Laplace Transforms

transcript

Laplace Transforms

Nirav B. Vyas

Department of MathematicsAtmiya Institute of Technology and Science

Yogidham, Kalavad roadRajkot - 360005 . Gujarat

N. B. Vyas Laplace Transforms

Laplace Transforms

Definition:

Let f(t) be a function of t defined for all t ≥ 0 then Laplacetransform of f(t) is denoted by L{f(t)} or f(s) and isdefined as

L {f (t)} = f (s) =

∞∫0

e−stf (t) dt

provided the integral exists where s is a parameter ( real orcomplex).

Laplace Transforms

NOTATIONS:

The original functions are denoted by lowercase letters suchas f(t), g(t), ...

Laplace transforms by the same letters with bars such asf(s)g(s), ...

Linearity of the Laplace Transforms

Theorem 1:

If L {f (t)} = f (s) and L {g (t)} = g (s) then for any constants a and bL {af (t) + bg (t)} = aL {f (t)}+ bL {g (t)}

Corollary 1:

Putting a = 0 and b = 0, we get L[0] = 0

Corollary 2:

Putting b = 0, we get L[af(t)] = aL[f(t)]

Corollary 3:

L [a1f1 (t) + a2f2 (t) + ...+ anfn (t)]

= a1L [f1(t)] + a2L [f2(t)] + ...+ anL [fn(t)]

Theorem 1:

Corollary 1:

Corollary 2:

Corollary 3:

L [a1f1 (t) + a2f2 (t) + ...+ anfn (t)]

= a1L [f1(t)] + a2L [f2(t)] + ...+ anL [fn(t)]

Theorem 1:

Corollary 1:

Corollary 2:

Corollary 3:

L [a1f1 (t) + a2f2 (t) + ...+ anfn (t)]

= a1L [f1(t)] + a2L [f2(t)] + ...+ anL [fn(t)]

Theorem 1:

Corollary 1:

Corollary 2:

Corollary 3:

L [a1f1 (t) + a2f2 (t) + ...+ anfn (t)]

= a1L [f1(t)] + a2L [f2(t)] + ...+ anL [fn(t)]

Laplace Transforms of some elementary functions

1 L(1) =1

2 L(eat) =1

s− a

cor.1 If a = 0⇒ L(1) =1

cor.2 L[e−at] =1

s+ aif s > −a

cor.3 L[cat] = L[eat log c] =1

s− a logcif s > a log c and c > 0

1 L(1) =1

2 L(eat) =1

s− a

cor.1 If a = 0⇒ L(1) =1

cor.2 L[e−at] =1

s+ aif s > −a

1 L(1) =1

2 L(eat) =1

s− a

cor.1 If a = 0⇒ L(1) =1

cor.2 L[e−at] =1

s+ aif s > −a

1 L(1) =1

2 L(eat) =1

s− a

cor.1 If a = 0⇒ L(1) =1

cor.2 L[e−at] =1

s+ aif s > −a

1 L(1) =1

2 L(eat) =1

s− a

cor.1 If a = 0⇒ L(1) =1

cor.2 L[e−at] =1

s+ aif s > −a

3 L[sinh at] =a

s2 − a2

4 L[cosh at] =s

s2 − a2

5 L[sin at] =a

s2 + a2

6 L[cos at] =s

s2 + a2, s > 0

cor.1 L[sin t] =1

s2 + 1, s > 0

cor.2 L[cos t] =s

s2 + 1, s > 0

3 L[sinh at] =a

s2 − a2

4 L[cosh at] =s

s2 − a2

5 L[sin at] =a

s2 + a2

6 L[cos at] =s

s2 + a2, s > 0

cor.1 L[sin t] =1

s2 + 1, s > 0

cor.2 L[cos t] =s

s2 + 1, s > 0

3 L[sinh at] =a

s2 − a2

4 L[cosh at] =s

s2 − a2

5 L[sin at] =a

s2 + a2

6 L[cos at] =s

s2 + a2, s > 0

cor.1 L[sin t] =1

s2 + 1, s > 0

cor.2 L[cos t] =s

s2 + 1, s > 0

3 L[sinh at] =a

s2 − a2

4 L[cosh at] =s

s2 − a2

5 L[sin at] =a

s2 + a2

6 L[cos at] =s

s2 + a2, s > 0

cor.1 L[sin t] =1

s2 + 1, s > 0

cor.2 L[cos t] =s

s2 + 1, s > 0

3 L[sinh at] =a

s2 − a2

4 L[cosh at] =s

s2 − a2

5 L[sin at] =a

s2 + a2

6 L[cos at] =s

s2 + a2, s > 0

cor.1 L[sin t] =1

s2 + 1, s > 0

cor.2 L[cos t] =s

s2 + 1, s > 0

3 L[sinh at] =a

s2 − a2

4 L[cosh at] =s

s2 − a2

5 L[sin at] =a

s2 + a2

6 L[cos at] =s

s2 + a2, s > 0

cor.1 L[sin t] =1

s2 + 1, s > 0

cor.2 L[cos t] =s

s2 + 1, s > 0

7 L[tn] =Γ(n+ 1)

sn+1, n = 0, 1, 2, ...

