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Laplace Transforms

ENGIN 211, Engineering Math

Why Laplace Transform ?

• Laplace transform converts a function in the time domain to its frequency domain.

• It is a powerful, systematic method in solving differential equations.

• It converts differential equations into algebraic equations.

• Initial or boundary conditions are automatically accounted for.

• In situations where other methods fail because the solution is discontinuous, it can succeed.

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Definition

ℒ 𝑓 𝑡 = 𝑒−𝑠𝑡𝑓 𝑡 𝑑𝑡∞

0

= 𝐹 𝑠

The parameter 𝑠 known as the complex

frequency is assumed to have positive and large

enough real part to ensure that the integral

converges.

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Useful Transforms

• 𝑓 𝑡 = 𝑎 → 𝐹 𝑠 =𝑎

𝑠

• 𝑓 𝑡 = 𝑒𝑎𝑡 → 𝐹 𝑠 =1

𝑠−𝑎

• 𝑓 𝑡 = sin 𝜔𝑡 → 𝐹 𝑠 =𝜔

𝑠2+𝜔2

• 𝑓 𝑡 = cos 𝜔𝑡 → 𝐹 𝑠 =𝑠

𝑠2+𝜔2

• 𝑓 𝑡 = 𝑡𝑛 → 𝐹 𝑠 =𝑛!

𝑠𝑛+1

• 𝑓 𝑡 = sinh 𝜔𝑡 → 𝐹 𝑠 =𝜔

𝑠2−𝜔2

• 𝑓 𝑡 = cosh 𝜔𝑡 → 𝐹 𝑠 =𝑠

𝑠2−𝜔2

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The Unit Impulse δ(t)

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The unit impulse function is defined as

𝛿(𝑡) =𝑑𝑢(𝑡)

𝑑𝑡

where 𝑢(𝑡) is the unit step function.

Total area underneath 𝛿(𝑡): 𝛿(𝑡 − 𝑡0)𝑑𝑡∞

−∞= 1

Sifting property of 𝛿(𝑡): 𝑓(𝑡)𝛿(𝑡 − 𝑡0)𝑑𝑡∞

−∞= 𝑓(𝑡0)

Laplace transform: ℒ 𝛿(𝑡) = 1, ℒ 𝛿(𝑡 − 𝑡0) = 𝑒−𝑠𝑡0

𝛿(𝑡 − 𝑡0)

Properties

Linearity

ℒ 𝑓 𝑡 ± 𝑔 𝑡 = ℒ 𝑓 𝑡 ± ℒ 𝑔 𝑡

ℒ 𝑘𝑓 𝑡 = 𝑘ℒ 𝑓 𝑡

But ℒ 𝑓 𝑡 𝑔 𝑡 ≠ ℒ 𝑓 𝑡 ℒ 𝑔 𝑡

Rather the convolution of two function,

ℒ 𝑓 𝑡 ∗ 𝑔 𝑡 = ℒ 𝑓 𝑡 ℒ 𝑔 𝑡

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Examples

•ℒ 2 sin 3𝑡 + 4 sinh 3𝑡 = 2

3

𝑠2+9+ 4

3

𝑠2−9

=18 𝑠2+3

𝑠4−81

•ℒ 5e4𝑡 + cosh 2𝑡 =

5

𝑠−4+

𝑠

𝑠2−4

=6𝑠2−4𝑠−20

𝑠−4 𝑠4−4

•ℒ 𝑡3 + 2𝑡2 − 4𝑡 + 1 =

3!

𝑠4 + 22!

𝑠3 − 41

𝑠2 +1

𝑠

=1

𝑠4 𝑠3 − 4𝑠2 + 4𝑠 + 6

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Main Theorems

Assume ℒ 𝑓 𝑡 = 𝐹 𝑠 :

• Similarity theorem ℒ 𝑓 𝑎𝑡 =1

𝑎𝐹

𝑠

𝑎

• Real Shifting theorem: for 𝑎 > 0

ℒ 𝑓 𝑡 − 𝑎 = 𝑒−𝑎𝑠𝐹 𝑠

• Complex Shifting theorem: for 𝑎 > 0

ℒ 𝑒−𝑎𝑡𝑓 𝑡 = 𝐹 𝑠 + 𝑎

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Main Theorems (Cont’d)

• Derivative theorem:

