At the decoder side, inverse TDLP filtering on the FDLP decoded signal gives the sub-band signal back. .
Substituting (A^{-1}) for A and (B^{-1}) for B in (4.8) and then taking the inverse of both sides, we get (4.9).
Applying the inverse to both sides of the last equality, it yields the following: - ζ 2 h ~ + = 1 b + P T + 1 b - g σ S - = 1 - ζ c a, φ ′ q P T + ζ ζ - 1 | c a, φ ′ | σ S - 2 U 0, 1.
Applying the inverse to both sides of (47) leads us to the following: - ζ 2 h + = 1 b + P T + g + b - = 1 b + P T + g b - = 1 - ζ c a, φ ′ q P T + ζ ζ - 1 | c a, φ ′ | 2 U 1. Open image in new window.
Applying the inverse to both sides of the last equality yields the following unique solution: - h ~ + = 1 b + P T + 1 b - g σ S - = 1 1 - ζ w c a, γ, δ ′ q P T + ζ - 1 ζ | c a, γ, δ ′ | σ S - 2 U 0, 1.
This operator is invertible, so applying its inverse to both sides of the equation, we rewrite it equivalently as follows: 1 ã + P T + 1 ã - P T + - ã + ζ ã - h + + ( T - 1 h + ( 0 ) ¯ = ( T - 1 g +, Open image in new window. that is - ζ h + + h + ( 0 ) ¯ T - 1 1 = 1 ã + P T + 1 ã - g Open image in new window (36).
Figure 2A plots the cueing effects (the cued-side performance advantage calculated by subtracting inverse efficiency for cued-side targets from that for targets on the uncued side) for each combination of target eccentricity and cue eccentricity.
On the other side, the inverse transform,, for the transformed triplets is defined as (8).
Subsequently, the decoded quantization indices are fed to the reconstruction module which performs inverse quantization using the side information and the correlation channel statistics.
At the Rx side, the inverse operations are done resulting in the following complexity begin{array}rcl@ C^{text{Tx}}_{text{FS-FB}} &=& 2C_{text{FFT} } KM + 8M_{f}(K - 1). end{array} (6).
