## Appendix C Review on the Frequency-Domain-Based Bit Flipping Control Strategy  
Denote the transfer function of the loop filter as \(F(z)\). Denote the noise transfer function and the signal transfer function as \(NTF(z)\) and \(STF(z)\), respectively. In order to achieve a high signal to noise ratio, it is desired to achieve \(NTF(z)\to 0\) and \(STF(z)\to 1\) in the signal band as well as \(NTF(z)\to 1\) and \(STF(z)\to 0\) in the noise band. Since \(NTF(z)=\frac{1}{1+F(z)}\) and \(STF(z)=\frac{F(z)}{1+F(z)}\), it is implied that \(\left|F(z)\right|\rightarrow+\infty\) in the signal band and \(F(z)=0\) in the noise band. This further implies that \(F(z)\) contains some unstable poles ((1) the poles outside the unit circle; or (2) the poles on the unit circle with the frequencies inside the signal band in which the resonance could occur). As the loop filter is unstable, the stability issue is very critical. As a result, the control of the sigma delta modulator is very important. Assume that the loop filter could be realized via the state space representation with the state space matrices \(\mathbf{A}\), \(\mathbf{B}\), \(\mathbf{C}\) and \(D\). Here, \(D=0\). Let the input sequence and the quantized output sequence of the sigma delta modulator be \(u(n)\) and \(s(n)\), respectively. Denote the state vector as \(\mathbf{x}(n)\). Then, it can be shown that:  
\[\mathbf{x}(n)=\mathbf{A}^{n}\mathbf{x}(0)+\sum\limits_{k=0}^{n-1}\mathbf{A}^{ n-1-k}\mathbf{B}(u(k)-s(k))\]  
Assume that \(\mathbf{A}\) is diagonalizable. Denote a matrix \(\mathbf{T}\) in such a way that the columns of \(\mathbf{T}\) are the eigenvectors of \(\mathbf{A}\). Moreover, denote a diagonal matrix \(\mathbf{D}\) in such a way that its diagonal elements are the corresponding eigenvalues of \(\mathbf{A}\). That is, \(\mathbf{A}=\mathbf{T}\mathbf{D}\mathbf{T}^{-1}\). Let the diagonal elements of \(\mathbf{D}\) be \(d_{i}\). Assume that the state vectors are \(N\) dimensional. Denote:  
\[\widetilde{\mathbf{x}}(n)=\left[ \begin{array}{ccc}\widetilde{x}_{0}(n)& \cdots&\widetilde{x}_{N-1}(n)\end{array} \right]^{T}=\mathbf{T}^{-1}\mathbf{x} (n)\]  
and  
\[\widetilde{\mathbf{B}}=\left[ \begin{array}{ccc}\widetilde{b}_{0}&\cdots& \widetilde{b}_{N-1}\end{array} \right]^{T}=\mathbf{T}^{-1}\mathbf{B}\]  
Then, we have:  
\[\mathbf{x}(n)=\mathbf{T}\mathbf{D}^{n}\mathbf{T}^{-1}\mathbf{x}(0)+\sum \limits_{k=0}^{n-1}\mathbf{T}\mathbf{D}^{n-1-k}\mathbf{T}^{-1}\mathbf{B}(u(k) -s(k))\]  
\[\widetilde{\mathbf{x}}(n)=\mathbf{T}^{-1}\mathbf{x}(n)=\mathbf{D}^{n}\mathbf{ T}^{-1}\mathbf{x}(0)+\sum\limits_{k=0}^{n-1}\mathbf{D}^{n-1-k}\mathbf{T}^{-1} \mathbf{B}(u(k)-s(k))\]  
\[=\mathbf{D}^{n}\widetilde{\mathbf{x}}(0)+\sum\limits_{k=0}^{n-1}\mathbf{D}^{ n-1-k}\widetilde{\mathbf{B}}(u(k)-s(k))\]  
This implies that:  
\[\widetilde{x}_{i}(n)=d_{i}^{n}\widetilde{x}_{i}(0)+\sum\limits_{k=0}^{n-1}d_{i} ^{n-1-k}\widetilde{b}_{i}(u(k)-s(k))\]  
In other words, we have:  
\[\frac{\widetilde{x}_{i}(n)}{\widetilde{b}_{i}d_{i}^{n-1}}=\frac{d_{i} \widetilde{x}_{i}(0)}{\widetilde{b}_{i}}+\sum\limits_{k=0}^{n-1}\frac{u(k)-s(k )}{d_{i}^{k}}\]  
Denote the z-transform of the input sequence and that of the quantized output sequence of the sigma delta modulator as \(U(z)\) and \(S(z)\), respectively. Suppose that the eigenvectors in \(\mathbf{T}\) and the corresponding eigenvalues in \(\mathbf{D}\) are arranged in such a way that \(|d_{i}|\geq|d_{i}|\) for \(j>i\). Moreover, assume that there are \(P\) unstable eigenvalues in \(\mathbf{D}\). That is, \(|d_{i}|\geq 1\) for \(i=0,\cdots,P-1\). By flipping somevalues in the quantizer output, denote the z-transform of the new quantizer output as \(\hat{S}(z)\). In order to stabilize the sigma delta modulator, we have:  
\[\lim_{n\rightarrow+\infty}\frac{\overline{x}_{i}(n)}{\overline{b}_{i}d_{i}^{n-1 }}=0=\frac{d_{i}\overline{x}_{i}(0)}{\overline{b}_{i}}+\left.U(z)\right|_{z=d_ {i}}-\hat{S}(z)\right|_{z=d_{i}}\]  
for \(i=0,\cdots,P-1\).