cor.1 If n = 0, L[1] =1

cor.2 If n = −1

L(t−

Γ(12)

7 L[tn] =Γ(n+ 1)

sn+1, n = 0, 1, 2, ...

cor.1 If n = 0, L[1] =1

cor.2 If n = −1

L(t−

Γ(12)

7 L[tn] =Γ(n+ 1)

sn+1, n = 0, 1, 2, ...

cor.1 If n = 0, L[1] =1

cor.2 If n = −1

L(t−

Γ(12)

7 L[tn] =Γ(n+ 1)

sn+1, n = 0, 1, 2, ...

cor.1 If n = 0, L[1] =1

cor.2 If n = −1

L(t−

Γ(12)

7 L[tn] =Γ(n+ 1)

sn+1, n = 0, 1, 2, ...

cor.1 If n = 0, L[1] =1

cor.2 If n = −1

L(t−

Γ(12)

7 L[tn] =Γ(n+ 1)

sn+1, n = 0, 1, 2, ...

cor.1 If n = 0, L[1] =1

cor.2 If n = −1

L(t−

Γ(12)

Examples of Laplace Transform

{2t3 + e−2t + t

{A+B t

12 + C t−

{eat − 1

}4 L {sin(at+ b)}5 L {sin 2t cos 3t}6 L

{cos24t

{cos32t

{2t3 + e−2t + t

{A+B t

12 + C t−

{eat − 1

{cos24t

{cos32t

{2t3 + e−2t + t

{A+B t

12 + C t−

{eat − 1

4 L {sin(at+ b)}5 L {sin 2t cos 3t}6 L

{cos24t

{cos32t

{2t3 + e−2t + t

{A+B t

12 + C t−

{eat − 1

}4 L {sin(at+ b)}

5 L {sin 2t cos 3t}6 L

{cos24t

{cos32t

{2t3 + e−2t + t

{A+B t

12 + C t−

{eat − 1

}4 L {sin(at+ b)}5 L {sin 2t cos 3t}

cos24t}

cos32t}

{2t3 + e−2t + t

{A+B t

12 + C t−

{eat − 1

{cos24t

cos32t}

{2t3 + e−2t + t

{A+B t

12 + C t−

{eat − 1

{cos24t

{cos32t

First Shifting Theorem

If L {f (t)} = f (s) then L{eatf (t)

}= f (s− a)

Proof: By the def. of Laplace

L{eatf (t)

}=∞∫0

e−steatf (t) dt

=∞∫0

e−(s−a)tf (t) dt

=∞∫0

e−rtf (t) dt

= f(r)

= f(s− a)

Thus if we know the transformation f(s) of f(t) then we canwrite the transformation of eatf(t) simply replacing s by s− a toget F (s− a)

}= f (s− a)

L{eatf (t)

}=∞∫0

e−steatf (t) dt

=∞∫0

e−rtf (t) dt

= f(r)

= f(s− a)

}= f (s− a)

L{eatf (t)

}=∞∫0

e−steatf (t) dt

=∞∫0

e−rtf (t) dt

= f(r)

= f(s− a)

}= f (s− a)

L{eatf (t)

}=∞∫0

e−steatf (t) dt

=∞∫0

e−rtf (t) dt

= f(r)

= f(s− a)

}= f (s− a)

L{eatf (t)

}=∞∫0

e−steatf (t) dt

=∞∫0

e−rtf (t) dt

= f(r)

= f(s− a)

}= f (s− a)

L{eatf (t)

}=∞∫0

e−steatf (t) dt

=∞∫0

e−rtf (t) dt

= f(r)

= f(s− a)

}= f (s− a)

L{eatf (t)

}=∞∫0

e−steatf (t) dt

=∞∫0

e−rtf (t) dt

= f(r)

= f(s− a)

}= f (s− a)

L{eatf (t)

}=∞∫0

e−steatf (t) dt

=∞∫0

e−rtf (t) dt

= f(r)

= f(s− a)

