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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).
N. B. Vyas Laplace Transforms
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), ...
N. B. Vyas Laplace Transforms
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)]
N. B. Vyas Laplace Transforms
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)]
N. B. Vyas Laplace Transforms
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)]
N. B. Vyas Laplace Transforms
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)]
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
1 L(1) =1
s
2 L(eat) =1
s− a
cor.1 If a = 0⇒ L(1) =1
s
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
1 L(1) =1
s
2 L(eat) =1
s− a
cor.1 If a = 0⇒ L(1) =1
s
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
1 L(1) =1
s
2 L(eat) =1
s− a
cor.1 If a = 0⇒ L(1) =1
s
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
1 L(1) =1
s
2 L(eat) =1
s− a
cor.1 If a = 0⇒ L(1) =1
s
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
1 L(1) =1
s
2 L(eat) =1
s− a
cor.1 If a = 0⇒ L(1) =1
s
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
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
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
7 L[tn] =Γ(n+ 1)
sn+1
=n!
sn+1, n = 0, 1, 2, ...
cor.1 If n = 0, L[1] =1
s
cor.2 If n = −1
2
L(t−
12
)=
Γ(12)
s12
=
√π
s
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
7 L[tn] =Γ(n+ 1)
sn+1
=n!
sn+1, n = 0, 1, 2, ...
cor.1 If n = 0, L[1] =1
s
cor.2 If n = −1
2
L(t−
12
)=
Γ(12)
s12
=
√π
s
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
7 L[tn] =Γ(n+ 1)
sn+1
=n!
sn+1, n = 0, 1, 2, ...
cor.1 If n = 0, L[1] =1
s
cor.2 If n = −1
2
L(t−
12
)=
Γ(12)
s12
=
√π
s
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
7 L[tn] =Γ(n+ 1)
sn+1
=n!
sn+1, n = 0, 1, 2, ...
cor.1 If n = 0, L[1] =1
s
cor.2 If n = −1
2
L(t−
12
)=
Γ(12)
s12
=
√π
s
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
7 L[tn] =Γ(n+ 1)
sn+1
=n!
sn+1, n = 0, 1, 2, ...
cor.1 If n = 0, L[1] =1
s
cor.2 If n = −1
2
L(t−
12
)=
Γ(12)
s12
=
√π
s
N. B. Vyas Laplace Transforms
Laplace Transforms of some elementary functions
7 L[tn] =Γ(n+ 1)
sn+1
=n!
sn+1, n = 0, 1, 2, ...
cor.1 If n = 0, L[1] =1
s
cor.2 If n = −1
2
L(t−
12
)=
Γ(12)
s12
=
√π
s
N. B. Vyas Laplace Transforms
Examples of Laplace Transform
1 L
{2t3 + e−2t + t
43
}
2 L
{A+B t
12 + C t−
12
}3 L
{eat − 1
a
}4 L {sin(at+ b)}5 L {sin 2t cos 3t}6 L
{cos24t
}7 L
{cos32t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform
1 L
{2t3 + e−2t + t
43
}2 L
{A+B t
12 + C t−
12
}
3 L
{eat − 1
a
}4 L {sin(at+ b)}5 L {sin 2t cos 3t}6 L
{cos24t
}7 L
{cos32t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform
1 L
{2t3 + e−2t + t
43
}2 L
{A+B t
12 + C t−
12
}3 L
{eat − 1
a
}
4 L {sin(at+ b)}5 L {sin 2t cos 3t}6 L
{cos24t
}7 L
{cos32t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform
1 L
{2t3 + e−2t + t
43
}2 L
{A+B t
12 + C t−
12
}3 L
{eat − 1
a
}4 L {sin(at+ b)}
5 L {sin 2t cos 3t}6 L
{cos24t
}7 L
{cos32t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform
1 L
{2t3 + e−2t + t
43
}2 L
{A+B t
12 + C t−
12
}3 L
{eat − 1
a
}4 L {sin(at+ b)}5 L {sin 2t cos 3t}
6 L{
cos24t}
7 L{
cos32t}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform
1 L
{2t3 + e−2t + t
43
}2 L
{A+B t
12 + C t−
12
}3 L
{eat − 1
a
}4 L {sin(at+ b)}5 L {sin 2t cos 3t}6 L
{cos24t
}
7 L{
cos32t}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform
1 L
{2t3 + e−2t + t
