3t) - 1

Check by WolframAlpha:
https://www.wolframalpha.com/input/?i=inverse laplace transform ( (2s - 13) / [ s(s^2 - 4s + 13) ] ) =
(The 1st result under the "Alternate forms:" section)
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Soruların Yedeği: https://i.ibb.co/mbts8yp/inverse-laplace.jpg

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