1 L(eat) =1

s− a

2 L[eattn] =Γ(n+ 1)

(s− a)n+1

3 L[eatsinh bt] =b

(s− a)2 − b2

4 L[eatcosh bt] =s

(s− a)2 − b2

5 L[eatsin bt] =b

(s− a)2 + b2

6 L[eatcos bt] =s− a

(s− a)2 + b2, s > 0

1 L(eat) =1

s− a

(s− a)n+1

3 L[eatsinh bt] =b

(s− a)2 − b2

4 L[eatcosh bt] =s

(s− a)2 − b2

5 L[eatsin bt] =b

(s− a)2 + b2

(s− a)2 + b2, s > 0

1 L(eat) =1

s− a

(s− a)n+1

3 L[eatsinh bt] =b

(s− a)2 − b2

4 L[eatcosh bt] =s

(s− a)2 − b2

5 L[eatsin bt] =b

(s− a)2 + b2

(s− a)2 + b2, s > 0

1 L(eat) =1

s− a

(s− a)n+1

3 L[eatsinh bt] =b

(s− a)2 − b2

4 L[eatcosh bt] =s

(s− a)2 − b2

5 L[eatsin bt] =b

(s− a)2 + b2

(s− a)2 + b2, s > 0

1 L(eat) =1

s− a

(s− a)n+1

3 L[eatsinh bt] =b

(s− a)2 − b2

4 L[eatcosh bt] =s

(s− a)2 − b2

5 L[eatsin bt] =b

(s− a)2 + b2

(s− a)2 + b2, s > 0

1 L(eat) =1

s− a

(s− a)n+1

3 L[eatsinh bt] =b

(s− a)2 − b2

4 L[eatcosh bt] =s

(s− a)2 − b2

5 L[eatsin bt] =b

(s− a)2 + b2

(s− a)2 + b2, s > 0

Examples

1 Find out the Laplace transform of e−3t (2 cos 5t− 3 sin 5t)

2 L[e−atsinhbt]

3 L[t3e−3t]

4 L[(t+ 2)2et]

5 L[e−tsin2t]

6 L[cosh at sin at]

Examples

2 L[e−atsinhbt]

3 L[t3e−3t]

4 L[(t+ 2)2et]

5 L[e−tsin2t]

6 L[cosh at sin at]

Examples

2 L[e−atsinhbt]

3 L[t3e−3t]

4 L[(t+ 2)2et]

5 L[e−tsin2t]

6 L[cosh at sin at]

Examples

2 L[e−atsinhbt]

3 L[t3e−3t]

4 L[(t+ 2)2et]

5 L[e−tsin2t]

6 L[cosh at sin at]

Examples

2 L[e−atsinhbt]

3 L[t3e−3t]

4 L[(t+ 2)2et]

5 L[e−tsin2t]

6 L[cosh at sin at]

Examples

2 L[e−atsinhbt]

3 L[t3e−3t]

4 L[(t+ 2)2et]

5 L[e−tsin2t]

6 L[cosh at sin at]

Examples

Ex. Find the Laplace transform of the function which is defined as

f(t) =

{t/T 0 < t < T1 when t > T

Examples

Ex. Find Laplace transform of f(t) =

{sin t 0 < t < π

0 when t > π

Examples

Ex. Find Laplace transform of f(t) where f(t) =

{t 0 < t < 45 when t > 4

Change of Scale property

If L {f (t)} = f (s) then L{eatf (bt)

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

e−(s−a)tf (bt) dt

=∞∫0

e−( s−ab )btf (bt) dt

Let bt = u⇒ b dt = du

L{eatf (bt)

}=∞∫0

e−( s−ab )uf (u)

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

L{eatf (bt)

}=∞∫0

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

L{eatf (bt)

}=∞∫0

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

L{eatf (bt)

}=∞∫0

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

L{eatf (bt)

}=∞∫0

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

L{eatf (bt)

}=∞∫0

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

L{eatf (bt)

}=∞∫0

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

L{eatf (bt)

}=∞∫0

(s− ab

), b > 0

L {f (t)} =

∞∫0

e−stf (t) dt

L{eatf (bt)

∞∫0

e−steatf (bt) dt

=∞∫0

L{eatf (bt)

}=∞∫0

(s− ab

)N. B. Vyas Laplace Transforms

Inverse Laplace Transform

If L {f (t)} = f (s) then f(t) is called the inverse Laplacetransform of f(s) and it is denoted by L−1

}= f (t)

1 L−1(

2 L−1(

s− a

)= eat

3 L−1(

s2 + a2

asin at

4 L−1(

s2 + a2

)= cos at

5 L−1(

s2 − a2

asinh at

6 L−1(

s2 − a2

)= cosh at

7 L−1(

tn−1

(n− 1)!

}= f (t)

1 L−1(

2 L−1(

s− a

)= eat

3 L−1(

s2 + a2

asin at

4 L−1(

s2 + a2

)= cos at

5 L−1(

s2 − a2

asinh at

6 L−1(

s2 − a2

)= cosh at

7 L−1(

tn−1

(n− 1)!

}= f (t)

1 L−1(

2 L−1(

s− a

)= eat

3 L−1(

s2 + a2

asin at

4 L−1(

s2 + a2

)= cos at

5 L−1(

s2 − a2

asinh at

6 L−1(

s2 − a2

)= cosh at

7 L−1(

tn−1

(n− 1)!

}= f (t)

1 L−1(

2 L−1(

s− a

)= eat

3 L−1(

s2 + a2

asin at

4 L−1(

s2 + a2

)= cos at

5 L−1(

s2 − a2

asinh at

6 L−1(

s2 − a2

)= cosh at

7 L−1(

tn−1

(n− 1)!

