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.
\end{align} The last term on the right-hand side always equals the inverse of the left-hand side of the next equation.
On the other side, the inverse transform,, for the transformed triplets is defined as (8).
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).
Repeat the inverse on the opposite side.
where and means that as The left-hand side of (1.2) is the inverse moment and the right-hand side is the inverse of the moment.
Applying the inverse operator, of both sides of the above equation, we obtain (4.4).