43
}2 L
{A+B t
12 + C t−
12
}3 L
{eat − 1
a
}4 L {sin(at+ b)}5 L {sin 2t cos 3t}6 L
{cos24t
}7 L
{cos32t
}
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
First Shifting Theorem
Note:
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
N. B. Vyas Laplace Transforms
First Shifting Theorem
Note:
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
N. B. Vyas Laplace Transforms
First Shifting Theorem
Note:
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
N. B. Vyas Laplace Transforms
First Shifting Theorem
Note:
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
N. B. Vyas Laplace Transforms
First Shifting Theorem
Note:
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
N. B. Vyas Laplace Transforms
First Shifting Theorem
Note:
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
N. B. Vyas Laplace Transforms
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]
N. B. Vyas Laplace Transforms
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]
N. B. Vyas Laplace Transforms
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]
N. B. Vyas Laplace Transforms
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]
N. B. Vyas Laplace Transforms
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]
N. B. Vyas Laplace Transforms
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]
N. B. Vyas Laplace Transforms
Examples
Ex. Find the Laplace transform of the function which is defined as
f(t) =
{t/T 0 < t < T1 when t > T
N. B. Vyas Laplace Transforms
Examples
Ex. Find Laplace transform of f(t) =
{sin t 0 < t < π
0 when t > π
N. B. Vyas Laplace Transforms
Examples
Ex. Find Laplace transform of f(t) where f(t) =
{t 0 < t < 45 when t > 4
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b=
1
bf
(s− ab
)
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b=
1
bf
(s− ab
)
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b=
1
bf
(s− ab
)
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b=
1
bf
(s− ab
)
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b=
1
bf
(s− ab
)
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b=
1
bf
(s− ab
)
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b=
1
bf
(s− ab
)
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b
=1
bf
(s− ab
)
N. B. Vyas Laplace Transforms
Change of Scale property
If L {f (t)} = f (s) then L{eatf (bt)
}=
1
bf
(s− ab
), b > 0
Proof: By the def. of Laplace
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)
du
b=
1
bf
(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(s)
}= f (t)
1 L−1(
1
s
)= 1
2 L−1(
1
s− a
)= eat
3 L−1(
1
s2 + a2
)=
1
asin at
4 L−1(
s
s2 + a2
)= cos at
5 L−1(
1
s2 − a2
)=
1
asinh at
6 L−1(
s
s2 − a2
)= cosh at
7 L−1(
1
sn
)=
tn−1
(n− 1)!
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(s)
}= f (t)
1 L−1(
1
s
)= 1
2 L−1(
1
s− a
)= eat
3 L−1(
1
s2 + a2
)=
1
asin at
4 L−1(
s
s2 + a2
)= cos at
5 L−1(
1
s2 − a2
)=
1
asinh at
6 L−1(
s
s2 − a2
)= cosh at
7 L−1(
1
sn
)=
tn−1
(n− 1)!
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(s)
}= f (t)
1 L−1(
1
s
)= 1
2 L−1(
1
s− a
)= eat
3 L−1(
1
s2 + a2
)=
1
asin at
4 L−1(
s
s2 + a2
)= cos at
5 L−1(
1
s2 − a2
)=
1
asinh at
6 L−1(
s
s2 − a2
)= cosh at
7 L−1(
1
sn
)=
tn−1
(n− 1)!
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(s)
}= f (t)
1 L−1(
1
s
)= 1
2 L−1(
1
s− a
)= eat
3 L−1(
1
s2 + a2
)=
1
asin at
4 L−1(
s
s2 + a2
)= cos at
5 L−1(
1
s2 − a2
)=
1
asinh at
6 L−1(
s
s2 − a2
)= cosh at
7 L−1(
1
sn
)=
tn−1
(n− 1)!
