# Low Complexity Equalization of Orthogonal Chirp Division Multiplexing in Doubly-Selective Channels
## 3 Low Complexity Equalizers for UP-OCDM under Doubly Selective Channels
### Iterative LSQR Block Equalizer
#### 3.3.1 Iterative LSQR Block Equalization Algorithm  
We derive the special structure of the transform domain channel matrix first. We assume the channel tap gains are modeled by a basis expansion model, which means each channel tap gain \(\mathbf{h}_{l}\) is a linear combination of basis functions. With a given set of basis, \(h_{l}[n]\) can be represented as:  
\[h_{l}[n]=\sum_{m=0}^{M-1}c_{l,m}\varphi_{m}[n], \tag{21}\]  
where \(M\) is the BEM model order, \(b_{l,m}\) is the \(m\)th basis coefficient of the \(l\)th channel path. Moreover, \(\varphi_{m}[n]\) and \(h_{l}[n]\) denote the \(m\)th basis function and the channel tap gain of the \(l\)th path at sampling time \(nT_{s}\), respectively. With the increase of \(M\), the modeling error can be reduced. Combining the input-output relationship in (15) with the BEM model in (21), the time-domain receive signal \(\mathbf{r}\) can be expressed as  
\[\mathbf{r} =\sum_{m=0}^{M-1}\mathbf{P}_{m}\mathbf{G}_{m}\mathbf{s}+\mathbf{w} \tag{22}\] \[=\sum_{m=0}^{M-1}\mathbf{P}_{m}\mathbf{G}_{m}\mathbf{\Phi}d+\mathbf{w}\] \[=\sum_{m=0}^{M-1}\mathbf{P}_{m}\mathbf{\Theta}_{m}\mathbf{d}+\mathbf{w}\] \[=\mathbf{C}\mathbf{d}+\mathbf{w},\]  
where \(\mathbf{P}_{m}=\mathcal{D}([\varphi_{m}[0],\varphi_{m}[1],\ldots,\varphi_{m}[N- 1])\) is a diagonal matrix consisted of the BEM basis, \(\mathbf{G}_{m}\) is a circulant matrix with its first column equaling to \([c_{0,m},c_{1,m},\ldots,c_{L-1,m}]^{T}\). Note that \(\mathbf{\Phi}\) is also circulant, so that \(\mathbf{\Theta}_{m}=\mathbf{G}_{m}\mathbf{\Phi}\) is a circulant matrix. We define \(\mathbf{C}=\sum_{m=0}^{M-1}\mathbf{P}_{m}\mathbf{\Theta}_{m}\), which is the transform domain channel matrix between the receive signal \(\mathbf{r}\) and the transmitted symbol vector \(\mathbf{d}\). With the well-known circulant matrix construction method via FFT, it is easy to reconstruct \(\mathbf{C}\). After the construction of \(\mathbf{C}\), we employ the LSQR algorithm, which is a Krylov subspace method, for the regularized solution of linear systems. Specifically, for a linear system \(\mathbf{r}=\mathbf{H}\mathbf{x}\), the Krylov subspace for LSQR is spanned as  
\[\mathcal{K}(\mathbf{H}^{H}\mathbf{H},\mathbf{H}^{H}\mathbf{r},i)=Span\big{\{}\mathbf{H}^{H}\mathbf{r },\left(\mathbf{H}^{H}\mathbf{H}\right)\mathbf{H}^{H}\mathbf{r},\ldots,\left(\mathbf{H}^{H}\mathbf{H} \right)^{(i-1)}\mathbf{H}^{H}\mathbf{r}\big{\}}. \tag{23}\]  
Essentially, LSQR is an iterative algorithm equivalent to the conjugate gradient method. The main idea of LSQR is to minimize the norm of the residual error in each iteration. It can be concluded from Reference [24] that LSQR is one of the best regularization methods which can achieve the minimum error possible. It will be shown in the simulation section that the LSQR based equalizer is able to achieve performance close to that of the MMSE equalizer when the iteration number is suitably chosen. The details of the LSQR algorithm can be found in Reference [20]. In this paper, we assume the BEM coefficients are already known to the receiver. The first step of the proposed equalization algorithm is to reconstruct the transform domain channel matrix \(\mathbf{C}\). It is commonly known that a circulant matrix can be easily constructed by a diagonal matrix. In order to achieve a fast implement, we denote  
\[\mathbf{\Theta}_{m} =\mathbf{F}^{H}\mathbf{D}_{m}\mathbf{\Lambda}\mathbf{F}, \tag{24}\] \[=\mathbf{F}^{H}\mathbf{V}_{m}\mathbf{F}\]  
where \(\mathbf{V}_{m}=\mathbf{D}_{m}\mathbf{\Lambda}\), and \(\mathbf{D}_{m}\) is a diagonal matrix whose diagonal elements are the DFT of the BEM coefficients \(c_{i,m}\) that zero-padded to length N, i.e.,  
\[\mathbf{D}_{m}=\mathcal{D}(\mathbf{F}[c_{0,m},c_{1,m},\ldots,c_{L-1,m},0,\ldots,0]^{T}), \tag{25}\]and \(\mathbf{\Lambda}\) is also diagonal, with \(\mathbf{\Lambda}(k,k)=e^{j\pi(\frac{k^{2}}{N}-\frac{1}{4})}\).