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6.6 The Convolution Integral

The Laplace transform is linear:

\[ \mathcal{L} \{ f(t) + g(t) \} = \mathcal{L} \{ f(t) \} + \mathcal{L} \{ g(t) \} = F(s) + G(s) \]

but \( \mathcal{L} \{ f(t) g(t) \} \) is NOT \( F(s) G(s) \)

because

\[ \int_{0}^{\infty} f(t) g(t) e^{-st} dt \neq \int_{0}^{\infty} f(t) e^{-st} dt \int_{0}^{\infty} g(t) e^{-st} dt \]

it turns out the convolution of \( f(t) \) and \( g(t) \)

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

\( \tau \): variable of integration (NOT \( t \))

has a Laplace transform of \( F(s) G(s) \)

\[ \mathcal{L} \left\{ \int_{0}^{t} f(\tau) g(t-\tau) d\tau \right\} = \mathcal{L} \left\{ \int_{0}^{t} f(t-\tau) g(\tau) d\tau \right\} = F(s) G(s) \]
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a lot of systems in science/engineering are modeled by the convolution integral (hereditary systems)

  • the way the system behaves does not only depend on its present state, but also how it got there

Why is \( \mathcal{L} \left\{ \int_{0}^{t} f(\tau) g(t-\tau) d\tau \right\} = F(s) G(s) \)?

\[ = \int_{0}^{\infty} e^{-st} \left( \int_{0}^{t} f(\tau) g(t-\tau) d\tau \right) dt \]\[ = \int_{0}^{\infty} \int_{\tau}^{\infty} e^{-st} f(\tau) g(t-\tau) dt d\tau \]

double integral: swap the order of integration here

Coordinate graph with axes t and tau showing a shaded triangular region bounded by the line t=tau.

let \( u = t - \tau \rightarrow t = u + \tau \quad dt = du \)

\[ = \int_{0}^{\infty} f(\tau) \int_{0}^{\infty} e^{-s(u+\tau)} g(u) du d\tau \]
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\[ = \int_{0}^{\infty} f(\tau) \int_{0}^{\infty} e^{-su} e^{-s\tau} g(u) \, du \, d\tau \]
\[ = \int_{0}^{\infty} f(\tau) e^{-s\tau} \, d\tau \int_{0}^{\infty} e^{-su} g(u) \, du \]
\[ = F(s) G(s) \]

this gives us another option in dealing with Laplace transforms

for example, if \( H(s) = \frac{1}{s(s^2+1)} \)     \( h(t) = ? \)

option 1:

\[ \frac{1}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1} = \frac{1}{s} + \frac{s}{s^2+1} \]

\( h(t) = 1 + \cos(t) \)

option 2:

\[ \frac{1}{s(s^2+1)} = \frac{1}{s} \cdot \frac{1}{s^2+1} = F(s) G(s) \]
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it turns into convolution in \( t \)

\[ h(t) = \int_{0}^{t} f(\tau) g(t-\tau) \, d\tau \text{ or } \int_{0}^{t} f(t-\tau) g(\tau) \, d\tau \]

here, \( f(t) = 1 \)     \( g(t) = \sin(t) \)

\[ h(t) = \int_{0}^{t} f(t-\tau) g(\tau) \, d\tau = \int_{0}^{t} 1 \cdot \sin(\tau) \, d\tau = -\cos(\tau) \Big|_0^t \]
\[ = -\cos(t) + \cos(0) = 1 - \cos(t) \]

or, avoid doing a difficult convolution

for example,

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

by parts but must cut off after two rounds

side step by transforming it, then come back to \( t \)

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Convolution Integral Example