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(s)
}= f (t)
1 L−1(
1
s
)= 1
2 L−1(
1
s− a
)= eat
3 L−1(
1
s2 + a2
)=
1
asin at
4 L−1(
s
s2 + a2
)= cos at
5 L−1(
1
s2 − a2
)=
1
asinh at
6 L−1(
s
s2 − a2
)= cosh at
7 L−1(
1
sn
)=
tn−1
(n− 1)!
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(s)
}= f (t)
1 L−1(
1
s
)= 1
2 L−1(
1
s− a
)= eat
3 L−1(
1
s2 + a2
)=
1
asin at
4 L−1(
s
s2 + a2
)= cos at
5 L−1(
1
s2 − a2
)=
1
asinh at
6 L−1(
s
s2 − a2
)= cosh at
7 L−1(
1
sn
)=
tn−1
(n− 1)!
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(s)
}= f (t)
1 L−1(
1
s
)= 1
2 L−1(
1
s− a
)= eat
3 L−1(
1
s2 + a2
)=
1
asin at
4 L−1(
s
s2 + a2
)= cos at
5 L−1(
1
s2 − a2
)=
1
asinh at
6 L−1(
s
s2 − a2
)= cosh at
7 L−1(
1
sn
)=
tn−1
(n− 1)!
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(s)
}= f (t)
1 L−1(
1
s
)= 1
2 L−1(
1
s− a
)= eat
3 L−1(
1
s2 + a2
)=
1
asin at
4 L−1(
s
s2 + a2
)= cos at
5 L−1(
1
s2 − a2
)=
1
asinh at
6 L−1(
s
s2 − a2
)= cosh at
7 L−1(
1
sn
)=
tn−1
(n− 1)!
N. B. Vyas Laplace Transforms
Partial Fractions
N. B. Vyas Laplace Transforms
Partial Fractions
N. B. Vyas Laplace Transforms
Partial Fractions
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - I
1 L−1(s2 − 3s+ 4
s3
)
2 L−1[
3
2
(s4 − 2s2 + 1
s5
)]3 L−1
(s+ 7
(s+ 1)2 + 1
)4 L−1
(3s+ 5
(s+ 1)4
)5 L−1
(3s
s2 + 2s− 8
)6 L−1
(3s+ 7
s2 − 2s− 3
)7 L−1
(2s2 − 6s+ 5
s3 − 6s2 + 11s− 6
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - I
1 L−1(s2 − 3s+ 4
s3
)2 L−1
[3
2
(s4 − 2s2 + 1
s5
)]
3 L−1(
s+ 7
(s+ 1)2 + 1
)4 L−1
(3s+ 5
(s+ 1)4
)5 L−1
(3s
s2 + 2s− 8
)6 L−1
(3s+ 7
s2 − 2s− 3
)7 L−1
(2s2 − 6s+ 5
s3 − 6s2 + 11s− 6
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - I
1 L−1(s2 − 3s+ 4
s3
)2 L−1
[3
2
(s4 − 2s2 + 1
s5
)]3 L−1
(s+ 7
(s+ 1)2 + 1
)
4 L−1(
3s+ 5
(s+ 1)4
)5 L−1
(3s
s2 + 2s− 8
)6 L−1
(3s+ 7
s2 − 2s− 3
)7 L−1
(2s2 − 6s+ 5
s3 − 6s2 + 11s− 6
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - I
1 L−1(s2 − 3s+ 4
s3
)2 L−1
[3
2
(s4 − 2s2 + 1
s5
)]3 L−1
(s+ 7
(s+ 1)2 + 1
)4 L−1
(3s+ 5
(s+ 1)4
)
5 L−1(
3s
s2 + 2s− 8
)6 L−1
(3s+ 7
s2 − 2s− 3
)7 L−1
(2s2 − 6s+ 5
s3 − 6s2 + 11s− 6
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - I
1 L−1(s2 − 3s+ 4
s3
)2 L−1
[3
2
(s4 − 2s2 + 1
s5
)]3 L−1
(s+ 7
(s+ 1)2 + 1
)4 L−1
(3s+ 5
(s+ 1)4
)5 L−1
(3s
s2 + 2s− 8
)
