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Example: Find 𝑳 {𝒄𝒐𝒔 𝒂𝒕−𝒄𝒐𝒔 𝒃𝒕𝒕 }
Solution: 𝐿{𝑐𝑜𝑠 𝑎𝑡 − 𝑐𝑜𝑠 𝑏𝑡} = 𝐿{𝑐𝑜𝑠 𝑎𝑡} − 𝐿{𝑐𝑜𝑠 𝑏𝑡} = (^) 𝑠 (^2) +𝑎𝑠 2 − (^) 𝑠 (^2) +𝑏𝑠 2
𝑡 } = ∫^ [^
𝑠^2 + 𝑎^2 −^
𝑠^2 + 𝑏^2 ] 𝑑𝑠
∞ 𝑠 = [^12 𝑙𝑜𝑔 (𝑠^2 + 𝑎^2 ) − 12 𝑙𝑜𝑔 (𝑠^2 + 𝑏^2 )]𝑠
∞
= [^12 𝑙𝑜𝑔 [𝑠
(^2) +𝑎 2 𝑠^2 +𝑏^2 ]] 𝑠
∞ = 12 [𝑙𝑜𝑔 [1+𝑎
(^2) /𝑠 2 1+𝑏^2 /𝑠^2 ]] 𝑠
∞
= 12 [𝑙𝑜𝑔 1 − 𝑙𝑜𝑔 [
1+𝑎𝑠 22 1+𝑏𝑠^22
]]
2 𝑙𝑜𝑔 [
2 𝑠^2 1 + 𝑏
2 𝑠^2
] =
2 𝑙𝑜𝑔 [
𝑠^2 + 𝑏^2
𝑠^2 + 𝑎^2 ]
𝑡 } = ∫^ 𝑒
∞ 0
2 𝑙𝑜𝑔 [
𝑠^2 + 𝑏^2
𝑠^2 + 𝑎^2 ]
∴ ∫ 𝑒−𝑡^ {
∞ 0
2 𝑙𝑜𝑔 [
1 + 𝑏^2
1 + 𝑎^2 ]
Example: Find 𝐿 {∫ 𝑒 0 𝑡 −𝑡^ 𝑠𝑖𝑛 𝑡𝑡 𝑑𝑡}
Solution : We know that
𝐿{𝑠𝑖𝑛 𝑡} = (^) 𝑠 (^21) +1 = 𝑓(𝑠).
Now by division rule i.e. if 𝐿{𝑓(𝑡)} = 𝑓(𝑠), then 𝐿 {𝑓(𝑡)𝑡 } = ∫𝑠 ∞𝑓(𝑠)𝑑𝑠
∴ 𝐿 {
𝑡 } = ∫^ 𝑓(𝑠)𝑑𝑠
∞ 𝑠
𝑠^2 + 1 𝑑𝑠
∞ 𝑠
= [𝑡𝑎𝑛−1^ 𝑠]𝑠∞
∴ 𝐿 {𝑒−𝑡^ 𝑠𝑖𝑛 𝑡𝑡 } = 𝑐𝑜𝑡−1(𝑠 + 1) by first shifting property 𝐿{𝑒𝑎𝑡𝑓(𝑡)} = 𝑓(𝑠 − 𝑎)
∴ 𝐿 {∫ 𝑒 0 𝑡 −𝑡^ 𝑠𝑖𝑛 𝑡𝑡 𝑑𝑡} = (^1) 𝑠 𝑐𝑜𝑡−1(𝑠 + 1) by integral rule 𝐿 {∫ 𝑓(𝑡) 0 𝑡 𝑑𝑡} = (^1) 𝑠 𝑓(𝑠)
Example: Find 𝐿{(𝒆𝒂𝒕^ − 𝒄𝒐𝒔 𝒃𝒕)/𝒕}
Solution : We know that
𝐿{𝑓(𝑡)} = 𝐿{𝑒𝑎𝑡^ − 𝑐𝑜𝑠 𝑏𝑡} = 𝐿{𝑒𝑎𝑡^ } − 𝐿{𝑐𝑜𝑠 𝑏𝑡} =
𝑠 − 𝑎 −^
𝑠^2 + 𝑏^2 = 𝐹(𝑠)
Now, 𝑓(𝑡) = 2𝑡^ = 𝑒𝑙𝑜𝑔2𝑡 = 𝑒𝑡.𝑙𝑜𝑔2^ = 𝑒𝑙𝑜𝑔2.𝑡
∴ 𝐿{𝑓(𝑡)} = 𝐿{𝑒𝑙𝑜𝑔2.𝑡} =
Now, 𝑔(𝑡) = 𝑐𝑜𝑠 2𝑡−𝑐𝑜𝑠 3𝑡𝑡
𝐿{𝑐𝑜𝑠 2𝑡 − 𝑐𝑜𝑠 3𝑡} =
𝑠^2 + 4 −^
𝑠^2 + 9
𝑡 } = ∫^ [^
𝑠^2 + 4 −^
𝑠^2 + 9] 𝑑𝑠
∞ 𝑠
= [
2 [𝑙𝑜𝑔 (𝑠
2 + 4) − 𝑙𝑜𝑔(𝑠 2 + 9)]]