}= f (t)

1 L−1(

2 L−1(

s− a

)= eat

3 L−1(

s2 + a2

asin at

4 L−1(

s2 + a2

)= cos at

5 L−1(

s2 − a2

asinh at

6 L−1(

s2 − a2

)= cosh at

7 L−1(

tn−1

(n− 1)!

}= f (t)

1 L−1(

2 L−1(

s− a

)= eat

3 L−1(

s2 + a2

asin at

4 L−1(

s2 + a2

)= cos at

5 L−1(

s2 − a2

asinh at

6 L−1(

s2 − a2

)= cosh at

7 L−1(

tn−1

(n− 1)!

}= f (t)

1 L−1(

2 L−1(

s− a

)= eat

3 L−1(

s2 + a2

asin at

4 L−1(

s2 + a2

)= cos at

5 L−1(

s2 − a2

asinh at

6 L−1(

s2 − a2

)= cosh at

7 L−1(

tn−1

(n− 1)!

}= f (t)

1 L−1(

2 L−1(

s− a

)= eat

3 L−1(

s2 + a2

asin at

4 L−1(

s2 + a2

)= cos at

5 L−1(

s2 − a2

asinh at

6 L−1(

s2 − a2

)= cosh at

7 L−1(

tn−1

(n− 1)!

Partial Fractions

Examples of Inverse Laplace Transform - I

1 L−1(s2 − 3s+ 4

2 L−1[

(s4 − 2s2 + 1

)]3 L−1

(s+ 1)2 + 1

)4 L−1

(3s+ 5

(s+ 1)4

)5 L−1

s2 + 2s− 8

)6 L−1

(3s+ 7

s2 − 2s− 3

)7 L−1

(2s2 − 6s+ 5

s3 − 6s2 + 11s− 6

1 L−1(s2 − 3s+ 4

)2 L−1

(s4 − 2s2 + 1

3 L−1(

(s+ 1)2 + 1

)4 L−1

(3s+ 5

(s+ 1)4

)5 L−1

s2 + 2s− 8

)6 L−1

(3s+ 7

s2 − 2s− 3

)7 L−1

(2s2 − 6s+ 5

s3 − 6s2 + 11s− 6

1 L−1(s2 − 3s+ 4

)2 L−1

(s4 − 2s2 + 1

)]3 L−1

(s+ 1)2 + 1

4 L−1(

(s+ 1)4

)5 L−1

s2 + 2s− 8

)6 L−1

(3s+ 7

s2 − 2s− 3

)7 L−1

(2s2 − 6s+ 5

s3 − 6s2 + 11s− 6

1 L−1(s2 − 3s+ 4

)2 L−1

(s4 − 2s2 + 1

)]3 L−1

(s+ 1)2 + 1

)4 L−1

(3s+ 5

(s+ 1)4

5 L−1(

s2 + 2s− 8

)6 L−1

(3s+ 7

s2 − 2s− 3

)7 L−1

(2s2 − 6s+ 5

s3 − 6s2 + 11s− 6

1 L−1(s2 − 3s+ 4

)2 L−1

(s4 − 2s2 + 1

)]3 L−1

(s+ 1)2 + 1

)4 L−1

(3s+ 5

(s+ 1)4

)5 L−1

s2 + 2s− 8

6 L−1(

s2 − 2s− 3

)7 L−1

(2s2 − 6s+ 5

s3 − 6s2 + 11s− 6

1 L−1(s2 − 3s+ 4

)2 L−1

(s4 − 2s2 + 1

)]3 L−1

(s+ 1)2 + 1

)4 L−1

(3s+ 5

(s+ 1)4

)5 L−1

s2 + 2s− 8

)6 L−1

(3s+ 7

s2 − 2s− 3

7 L−1(

2s2 − 6s+ 5

s3 − 6s2 + 11s− 6

1 L−1(s2 − 3s+ 4

)2 L−1

(s4 − 2s2 + 1

)]3 L−1

(s+ 1)2 + 1

)4 L−1

(3s+ 5

(s+ 1)4

)5 L−1

s2 + 2s− 8

)6 L−1

(3s+ 7

s2 − 2s− 3

)7 L−1

(2s2 − 6s+ 5

s3 − 6s2 + 11s− 6

Examples of Inverse Laplace Transform - II

1 L−1(

(s+ 4)(s2 + 9)

2 L−1(

(s2 − 1)

)3 L−1

(4s+ 5

(s− 1)2(s+ 2)

)4 L−1

(2s2 − 1

(s2 + 1)(s2 + 4)

)5 L−1

s4 + s2 + 1

)6 L−1

s4 + 4a4

)7 L−1

s2 + 6s+ 13

1 L−1(

(s+ 4)(s2 + 9)

)2 L−1

(s2 − 1)

3 L−1(

(s− 1)2(s+ 2)

)4 L−1

(2s2 − 1

(s2 + 1)(s2 + 4)

)5 L−1

s4 + s2 + 1

)6 L−1

s4 + 4a4

)7 L−1

s2 + 6s+ 13

1 L−1(

(s+ 4)(s2 + 9)