ℒ𝑑𝑓 𝑡

𝑑𝑡= 𝑠𝐹 𝑠 − 𝑓 0 +

ℒ𝑑2𝑓 𝑡

𝑑𝑡2= 𝑠2𝐹 𝑠 − 𝑠𝑓 0 + − 𝑓′ 0 +

• Complex differentiation theorem:

ℒ 𝑡𝑛𝑓 𝑡 = −1 𝑛𝑑𝑛𝐹 𝑠

𝑑𝑠𝑛

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Main Theorems (Cont’d)

• Real integral theorem:

ℒ 𝑓 𝜏𝑡

0

𝑑𝜏 =𝐹 𝑠

𝑠

• Complex integral theorem:

ℒ𝑓 𝑡

𝑡= 𝐹 𝜎

∞

𝑠

𝑑𝜎 if lim𝑡→0

𝑓 𝑡

𝑡exists

• Convolution theorem: if 𝑓1 ∗ 𝑓2 = 𝑓1 𝜏𝑡

0𝑓2 𝑡 − 𝜏 𝑑𝜏

ℒ 𝑓1 ∗ 𝑓2 = 𝐹1 𝑠 ∗ 𝐹2 𝑠

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Main Theorems (Cont’d)

Convolution theorem:

𝑓1 ∗ 𝑓2 = 𝑓1 𝜏𝑡

0

𝑓2 𝑡 − 𝜏 𝑑𝜏

ℒ 𝑓1 ∗ 𝑓2 = 𝐹1 𝑠 ∗ 𝐹2 𝑠

Initial theorem:

𝑓 0 + = lim𝑠→∞

𝑠𝐹 𝑠 if lim𝑡→0

𝑓 𝑡 exists

Final theorem:

𝑓 ∞ = lim𝑠→0

𝑠𝐹 𝑠 if lim𝑡→∞

𝑓 𝑡 exists

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Example (Shifting theorem)

The real shifting theorem states that if ℒ 𝑓 𝑡 = 𝐹 𝑠 , then

ℒ 𝑒−𝑎𝑡𝑓 𝑡 = 𝐹 𝑠 + 𝑎

• Because ℒ sin 3𝑡 =3

𝑠2+9, then

ℒ 𝑒−2𝑡 sin 3𝑡 =3

𝑠 + 2 2 + 9=

3

𝑠2 + 4𝑠 + 13

• Because ℒ 𝑡3 =3!

𝑠4, then

ℒ 𝑡3𝑒−4𝑡 =3!

𝑠 + 4 4=

6

𝑠 + 4 4

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Example (Complex Differentiation)

Complex differentiation theorem:

ℒ 𝑡𝑛𝑓 𝑡 = −1 𝑛𝑑𝑛𝐹 𝑠

𝑑𝑠𝑛

If ℒ cosh 3𝑡 =𝑠

𝑠2−9, then

ℒ 𝑡2 cosh 3𝑡 =𝑑2

𝑑𝑠2

𝑠

𝑠2 − 9=

2𝑠 𝑠2 + 27

𝑠2 − 9 3

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Example (Complex Integral)

• ℒ𝑓 𝑡

𝑡= 𝐹 𝜎

∞

𝑠𝑑𝜎 if lim

𝑡→0

𝑓 𝑡

𝑡exists

Example: find ℒ1−cos 2𝑡

𝑡=?

Since lim𝑡→0

1−cos 2𝑡

𝑡= lim

𝑡→0

2 sin2 𝑡

𝑡= 0 does exist, we obtain

ℒ1 − cos 2𝑡

𝑡=

1

𝜎−

𝜎

𝜎2 + 4𝑑𝜎

∞

𝑠

= ln𝜎 −1

2ln 𝜎2 + 4

𝑠

∞

= ln𝜎

𝜎2 + 4 𝑠

∞

= 0 − ln𝑠

𝑠2 + 4= ln

𝑠2 + 4

𝑠

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Inverse Laplace Transform

• Given Laplace transform 𝐹 𝑠 in 𝑠-domain, find its function

𝑓 𝑡 in 𝑡-domain.

ℒ−1 𝐹 𝑠 = 𝑓(𝑡)

• Useful pairs

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𝐹 𝑠 =𝑎

𝑠→ 𝑓 𝑡 = 𝑎

𝐹 𝑠 =1

𝑠 − 𝑎→ 𝑓 𝑡 = 𝑒𝑎𝑡

𝐹 𝑠 =𝜔

𝑠2 + 𝜔2→ 𝑓 𝑡 = sin𝜔𝑡

𝐹 𝑠 =𝑠

𝑠2 + 𝜔2→ 𝑓 𝑡 = cos𝜔𝑡

𝐹 𝑠 =𝑛!

