## 3 Single Channel Modulated Wideband converter
### Power Spectrum Sensing  
We aim to derive the relation between the sample sequences \(y\left[n\right]\) and the unknown power spectrum \(P_{x}(f)\). Since \(x(t)\) is a wide-sense stationary process, in T\({}_{i}\), the relation between \(X_{i}(f)\) and \(y(f)\) can be expressed as follows  
\[E\left[\left|X_{i}(f)\right|^{2}\right]=E\left[\left|X(f)\right|^{2}\right]=P_ {x}(f),\ 1\leq i\leq m \tag{9}\]  
Figure 3: The equivalent of proposed single channel sub-Nyquist sampling scheme. where \(E\left[\cdot\right]\) denotes the expectation operator. \(X_{i}(f)\)is the Fourier transform of \(x_{i}(t)\), and \(P_{x}(f)\) is given by Equation (2). Then we have  
\[E\left[X_{i}(f_{1})X_{i}^{*}(f_{2})\right]=P_{x}(f_{1})\delta(f_{1}-f_{2}),\ 1 \leq i\leq m \tag{10}\]  
where \(f_{1}\) and \(f_{2}\) are any frequencies within \(f_{nyq}\). We define the autocorrelation matrix  
\[\mathbf{R}_{z}(f)=E\left[z(f)z^{H}(f)\right]\, \tag{11}\]  
where \((\cdot)^{H}\) denotes the Hermitian operation. From Equation (10), we know that \(\mathbf{R}_{z}(f)\) is a block diagonal matrix with block diagonal element \(\mathbf{R}_{z}^{i}(f)=E[z_{i}(f)z_{i}^{H}(f)]\), \(\mathbf{R}_{z}^{i}(f)\) is also a diagonal matrix with diagonal line element \(\mathbf{R}_{z}^{i}(j,j,f)=P_{x}(f+l_{j}f_{s})\), where \(l_{j}=-L_{0}+j-1\), \(1\leq j\leq L\). Obviously, our aim can be described as recovery of \(\mathbf{R}_{z}(f)\), because once \(\mathbf{R}_{z}(f)\) is known, \(P_{x}(f)\) follows for all \(f\). Now, we relate \(\mathbf{R}_{z}(f)\) to the correlation of the sub-Nyquist samples. From Equation (8), we have the relationship in matrix form as  
\[\mathbf{R}_{y}(f)=\mathbf{A}\mathbf{R}_{z}(f)\mathbf{A}^{H},\ f\in F_{s} \tag{12}\]  
Besides, \(\mathbf{R}_{y}(f)\) can be expressed  
\[\mathbf{R}_{y}(f)=E[y(f)y^{H}(f)]=\left[ \begin{array}{ccc}E[Y_{1}(e^{j2\pi fT_{ s}})Y_{1}^{H}(e^{j2\pi fT_{s}})]&&0\\ &\ddots&\\ 0&&E[Y_{m}(e^{j2\pi fT_{s}})Y_{m}^{H}(e^{j2\pi fT_{s}})]\end{array} \right] \tag{13}\]  
where the diagonal line element of \(\mathbf{R}_{y}(f)\) is  
\[\mathbf{R}_{y}(i,i,f)=E\left[Y_{i}(e^{2\pi fT_{s}})Y_{i}^{H}(e^{j2\pi fT_{s}}) \right]=C_{i}R_{z}^{i}(f)C_{i}^{H}=\sum_{l_{i}=-L_{0}}^{L_{0}}|c_{i,l_{i}}|^{ 2}P_{x}(f-l_{i}f_{s}),\ f\in F_{s} \tag{14}\]  
For ease of description, we define  
\[\mathbf{r}_{y}(f)=\left[\mathbf{R}_{y}(1,1,f)\quad\mathbf{R}_{y}(2,2,f),\cdots,\mathbf{R}_{y} (m,m,f)\right]^{T}\!,\ f\in F_{s} \tag{15}\]  
\[\mathbf{\Phi}=\mathbf{A}.*\bar{\mathbf{A}} \tag{16}\]  
and  
\[\mathbf{r}_{x}(f)=\left[P_{x}(f+L_{0}f_{s})\ P_{x}(f+(L_{0}-1)f_{s}),\cdots,P_{x}( f-L_{0}f_{s})\right]^{T} \tag{17}\]  
where \((. *)\) denotes pointwise matrix multiplication and \((\cdot)\) denotes complex conjugacy. From Equation (14), finally we can get the matrix equation between the power spectrum \(P_{x}(f)\) and samples as follows  
\[\mathbf{r}_{y}(f)=\mathbf{\Phi}\ \mathbf{r}_{x}(f),\ f\in F_{s} \tag{18}\]  
Then we can recover the entire power spectrum of \(x(t)\) by recovering \(\mathbf{r}_{x}(f)\) for all \(f\in F_{s}\). Because of the larger of \(m\), the longer of sensing time needed. In general, the Equation (18) is a underdetermined function (\(m<L\)). So, next we should solve the undetermined function to detect the support. For this purpose, we adopt a operation called continuous-to-finite (CTF) detailed in [11]. From Equation (18), we have  
\[\mathbf{Q}=\mathbf{\Phi}\ \mathbf{Z}\ \mathbf{\Phi}^{\mathrm{H}},\ f\in F_{s} \tag{19}\]where \(\mathbf{Q}=\int_{f\in F_{s}}\mathbf{r}_{y}(f)\mathbf{r}_{y}^{H}(f)df\) and \(\mathbf{Z}=\int_{f\in F_{s}}r_{x}(f)r_{x}^{H}(f)df\). Then, any matrix \(\mathbf{V}\), for which \(\mathbf{Q}=\mathbf{V}\mathbf{V}^{H}\), is a frame for \(\mathbf{r}_{y}(F_{s})\)[17]. We construct it by performing an eigendecomposition of \(\mathbf{Q}\) and choosing \(\mathbf{V}\) as the matrix of eigenvectors corresponding to the non-zeros eigenvalues.