the laplace transform - wordpress.com filesolve ๐ฟsin2 using laplace transform ๐ฟsin2 = ......
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Definition
Let ๐ be a function defined for ๐ก โฅ 0. Then the integral
๐ฟ ๐ ๐ก = ๐โ๐ ๐ก๐ ๐ก ๐๐กโ
0
Let see some examples ๐ฟ 1
Solve ๐ฟ 1 using Laplace Transform
๐ฟ 1 = ๐โ๐ ๐ก(1)๐๐กโ
0
= ๐โ๐ ๐ก๐๐กโ
0
๐โ๐ ๐ก๐๐กโ
0
= โ๐โ๐ ๐ก
๐ โ0
= โ๐โ๐ โ
๐ โ โ
๐0
๐
= 0 +1
๐ =
1
๐
Let see some examples ๐ฟ ๐ก
Solve ๐ฟ ๐ก using Laplace Transform
๐ฟ ๐ก = ๐โ๐ ๐ก(๐ก)๐๐กโ
0
= ๐ก๐โ๐ ๐ก๐๐กโ
0
๐ก๐โ๐ ๐ก๐๐กโ
0
=๐ก๐โ๐ ๐ก
โ๐ โ
๐โ๐ ๐ก
๐ 2 โ0
=(โ)๐โ๐ (โ)
โ๐ โ
๐โ๐ (โ)
๐ 2
โ0 ๐โ๐ 0
โ๐ โ
๐โ๐ 0
๐ 2=
1
๐ 2
DIff Integr
+ ๐ก ๐โ๐ ๐ก
- 1 ๐โ๐ ๐ก
โ๐
+
0 ๐โ๐ ๐ก
๐ 2
Let see some examples ๐ฟ ๐โ5๐ก
Solve ๐ฟ ๐โ5๐ก using Laplace Transform
๐ฟ ๐โ5๐ก = ๐โ๐ ๐ก(๐โ5๐ก)๐๐กโ
0
= ๐โ(๐ +5)๐ก๐๐กโ
0
๐โ(๐ +5)๐ก๐๐กโ
0
=๐โ(๐ +5)๐ก
โ(๐ + 5) โ0
=๐โ(๐ +5)โ
โ(๐ + 5)โ
๐โ ๐ +5 0
โ(๐ + 5)=
1
๐ + 5
Let see some examples ๐ฟ sin 2๐ก
Solve ๐ฟ sin 2๐ก using Laplace Transform
๐ฟ sin 2๐ก = ๐โ๐ ๐ก(sin 2๐ก)๐๐กโ
0
= (sin 2๐ก)๐โ๐ ๐ก๐๐กโ
0
(sin 2๐ก)๐โ๐ ๐ก๐๐กโ
0
= โ sin 2๐ก๐โ๐ ๐ก
๐ โ 2 cos 2๐ก
๐โ๐ ๐ก
๐ 2
โ 4sin 4๐ก ๐โ๐ ๐ก
๐ 2 ๐๐กโ
0
1 +4
๐ 2 (sin 2๐ก)๐โ๐ ๐ก๐๐กโ
0
= โ sin 2๐ก๐โ๐ ๐ก
๐ โ 2 cos 2๐ก
๐โ๐ ๐ก
๐ 2
โ0
DIff Integr
+ sin 2๐ก ๐โ๐ ๐ก
- 2 cos 2๐ก ๐โ๐ ๐ก
โ๐
+
โ4 sin 2๐ก ๐โ๐ ๐ก
๐ 2
Let see some examples ๐ฟ sin 2๐ก cont
1 +4
๐ 2 (sin 2๐ก)๐โ๐ ๐ก๐๐ก
โ
0
= โsin 2๐ก๐โ๐ ๐ก
๐ โ 2 cos 2๐ก
๐โ๐ ๐ก
๐ 2
โ0
0 โ 0 โ 2 =2
s2
Thus
1 +4
๐ 2 (sin 2๐ก)๐โ๐ ๐ก๐๐ก
โ
0
=2
๐ 2
And
(sin 2๐ก)๐โ๐ ๐ก๐๐กโ
0
=2
๐ 2 + 4
Transform of Derivatives
Contoh:- ๐ฆโฒโฒ + 2๐ฆโฒ + ๐ฆ = 0
Then ๐ฟ*๐ฆโฒโฒ+ = ๐ 2๐ ๐ โ ๐ ๐ฆ 0 โ ๐ฆโฒ 0
And ๐ฟ*๐ฆโฒ+ = ๐ ๐ ๐ โ ๐ฆ 0
And ๐ฟ*๐ฆ+ = ๐(๐ )
Thus ,๐ 2๐ ๐ โ ๐ ๐ฆ 0 โ ๐ฆโฒ 0 - + 2 ๐ ๐ ๐ โ ๐ฆ 0 + ๐ ๐ = 0
Inverse Laplace transform
Solve ๐ฟโ1 2
๐ โ
1
๐ 3
2
First, try to expand it
๐ฟโ14
๐ 2โ
4
๐ 4+
1
๐ 6
= ๐ฟโ14
๐ 2โ ๐ฟโ1
4
๐ 4+ ๐ฟโ1
1
๐ 6
Inverse Laplace transform
Step 2nd
๐ฟโ14
๐ 2โ ๐ฟโ1
4
๐ 4+ ๐ฟโ1
1
๐ 6
= 4๐ฟโ11
๐ 2โ 4๐ฟโ1
1
๐ 4+ ๐ฟโ1
1
๐ 6
Now, let solve one by one
Inverse Laplace transform
Given 4๐ฟโ1 1
๐ 2 , we need to find which theorem
in your laplace table match this Laplace Transform
So, we know that ๐ก๐ =๐!