𝑠
∞
𝑠^2 + 9
𝑠^2 + 4)
Now, ℎ(𝑡) = 𝑡 𝑠𝑖𝑛 𝑡
𝐿{𝑠𝑖𝑛 𝑡} = (^) 𝑠 (^21) +1,
Then 𝐿{ 𝑡 𝑠𝑖𝑛 𝑡} = − (^) 𝑑𝑠𝑑 ( (^) 𝑠 (^21) +1) = (^) (𝑠 2 2𝑠+1) 2
∴ 𝐿 {2𝑡^ +
𝑡 + 𝑡 𝑠𝑖𝑛 𝑡} =^
𝑠^2 + 9
𝑠^2 + 4) +^
(𝑠^2 + 1)^2
Example. Find inverse Laplace Transform of 𝑓(𝑠) = 𝑠
2 (𝑠−2)^3.
Solution. Clearly,
𝑠^2 (𝑠 − 2)^3 =^
(𝑠 − 2) +^
(𝑠 − 2)^2 +^
(𝑠 − 2)^3
𝑠^2
(𝑠 − 2)^3 =
𝐴(𝑠 − 2)^2 + 𝐵(𝑠 − 2) + 𝐶
(𝑠 − 2)^3
⇒ 𝑠^2 = 𝐴(𝑠 − 2)^2 + 𝐵(𝑠 − 2) + 𝐶
⇒ 𝑠^2 = 𝐴(𝑠^2 − 4𝑠 + 4) + 𝐵(𝑠 − 2) + 𝐶
On comparing the coefficients of different powers of 𝑠, we get
⇒ 𝐴 = 1 −4𝐴 + 𝐵 = 0 ⇒ 𝐵 = 4 4𝐴 − 2𝐵 + 𝐶 = 0 ⇒ 𝐶 = 4
⇒
𝑠^2
(𝑠 − 2)^3 =^
(𝑠 − 2) +^
(𝑠 − 2)^2 +^
(𝑠 − 2)^3
Now, we know that
𝐿{sin 𝑎𝑡} =
𝑠^2 + 𝑎^2
⇒ 𝐿−1^ {
𝑠^2 + 𝑎^2 } =
𝑎 sin 𝑎𝑡
⇒ 𝐿−1^ {
𝑑𝑎 (^
𝑠^2 + 𝑎^2 )} =
𝑎 sin 𝑎𝑡)
⇒ 𝐿−1^ {
−1 𝑑𝑎 𝑑 (𝑠^2 + 𝑎^2 )
(𝑠^2 + 𝑎^2 )^2 } = (
𝑎. 𝑡 cos 𝑎𝑡 − 1. sin 𝑎𝑡 𝑎^2 )
⇒ 𝐿−1^ {
(𝑠^2 + 𝑎^2 )^2 } = (
𝑎. 𝑡 cos 𝑎𝑡 − 1. sin 𝑎𝑡 𝑎^2 )
⇒ 𝑳−𝟏^ {
(𝒔𝟐^ + 𝒂𝟐)𝟐} = (
𝟐𝒂𝟑^ )
∴ 𝐿−1^ {
𝑠^2
(𝑠^2 + 4)^2 } = 𝐿
𝑠^2 + 4} − 𝐿
(𝑠^2 + 4)^2 }
⇒ 𝐿−1^ {
𝑠^2
(𝑠^2 + 4)^2 } = 𝐿
𝑠^2 + 4} − 4𝐿
(𝑠^2 + 4)^2 }
⇒ 𝐿−1^ {
𝑠^2
(𝑠^2 + 4)^2 } =
4 sin 4𝑡 − 4 (
sin 4𝑡 − 4𝑡 cos 4𝑡
- 4^3 )
⇒ 𝐿−1^ {
𝑠^2
(𝑠^2 + 4)^2 } =
4 sin 4𝑡 − (
sin 4𝑡 − 4𝑡 cos 4𝑡 32 )
⇒ 𝐿−1^ {
𝑠^2
(𝑠^2 + 4)^2 } =
32 sin 4𝑡 +
8 𝑡 cos 4𝑡
Example. Find inverse Laplace Transform of 𝑓(𝑠) = (^) (𝒔−𝟏)(𝒔𝟓𝒔+𝟑𝟐+𝟐𝒔+𝟓).
