The Laplace Transform - WordPress.com fileSolve 𝐿sin2 using Laplace Transform 𝐿sin2 = ......
Transcript of The Laplace Transform - WordPress.com fileSolve 𝐿sin2 using Laplace Transform 𝐿sin2 = ......
Let see some examples 𝐿 1
Solve 𝐿 1 using Laplace Transform
𝐿 1 = 𝑒−𝑠𝑡(1)𝑑𝑡∞
0
= 𝑒−𝑠𝑡𝑑𝑡∞
0
𝑒−𝑠𝑡𝑑𝑡∞
0
= −𝑒−𝑠𝑡
𝑠 ∞0
= −𝑒−𝑠 ∞
𝑠− −
𝑒0
𝑠
= 0 +1
𝑠=
1
𝑠
Let see some examples 𝐿 𝑡
Solve 𝐿 𝑡 using Laplace Transform
𝐿 𝑡 = 𝑒−𝑠𝑡(𝑡)𝑑𝑡∞
0
= 𝑡𝑒−𝑠𝑡𝑑𝑡∞
0
𝑡𝑒−𝑠𝑡𝑑𝑡∞
0
=𝑡𝑒−𝑠𝑡
−𝑠−
𝑒−𝑠𝑡
𝑠2 ∞0
=(∞)𝑒−𝑠(∞)
−𝑠−
𝑒−𝑠(∞)
𝑠2
−0 𝑒−𝑠 0
−𝑠−
𝑒−𝑠 0
𝑠2=
1
𝑠2
DIff Integr
+ 𝑡 𝑒−𝑠𝑡
- 1 𝑒−𝑠𝑡
−𝑠
+
0 𝑒−𝑠𝑡
𝑠2
Let see some examples 𝐿 𝑒−5𝑡
Solve 𝐿 𝑒−5𝑡 using Laplace Transform
𝐿 𝑒−5𝑡 = 𝑒−𝑠𝑡(𝑒−5𝑡)𝑑𝑡∞
0
= 𝑒−(𝑠+5)𝑡𝑑𝑡∞
0
𝑒−(𝑠+5)𝑡𝑑𝑡∞
0
=𝑒−(𝑠+5)𝑡
−(𝑠 + 5) ∞0
=𝑒−(𝑠+5)∞
−(𝑠 + 5)−
𝑒− 𝑠+5 0
−(𝑠 + 5)=
1
𝑠 + 5
Let see some examples 𝐿 sin 2𝑡
Solve 𝐿 sin 2𝑡 using Laplace Transform
𝐿 sin 2𝑡 = 𝑒−𝑠𝑡(sin 2𝑡)𝑑𝑡∞
0
= (sin 2𝑡)𝑒−𝑠𝑡𝑑𝑡∞
0
(sin 2𝑡)𝑒−𝑠𝑡𝑑𝑡∞
0
= − sin 2𝑡𝑒−𝑠𝑡
𝑠− 2 cos 2𝑡
𝑒−𝑠𝑡
𝑠2
− 4sin 4𝑡 𝑒−𝑠𝑡
𝑠2 𝑑𝑡∞
0
1 +4
𝑠2 (sin 2𝑡)𝑒−𝑠𝑡𝑑𝑡∞
0
= − sin 2𝑡𝑒−𝑠𝑡
𝑠− 2 cos 2𝑡
𝑒−𝑠𝑡
𝑠2
∞0
DIff Integr
+ sin 2𝑡 𝑒−𝑠𝑡
- 2 cos 2𝑡 𝑒−𝑠𝑡
−𝑠
+
−4 sin 2𝑡 𝑒−𝑠𝑡
𝑠2
Let see some examples 𝐿 sin 2𝑡 cont
1 +4
𝑠2 (sin 2𝑡)𝑒−𝑠𝑡𝑑𝑡
∞
0
= −sin 2𝑡𝑒−𝑠𝑡
𝑠− 2 cos 2𝑡
𝑒−𝑠𝑡
𝑠2
∞0
0 − 0 − 2 =2
s2
Thus
1 +4
𝑠2 (sin 2𝑡)𝑒−𝑠𝑡𝑑𝑡
∞
0
=2
𝑠2
And
(sin 2𝑡)𝑒−𝑠𝑡𝑑𝑡∞
0
=2
𝑠2 + 4
Transform of Derivatives
Contoh:- 𝑦′′ + 2𝑦′ + 𝑦 = 0
Then 𝐿*𝑦′′+ = 𝑠2𝑌 𝑠 − 𝑠𝑦 0 − 𝑦′ 0
And 𝐿*𝑦′+ = 𝑠𝑌 𝑠 − 𝑦 0
And 𝐿*𝑦+ = 𝑌(𝑠)
Thus ,𝑠2𝑌 𝑠 − 𝑠𝑦 0 − 𝑦′ 0 - + 2 𝑠𝑌 𝑠 − 𝑦 0 + 𝑌 𝑠 = 0
Inverse Laplace transform
Solve 𝐿−1 2
𝑠−
1
𝑠3
2
First, try to expand it
𝐿−14
𝑠2−
4
𝑠4+
1
𝑠6
= 𝐿−14
𝑠2− 𝐿−1
4
𝑠4+ 𝐿−1
1
𝑠6
Inverse Laplace transform
Step 2nd
𝐿−14
𝑠2− 𝐿−1
4
𝑠4+ 𝐿−1
1
𝑠6
= 4𝐿−11
𝑠2− 4𝐿−1
1
𝑠4+ 𝐿−1
1
𝑠6
Now, let solve one by one
Inverse Laplace transform
Given 4𝐿−1 1
𝑠2 , we need to find which theorem
in your laplace table match this Laplace Transform
So, we know that 𝑡𝑛 =𝑛!