6 L−1(
3s+ 7
s2 − 2s− 3
)7 L−1
(2s2 − 6s+ 5
s3 − 6s2 + 11s− 6
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - I
1 L−1(s2 − 3s+ 4
s3
)2 L−1
[3
2
(s4 − 2s2 + 1
s5
)]3 L−1
(s+ 7
(s+ 1)2 + 1
)4 L−1
(3s+ 5
(s+ 1)4
)5 L−1
(3s
s2 + 2s− 8
)6 L−1
(3s+ 7
s2 − 2s− 3
)
7 L−1(
2s2 − 6s+ 5
s3 − 6s2 + 11s− 6
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - I
1 L−1(s2 − 3s+ 4
s3
)2 L−1
[3
2
(s4 − 2s2 + 1
s5
)]3 L−1
(s+ 7
(s+ 1)2 + 1
)4 L−1
(3s+ 5
(s+ 1)4
)5 L−1
(3s
s2 + 2s− 8
)6 L−1
(3s+ 7
s2 − 2s− 3
)7 L−1
(2s2 − 6s+ 5
s3 − 6s2 + 11s− 6
)N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - II
1 L−1(
s+ 29
(s+ 4)(s2 + 9)
)
2 L−1(
s
(s2 − 1)
)3 L−1
(4s+ 5
(s− 1)2(s+ 2)
)4 L−1
(2s2 − 1
(s2 + 1)(s2 + 4)
)5 L−1
(s
s4 + s2 + 1
)6 L−1
(s
s4 + 4a4
)7 L−1
(s+ 3
s2 + 6s+ 13
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - II
1 L−1(
s+ 29
(s+ 4)(s2 + 9)
)2 L−1
(s
(s2 − 1)
)
3 L−1(
4s+ 5
(s− 1)2(s+ 2)
)4 L−1
(2s2 − 1
(s2 + 1)(s2 + 4)
)5 L−1
(s
s4 + s2 + 1
)6 L−1
(s
s4 + 4a4
)7 L−1
(s+ 3
s2 + 6s+ 13
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - II
1 L−1(
s+ 29
(s+ 4)(s2 + 9)
)2 L−1
(s
(s2 − 1)
)3 L−1
(4s+ 5
(s− 1)2(s+ 2)
)
4 L−1(
2s2 − 1
(s2 + 1)(s2 + 4)
)5 L−1
(s
s4 + s2 + 1
)6 L−1
(s
s4 + 4a4
)7 L−1
(s+ 3
s2 + 6s+ 13
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - II
1 L−1(
s+ 29
(s+ 4)(s2 + 9)
)2 L−1
(s
(s2 − 1)
)3 L−1
(4s+ 5
(s− 1)2(s+ 2)
)4 L−1
(2s2 − 1
(s2 + 1)(s2 + 4)
)
5 L−1(
s
s4 + s2 + 1
)6 L−1
(s
s4 + 4a4
)7 L−1
(s+ 3
s2 + 6s+ 13
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - II
1 L−1(
s+ 29
(s+ 4)(s2 + 9)
)2 L−1
(s
(s2 − 1)
)3 L−1
(4s+ 5
(s− 1)2(s+ 2)
)4 L−1
(2s2 − 1
(s2 + 1)(s2 + 4)
)5 L−1
(s
s4 + s2 + 1
)
6 L−1(
s
s4 + 4a4
)7 L−1
(s+ 3
s2 + 6s+ 13
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - II
1 L−1(
s+ 29
(s+ 4)(s2 + 9)
)2 L−1
(s
(s2 − 1)
)3 L−1
(4s+ 5
(s− 1)2(s+ 2)
)4 L−1
(2s2 − 1
(s2 + 1)(s2 + 4)
)5 L−1
(s
s4 + s2 + 1
)6 L−1
(s
s4 + 4a4
)
7 L−1(
s+ 3
s2 + 6s+ 13
)
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform - II
1 L−1(
s+ 29
(s+ 4)(s2 + 9)
)2 L−1
(s
(s2 − 1)
)3 L−1
(4s+ 5
(s− 1)2(s+ 2)
)4 L−1
(2s2 − 1
(s2 + 1)(s2 + 4)
)5 L−1
(s
s4 + s2 + 1
)6 L−1
(s
s4 + 4a4
)7 L−1
(s+ 3
s2 + 6s+ 13
)N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
Examples of Transformation of Derivatives
