# Convolution - Can someone explain this solution?

Convolution & Inverses

Given an impulse response h[n] to a system, and the impulse response g[n] of the inverse system, why is $$h[n] * g[n] = \delta[n]$$? Where the * sign is used to denote convolution.

Last edited:

Related Introductory Physics Homework Help News on Phys.org
xanthym
Lomion said:
Given an impulse response h[n] to a system, and the impulse response g[n] of the inverse system, why is $$h[n] * g[n] = \delta[n]$$? Where the * sign is used to denote convolution.
This result follows from the properties of Laplace Transforms.
First, define the Laplace Transforms of the Impulse Responses h(n) and g(n):
H(s) = L{h(n)}
G(s) = L{g(n)}
Next, find the Laplace Transform of the given Convolution, remembering that the Laplace Transform of a Convolution is the product of the Laplace Transforms of the convolved functions:
L{h(n)*g(n)} = L{h(n)}L{g(n)} = H(s)G(s)
However, since both h(n) and g(n) are Impulse Response functions, their Laplace Transforms are their system Transfer Functions. Moreover, since we are given that h(n) represents the Inverse system to g(n), their TRANSFER FUNCTIONS must be be reciprocal to each other:
H(s)G(s) = 1
-----> L{h(n)*g(n)} = 1
-----> h(n)*g(n) = DIRAC-DELTA(n)
where we used the result that L^(-1)(1)=DIRAC-DELTA(n).
~

Last edited: