## Appendix B  
In this appendix we derive the Fisher information matrix needed to obtain the CRLBs of proposed estimators. We need to calculate the expectations of the first and second partial derivatives of the natural logarithm of the following conditional pdf  
\[f\left(y_{0},\ldots,y_{K-1}|S,N\right)=\frac{1}{\left(2\pi N\right)^{K/2}}\exp \left\{-\frac{1}{2N}\sum_{k=0}^{K-1}\left(y_{k}-\sqrt{2S}x_{k}\right)^{2} \right\}, \tag{11}\]  
where \(S=A^{2}/2\) and \(N=\sigma^{2}\), to get each element of thee matrix.  
We start from the derivative of \(\ln f\) with respect to \(S\). We have  
\[\frac{\partial\ln f}{\partial S} =-\frac{1}{2N}\sum_{k=0}^{K-1}\frac{\partial}{\partial S}\left(y _{k}^{2}-2\sqrt{2S}x_{k}y_{k}+2 Sx_{k}^{2}\right). \tag{12}\] \[=-\frac{1}{2N}\sum_{k=0}^{K-1}\left(-\sqrt{\frac{2}{S}}x_{k}y_{k }+2x_{k}^{2}\right). \tag{13}\]  
Differentiating again with respect to \(S\), we get  
\[\frac{\partial^{2}\ln f}{\partial S^{2}} =\frac{1}{2N}\sum_{k=0}^{K-1}\sqrt{2}\left(-\frac{1}{2\sqrt{S^{3} }}\right)x_{k}y_{k} \tag{14}\] \[=\frac{1}{2\sigma^{2}}\sum_{k=0}^{K-1}\sqrt{2}\left(-\frac{2\sqrt {2}}{2A^{3}}\right)x_{k}y_{k}\] (15) \[=-\frac{1}{A^{3}\sigma^{2}}\sum_{k=0}^{K-1}x_{k}y_{k}. \tag{16}\]  
Since \(E[y_{k}]=Ax_{k}\), the corresponding element in the Fisher matrix is given by  
\[-E\left[\frac{\partial^{2}\ln f}{\partial S^{2}}\right]=\frac{1}{A^{3}\sigma ^{2}}\sum_{k=0}^{K-1}Ax_{k}^{2}. \tag{17}\]  
Now, consider the term  
\[\frac{1}{K}\sum_{k=0}^{K-1}x_{k}^{2}=\frac{1}{K}\sum_{k=0}^{K-1}\cos^{2}\left( 2\pi\
u k+\phi\right). \tag{18}\]  
Under the assumption that the frequency \(\
u\) is not close to \(0\) or \(1/2\) and for \(K\) large enough, we can write [7] (p. 33)  
\[\frac{1}{K}\sum_{k=0}^{K-1}\cos\left(4\pi\
u k+\phi\right)\simeq 0. \tag{19}\]  
Hence,  
\[\frac{1}{K}\sum_{k=0}^{K-1}x_{k}^{2}\simeq\frac{1}{2} \tag{20}\]  
and the first element of Fisher matrix is then given by  
\[-E\left[\frac{\partial^{2}\ln f}{\partial S^{2}}\right]=\frac{K}{2A^{2}\sigma ^{2}}. \tag{21}\]Similarly, we obtain the first and second partial derivatives with respect to \(N\). The first derivative is given by  
\[\frac{\partial\ln f}{\partial N} =\frac{\partial}{\partial N}\left(-\frac{K}{2}\ln\left(2\pi N \right)\right)-\frac{\partial}{\partial N}\left(\frac{1}{2N}\sum_{k=0}^{K-1} \left|y_{k}-Ax_{k}\right|^{2}\right)\] \[=-\frac{K}{2N}+\frac{1}{2N^{2}}\sum_{k=0}^{K-1}\left|y_{k}-Ax_{k }\right|^{2}.\]  
Differentiating again with respect to \(N\) we get  
\[\frac{\partial^{2}\ln f}{\partial N^{2}}=\frac{K}{2N^{2}}-\frac{1}{N^{3}}\sum_ {k=0}^{K-1}\left|y_{k}-Ax_{k}\right|^{2}. \tag{104}\]  
Hence, taking the expectation  
\[-E\left[\frac{\partial^{2}\ln f}{\partial N^{2}}\right]=-\frac{K}{2N^{2}}+ \frac{1}{N^{3}}E\left[\sum_{k=0}^{K-1}\left|y_{k}-Ax_{k}\right|^{2}\right]. \tag{105}\]  
Consider, now, the term  
\[E\left[\sum_{k=0}^{K-1}\left|y_{k}-Ax_{k}\right|^{2}\right]=\sum_{k=0}^{K-1}E \left[y_{k}^{2}\right]-2 Ax_{k}E\left[y_{k}\right]+A^{2}x_{k}^{2}. \tag{106}\]  
Since \(E\left[y_{k}^{2}\right]=A^{2}x_{k}^{2}+N\) and \(E\left[y_{k}\right]=Ax_{k}\), the above term can be rewritten as  
\[E\left[\sum_{k=0}^{K-1}\left|y_{k}-Ax_{k}\right|^{2}\right]=\sum_{k=0}^{K-1}A^ {2}x_{k}^{2}+N-2A^{2}x_{k}^{2}+A^{2}x_{k}^{2}=KN. \tag{107}\]  
The second derivative is then  
\[-E\left[\frac{\partial^{2}\ln f}{\partial N^{2}}\right]=-\frac{K}{2N^{2}}+ \frac{K}{N^{2}}=\frac{K}{\sigma^{4}}. \tag{108}\]  
The off-diagonal terms of the Fisher matrix are both equal to zero. In detail, we have  
\[\frac{\partial^{2}\ln f}{\partial S\partial N}=\frac{\partial^{2}\ln f}{ \partial N\partial S}=-\frac{1}{\sigma^{4}}\sum_{k=0}^{K-1}\left[\frac{x_{k}y _{k}}{A}-x_{k}^{2}\right]. \tag{109}\]  
By taking the expectation, we immediately obtain that  
\[-E\left[\frac{\partial^{2}\ln f}{\partial S\partial N}\right]=0. \tag{110}\]