Ex. Derive the Laplace transform of sin at and cos at
Ex. Obtain L {tn} from L(1) =1
s
Ex. Find L(t sin at)
Ex. Find L(t cos at)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
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)
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}
∴1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}
∴ L−1{
1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Transformation of Integrals
Thm: If L [f(t)] = f(s) then L
{∫ t
0f(u)du
}=
1
sf(s)
Proof: Let I(t) =
∫ t
0f(u)du
∴ I ′(t) =d
dt
[∫ 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
}∴
1
sf(s) = L
{∫ t
0f(u)du
}∴ L−1
{1
sf(s)
}=
∫ t
0f(u)du
N. B. Vyas Laplace Transforms
Examples of Transformation of Integrals
Ex. Prove that: L−1(
1
s2 + 1
)= sin t
Ex. Prove that: L−1(
1
s(s2 + 1)
)= 1− cos t
Ex. Find inverse Laplace transform of1
s3(s2 + a2)
N. B. Vyas Laplace Transforms
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
ds
(f(s)
)then L−1
{f ′(s)
}= −tf(t)
N. B. Vyas Laplace Transforms
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
d
ds
(f(s)
)then L−1
{f ′(s)
}= −tf(t)
N. B. Vyas Laplace Transforms
Examples of Laplace transform when tn is inmultiplication
1 L{t2eat
}
2 L{t3e−3t
}3 L {tcos at}4 L
{tsin2t
}5 L
{te2tcos 3t
}6 L {tcos(4t+ 3)}
N. B. Vyas Laplace Transforms
Examples of Laplace transform when tn is inmultiplication
1 L{t2eat
}2 L
{t3e−3t
}
3 L {tcos at}4 L
{tsin2t
}5 L
{te2tcos 3t
}6 L {tcos(4t+ 3)}
N. B. Vyas Laplace Transforms
Examples of Laplace transform when tn is inmultiplication
1 L{t2eat
}2 L
{t3e−3t
}3 L {tcos at}
4 L{tsin2t
}5 L
{te2tcos 3t
}6 L {tcos(4t+ 3)}
N. B. Vyas Laplace Transforms
Examples of Laplace transform when tn is inmultiplication
1 L{t2eat
}2 L
{t3e−3t
}3 L {tcos at}4 L
{tsin2t
}
5 L{te2tcos 3t
}6 L {tcos(4t+ 3)}
N. B. Vyas Laplace Transforms
Examples of Laplace transform when tn is inmultiplication
1 L{t2eat
}2 L
{t3e−3t
}3 L {tcos at}4 L
{tsin2t
}5 L
{te2tcos 3t
}
6 L {tcos(4t+ 3)}
N. B. Vyas Laplace Transforms
Examples of Laplace transform when tn is inmultiplication
1 L{t2eat
}2 L
{t3e−3t
}3 L {tcos at}4 L
{tsin2t
}5 L
{te2tcos 3t
}6 L {tcos(4t+ 3)}
N. B. Vyas Laplace Transforms
Division by t
Thm: If L [f(t)] = f(s) then L
{1
tf(t)
}=
∫ ∞s
f(s) provided the
integral exists.
