Transforms and its Applications: UNIT 1: Laplace Transforms

Inverse Laplace Transform

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Laplace Transforms: Important formulae and Example Important Solved Problems with formula, steps, derivation and answer based on Inverse Laplace Transform

INVERSE LAPLACE TRANSFORM

Now, we obtain f(t) when F(s) is given, then we say that inverse Laplace transform of F(s) is f(t).

(1) If L [f(t)] = F(s), then L^‒1[F(s)] = f(t),

where L^‒1 is called the inverse Laplace transform operator.

(2) If F₁(s) and F₂(s) are Laplace transform of f(t) and g(t) respectively, then L^‒1[C₁F₁(s) + C₂ F₂(s)] = C₁L^‒1 [F₁(s)] + C₂L^‒1[F₂ (s)]

Proof:

Given: L[f(t)] = F₁(s)

f(t) = L^‒1[F₁(s)]

L[g(t)] = F₂(s)

g(t) = L^‒1[F₂(s)]

We know that, L[C₁f(t) + C₂g(t)]

= C₁L[f(t)] + C₂L[g (t)]

= C₁F₁(s) + C₂F₂(s)

C₁f(t) + C₂g(t) = L^‒1[C₁F₁(s) + C₂ F₂ (s)]

i.e., L^‒1[C₁ F₁(s) + C₂F₂(s)] = C₁f(t) + C₂F₂(s)

= C₁L^‒1F₁(s) + C₂L^‒1F₂(s)

Note: (1) If L[f(t)] = F(s), then L[e^at f(t)] = F(s − a)

i.e., If L^‒1[F(s)] = f(t), then

L^‒1[F(s‒a)] = e^atf(t) = e^at f(t) = e^at L^‒1 [F(s)]

Note: (2) If L[f(t)] = F(s), then L[e^‒at f(t)] = F(s + a)

i.e., If L^‒1[F(s)] = f(t), then

L^‒1[F(s+a)] = e^‒atf(t) = e^‒at f(t) = e^‒at L^‒1 [F(s)]

IMPORTANT FORMULAE - INVERSE LAPLACE TRANSFORM

PROBLEMS BASED ON INVERSE LAPLACE TRANSFORM

Transforms and its Applications: UNIT 1: Laplace Transforms : Tag: Engineering mathematics, Maths : - Inverse Laplace Transform

Home | Department: EEE Department | Ist Year | Subject: Transforms and its Applications |

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