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Textbook by: Carlos E. Davila. E-mail the author

# Parallel and Cascaded Filters

Module by: Carlos E. Davila. E-mail the author

In some applications, such as graphic equalizers, it is useful to place filters in parallel as shown in Figure 1. Can the parallel combination of filters be characterized by a single equivalent filter heq(t)heq(t)? The answer is yes and results by noting that

y ( t ) = i = 1 N x ( t ) * h i ( t ) = i = 1 N - x ( t - τ ) h i ( τ ) d τ = - x ( t - τ ) i = 1 N h i ( τ ) d τ y ( t ) = i = 1 N x ( t ) * h i ( t ) = i = 1 N - x ( t - τ ) h i ( τ ) d τ = - x ( t - τ ) i = 1 N h i ( τ ) d τ
(1)

Therefore, the last equation in Equation 1 shows that

h e q ( t ) = i = 1 N h i ( t ) h e q ( t ) = i = 1 N h i ( t )
(2)

The equivalent transfer function for the parallel filter structure is given by

H e q ( j Ω ) = i = 1 N H i ( j Ω ) H e q ( j Ω ) = i = 1 N H i ( j Ω )
(3)

Next we wish to find an equivalent filter for the cascaded structure shown in Figure 2.

This can be done by finding an expression for the intermediate value y1(t)y1(t):

y 1 ( t ) = - x ( t - τ ) h 1 ( τ ) d τ y 1 ( t ) = - x ( t - τ ) h 1 ( τ ) d τ
(4)

The output of the cascaded structure is given by

y ( t ) = - y 1 ( t - γ ) h 2 ( γ ) d γ y ( t ) = - y 1 ( t - γ ) h 2 ( γ ) d γ
(5)

substituting Equation 4 into Equation 5 gives

y ( t ) = - - x ( t - γ - τ ) h 1 ( τ ) d τ h 2 ( γ ) d γ y ( t ) = - - x ( t - γ - τ ) h 1 ( τ ) d τ h 2 ( γ ) d γ
(6)

Reversing the order of integration and rearranging slightly gives

y ( t ) = - - x ( t - γ - τ ) h 1 ( τ ) h 2 ( γ ) d γ d τ y ( t ) = - - x ( t - γ - τ ) h 1 ( τ ) h 2 ( γ ) d γ d τ
(7)

Now let ξ=γ+τξ=γ+τ, solving for ττ gives τ=ξ-γτ=ξ-γ and dξ=dτdξ=dτ. Substituting these quantities into Equation 7 leads to

y ( t ) = - x ( t - ξ ) - h 1 ( ξ - γ ) h 2 ( γ ) d γ d ξ y ( t ) = - x ( t - ξ ) - h 1 ( ξ - γ ) h 2 ( γ ) d γ d ξ
(8)

Notice that we can factor x(t-ξ)x(t-ξ) from the inner integral since x(t-ξ)x(t-ξ) does not depend on γγ. The integral in the brackets is recognized as h1(t)*h2(t)h1(t)*h2(t) evaluated at ξξ. Therefore for the cascaded system, the equivalent impulse response is given by

h e q ( t ) = - h 1 ( t - γ ) h 2 ( γ ) d γ h e q ( t ) = - h 1 ( t - γ ) h 2 ( γ ) d γ
(9)

This can be generalized to any number of cascaded filters giving

h e q ( t ) = h 1 ( t ) * h 1 ( t ) * * h N ( t ) h e q ( t ) = h 1 ( t ) * h 1 ( t ) * * h N ( t )
(10)

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