N. B. Vyas Laplace Transforms
Examples of Laplace Transform when t is in division
1 L
{sin t
t
}
2 L
{1− cos 2t
t
}3 L
{e−at − e−bt
t
}4 L
{cos 2t− cos 3t
t
}5 L
{1− et
t
}6 L
{cos at− cos bt
t
}7 L
{e−tsin t
t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform when t is in division
1 L
{sin t
t
}2 L
{1− cos 2t
t
}
3 L
{e−at − e−bt
t
}4 L
{cos 2t− cos 3t
t
}5 L
{1− et
t
}6 L
{cos at− cos bt
t
}7 L
{e−tsin t
t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform when t is in division
1 L
{sin t
t
}2 L
{1− cos 2t
t
}3 L
{e−at − e−bt
t
}
4 L
{cos 2t− cos 3t
t
}5 L
{1− et
t
}6 L
{cos at− cos bt
t
}7 L
{e−tsin t
t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform when t is in division
1 L
{sin t
t
}2 L
{1− cos 2t
t
}3 L
{e−at − e−bt
t
}4 L
{cos 2t− cos 3t
t
}
5 L
{1− et
t
}6 L
{cos at− cos bt
t
}7 L
{e−tsin t
t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform when t is in division
1 L
{sin t
t
}2 L
{1− cos 2t
t
}3 L
{e−at − e−bt
t
}4 L
{cos 2t− cos 3t
t
}5 L
{1− et
t
}
6 L
{cos at− cos bt
t
}7 L
{e−tsin t
t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform when t is in division
1 L
{sin t
t
}2 L
{1− cos 2t
t
}3 L
{e−at − e−bt
t
}4 L
{cos 2t− cos 3t
t
}5 L
{1− et
t
}6 L
{cos at− cos bt
t
}
7 L
{e−tsin t
t
}
N. B. Vyas Laplace Transforms
Examples of Laplace Transform when t is in division
1 L
{sin t
t
}2 L
{1− cos 2t
t
}3 L
{e−at − e−bt
t
}4 L
{cos 2t− cos 3t
t
}5 L
{1− et
t
}6 L
{cos at− cos bt
t
}7 L
{e−tsin t
t
}N. B. Vyas Laplace Transforms
Examples of infinite integral using Laplace Transform
1 Find
∫ ∞0
te−2tsin t dt
2 Find
∫ ∞0
sinmt
tdt
3 Find
∫ ∞0
e−t − e−3t
tdt
4 Find
∫ ∞0
e−tsin2 t
tdt
N. B. Vyas Laplace Transforms
Examples of infinite integral using Laplace Transform
1 Find
∫ ∞0
te−2tsin t dt
2 Find
∫ ∞0
sinmt
tdt
3 Find
∫ ∞0
e−t − e−3t
tdt
4 Find
∫ ∞0
e−tsin2 t
tdt
N. B. Vyas Laplace Transforms
Examples of infinite integral using Laplace Transform
1 Find
∫ ∞0
te−2tsin t dt
2 Find
∫ ∞0
sinmt
tdt
3 Find
∫ ∞0
e−t − e−3t
tdt
4 Find
∫ ∞0
e−tsin2 t
tdt
N. B. Vyas Laplace Transforms
Examples of infinite integral using Laplace Transform
1 Find
∫ ∞0
te−2tsin t dt
2 Find
∫ ∞0
sinmt
tdt
3 Find
∫ ∞0
e−t − e−3t
tdt
4 Find
∫ ∞0
e−tsin2 t
tdt
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform
1 L−1(
s
(s2 + a2)2
)
2 L−1(cot−1
s
a
)3 L−1
(log
(s+ 1
s− 1
))
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform
1 L−1(
s
(s2 + a2)2
)2 L−1
(cot−1
s
a
)
3 L−1(log
(s+ 1
s− 1
))
N. B. Vyas Laplace Transforms
Examples of Inverse Laplace Transform
1 L−1(
s
(s2 + a2)2
)2 L−1
(cot−1
s
a
)3 L−1
(log
(s+ 1
s− 1
))
N. B. Vyas Laplace Transforms
Convolution
Defn:Convolution of function f(t) and g(t) is denoted f(t) ∗ g(t) and
defined as f(t) ∗ g(t) =
∫ 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)
)=
∫ t
0f(u)g(t− u) du
N. B. Vyas Laplace Transforms
Convolution
Defn:Convolution of function f(t) and g(t) is denoted f(t) ∗ g(t) and
defined as f(t) ∗ g(t) =
∫ 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)
)=
∫ t
0f(u)g(t− u) du
N. B. Vyas Laplace Transforms
Convolution
Proof: Let φ(t) =
∫ t
0f(u)g(t− u) du
then L(φ(t)) =
∫ ∞0
e−st{∫ t
0f(u)g(t− u) du
}dt
=
∫ ∞0
∫ t
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
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−s(t−u)g(t− u) dt
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−svg(v) dv
}du, Putting t− u = v
N. B. Vyas Laplace Transforms
Convolution
Proof: Let φ(t) =
∫ t
0f(u)g(t− u) du
then L(φ(t)) =
∫ ∞0
e−st{∫ t
0f(u)g(t− u) du
}dt
=
∫ ∞0
∫ t
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
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−s(t−u)g(t− u) dt
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−svg(v) dv
}du, Putting t− u = v
N. B. Vyas Laplace Transforms
Convolution
Proof: Let φ(t) =
∫ t
0f(u)g(t− u) du