𝑠𝑛+1→ 𝑓 𝑡 = 𝑡𝑛

𝐹 𝑠 =𝜔

𝑠2 − 𝜔2→ 𝑓 𝑡 = sinh𝜔𝑡

𝐹 𝑠 =𝑠

𝑠2 − 𝜔2→ 𝑓 𝑡 = cosh𝜔𝑡

Example

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Find ℒ−1 4𝑠2−5𝑠+6

𝑠+1 𝑠2+4=?

Solution: 4𝑠2−5𝑠+6

𝑠+1 𝑠2+4=

𝐴

𝑠+1+

𝐵𝑠+𝐶

𝑠2+4

4𝑠2 − 5𝑠 + 6 = 𝐴 𝑠2 + 4 + 𝐵𝑠 + 𝐶 𝑠 + 1

• 𝑠 + 1 = 0 → 𝐴 = 3

• Equating coefficients of various powers 𝐵 = 1, and 𝐶 = −6

ℒ−13

𝑠 + 1+

𝑠 − 6

𝑠2 + 4= ℒ−1

3

𝑠 + 1+

𝑠

𝑠2 + 4− 3

2

𝑠2 + 4

= 3𝑒−𝑡 + cos 2𝑡 − 3 sin 2𝑡

Partial Fractions

• Order in numerator less than denominator

• Factorize the denominator into prime factors

• A linear factor 𝑠 + 𝑎 gives 𝐴

𝑠+𝑎

• A repeated factor 𝑠 + 𝑎 2 gives 𝐴

𝑠+𝑎+

𝐵

𝑠+𝑎 2

• A quadratic factor 𝑠2 + 𝑝𝑠 + 𝑞 gives 𝑃𝑠+𝑄

𝑠2+𝑝𝑠+𝑞

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Example (Partial Fractions) 𝑠2 − 9𝑠 + 7

𝑠2 + 𝑠 − 2 𝑠2 + 2𝑠 + 2=

𝐴

𝑠 + 2+

𝐵

𝑠 − 1+

𝐶𝑠 + 𝐷

𝑠2 + 2𝑠 + 2

• 𝐴 =𝑠2−9𝑠+7

𝑠−1 𝑠2+2𝑠+2 𝑠=−2

= −4 and 𝐵 =𝑠2−9𝑠+7

𝑠+2 𝑠2+2𝑠+2 𝑠=1

= −2

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• 𝑠2 − 9𝑠 + 7 = 𝐴 𝑠 − 1 𝑠2 + 2𝑠 + 2 + 𝐵 𝑠 + 2 𝑠2 + 2𝑠 + 2 + 𝐶𝑠 + 𝐷 𝑠2 + 𝑠 − 2

• Equating highest-order 𝑠3 coefficient 𝐴 + 𝐵 + 𝐶 = 0 → 𝐶 =22

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• Equating lowest-order 𝑠0 coefficient −2𝐴 + 4𝐵 − 2𝐷 = 0 → 𝐷 =16

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ℒ−1𝐴

𝑠 + 2+

𝐵

𝑠 − 1+

𝐶𝑠 + 𝐷

𝑠2 + 2𝑠 + 2= ℒ−1

−4

𝑠 + 2−

2

5

1

𝑠 − 1+

2

5

11𝑠 + 8

𝑠 + 1 2 + 1

= ℒ−1−4

𝑠 + 2−

2

5

1

𝑠 − 1+

2

5

11 𝑠 + 1 − 3

𝑠 + 1 2 + 1

= ℒ−1−4

𝑠 + 2−

2

5

1

𝑠 − 1+

22

5

𝑠 + 1

𝑠 + 1 2 + 1−

6

5

1

𝑠 + 1 2 + 1

= −4𝑒−2𝑡 −2

5𝑒𝑡 +

22

5𝑒−𝑡 sin 𝑡 −

6

5𝑒−𝑡 cos 𝑡

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Applying to Differential Equations

With the use of Derivative theorems:

ℒ𝑑𝑓 𝑡

𝑑𝑡= 𝑠𝐹 𝑠 − 𝑓 0 +

ℒ𝑑2𝑓 𝑡

𝑑𝑡2= 𝑠2𝐹 𝑠 − 𝑠𝑓 0 + − 𝑓′ 0 +

We can convert differential equations into algebraic equations

where all initial conditions are taken into account. Effectively, we

are converting problems in 𝑡-domain into 𝑠-domain. In the end,

we need to perform inverse transform to go back to 𝑡-domain.