Solution. Clearly,
5𝑠 + 3 (𝑠 − 1)(𝑠^2 + 2𝑠 + 5) =^
𝑠 − 1 +^
𝑠^2 + 2𝑠 + 5
5𝑠 + 3 = (𝑠^2 + 2𝑠 + 5)𝐴 + (𝑠 − 1)( 𝐵𝑠 + 𝐶)
On comparing the coefficients of different powers of 𝑠, we get
𝐴 + 𝐵 = 0
2𝐴 − 𝐵 + 𝐶 = 5
⇒ 3𝐴 + 𝐶 = 5
5𝐴 − 𝐶 = 3
⇒ 𝐿−1^ {
(𝑠 − 1)(𝑠^2 + 2𝑠 + 5)} = 𝑒
𝑡 − 𝑒−𝑡𝑐𝑜𝑠 2𝑡 +^3
Example. Find inverse Laplace Transform of 𝑓(𝑠) = (^) 𝒔𝟒+𝟒𝒂𝒔 𝟒.
Solution. Clearly, 𝑠 𝑠^4 + 4𝑎^4 =^
(𝑠^2 + 2𝑎^2 )^2 − 4𝑎^2 𝑠^2
⇒ (𝑠^2 + 2𝑎^2 )^2 − (2𝑎𝑠)^2 = (𝑠^2 + 2𝑎^2 + 2𝑎𝑠)(𝑠^2 + 2𝑎^2 − 2𝑎𝑠)
𝑠^4 + 4𝑎^4 =^
(𝑠^2 + 2𝑎^2 + 2𝑎𝑠)(𝑠^2 + 2𝑎^2 − 2𝑎𝑠)
(𝑠^2 + 2𝑎^2 + 2𝑎𝑠)(𝑠^2 + 2𝑎^2 − 2𝑎𝑠) =^
(𝑠^2 + 2𝑎^2 + 2𝑎𝑠) +^
(𝑠^2 + 2𝑎^2 − 2𝑎𝑠)
⇒ (^) (𝑠 (^2) + 2𝑎 (^2) + 2𝑎𝑠)(𝑠𝑠 (^2) + 2𝑎 (^2) − 2𝑎𝑠) =
(𝑠^2 + 2𝑎^2 − 2𝑎𝑠)(𝐴𝑠 + 𝐵) + (𝑠^2 + 2𝑎^2 + 2𝑎𝑠)(𝐶𝑠 + 𝐷) (𝑠^2 + 2𝑎^2 + 2𝑎𝑠)(𝑠^2 + 2𝑎^2 − 2𝑎𝑠)
⇒ 𝑠 = (𝑠^2 + 2𝑎^2 − 2𝑎𝑠)(𝐴𝑠 + 𝐵) + (𝑠^2 + 2𝑎^2 + 2𝑎𝑠)(𝐶𝑠 + 𝐷)
On comparing the coefficients of different powers of 𝑠, we get
(1) (^) 0 = 𝐴 + 𝐶, (2) 0 = −2𝑎𝐴 + 𝐵 + 2𝑎𝐶 + 𝐷
(3) (^) 0 = 2𝑎^2 𝐴 − 2𝑎𝐵 + 2𝑎^2 𝐶 + 2𝑎𝐷
(4) (^) 1 = 2𝑎^2 𝐵 + 2𝑎^2 𝐷 ⇒ 𝐵 + 𝐷 = 1 2𝑎^2
From (2) and (4), we get
(5) (^) 0 = −𝐴 + 𝐶
Again from (1) and (5), we get
𝐴 = 𝐶 = 0
Now, from (3), we get
0 = −𝐵 + 𝐷 ⇒ 𝐵 = 𝐷
Thus, from (4), we get
2𝐵 = (^) 2𝑎^12 ⇒ 𝐵 = (^) 4𝑎^12 and so 𝐷 = (^) 4𝑎^12.