then L(φ(t)) =
∫ ∞0
e−st{∫ t
0f(u)g(t− u) du
}dt
=
∫ ∞0
∫ t
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
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−s(t−u)g(t− u) dt
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−svg(v) dv
}du, Putting t− u = v
N. B. Vyas Laplace Transforms
Convolution
Proof: Let φ(t) =
∫ t
0f(u)g(t− u) du
then L(φ(t)) =
∫ ∞0
e−st{∫ t
0f(u)g(t− u) du
}dt
=
∫ ∞0
∫ t
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
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−s(t−u)g(t− u) dt
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−svg(v) dv
}du, Putting t− u = v
N. B. Vyas Laplace Transforms
Convolution
Proof: Let φ(t) =
∫ t
0f(u)g(t− u) du
then L(φ(t)) =
∫ ∞0
e−st{∫ t
0f(u)g(t− u) du
}dt
=
∫ ∞0
∫ t
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
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−s(t−u)g(t− u) dt
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−svg(v) dv
}du, Putting t− u = v
N. B. Vyas Laplace Transforms
Convolution
Proof: Let φ(t) =
∫ t
0f(u)g(t− u) du
then L(φ(t)) =
∫ ∞0
e−st{∫ t
0f(u)g(t− u) du
}dt
=
∫ ∞0
∫ t
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
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−s(t−u)g(t− u) dt
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−svg(v) dv
}du, Putting t− u = v
N. B. Vyas Laplace Transforms
Convolution
Proof: Let φ(t) =
∫ t
0f(u)g(t− u) du
then L(φ(t)) =
∫ ∞0
e−st{∫ t
0f(u)g(t− u) du
}dt
=
∫ ∞0
∫ t
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
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−s(t−u)g(t− u) dt
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−svg(v) dv
}du, Putting t− u = v
N. B. Vyas Laplace Transforms
Convolution
Proof: Let φ(t) =
∫ t
0f(u)g(t− u) du
then L(φ(t)) =
∫ ∞0
e−st{∫ t
0f(u)g(t− u) du
}dt
=
∫ ∞0
∫ t
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
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−s(t−u)g(t− u) dt
}du
=
∫ ∞0
e−suf(u)
{∫ ∞u
e−svg(v) dv
}du, Putting t− u = v
N. B. Vyas Laplace Transforms
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) =
∫ t
0f(u)g(t− u) du
N. B. Vyas Laplace Transforms
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) =
∫ t
0f(u)g(t− u) du
N. B. Vyas Laplace Transforms
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) =
∫ t
0f(u)g(t− u) du
N. B. Vyas Laplace Transforms
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) =
∫ t
0f(u)g(t− u) du
N. B. Vyas Laplace Transforms
Examples of Convolution theorem
Apply convolution theorem to evaluate:
Ex. L−1{
1
s2(s− 1)
}
Ex. L−1{
s
(s2 + 4)2
}Ex. L−1
{1
(s+ a)(s+ b)
}Ex. L−1
{1
s(s2 + 4)
}
N. B. Vyas Laplace Transforms
Examples of Convolution theorem
Apply convolution theorem to evaluate:
Ex. L−1{
1
s2(s− 1)
}Ex. L−1
{s
(s2 + 4)2
}
Ex. L−1{
1
(s+ a)(s+ b)
}Ex. L−1
{1
s(s2 + 4)
}
N. B. Vyas Laplace Transforms
Examples of Convolution theorem
Apply convolution theorem to evaluate:
Ex. L−1{
1
s2(s− 1)
}Ex. L−1
{s
(s2 + 4)2
}Ex. L−1
{1
(s+ a)(s+ b)
}
Ex. L−1{
1
s(s2 + 4)
}
N. B. Vyas Laplace Transforms
Examples of Convolution theorem
Apply convolution theorem to evaluate:
Ex. L−1{
1
s2(s− 1)
}Ex. L−1
{s
(s2 + 4)2
}Ex. L−1
{1
(s+ a)(s+ b)
}Ex. L−1
{1
s(s2 + 4)
}
N. B. Vyas Laplace Transforms
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
N. B. Vyas Laplace Transforms
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
N. B. Vyas Laplace Transforms
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
∫ p
0
e−stf(t)dt, (s > 0)
N. B. Vyas Laplace Transforms
Laplace transform of Periodic function
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
N. B. Vyas Laplace Transforms
Laplace transform of Periodic function
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
N. B. Vyas Laplace Transforms
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..