𝑠𝑛+1
Then since we knew that 𝑠𝑛+1 = 𝑠2
We can conclude that 𝑛 + 1 = 2, 𝑡𝑢𝑠 𝑛 = 1
Inverse Laplace transform
Now, we know n=1
So,
𝑡1 =1!
𝑠2=
1
𝑠2
Eh!, the laplace that we are trying to solve is 4
𝑠2 , so we need to modify a bit.
Inverse Laplace transform
Try to solve others:-
4𝐿−11
𝑠4
𝑛 + 1 = 4, 𝑡𝑢𝑠 𝑛 = 3
𝛽𝑡3 =𝛽3!
𝑠4=
6𝛽
𝑠4
4𝐿−11
𝑠4=
6𝛽
𝑠4
𝛽 =2
3
Thus
4𝐿−11
𝑠4=
2
3𝑡3
Inverse Laplace transform
Try to solve others:-
𝐿−11
𝑠6
𝑛 + 1 = 6, 𝑡𝑢𝑠 𝑛 = 5
𝛾𝑡5 =𝛾5!
𝑠6=
120𝛾
𝑠6
𝐿−11
𝑠6=
120𝛾
𝑠6
𝛾 =1
120
Thus
4𝐿−11
𝑠4=
1
120𝑡5
Application
Solve the system Figure 1 under the conditions E(t) = 60 V, L = 1 h, R = 50 Ohm, C = 10−4 f, and the currents 𝑖1 𝑎𝑛𝑑 𝑖2 are initially zero.
Given that:-
𝐿𝑑𝑖1𝑑𝑡
+ 𝑅𝑖2 = 𝐸 𝑡
𝑅𝐶𝑑𝑖2𝑑𝑡
+ 𝑖2 − 𝑖1 = 0
Application
How to solve it?
1st step
𝐿𝑑𝑖1𝑑𝑡
+ 𝑅𝑖2 = 𝐸 𝑡
𝑑𝑖1𝑑𝑡
+ 50𝑖2 = 60
And
𝑅𝐶𝑑𝑖2𝑑𝑡
+ 𝑖2 − 𝑖1 = 0
50 10−4𝑑𝑖2𝑑𝑡
+ 𝑖2 − 𝑖1 = 0
Application
How to solve it? 2nd Step Applying the Laplace transform to each equation of the system and simplifying gives
𝑑𝑖1𝑑𝑡
+ 50𝑖2 = 60
=> ,𝑠𝐼1 𝑠 − 𝑖1(0)- + 50𝐼2 𝑠 =60
𝑠
50 10−4𝑑𝑖2𝑑𝑡
+ 𝑖2 − 𝑖1 = 0
0.005,𝑠𝐼2 𝑠 − 𝑖2(0)- + 𝐼2 𝑠 − 𝐼1 𝑠 = 0
Application
How to solve it? 2nd Step (Cari 𝐼2 𝑠 )
𝑠𝐼1 𝑠 + 50𝐼2 𝑠 =60
𝑠
0.005𝑠𝐼2 𝑠 + 𝐼2 𝑠 = 𝐼1 𝑠
𝑠 0.005𝑠𝐼2 𝑠 + 𝐼2 𝑠 + 50𝐼2 𝑠 =60
𝑠
𝐼2 𝑠𝑠2 + 200𝑠 + 10000
200=
60
𝑠
𝐼2 𝑠 =12000
𝑠 𝑠 + 100 2
Application
How to solve it?
2nd Step (Cari 𝐼1 𝑠 ) 0.005𝑠𝐼2 𝑠 + 𝐼2 𝑠 = 𝐼1 𝑠
0.005𝑠12000
𝑠 𝑠 + 100 2+
12000
𝑠 𝑠 + 100 2= 𝐼1 𝑠
𝐼1 𝑠 =60𝑠
𝑠 𝑠 + 100 2+
12000
𝑠 𝑠 + 100 2
𝐼1 𝑠 =60𝑠 + 12000
𝑠 𝑠 + 100 2
Application
How to solve it?
3rd Step
Solving the system for 𝐼1 and 𝐼2 and decomposing the results into partial fractions gives
𝐼1 𝑠 =60𝑠 + 12000
𝑠 𝑠+100 2 =6
5𝑠−
6
5 𝑠+100−
60
𝑠+100 2
𝐼2 𝑠 =12000
𝑠 𝑠 + 100 2=
6
5𝑠−
6
5 𝑠 + 100−
120
𝑠 + 100 2