## 2 IIr-Ofdm  
Assume that the number of subcarriers is \(N\) and the number of transmitted symbols is \(V\). \(d_{n,v}\) is the \(v\)th transmitted symbol at the \(n\)th subcarrier (\(n=0,\cdots,N-1\)). Let \(\mathbf{d}_{v}\) represent the transmitted symbol matrix, which can be written as  
\[\mathbf{d}_{v}=[d_{0,v},\cdots,d_{N-1,v}]^{\mathrm{T}}. \tag{1}\]  
After the \(N\)-point inverse discrete Fourier transform (IDFT), the equivalent baseband transmitted signal can be written as  
\[s_{v}(k)=\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}d_{n,v}e^{i2\pi mk/N},k\in[0,N-1], \tag{2}\]  
where \(\frac{1}{\sqrt{N}}\) is the power normalization factor.  
Assume that the channel consists of a forward path and several loops, which may be due to reflections or loop interference. The transmitted signal passing through a channel with loops can be indicated as  
\[r_{v}(k)=h(0)[s_{v}(k)+h(1)r_{v}(k-1)+\cdots+h(l_{g})r_{v}(k-l_{g})], \tag{3}\]  
where \(r_{v}(k)\) is the received signal corresponding to \(s_{v}(k)\) at the receiver and \(l_{g}\) is the number of loops. Under this channel, only \(h(0)\) is the forward path.  
Suppose that the transmitted signal passes through a pure IIR channel \(H(z)=1/A(z)\), we have  
\[A(z)=\sum_{i=0}^{l_{g}}a(i)z^{-i}, \tag{4}\]  
where \(a(i)\) is the coefficient of the IIR channel. \(H(z)\) and \(A(z)\) are the \(z\)-transforms of the channel.  
Linking Equation (4) to the channel, we have  
\[a(i)=\begin{cases}\frac{1}{n_{i}},&i=0,\\ -h_{i},&i=[1,\cdots,l_{g}].\end{cases} \tag{5}\]  
Therefore, the differential Equation (3) for the system of the IIR channel can be expressed as  
\[s_{v}(k)=\sum_{i=0}^{l_{g}}a(i)r_{v}(k-i), \tag{6}\]  
where received \(r_{v}(k)\) and transmitted signal \(s_{v}(k)\) satisfy linear convolution, like classical OFDM.  
We consider \(\mathbf{r}_{v}=[r_{v}(0),\cdots,r_{v}(N-1)]^{\mathrm{T}}\) as the received signal vector. Then, the transmitted signal vector \(\mathbf{s}_{v}=[s_{v}(0),\cdots,s_{v}(N-1)]^{\mathrm{T}}\) can be expressed as  
\[\mathbf{s}_{v}=\mathbf{A}\mathbf{r}_{v}, \tag{7}\]  
where \(\mathbf{A}\) represents the IIR channel matrix, which can be written as\[\mathbf{A}=\sum_{i=0}^{l_{g}}a(i)\Lambda^{i}=\begin{bmatrix}a(0)&&&&\\ \vdots&\ddots&&\\ a(l_{g})&\cdots&a(0)&&\\ &a(l_{g})&\cdots&a(0)&&\\ &&\ddots&&\ddots&\\ &&&a(l_{g})&\cdots&a(0)\end{bmatrix} \tag{8}\]  
with  
\[\Lambda=\begin{bmatrix}0&\cdots&0&0\\ 1&\ddots&0&0\\ \vdots&\ddots&\ddots&\vdots\\ 0&\cdots&1&0\end{bmatrix}_{N\times N},\Gamma=\begin{bmatrix}0&\cdots&0&1\\ 1&\ddots&0&0\\ \vdots&\ddots&\ddots&\vdots\\ 0&\cdots&1&0\end{bmatrix}_{N\times N}, \tag{9}\]  
note that \(\Lambda^{0}\) and \(\Gamma^{0}\) are the identity matrix.  
However, it can be anticipated from (7) that when \(k>N-1\), the received signal will continue to be influenced by the loops of the IIR channel. It can be obtained by  
\[a(0)r_{v}(k)=-\sum_{i=1}^{l_{g}}a(i)r_{v}(k-i). \tag{10}\]  
The ISI signal can be obtained by  
\[\mathbf{s}_{v}^{\mathbf{ISI}}=\mathbf{A}^{\mathbf{ISI}}\mathbf{r}_{v}, \tag{11}\]  
where \(\mathbf{A}^{\mathbf{ISI}}\) represents the ISI matrix of IIR channel, which can be written as  
\[\mathbf{A}^{\mathbf{ISI}}=\sum_{i=0}^{l_{g}}a(i)(\Gamma^{i}-\Lambda^{i})= \begin{bmatrix}0&&a(l_{g})&\cdots&a(1)\\ \vdots&\ddots&&\ddots&\vdots\\ 0&\cdots&0&&a(l_{g})\\ &0&\cdots&0\\ \end{bmatrix} \tag{12}\]  
where we consider \(r_{v}(k)\) as the transmitted signal and \(s_{v}(k)\) as the received signal; the ISI signal under the IIR channel is similar to the transmitted signal under the FIR channel.