N. B. Vyas Laplace Transforms
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..
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
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
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
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
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
Laplace Transform of Unit Step Function:
By definition of Laplace transform
L{u(t− a)} =
∫ ∞0
e−stu(t− a)dt
=
∫ a
0
e−st(0)dt+
∫ ∞a
e−st(1)dt
=
∫ ∞a
e−stdt =
[e−st
−s
]∞a
=1
se−as
∴ L−1
[1
se−as
]= u(t− a)
In particular, if a = 0
L(u(t)) =1
s⇒ L−1
(1
s
)= u(t)
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
Laplace Transform of Unit Step Function:
By definition of Laplace transform
L{u(t− a)} =
∫ ∞0
e−stu(t− a)dt
=
∫ a
0
e−st(0)dt+
∫ ∞a
e−st(1)dt
=
∫ ∞a
e−stdt =
[e−st
−s
]∞a
=1
se−as
∴ L−1
[1
se−as
]= u(t− a)
In particular, if a = 0
L(u(t)) =1
s⇒ L−1
(1
s
)= u(t)
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
Laplace Transform of Unit Step Function:
By definition of Laplace transform
L{u(t− a)} =
∫ ∞0
e−stu(t− a)dt
=
∫ a
0
e−st(0)dt+
∫ ∞a
e−st(1)dt
=
∫ ∞a
e−stdt =
[e−st
−s
]∞a
=1
se−as
∴ L−1
[1
se−as
]= u(t− a)
In particular, if a = 0
L(u(t)) =1
s⇒ L−1
(1
s
)= u(t)
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
Laplace Transform of Unit Step Function:
By definition of Laplace transform
L{u(t− a)} =
∫ ∞0
e−stu(t− a)dt
=
∫ a
0
e−st(0)dt+
∫ ∞a
e−st(1)dt
=
∫ ∞a
e−stdt =
[e−st
−s
]∞a
=1
se−as
∴ L−1
[1
se−as
]= u(t− a)
In particular, if a = 0
L(u(t)) =1
s⇒ L−1
(1
s
)= u(t)
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
Laplace Transform of Unit Step Function:
By definition of Laplace transform
L{u(t− a)} =
∫ ∞0
e−stu(t− a)dt
=
∫ a
0
e−st(0)dt+
∫ ∞a
e−st(1)dt
=
∫ ∞a
e−stdt =
[e−st
−s
]∞a
=1
se−as
∴ L−1
[1
se−as
]= u(t− a)
In particular, if a = 0
L(u(t)) =1
s⇒ L−1
(1
s
)= u(t)
N. B. Vyas Laplace Transforms
Unit Step function or Heaviside’s unit function
Laplace Transform of Unit Step Function:
By definition of Laplace transform
L{u(t− a)} =
∫ ∞0
e−stu(t− a)dt
=
∫ a
0
e−st(0)dt+
∫ ∞a
e−st(1)dt
=
∫ ∞a
e−stdt =
[e−st
−s
]∞a
=1
se−as
∴ L−1
[1
se−as
]= u(t− a)
In particular, if a = 0
L(u(t)) =1
s⇒ L−1
(1
s
)= u(t)
N. B. Vyas Laplace Transforms
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)}
N. B. Vyas Laplace Transforms
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)}
N. B. Vyas Laplace Transforms
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
s
Corollary:L{f(t) [H(t− a)−H(t− b)]} = e−asL{f(t+a)}−e−bsL{f(t+b)}
N. B. Vyas Laplace Transforms
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)}
N. B. Vyas Laplace Transforms