)2 L−1

(s2 − 1)

)3 L−1

(4s+ 5

(s− 1)2(s+ 2)

4 L−1(

2s2 − 1

(s2 + 1)(s2 + 4)

)5 L−1

s4 + s2 + 1

)6 L−1

s4 + 4a4

)7 L−1

s2 + 6s+ 13

1 L−1(

(s+ 4)(s2 + 9)

)2 L−1

(s2 − 1)

)3 L−1

(4s+ 5

(s− 1)2(s+ 2)

)4 L−1

(2s2 − 1

(s2 + 1)(s2 + 4)

5 L−1(

s4 + s2 + 1

)6 L−1

s4 + 4a4

)7 L−1

s2 + 6s+ 13

1 L−1(

(s+ 4)(s2 + 9)

)2 L−1

(s2 − 1)

)3 L−1

(4s+ 5

(s− 1)2(s+ 2)

)4 L−1

(2s2 − 1

(s2 + 1)(s2 + 4)

)5 L−1

s4 + s2 + 1

6 L−1(

s4 + 4a4

)7 L−1

s2 + 6s+ 13

1 L−1(

(s+ 4)(s2 + 9)

)2 L−1

(s2 − 1)

)3 L−1

(4s+ 5

(s− 1)2(s+ 2)

)4 L−1

(2s2 − 1

(s2 + 1)(s2 + 4)

)5 L−1

s4 + s2 + 1

)6 L−1

s4 + 4a4

7 L−1(

s2 + 6s+ 13

1 L−1(

(s+ 4)(s2 + 9)

)2 L−1

(s2 − 1)

)3 L−1

(4s+ 5

(s− 1)2(s+ 2)

)4 L−1

(2s2 − 1

(s2 + 1)(s2 + 4)

)5 L−1

s4 + s2 + 1

)6 L−1

s4 + 4a4

)7 L−1

s2 + 6s+ 13

Transformation of Derivatives

Thm: If f ′(t) be continuous and L [f(t)] = f(s) thenL {f ′(t)} = sf(s)− f(0) provided lim

t→∞e−stf(t) = 0

i.e. L {f ′(t)} = sL {f(t)} − f(0)

Simillarly L {f ′′(t)} = sL {f ′(t)} − f ′(0)

= s [sL {f(t)} − f(0)]− f ′(0)

= s2L {f(t)} − sf(0)− f ′(0)

= s2f(s)− sf(0)− f ′(0)

In generalL {fn(t)} = snf(s)− sn−1f(0)− sn−2f ′(0)− . . .− fn−1(0)

i.e. L {f ′(t)} = sL {f(t)} − f(0)

= s [sL {f(t)} − f(0)]− f ′(0)

= s2L {f(t)} − sf(0)− f ′(0)

= s2f(s)− sf(0)− f ′(0)

i.e. L {f ′(t)} = sL {f(t)} − f(0)

= s [sL {f(t)} − f(0)]− f ′(0)

= s2L {f(t)} − sf(0)− f ′(0)

= s2f(s)− sf(0)− f ′(0)

i.e. L {f ′(t)} = sL {f(t)} − f(0)

= s [sL {f(t)} − f(0)]− f ′(0)

= s2L {f(t)} − sf(0)− f ′(0)

= s2f(s)− sf(0)− f ′(0)

i.e. L {f ′(t)} = sL {f(t)} − f(0)

= s [sL {f(t)} − f(0)]− f ′(0)

= s2L {f(t)} − sf(0)− f ′(0)

= s2f(s)− sf(0)− f ′(0)

i.e. L {f ′(t)} = sL {f(t)} − f(0)

= s [sL {f(t)} − f(0)]− f ′(0)

= s2L {f(t)} − sf(0)− f ′(0)

= s2f(s)− sf(0)− f ′(0)

i.e. L {f ′(t)} = sL {f(t)} − f(0)

= s [sL {f(t)} − f(0)]− f ′(0)

= s2L {f(t)} − sf(0)− f ′(0)

= s2f(s)− sf(0)− f ′(0)

i.e. L {f ′(t)} = sL {f(t)} − f(0)

= s [sL {f(t)} − f(0)]− f ′(0)

= s2L {f(t)} − sf(0)− f ′(0)

= s2f(s)− sf(0)− f ′(0)

Examples of Transformation of Derivatives

Ex. Derive the Laplace transform of sin at and cos at

Ex. Obtain L {tn} from L(1) =1

sEx. Find L(t sin at)

Ex. Find L(t cos at)

Ex. Find L(t sin at)

Transformation of Integrals

Thm: If L [f(t)] = f(s) then L

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t)

and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

∴ L−1{

0f(u)du

{∫ t

0f(u)du

Proof: Let I(t) =

0f(u)du

∴ I ′(t) =d

[∫ t

0f(u)du

]= f(t) and I(0) = 0

∴ L {f(t)} = L {I ′(t)} = sI(s)− I(0) = sI(s)

∴ L {f(t)} = sI(s)

