## Appendix A  
To investigate \(L(s)\), substitute \(e^{-\alpha s}=\cos\left(\alpha w\right)-j\sin\left(\alpha\omega\right)\) into Equation (16),  
\[L(j\omega)=\frac{j(k\omega+T\omega)+1}{1-\cos\left(\alpha\omega\right)+j[T \omega+\sin\left(\alpha\omega\right)]} \tag{18}\]  
Figure 17: The residual error with the given signals of the target and disturbances.  
Because of the effect of the trigonometric functions, the phase function and magnitude function of \(L(j\omega)\) will oscillate in each cycle. Assume \(\phi(\omega)\) is the phase of \(L(j\omega)\), and we could get  
\[\phi(\omega)=ac\ \tan\left(k\omega+T\omega\right)-ac\ \tan\frac{T\omega+\sin\left( \alpha\omega\right)}{1-\cos\left(\alpha\omega\right)} \tag{2}\]  
If \(\alpha=0\), we have \(\phi(\omega)=ac\ \tan\left(k\omega+T\omega\right)-0.5\pi<0\). Hence, \(T\) should be big enough to make the loss of phase small at \(\omega_{c}\). Nevertheless, if \(T\) is too big, it will lead to the decreasing of the promotion for the system performance. Concentrating on the derivative function of \(\phi(\omega)\) as follows,  
\[\phi^{\prime}(\omega)=\frac{k+T}{1+\left(k+T\right)^{2}\omega^{2}}-\frac{(T- \alpha)[1-\cos\left(\alpha\omega\right)]-T\alpha\omega\sin\left(\alpha\omega \right)}{\left[1-\cos\left(\alpha\omega\right)\right]^{2}+\left[T\omega+\sin \left(\alpha\omega\right)\right]^{2}} \tag{3}\]  
Obviously, if \(T<\alpha\), then \(\phi^{\prime}(\omega)>0\) when \(T\omega\in\left(0,\ \frac{\pi}{4}\right)\). However, since \(\alpha\) (the CCD's delay) cannot be too small, commonly, \(T\) has to be much bigger than \(\alpha\) or it will result in instability. Actually, when \(T>\alpha\) and \(T\omega\in\left(0,\ \frac{\pi}{4}\right)\), we can still get \(\phi^{\prime}(\omega)>0\).  
Proof.: \(T\omega\in\left(0,\ \frac{\pi}{4}\right)\).  
Define \(\psi(\omega)=(T-\alpha)[1-\cos\left(\alpha\omega\right)]-T\alpha\omega\sin \left(\alpha\omega\right)\).  
If \(\psi(\omega)<0\), obviously, \(\phi^{\prime}(\omega)>0\).  
Since \(\psi(0)=0\) and \(\psi^{\prime}(\omega)=-\alpha^{2}[\sin\left(\alpha\omega\right)+T\omega\cos \left(\alpha\omega\right)]<0\), we get \(\psi(\omega)<0\). Then, \(\phi^{\prime}(\omega)>0\).  
From the above discussion, \(\phi(\omega)\) is always increasing when \(\omega\) is from \(0\) to \(\frac{\pi}{4\pi}\). Note that  
\[\left\{\begin{array}{c}\phi(0)=-0.5\pi\\ \phi(\omega_{c})=ac\ \tan\left(1+T\omega_{c}\right)-ac\ \tan\frac{T\omega_{c}+ \sqrt{2}/2}{1-\sqrt{2}/2}<0\end{array}\right. \tag{4}\]  
It means that no matter what \(T\) is, the phase loss is inevitable. If we consider the amplitude of \(L(j\omega)\), the same result will be obtained. The gain function of \(L(j\omega)\) is as follows.  
\[-20\log\lvert L(j\omega)\rvert=-10\log\Biggl{[}\frac{\left(k\omega+T\omega \right)^{2}+1}{\left[1-\cos\left(\alpha\omega\right)\right]^{2}+\left[T\omega +\sin\left(\alpha\omega\right)\right]^{2}}\Biggr{]} \tag{5}\]  
It is obvious that \(-20\log\lvert L(j\infty)\rvert=-20\log\left(\frac{k}{T}+1\right)\), \(T>>k\) is hoped for reducing the sacrificing of GM. But,  
\[-20\log\lvert L(j\omega_{g})\rvert=-10\log\Biggl{[}1+\frac{4T\omega_{g}+1}{4 +T^{2}\omega_{g}^{2}}\Biggr{]}<0 \tag{6}\]  
From Equation (6), it can be concluded that the GM will also suffer losses. \(\square\)