(𝑠^2 + 2𝑎^2 + 2𝑎𝑠)(𝑠^2 + 2𝑎^2 − 2𝑎𝑠) =^
4𝑎^2
(𝑠^2 + 2𝑎^2 + 2𝑎𝑠) +^
4𝑎^2
(𝑠^2 + 2𝑎^2 − 2𝑎𝑠)
Example. Find the inverse Laplace transform (^) 𝑠 (^3) −𝑎^13.
Now , 𝑠^3 − 𝑎^3 = (𝑠 − 𝑎)(𝑠^2 + 𝑎𝑠 + 𝑎^2 )
Given, 𝑠 3 −𝑎^13 = (𝑠−𝑎)(𝑠 21 +𝑎𝑠+𝑎 2 ) = (𝑠−𝑎)𝐴 + (𝑠 2 +𝑎𝑠+𝑎𝐵𝑠+𝐶 2 )
(𝑠 − 𝑎)(𝑠^2 + 𝑎𝑠 + 𝑎^2 ) =
𝐴(𝑠^2 + 𝑎𝑠 + 𝑎^2 ) + (𝐵𝑠 + 𝐶)(𝑠 − 𝑎)
(𝑠 − 𝑎)(𝑠^2 + 𝑎𝑠 + 𝑎^2
⇒ 1 = 𝐴(𝑠^2 + 𝑎𝑠 + 𝑎^2 ) + (𝐵𝑠 + 𝐶)(𝑠 − 𝑎)
Comparing the different powers of ‘s’.
𝐴 + 𝐵 = 0 (1)
𝑎𝐴 − 𝑎𝐵 + 𝐶 = 0 (2)
𝑎^2 𝐴 − 𝑎𝐶 = 1 (3)
From (1), 𝐵 = −𝐴
Again from (2), 2𝑎𝐴 + 𝐶 = 0 ⇒ 𝐶 = −2𝑎𝐴
Now from (3), 𝑎^2 𝐴 + 2𝑎^2 𝐴 = 1 ⇒ 𝐴 = (^) 3𝑎^12
∴ 𝐵 = − (^) 3𝑎^12 and 𝐶 = − (^) 3𝑎^2
Thus,
𝑠^3 − 𝑎^3 =^
3𝑎^2
− 3𝑎^12 𝑠 − 3𝑎^2
(𝑠^2 + 𝑎𝑠 + 𝑎^2 )
𝑠^3 − 𝑎^3 =^
3𝑎^2
(𝑠 − 𝑎) −^
3𝑎^2
(𝑠^2 + 𝑎𝑠 + 𝑎^2 )
∴ 𝐿−1^ {
𝑠^3 − 𝑎^3 } =^
3𝑎^2 𝐿
(𝑠 − 𝑎)} −^
3𝑎^2 𝐿
(𝑠^2 + 𝑎𝑠 + 𝑎^2 )}
⇒ 𝐿−1^ {
𝑠^3 − 𝑎^3 } =^
3𝑎^2 𝑒
3𝑎^2 𝐿
(𝑠^2 + 𝑎𝑠 + 𝑎^2 )}
Now, 𝐿−1^ { (^) (𝑠 (^2) +𝑎𝑠+𝑎𝑠+2𝑎 (^2) )} = 𝐿−1^ {
𝑠+𝑎 2 +3𝑎 2 (𝑠+𝑎 2 )^2 +3𝑎 42
= 𝐿−1^ {
(𝑠 + 𝑎/2)^2 + (√3 𝑎/2)^2
(𝑠 + 𝑎/2)^2 + (√3 𝑎/2)^2
= 𝑒−𝑎𝑡/2𝐿−1^ {
𝑠^2 + (√3 𝑎/2)^2
𝑠^2 + (√3 𝑎/2)^2
2.^
𝑡 0
=
𝑡 0
We know that 𝟐 𝐜𝐨𝐬 𝑨 𝐬𝐢𝐧 𝑩 = 𝒔𝒊𝒏 (𝑨 + 𝑩) − 𝐬𝐢𝐧(𝑨 − 𝑩)
∴ 2 𝑐𝑜𝑠 𝒂𝒖 𝑠𝑖𝑛 𝒂(𝒕 − 𝒖) = 𝑠𝑖𝑛(𝑎𝑢 + 𝑎𝑡 − 𝑎𝑢) − sin(𝑎𝑢 − 𝑎𝑡 + 𝑎𝑢) = sin 𝑎𝑡 − sin(2𝑎𝑢 − 𝑎𝑡)