∴ L {f(t)} = sL {I(t)}

∴ f(s) = sL

{∫ t

0f(u)du

sf(s) = L

{∫ t

0f(u)du

}∴ L−1

0f(u)du

Examples of Transformation of Integrals

Ex. Prove that: L−1(

s2 + 1

)= sin t

Ex. Prove that: L−1(

s(s2 + 1)

)= 1− cos t

Ex. Find inverse Laplace transform of1

s3(s2 + a2)

Multiplication by tn

Thm: If L [f(t)] = f(s) then L {tnf(t)} = (−1)ndn

dsn[f(s)

if L {tf(t)} = (−1)1d

)then L−1

{f ′(s)

}= −tf(t)

Multiplication by tn

Thm: If L [f(t)] = f(s) then L {tnf(t)} = (−1)ndn

dsn[f(s)

]if L {tf(t)} = (−1)1

)then L−1

{f ′(s)

}= −tf(t)

Examples of Laplace transform when tn is inmultiplication

1 L{t2eat

2 L{t3e−3t

}3 L {tcos at}4 L

{tsin2t

{te2tcos 3t

}6 L {tcos(4t+ 3)}

1 L{t2eat

{t3e−3t

3 L {tcos at}4 L

{tsin2t

{te2tcos 3t

}6 L {tcos(4t+ 3)}

1 L{t2eat

{t3e−3t

}3 L {tcos at}

4 L{tsin2t

{te2tcos 3t

}6 L {tcos(4t+ 3)}

1 L{t2eat

{t3e−3t

}3 L {tcos at}4 L

{tsin2t

5 L{te2tcos 3t

}6 L {tcos(4t+ 3)}

1 L{t2eat

{t3e−3t

}3 L {tcos at}4 L

{tsin2t

{te2tcos 3t

6 L {tcos(4t+ 3)}

1 L{t2eat

{t3e−3t

}3 L {tcos at}4 L

{tsin2t

{te2tcos 3t

}6 L {tcos(4t+ 3)}

Division by t

∫ ∞s

f(s) provided the

integral exists.

Examples of Laplace Transform when t is in division

{sin t

{1− cos 2t

{e−at − e−bt

{cos 2t− cos 3t

{1− et

{cos at− cos bt

{e−tsin t

{sin t

{1− cos 2t

{e−at − e−bt

{cos 2t− cos 3t

{1− et

{cos at− cos bt

{e−tsin t

{sin t

{1− cos 2t

{e−at − e−bt

{cos 2t− cos 3t

{1− et

{cos at− cos bt

{e−tsin t

{sin t

{1− cos 2t

{e−at − e−bt

{cos 2t− cos 3t

{1− et

{cos at− cos bt

{e−tsin t

{sin t

{1− cos 2t

{e−at − e−bt

{cos 2t− cos 3t

{1− et

{cos at− cos bt

{e−tsin t

{sin t

{1− cos 2t

{e−at − e−bt

{cos 2t− cos 3t

{1− et

{cos at− cos bt

{e−tsin t

{sin t

{1− cos 2t

{e−at − e−bt

{cos 2t− cos 3t

{1− et

{cos at− cos bt

{e−tsin t

}N. B. Vyas Laplace Transforms

Examples of infinite integral using Laplace Transform

1 Find

∫ ∞0

te−2tsin t dt

2 Find

∫ ∞0

3 Find

∫ ∞0

e−t − e−3t

4 Find

∫ ∞0

e−tsin2 t

1 Find

∫ ∞0

te−2tsin t dt

2 Find

∫ ∞0

3 Find

∫ ∞0

e−t − e−3t

4 Find

∫ ∞0

e−tsin2 t

1 Find

∫ ∞0

te−2tsin t dt

2 Find

∫ ∞0

3 Find

∫ ∞0

e−t − e−3t

4 Find

∫ ∞0

e−tsin2 t

1 Find

∫ ∞0

te−2tsin t dt

2 Find

∫ ∞0

3 Find

∫ ∞0

e−t − e−3t

4 Find

∫ ∞0

e−tsin2 t

Examples of Inverse Laplace Transform

1 L−1(

(s2 + a2)2

2 L−1(cot−1

)3 L−1

s− 1

1 L−1(

(s2 + a2)2

)2 L−1

(cot−1

3 L−1(log

s− 1

1 L−1(

(s2 + a2)2

)2 L−1

(cot−1

)3 L−1

s− 1

Convolution

Defn:Convolution of function f(t) and g(t) is denoted f(t) ∗ g(t) and

defined as f(t) ∗ g(t) =

0f(u)g(t− u) du

Theorem: Convolution theorem

If L−1{f(s)

}= f(t) and L−1 {g(s)} = g(t) then

L−1(f(s)g(s)

0f(u)g(t− u) du

Convolution

Defn:Convolution of function f(t) and g(t) is denoted f(t) ∗ g(t) and

defined as f(t) ∗ g(t) =

0f(u)g(t− u) du

Theorem: Convolution theorem

If L−1{f(s)