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1st Order Differential Equation

Example: solve 𝑑𝑥

𝑑𝑡+ 4𝑥 = 2𝑒−2𝑡 + 4𝑒−4𝑡 , at 𝑡 = 0, 𝑥 = 2.

Taking Laplace on both sides to s-domain:

𝑠𝑋 − 2 + 4𝑋 =2

𝑠 + 2+

4

𝑠 + 4

Then 𝑋 =2

𝑠+4+

2

𝑠+2 𝑠+4+

4

𝑠+4 2 =2

𝑠+4+

1

𝑠+2−

1

𝑠+4+

4

𝑠+4 2

𝑋 =1

𝑠 + 2+

1

𝑠 + 4+

4

𝑠 + 4 2

Inverse Laplace back to t-domain

𝑥 𝑡 = ℒ−1 𝑋 = 𝑒−2𝑡 + 𝑒−4𝑡 + 4𝑡𝑒−4𝑡

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2nd Order Differential Equation Example: 𝒙" + 𝟓𝒙′ + 𝟔𝒙 = 𝟐𝟒 𝐬𝐢𝐧 𝟐𝒕 , 𝐚𝐭 𝒕 = 𝟎, 𝒙 = 𝟎 𝐚𝐧𝐝 𝒙′ = 𝟎

Taking Laplace transform on both sides:

𝑠2𝑋 − 𝑠𝑥 0 − 𝑥′ 0 + 5 𝑠𝑋 − 𝑥(0) + 6𝑋 = 242

𝑠2 + 4

Plug in initial conditions:

𝑠2 + 5𝑠 + 6 𝑋 =48

𝑠2 + 4

So 𝑋 =48

𝑠2+5𝑠+6 𝑠2+4=

𝐴

𝑠+2+

𝐵

𝑠+3+

𝐶𝑠+𝐷

𝑠2+4

where 𝐴 =48

(𝑠+3) 𝑠2+4 𝑠=−2

= 6, 𝐵 =48

(𝑠+2) 𝑠2+4 𝑠=−3

= −48

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And

𝐴 𝑠 + 3 𝑠2 + 4 + 𝐵 𝑠 + 2 𝑠2 + 4 + 𝐶𝑠 + 𝐷 𝑠 + 2 𝑠 + 3 = 48

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2nd Order Differential Equation (Cont’d)

Equating the coefficients for 𝑠3 and 𝑠0, we obtain

𝐴 + 𝐵 + 𝐶 = 0, and 12𝐴 + 8𝐵 + 6𝐷 = 48

Thus,

𝐶 = − 𝐴 + 𝐵 = −30

13, and 𝐷 =

24 − 6𝐴 − 4𝐵

3=

12

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Taking inverse Laplace:

𝑥 𝑡 = ℒ−1 𝑋 = ℒ−1𝐴

𝑠 + 2+

𝐵

𝑠 + 3+

𝐶𝑠 + 𝐷

𝑠2 + 4

= 6ℒ−11

𝑠 + 2−

48

13ℒ−1

1

𝑠 + 3−

30

13ℒ−1

𝑠

𝑠2 + 4+

6

13 ℒ−1

2

𝑠2 + 4

= 6𝑒−2𝑡 −48

13𝑒−3𝑡 −

30

13cos 2𝑡 +

6

13sin 2𝑡

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System of Differential Equations

𝑥" + 2𝑥 − 𝑦 = 0𝑦" + 2𝑦 − 𝑥 = 0

, with initial conditions 𝑥 0 = 4𝑦 0 = 2

and 𝑥′ 0 = 0

𝑦′ 0 = 0

Taking Laplace transforms using ℒ 𝑓" = 𝑠2𝐹 𝑠 − 𝑠𝑓 0 − 𝑓′ 0

𝑠2𝑋 − 𝑠𝑥 0 − 𝑥′ 0 + 2𝑋 − 𝑌 = 0

𝑠2𝑌 − 𝑠𝑦 0 − 𝑦′ 0 + 2𝑌 − 𝑋 = 0 or

𝑠2 + 2 𝑋 − 𝑌 = 4𝑠

−𝑋 + 𝑠2 + 2 𝑌 = 2𝑠

In matrix form

𝑠2 + 2 −1−1 𝑠2 + 2

𝑋𝑌

=4𝑠2𝑠

Using inverse matrix, we obtain

𝑋𝑌

=1

𝑠2 + 2 2 − 1𝑠2 + 2 1

1 𝑠2 + 24𝑠2𝑠

=2𝑠

𝑠2 + 2 2 − 12𝑠2 + 5𝑠2 + 4

Thus

𝑋 =2𝑠 2𝑠2+5

𝑠2+2 2−1=

2𝑠 2𝑠2+5

𝑠2+1 𝑠2+3, and 𝑌 =

2𝑠 𝑠2+4

𝑠2+2 2−1=

2𝑠 𝑠2+4

𝑠2+1 𝑠2+3

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System of Differential Equations (Cont’d)

