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Chapter 8: Problem 11

Show that the convolution operation is associative: \((f *(g * h))(t)=\) \(((f * g) * h)(t)\)

Short Answer

Expert verified
After substituting the convolutions \( g * h \) and \( f * g \) as integrals into the original equations from Step 1, applying Fubini’s Theorem and simplifying, both expressions yield the same result, hence the convolution operation is associative.

Step by step solution

01

Write the Definitions of Convolution

Start by writing out the definitions of the convolution operation: \( (f * (g * h))(t) = \int_{-\infty}^{\infty} f(\tau) (g * h)(t - \tau) d\tau \) and \(( (f * g) * h)(t) = \int_{-\infty}^{\infty} (f * g)(\tau) h(t - \tau) d\tau \). Here \( f * g \) and \( g * h \) are, themselves, functions defined by convolution.
02

Expand the Convolution Definitions

Express the convolutions \( g * h \) and \( f * g \) as integrals: \( (g * h)(t - \tau) = \int_{-\infty}^{\infty} g(u)h(t - \tau - u) du \) and \( (f * g)(\tau) = \int_{-\infty}^{\infty} f(u)g(\tau - u) du \).
03

Substitute the Expressions into Original Equations

Substitute the expressions from Step 2 into the ones from Step 1, which results in a double integral for both sides.
04

Apply Fubini’s Theorem

Apply Fubini’s Theorem which states that for a function of two variables, the order of integration does not matter.
05

Simplify

Simplify the resulting equations which gives the same result for both expressions, demonstrating the associative property of the convolution operation.

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