}= f(t) and L−1 {g(s)} = g(t) then

L−1(f(s)g(s)

0f(u)g(t− u) du

Convolution

Proof: Let φ(t) =

0f(u)g(t− u) du

then L(φ(t)) =

∫ ∞0

e−st{∫ t

0f(u)g(t− u) du

∫ ∞0

0e−stf(u)g(t− u) du dt

The region integration for this double integration is entire arealying between the lines u = 0 and u = t. On changing the orderof integration, we get

L(φ(t)) =

∫ ∞0

∫ ∞u

e−stf(u)g(t− u) dt du

∫ ∞0

e−suf(u)

{∫ ∞u

e−st+sug(t− u) dt

∫ ∞0

e−suf(u)

{∫ ∞u

e−s(t−u)g(t− u) dt

∫ ∞0

e−suf(u)

{∫ ∞u

e−svg(v) dv

}du, Putting t− u = v

Convolution

Proof: Let φ(t) =

0f(u)g(t− u) du

then L(φ(t)) =

∫ ∞0

e−st{∫ t

0f(u)g(t− u) du

∫ ∞0

L(φ(t)) =

∫ ∞0

∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

e−svg(v) dv

Convolution

Proof: Let φ(t) =

0f(u)g(t− u) du

then L(φ(t)) =

∫ ∞0

e−st{∫ t

0f(u)g(t− u) du

∫ ∞0

L(φ(t)) =

∫ ∞0

∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

e−svg(v) dv

Convolution

Proof: Let φ(t) =

0f(u)g(t− u) du

then L(φ(t)) =

∫ ∞0

e−st{∫ t

0f(u)g(t− u) du

∫ ∞0

L(φ(t)) =

∫ ∞0

∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

e−svg(v) dv

Convolution

Proof: Let φ(t) =

0f(u)g(t− u) du

then L(φ(t)) =

∫ ∞0

e−st{∫ t

0f(u)g(t− u) du

∫ ∞0

L(φ(t)) =

∫ ∞0

∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

e−svg(v) dv

Convolution

Proof: Let φ(t) =

0f(u)g(t− u) du

then L(φ(t)) =

∫ ∞0

e−st{∫ t

0f(u)g(t− u) du

∫ ∞0

L(φ(t)) =

∫ ∞0

∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

e−svg(v) dv

Convolution

Proof: Let φ(t) =

0f(u)g(t− u) du

then L(φ(t)) =

∫ ∞0

e−st{∫ t

0f(u)g(t− u) du

∫ ∞0

L(φ(t)) =

∫ ∞0

∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

e−svg(v) dv

Convolution

Proof: Let φ(t) =

0f(u)g(t− u) du

then L(φ(t)) =

∫ ∞0

e−st{∫ t

0f(u)g(t− u) du

∫ ∞0

L(φ(t)) =

∫ ∞0

∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

∫ ∞0

e−suf(u)

{∫ ∞u

e−svg(v) dv

Convolution

∫ ∞0

e−suf(u)g(s)du

= g(s)

∫ ∞0

e−suf(u)du

∴ L(φ(t)) = g(s)f(s)

L−1{g(s)f(s)

}= φ(t) =

0f(u)g(t− u) du

Convolution

∫ ∞0

e−suf(u)g(s)du

= g(s)

∫ ∞0

e−suf(u)du

∴ L(φ(t)) = g(s)f(s)

L−1{g(s)f(s)

}= φ(t) =

0f(u)g(t− u) du

Convolution

∫ ∞0

e−suf(u)g(s)du

= g(s)

∫ ∞0

e−suf(u)du

∴ L(φ(t)) = g(s)f(s)

L−1{g(s)f(s)

}= φ(t) =

0f(u)g(t− u) du

Convolution

∫ ∞0

e−suf(u)g(s)du

= g(s)

∫ ∞0

e−suf(u)du

∴ L(φ(t)) = g(s)f(s)

L−1{g(s)f(s)

}= φ(t) =

0f(u)g(t− u) du

Examples of Convolution theorem

Apply convolution theorem to evaluate:

Ex. L−1{

s2(s− 1)

Ex. L−1{

(s2 + 4)2

}Ex. L−1

(s+ a)(s+ b)

}Ex. L−1

s(s2 + 4)

Ex. L−1{

s2(s− 1)

}Ex. L−1

(s2 + 4)2

Ex. L−1{

(s+ a)(s+ b)

}Ex. L−1

s(s2 + 4)

Ex. L−1{

s2(s− 1)

}Ex. L−1

(s2 + 4)2

}Ex. L−1

(s+ a)(s+ b)

Ex. L−1{

s(s2 + 4)

Ex. L−1{

s2(s− 1)

}Ex. L−1

(s2 + 4)2

}Ex. L−1

(s+ a)(s+ b)

}Ex. L−1

s(s2 + 4)

