Convolution

Author: John J Weber III, PhD Corresponding Textbook Sections:

Expected Educational Results

Convolution

Definition: Convolution

Let f(t) and g(t) be piecewise continuous on [0,). The convolution of f(t) and g(t), (fg)(t), is defined by (fg)(t)=0tf(tv)g(v)dv

NOTE: In the definition of convolution, the “first” function is translated by v where v is the “dummy” variable for the integral.

Example 01: Evaluate e2tt.

Solution:

Use convolution definition:

e2tt=0te2(tv)vdv

Rewrite using algebra:

e2tt=0te2t2vvdv

Find antiderivative using integration by parts:

e2tt=(12ve2t2v14e2t2v)|v=0v=t

Use FTC-II:

e2tt=12te2t2t14e2t2t(12(0)e2t2(0)14e2t2(0))

Simplify:

e2tt=12t14+0+14e2t

Investigation 01

Evaluate the following convolutions.

  1. tt2

  2. tsin(t)

Properties of Convolution

Let f(t), g(t), and h(t) be piecewise continuous on [0,)

  1. (fg)=(gf)

  2. (f(g+h))=(fg)+(fh)

  3. (fg)h=f(gh)

  4. (f0)=0

Investigation 02

Prove the above properties of the convolution.

Theorem: Convolution Theorem

Let f(t) and g(t) be piecewise continuous on [0,) and of exponential order α. Let F(s)=L{f(t)}(s) and G(s)=L{g(t)}(s), then L{f(t)g(t)}(s)=F(s)G(s) or, equivalently, L1{F(s)G(s)}(t)=f(t)g(t).

NOTE: The convolution theorem helps find inverse Laplace transforms when the method of partial fractions fails.

Investigation 03

Prove the Convolution Theorem.

Hint: You will need to switch the order of integration and use u-substitution.

Example 02: Evaluate L1{1s23s10}(t).

Solution:

Rewrite by factoring:

L1{1s23s10}(t)=L1{1(s5)(s+2)}(t)

Use Convolution Theorem:

L1{1s23s10}(t)=e5te2t

Use the definition of convolution:

L1{1s23s10}(t)=0te5(tv)e2vdv

Simplify using algebra:

L1{1s23s10}(t)=0te5t7vdv

Find antiderivative:

L1{1s23s10}(t)=(17e5t7v)|v=0v=t

Use FTC-II:

L1{1s23s10}(t)=17e5t7t(15e5t7(0))

Simplify using algebra:

L1{1s23s10}(t)=17e2t+17e5t

NOTE: Compare this result with the solution to Example 01 in CPT_22.

Investigation 04

Use the convolution to find the following inverse Laplace transform:

  1. Find L1{1s210s+21}(t)

  2. Find L1{1s(s2+1)}(t)

  3. Find L1{s+1(s2+1)2}(t)

Solving IVPs Using Convolution

Investigation 05

Solve the following IVPs:

  1. y(t)5y(t)+6y(t)=cos(2t), y(0)=0, y(0)=0.

  2. y(t)+2y(t)+2y(t)=sin(2t), y(0)=1, y(0)=1.

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Last Modified: Sunday, 8 November 2020 22:22 EDT