Now let’s find the inverse Laplace of 𝑋,

𝑋 =2𝑠 2𝑠2 + 5

𝑠2 + 1 𝑠2 + 3=

𝐴1𝑠 + 𝐵1

𝑠2 + 1+

𝐶1𝑠 + 𝐷1

𝑠2 + 3

(a) Multiply 𝑠2 + 1 and then let 𝑠2 = −1,

𝐴1𝑠 + 𝐵1 =2𝑠 2𝑠2 + 5

𝑠2 + 3 𝑠2=−1

= 3𝑠

Thus, 𝐴1 = 3, and 𝐵1 = 0

(b) Multiply 𝑠2 + 3 and then let 𝑠2 = −3,

𝐶1𝑠 + 𝐷1 =2𝑠 2𝑠2 + 5

𝑠2 + 1 𝑠2=−3

= 𝑠

Thus, 𝐶1 = 1, and 𝐷1 = 0

So we have 𝑋 =3𝑠

𝑠2+1+

𝑠

𝑠2+3

Inverse Laplace: 𝑥 𝑡 = ℒ−1 𝑋 = 3 cos 𝑡 + cos 3𝑡

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System of Differential Equations (Cont’d)

Next let’s find the inverse Laplace of 𝑌,

𝑌 =2𝑠 𝑠2 + 4

𝑠2 + 1 𝑠2 + 3=

𝐴2𝑠 + 𝐵2

𝑠2 + 1+

𝐶2𝑠 + 𝐷2

𝑠2 + 3

(a) Multiply 𝑠2 + 1 and then let 𝑠2 = −1,

𝐴2𝑠 + 𝐵2 =2𝑠 𝑠2 + 4

𝑠2 + 3 𝑠2=−1

= 3𝑠

Thus, 𝐴2 = 3, and 𝐵2 = 0

(b) Multiply 𝑠2 + 3 and then let 𝑠2 = −3,

𝐶2𝑠 + 𝐷2 =2𝑠 𝑠2 + 4

𝑠2 + 1 𝑠2=−3

= −𝑠

Thus 𝐶2 = −1, and 𝐷2 = 0

We have 𝑌 =3𝑠

𝑠2+1−

𝑠

𝑠2+3

Inverse Laplace: 𝑦 𝑡 = ℒ−1 𝑌 = 3 cos 𝑡 − cos 3𝑡

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Solution: 𝑥 = 3 cos 𝑡 + cos 3𝑡

𝑦 = 3 cos 𝑡 − cos 3𝑡

Impulse Response

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Network

Network

Network

Network

Convolution and Laplace

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If we define the impulse response as h(t), then the output y(t) is related to the input x(t) via the convolution integral:

Convolution in the time domain is multiplication in the frequency domain:

Convolution Theorem for Inverse Laplace

This theorem can be used to find inverse Laplace transform ℒ−1 1

𝑠2 𝑠−3=?

𝐹1 𝑠 =1

𝑠2, and 𝐹2 𝑠 =

1

𝑠 − 3

Their inverse: 𝑓1 𝑡 = 𝑡, and 𝑓2 𝑡 = 𝑒3𝑡

Using Convolution Theorem

ℒ−11

𝑠2 𝑠 − 3= 𝑥𝑒3(𝑡−𝑥)𝑑𝑥

𝑡

0

= 𝑒3𝑡 𝑥𝑒−3𝑥𝑑𝑥𝑡

0

= 𝑒3𝑡 𝑥𝑒−3𝑥𝑑𝑥𝑡

0

=1

9𝑒3𝑡 − 3𝑡 − 1

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Summary

Key points:

Useful Laplace transform pairs

δ-function and its properties

Laplace transform of derivatives

Inverse Laplace using partial fractions

Solving differential equations with Laplace

Supplemental:

Convolution theorem

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