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\[ \int_{0}^{t} e^{-(t-\tau)} \sin(\tau) d\tau = \int_{0}^{t} f(t-\tau) g(\tau) d\tau \]
\[ f(t) = e^{-t} \]
\[ g(t) = \sin(t) \]
\[ \mathcal{L} \left\{ \int_{0}^{t} e^{-(t-\tau)} \sin(\tau) d\tau \right\} = \mathcal{L} \{ e^{-t} \} \mathcal{L} \{ \sin(t) \} \] \[ = \frac{1}{s+1} \cdot \frac{1}{s^2+1} \] \[ = \frac{1}{2} \frac{1}{s+1} + \frac{1}{2} \frac{1}{s^2+1} - \frac{1}{2} \frac{s}{s^2+1} \]

back to t domain:

\[ \frac{1}{2} e^{-t} + \frac{1}{2} \sin(t) - \frac{1}{2} \cos(t) \]
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Applied to Differential Equations

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\[ y'' + y = -4 \sin(2t) \quad y(0)=0, \, y'(0)=2 \]
\[ s^2 Y - s y(0) - y'(0) + Y = -4 \cdot \frac{2}{s^2+4} \]

Note: \( y(0)=0 \) and \( y'(0)=2 \)

\[ (s^2+1) Y = 2 - \frac{8}{s^2+4} \] \[ Y = \frac{2}{s^2+1} - \frac{8}{(s^2+1)(s^2+4)} \]

Instead of partial fraction here, let's choose to express answer in terms of convolution

\[ = \frac{2}{s^2+1} - \frac{4}{s^2+1} \cdot \frac{2}{s^2+4} \] \[ y = \mathcal{L}^{-1} \left\{ \frac{2}{s^2+1} \right\} - \mathcal{L}^{-1} \left\{ \left( \frac{4}{s^2+1} \right) \left( \frac{2}{s^2+4} \right) \right\} \]

Mapping to convolution terms:

  • \( \frac{4}{s^2+1} \rightarrow 4 \sin(t) = f(t) \)
  • \( \frac{2}{s^2+4} \rightarrow \sin(2t) = g(t) \)
\[ y = 2 \sin(t) - \int_{0}^{t} 4 \sin(t-\tau) \sin(2\tau) d\tau \]

\( f(t-\tau) \cdot g(\tau) \)

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Application of Convolution in Differential Equations

Useful if right side not explicitly stated.

Example: Mass-Spring System

For example, consider the following second-order differential equation with initial conditions:

\[my'' + ky = f(t), \quad y(0) = y'(0) = 0\]

Dividing by \(m\):

\[y'' + \frac{k}{m}y = \frac{f(t)}{m}\]

Taking the Laplace Transform:

\[(s^2 + \frac{k}{m})Y = \frac{1}{m}F(s)\]
\[Y = \frac{1}{m} \cdot \left( \frac{1}{s^2 + \frac{k}{m}} \right) F(s)\]

Note: The term \(\frac{1}{s^2 + \frac{k}{m}}\) corresponds to the inverse transform \(\sin\left(\sqrt{\frac{k}{m}}t\right)\).

\[y = \frac{1}{m} \int_{0}^{t} \sin\left(\sqrt{\frac{k}{m}}(t-\tau)\right) f(\tau) d\tau\]

Transfer Function

\[\frac{Y}{F} = \frac{1}{m} \frac{1}{s^2 + \frac{k}{m}}\]

\(\frac{Y}{F}\) is called the "transfer function". It describes how the system behaves based on an input force.

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\(\frac{Y}{F}\) is also the impulse response of the system.

  • Solution if \(f(t)\) is impulse at \(t=0 \rightarrow \delta(t)\)
\[\mathcal{L}\{\delta(t-t_0)\} = e^{-t_0 s}\]\[\mathcal{L}\{\delta(t)\} = 1\]

Understanding Convolution

This explains what convolution is doing:

\[y = \frac{1}{m} \int_{0}^{t} \sin\left(\sqrt{\frac{k}{m}}(t-\tau)\right) f(\tau) d\tau \rightarrow \text{sum up the impulse response from } \tau=0 \text{ to } \tau=t\]

\(y(t) = ?\)

Three coordinate graphs showing sine waves starting at t=0, a little later, and another a little later.