𝐿−1^ {
(𝑠^2 + 𝑎^2 ).^
(𝑠^2 + 𝑎^2 )} =
2𝑎 ∫ [sin 𝑎𝑡 − sin(2𝑎𝑢 − 𝑎𝑡)]𝑑𝑢
𝑡 0 = (^) 2𝑎^1 [∫ sin 𝑎𝑡 𝑑𝑢 − ∫ sin(2𝑎𝑢 − 𝑎𝑡)𝑑𝑢 0 𝑡 0 𝑡 ]
= (^) 2𝑎^1 [sin 𝑎𝑡 (𝑢) 0 𝑡^ − (^) 2𝑎^1 (cos(2𝑎𝑢 − 𝑎𝑡)) 0 𝑡^ ]
= (^) 2𝑎^1 [t sin 𝑎𝑡 − (^) 2𝑎^1 (cos 𝑎𝑡 − cos 𝑎𝑡)] = (^) 𝟐𝒂𝟏 𝐭 𝐬𝐢𝐧 𝒂𝒕
Verification:
Again, 𝐿{sin 𝑎𝑡} = (^) 𝑠 (^2) +𝑎𝑎 2 ⇒ 𝐿{𝑡 sin 𝑎𝑡} = − (^) 𝑑𝑠𝑑 { (^) 𝑠 (^2) +𝑎𝑎 2 } = (^) (𝑠 2 2𝑎𝑠+𝑎 (^2) ) 2
∴ 𝐿 {
2𝑎 t sin 𝑎𝑡} =
2𝑎 𝐿{t sin 𝑎𝑡} =
2𝑎.^
(𝑠^2 + 𝑎^2 )^2 =^
(𝑠^2 + 𝑎^2 )^2
Example. Apply Convolution theorem to evaluate 𝐿−1^ {(𝒔 (^) 𝟐+𝟐𝟓)(𝒔𝒔 𝟐+𝟏𝟔)}.
Solution. We have
𝐿−1^ {
(𝒔𝟐^ + 𝟐𝟓)(𝒔𝟐^ + 𝟏𝟔)} = 𝐿
(𝑠^2 + 5^2 ).^
(𝑠^2 + 4^2 )} = 𝐿
Thus we have
𝐿−1{𝑓(𝑠)} = 𝐿−1^ { (^) (𝑠 (^2) +5𝑠 (^2) )} = 𝑐𝑜𝑠 5𝑡 = 𝑓(𝑡) and 𝐿−1{𝑔(𝑠)} = 𝐿−1^ { (^) (𝑠 (^2) +4^12 )} = 14 𝑠𝑖𝑛 4𝑡 = 𝑔(𝑡)
∴ 𝑓(𝑢) = 𝑐𝑜𝑠 5𝑢 and 𝑔(𝑡 − 𝑢) = 14 sin 4(𝑡 − 𝑢)
From Convolution theorem,
𝐿−1{𝑓(𝑠). 𝑔(𝑠)} = 𝑓(𝑡) ∗ 𝑔(𝑡) = ∫ 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢
𝑡 0
⇒ 𝐿−1^ {
(𝑠^2 + 5^2 ).^
(𝑠^2 + 4^2 )} = 𝑓(𝑡) ∗ 𝑔(𝑡) = ∫ 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢
𝑡 0
= ∫ 𝑐𝑜𝑠 5𝑢
𝑡 0
=
𝑡 0
We know that 𝟐 𝐜𝐨𝐬 𝑨 𝐬𝐢𝐧 𝑩 = 𝒔𝒊𝒏 (𝑨 + 𝑩) − 𝐬𝐢𝐧(𝑨 − 𝑩)
Example. Apply Convolution theorem to evaluate 𝐿−1^ { (^) (𝒔𝟐+𝟒𝒔+𝟏𝟑)𝟏 𝟐}.
