Transforms and its Applications: UNIT 1: Laplace Transforms

Laplace Transforms using Convolution Theorem

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Laplace Transforms: Example Important Solved Problems with formula, steps, derivation and answer based on Convolution Theorem.

CONVOLUTION THEOREM

"Reverse, shift, multiply and integrate" technique is called CONVOLUTION.

Reverse : [L^‒1F(s)]_t=u = f(u)

Shift : g (t‒u)

Multiply : f(u) g (t‒u)

Integrate : ₀ʃ^t f(u) g(t‒u) du = f*g

= Convolution of f and g.

Example 1: Define convolution.

Solution: The convolution of two functions f(t) and g (t) is defined as

f(t) * g (t) = ₀ʃ^t f(u) g (t − u) du

Example 2: (a) Prove that f(t) * g(t) = g(t) * f(t).

Solution:

f(t) * g (t) = ₀ʃ^t f(u) g (t − u) du

We know that,

₀ʃ^a f(x) dx = ₀ʃ^a f(a−x) dx

= ₀ʃ^t f(t−u) g [t−( t−u)] du = ₀ʃ^t f(t−u) g (u) du

= ₀ʃ^t g(u) f(t−u) du

= f(t) * g (t)

Note: Convolution Integral or Falting integral.

Example 2: (b) Write the value of t*e^t.

Solution:

By definition of convolution, we have

Example 3: State and prove convolution theorem.

Statement:

If f(t) and g (t) are functions defined for t ≥ 0,

then L [f(t) * g (t)] = L [f(t)] . L [g (t)], where

f(t) * g (t) = ₀ʃ^t f(u) g (t − u) du (OR) ₀ʃ^t f(t − u) g (u) du

Proof: We know that, L [f(t)] = ₀ʃ^∞ e^‒st f(t) dt

i.e., L[f(t) * g (t)] = L[f(t)]. L[g (t)] = F(s) . G(s),

where L[f(t)] = F(s)

L[g (t)] = G(s)

L^‒1[F(s) G(s)] = f(t)*g(t)

= L^‒1[F(s)] * L^‒1[G(s)]

PROBLEMS BASED ON CONVOLUTION THEOREM

Transforms and its Applications: UNIT 1: Laplace Transforms : Tag: Engineering mathematics, Maths : - Laplace Transforms using Convolution Theorem

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

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