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PAGE 1

6.6 The Convolution Integral

We know:

\[ \mathcal{L} \{ f(t) \pm g(t) \} = \mathcal{L} \{ f(t) \} \pm \mathcal{L} \{ g(t) \} \]
\[ \mathcal{L}^{-1} \{ F(s) \pm G(s) \} = \mathcal{L}^{-1} \{ F(s) \} \pm \mathcal{L}^{-1} \{ G(s) \} \]

BUT

\[ \mathcal{L} \{ f(t) \cdot g(t) \} \neq \mathcal{L} \{ g(t) \} \mathcal{L} \{ f(s) \} \]
\[ \mathcal{L}^{-1} \{ F(s) \cdot G(s) \} \neq \mathcal{L}^{-1} \{ F(s) \} \cdot \mathcal{L}^{-1} \{ G(s) \} \]

e.g.

\[ \mathcal{L}^{-1} \left\{ \frac{1}{s(s-1)} \right\} \neq \mathcal{L}^{-1} \left\{ \frac{1}{s} \right\} \mathcal{L}^{-1} \left\{ \frac{1}{s-1} \right\} \]
\[ = \mathcal{L}^{-1} \left\{ -\frac{1}{s} + \frac{1}{s-1} \right\} \]
\[ = -1 + e^t \]

so what is \( \mathcal{L}^{-1} \{ F(s) \cdot G(s) \} \) ?

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\[ \mathcal{L}^{-1} \{ F(s) \cdot G(s) \} = \int_{0}^{t} f(t-\tau) g(\tau) d\tau \]
\[ = \int_{0}^{t} f(\tau) g(t-\tau) d\tau \]
\[ = f(t) * g(t) \]

Convolution Integral

\( \tau \): dummy variable → integration variable

example

\[ \mathcal{L}^{-1} \left\{ \frac{1}{(s+1)(s+2)} \right\} \]

two ways: first with partial fraction expansion

\[ \mathcal{L}^{-1} \left\{ \frac{1}{s+1} - \frac{1}{s+2} \right\} = e^{-t} - e^{-2t} \]

now with convolution integral

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Inverse Laplace Transform via Convolution

\[ \mathcal{L}^{-1} \left\{ \frac{1}{(s+1)(s+2)} \right\} = \mathcal{L}^{-1} \left\{ \frac{1}{s+1} \cdot \frac{1}{s+2} \right\} \]

\( F(s) \)

\( f(t) = e^{-t} \)

\( G(s) \)

\( g(t) = e^{-2t} \)

Applying the convolution theorem, we have two equivalent integral forms:

\[ = \int_{0}^{t} e^{-(t-\tau)} e^{-2\tau} d\tau \]

or

\[ = \int_{0}^{t} e^{-\tau} e^{-2(t-\tau)} d\tau \]

\( \tau \) is variable

\( t \) is "constant" (for integration purposes)

Evaluating the first integral:

\[ = \int_{0}^{t} e^{-t} e^{\tau} e^{-2\tau} d\tau \] \[ = \int_{0}^{t} e^{-t} e^{-\tau} d\tau \]

Note: \( e^{-t} \) is treated as a "constant" relative to \( \tau \).

\[ = e^{-t} \left( -e^{-\tau} \right) \bigg|_{\tau=0}^{\tau=t} \] \[ = e^{-t} (-e^{-t} + 1) = e^{-t} - e^{-2t} \]

Matches answer of partial fraction way.

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What is the Convolution Integral Doing?

\[ \int_{0}^{t} f(t-\tau) g(\tau) d\tau \]

Let \( f(t) = e^{-t} \), \( g(t) = e^{-2t} \), and \( t = 3 \):

\[ \int_{0}^{3} e^{-(3-\tau)} e^{-2\tau} d\tau \]

from end of first

from beginning of second

Conceptual Interpretation

  • Integration steps through functions in opposite directions.
  • Integrate the product.

When \( \tau = 1 \), the integrand is \( e^{-2} \cdot e^{-2} \).

Graph of the function e^-t showing a decreasing curve with a vertical slice highlighted at tau=2.
Graph of the function e^-2t showing a steeper decreasing curve with a vertical slice highlighted at tau=2.

At \( \tau = 2 \), the product is \( e^{-(3-2)} \cdot e^{-2(2)} \).

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Convolution Visualization

This page illustrates the process of convolution between two functions, \( f(\tau) \) and \( g(t-\tau) \), by showing the functions individually, the time-reversal and shifting of \( g \), and their overlapping area.

Initial Functions

The functions \( g(t) \) and \( f(t) \) are defined on the time axis \( t \). In the convolution integral, these are transformed into the \( \tau \) domain.

Graphs of g(t) and f(t) showing exponential-like curves on a coordinate system with t-axis.

Time Reversal and Shifting

The function \( g(-\tau) \) is a reflection of \( g(\tau) \) across the vertical axis. As \( t \) increases, the function \( g(t-\tau) \) shifts to the right along the \( \tau \) axis.

Graphs showing g(-tau) and the shifted version g(t-tau) relative to the tau axis.

Integration of Intersection

integrate intersection bar

The convolution result at time \( t \) is the area of the product of the two functions, represented by the shaded region where \( f(\tau) \) and \( g(t-\tau) \) overlap.

Graphs showing the overlap of f(tau) and g(t-tau) with a shaded intersection area.
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Example: Solving a Second-Order ODE with Convolution

Problem Statement

Solve the following differential equation using Laplace transforms and the convolution theorem:

\[ y'' - 2y' + y = 4t^2, \quad y(0) = y'(0) = 0 \]

Laplace Transform Step

Taking the Laplace transform of both sides:

\[ s^2 Y - 2sY + Y = 4 · \frac{2}{s^3} = \frac{8}{s^3} \]

Factoring out \( Y \):

\[ Y(s^2 - 2s + 1) = \frac{8}{s^3} \]

Solving for \( Y \):

\[ Y = \frac{8}{s^3(s^2 - 2s + 1)} = \frac{8}{s^3(s-1)^2} \]

Inverse Laplace Transform via Convolution

We can express \( y(t) \) as the inverse transform of a product:

\[ y = \mathcal{L}^{-1} \left\{ \underbrace{\frac{8}{s^3}}_{4t^2} \cdot \underbrace{\frac{1}{(s-1)^2}}_{te^t} \right\} \]

Applying the convolution theorem:

\[ y = \int_{0}^{t} 4(t-\tau)^2 · \tau e^{\tau} d\tau = \int_{0}^{t} 4\tau^2 · (t-\tau) e^{(t-\tau)} d\tau \]

the one w/ \( t-\tau \) is the one we integrate backwards (end first)