๐ ๐+1
Then since we knew that ๐ ๐+1 = ๐ 2
We can conclude that ๐ + 1 = 2, ๐ก๐๐ข๐ ๐ = 1
Inverse Laplace transform
Now, we know n=1
So,
๐ก1 =1!
๐ 2=
1
๐ 2
Eh!, the laplace that we are trying to solve is 4
๐ 2 , so we need to modify a bit.
Inverse Laplace transform
Try to match:-
๐ฟ ๐ผ๐ก =๐ผ
๐ 2=
4
๐ 2
Wow, obviously, ๐ผ = 4,
So
4๐ฟโ11
๐ 2= 4๐ก
Inverse Laplace transform
Try to solve others:-
4๐ฟโ11
๐ 4
๐ + 1 = 4, ๐ก๐๐ข๐ ๐ = 3
๐ฝ๐ก3 =๐ฝ3!
๐ 4=
6๐ฝ
๐ 4
4๐ฟโ11
๐ 4=
6๐ฝ
๐ 4
๐ฝ =2
3
Thus
4๐ฟโ11
๐ 4=
2
3๐ก3
Inverse Laplace transform
Try to solve others:-
๐ฟโ11
๐ 6
๐ + 1 = 6, ๐ก๐๐ข๐ ๐ = 5
๐พ๐ก5 =๐พ5!
๐ 6=
120๐พ
๐ 6
๐ฟโ11
๐ 6=
120๐พ
๐ 6
๐พ =1
120
Thus
4๐ฟโ11
๐ 4=
1
120๐ก5
Application
Solve the system Figure 1 under the conditions E(t) = 60 V, L = 1 h, R = 50 Ohm, C = 10โ4 f, and the currents ๐1 ๐๐๐ ๐2 are initially zero.
Given that:-
๐ฟ๐๐1๐๐ก
+ ๐ ๐2 = ๐ธ ๐ก
๐ ๐ถ๐๐2๐๐ก
+ ๐2 โ ๐1 = 0
Application
How to solve it?
1st step
๐ฟ๐๐1๐๐ก
+ ๐ ๐2 = ๐ธ ๐ก
๐๐1๐๐ก
+ 50๐2 = 60
And
๐ ๐ถ๐๐2๐๐ก
+ ๐2 โ ๐1 = 0
50 10โ4๐๐2๐๐ก
+ ๐2 โ ๐1 = 0
Application
How to solve it? 2nd Step Applying the Laplace transform to each equation of the system and simplifying gives
๐๐1๐๐ก
+ 50๐2 = 60
=> ,๐ ๐ผ1 ๐ โ ๐1(0)- + 50๐ผ2 ๐ =60
๐
50 10โ4๐๐2๐๐ก
+ ๐2 โ ๐1 = 0
0.005,๐ ๐ผ2 ๐ โ ๐2(0)- + ๐ผ2 ๐ โ ๐ผ1 ๐ = 0
Application
How to solve it? 2nd Step (Cari ๐ผ2 ๐ )
๐ ๐ผ1 ๐ + 50๐ผ2 ๐ =60
๐
0.005๐ ๐ผ2 ๐ + ๐ผ2 ๐ = ๐ผ1 ๐
๐ 0.005๐ ๐ผ2 ๐ + ๐ผ2 ๐ + 50๐ผ2 ๐ =60
๐
๐ผ2 ๐ ๐ 2 + 200๐ + 10000
200=
60
๐
๐ผ2 ๐ =12000
๐ ๐ + 100 2
Application
How to solve it?
2nd Step (Cari ๐ผ1 ๐ ) 0.005๐ ๐ผ2 ๐ + ๐ผ2 ๐ = ๐ผ1 ๐
0.005๐ 12000
๐ ๐ + 100 2+
12000
๐ ๐ + 100 2= ๐ผ1 ๐
๐ผ1 ๐ =60๐
๐ ๐ + 100 2+
12000
๐ ๐ + 100 2
๐ผ1 ๐ =60๐ + 12000
๐ ๐ + 100 2
Application
How to solve it?
3rd Step
Solving the system for ๐ผ1 and ๐ผ2 and decomposing the results into partial fractions gives
๐ผ1 ๐ =60๐ + 12000
๐ ๐ +100 2 =6
5๐ โ
6
5 ๐ +100โ
60
๐ +100 2
๐ผ2 ๐ =12000
๐ ๐ + 100 2=
6
5๐ โ
6
5 ๐ + 100โ
120
๐ + 100 2
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