Application to Differential Equations

Ex. Use transform method to solve y′′ + 3y′ + 2y = et, y(0) = 1 ,y′(0) = 0

Ex. Solve the equation x′′ + 2x′ + 5x = e−t sin t, x(0) = 0 , x′(0) = 1

Application to Differential Equations

Ex. Use transform method to solve y′′ + 3y′ + 2y = et, y(0) = 1 ,y′(0) = 0

Ex. Solve the equation x′′ + 2x′ + 5x = e−t sin t, x(0) = 0 , x′(0) = 1

Laplace transform of Periodic function

If f(t) is sectionally continuous function over aninterval of length p (0 ≤ t ≤ p) and f(t) is aperiodic function with period p (p > 0), that isf(t+ p) = f(t), then its Laplace transform existsand

L{f(t)} =1

1− e−ps

e−stf(t)dt, (s > 0)

Periodic Square Wave

Ex. Find the Laplace transform of the square wavefunction of period 2a defined as

f(t) =

{k if 0 ≤ t < a−k if a < t ≤ 2a

Periodic Triangular Wave

Ex. Find the Laplace transform of periodic function

f(t) =

{t if 0 < t < a2a− t if a < t < 2a

with period 2a

Unit Step function or Heaviside’s unit function

The Heaviside step function, or the unit stepfunction, usually denoted by H (but sometimes uor θ), is a discontinuous function whose value iszero for negative argument and one for positiveargument.

The function is used in the mathematics of controltheory, signal processing, structural mechanics,etc..

The Heaviside step function, or the unit stepfunction, usually denoted by H (but sometimes uor θ), is a discontinuous function whose value iszero for negative argument and one for positiveargument.

The function is used in the mathematics of controltheory, signal processing, structural mechanics,etc..

It is denoted by ua(t) or u(t− a) or H(t− a) and

is defined as H(t− a) =

{0 t < a1 t ≥ a

In particular, when a = 0

H(t) =

{0 t < 01 t ≥ 0

It is denoted by ua(t) or u(t− a) or H(t− a) and

is defined as H(t− a) =

{0 t < a1 t ≥ a

In particular, when a = 0

H(t) =

{0 t < 01 t ≥ 0

Laplace Transform of Unit Step Function:

By definition of Laplace transform

L{u(t− a)} =

∫ ∞0

e−stu(t− a)dt

e−st(0)dt+

∫ ∞a

e−st(1)dt

∫ ∞a

e−stdt =

[e−st

se−as

∴ L−1

se−as

]= u(t− a)

In particular, if a = 0

L(u(t)) =1

s⇒ L−1

)= u(t)

L{u(t− a)} =

∫ ∞0

e−stu(t− a)dt

e−st(0)dt+

∫ ∞a

e−st(1)dt

∫ ∞a

e−stdt =

[e−st

se−as

∴ L−1

se−as

]= u(t− a)

L(u(t)) =1

s⇒ L−1

)= u(t)

L{u(t− a)} =

∫ ∞0

e−stu(t− a)dt

e−st(0)dt+

∫ ∞a

e−st(1)dt

∫ ∞a

e−stdt =

[e−st

se−as

∴ L−1

se−as

]= u(t− a)

L(u(t)) =1

s⇒ L−1

)= u(t)

L{u(t− a)} =

∫ ∞0

e−stu(t− a)dt

e−st(0)dt+

∫ ∞a

e−st(1)dt

∫ ∞a

e−stdt =

[e−st

se−as

∴ L−1

se−as

]= u(t− a)

L(u(t)) =1

s⇒ L−1

)= u(t)

L{u(t− a)} =

∫ ∞0

e−stu(t− a)dt

e−st(0)dt+

∫ ∞a

e−st(1)dt

∫ ∞a

e−stdt =

[e−st

se−as

∴ L−1

se−as

]= u(t− a)

L(u(t)) =1

s⇒ L−1

)= u(t)

L{u(t− a)} =

∫ ∞0

e−stu(t− a)dt

e−st(0)dt+

∫ ∞a

e−st(1)dt

∫ ∞a

e−stdt =

[e−st

se−as

∴ L−1

se−as

]= u(t− a)

L(u(t)) =1

s⇒ L−1

)= u(t)

Second Shifting Theorem

Second Shifting Theorem:If L{f(t)} = f(s), then L{f(t− a)u(t− a)} = e−asf(s)

∴ L−1[e−asf(s)] = f(t− a)u(t− a)

Corollary: L{f(t)H(t− a)} = e−asL{f(t+ a)}

Corollary: L{H(t− a)−H(t− b)} =e−as − e−bs

sCorollary:L{f(t) [H(t− a)−H(t− b)]} = e−asL{f(t+a)}−e−bsL{f(t+b)}

Second Shifting Theorem:If L{f(t)} = f(s), then L{f(t− a)u(t− a)} = e−asf(s)∴ L−1[e−asf(s)] = f(t− a)u(t− a)

Corollary:L{f(t) [H(t− a)−H(t− b)]} = e−asL{f(t+a)}−e−bsL{f(t+b)}

Laplace Transforms

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