Solution. We have
(𝑠^2 + 4𝑠 + 13)^2 = [(𝑠 + 2)^2 + 3^2 ]^2
Now, 𝐿−1^ { (^) (𝑠 (^2) +4𝑠+13)^12 } = 𝐿−1^ { (^) [(𝑠+2)^12 +3 (^2) ] 2 } = 𝑒−2𝑡𝐿−1^ { (^) (𝑠 (^2) +3^12 ) 2 } = 𝑒−2𝑡𝐿−1^ {(𝑠 (^2) +3^12 ). (^) (𝑠 (^2) +3^12 )}
Clearly, 𝐿−1^ { (^) (𝑠 (^2) +3^12 )} = 13 𝑠𝑖𝑛 3𝑡 = 𝑓(𝑡) = 𝑔(𝑡)
∴ 𝑓(𝑢) = 13 𝑠𝑖𝑛 3𝑢 and 𝑔(𝑡 − 𝑢) = 13 𝑠𝑖𝑛 3(𝑡 − 𝑢)
From Convolution theorem,
𝐿−1^ { (^) (𝑠 (^2) +3^12 ). (^) (𝑠 (^2) +3^12 )} = 𝑓(𝑡) ∗ 𝑔(𝑡) = ∫ 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢 0 𝑡 =∫ 0 𝑡^13 𝑠𝑖𝑛 3𝑢 13 𝑠𝑖𝑛 3(𝑡 − 𝑢)𝑑𝑢
𝑡 0
𝑡 0
We know that 𝟐 𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝑩 = 𝒄𝒐𝒔 (𝑨 − 𝑩) − 𝐜𝐨𝐬(𝑨 + 𝑩)
∴ 2 𝑠𝑖𝑛 3𝑢 𝑠𝑖𝑛 3(𝑡 − 𝑢) = cos (3𝑢 − 3𝑡 + 3𝑢) − cos(3𝑢 + 3𝑡 − 3𝑢) = cos(6𝑢 − 3𝑡) − cos 3𝑡
∴ 𝐿−1^ { (^) (𝑠 (^2) + 3^12 ). (^) (𝑠 (^2) + 3^12 )} = 18 1 ∫ [cos(6𝑢 − 3𝑡) − cos 3𝑡]𝑑𝑢 = 18 1 [∫ cos(6𝑢 − 3𝑡)
𝑡 0
du − ∫ cos 3𝑡
𝑡 0
𝑑𝑢]
𝑡 0 = 181 [sin(6𝑢−3𝑡) 6 − cos 3t. 𝑢] 0
𝑡 = 181 [{sin 3𝑡 6 − cos 3𝑡. 𝑡} − {− sin 3𝑡 6 − cos 3𝑡 .0}]
=
18 [
sin 3𝑡 6 − cos 3𝑡. 𝑡 +
sin 3𝑡 6 ] =
sin 3𝑡 3 − 𝑡 cos 3𝑡)
=
(sin 3𝑡 − 3𝑡 cos 3𝑡)
𝐿−1^ {
(𝑠^2 + 4𝑠 + 13)^2 } = 𝑒
(𝑠^2 + 3^2 )^2 } =
(sin 3𝑡 − 3𝑡 cos 3𝑡)
Example. Apply Convolution theorem to evaluate 𝐿−1^ {(𝒔+𝒂)(𝒔+𝒃) 𝟏 }
Solution. We have
𝐿−1^ {
where 𝑓(𝑠) = (^) (𝑠+𝑎)^1 , 𝑔(𝑠) = (^) (𝑠+𝑏)^1
Then, we have
𝐿−1{𝑓(𝑠)} = 𝐿−1^ { (^) (𝑠+𝑎)^1 } = 𝑒−𝑎𝑡^ = 𝑓(𝑡) and 𝐿−1{𝑔(𝑠)} = 𝐿−1^ { (^) (𝑠+𝑏)^1 } = 𝑒−𝑏𝑡^ = 𝑔(𝑡)
