## Theory
### The Influence of the Erroneous Sampling Points in the Auxiliary Interferometer  
In the paper, we take the extreme points of the auxiliary interference signal as the sampling points. Figure 2 shows the extreme points of the auxiliary interferometer output in the experiments. Obviously, because of the influence of noises and mode-hopping signals, some of the extreme points are incorrect (as shown in the red circle section). Thus, the parameters \(N\) and \(P\) in Equation (5) are different from their ideal values.  
When there are noises and mode-hopping signals in the auxiliary interferometer, Equation (3) should be rewritten as:  
\[2\pi\Delta f(k)\tau_{a}=\ (\pi+\Delta\ \Omega(k_{1}))k,\ k=1,\ 2,\ldots,\ N,\ k_{1}= 1,\ 2,\ \ldots,\ N_{1} \tag{8}\]  
where \(\Delta\Omega(k_{1})\) is the phase deviation caused by all of the sampling points of the auxiliary interferometer output in the actual situation, and only part of the \(\Delta\Omega(k_{1})\) is caused by the erroneous sampling points is not equal to zero, while \(N\) is the number of the ideal sampling points, and \(N_{1}\) is the total number of the sampling points in the actual situation. Thus, Equation (4) turns into:  
\[I_{m}(k)=A_{m}\cos(2\pi f_{0}\tau_{m}+\pi k\tau_{m}/\tau_{a}+\Delta\Omega(k_{1} )k\tau_{m}/\tau_{a}) \tag{9}\]  
Equation (9) by FFT yields:  
\[(1+\sum_{k_{2}=1}^{N_{2}}\frac{\Delta\Omega(k_{2})}{\pi N})\frac{\tau_{m}}{2 \tau_{a}}=\frac{P_{1}}{N_{1}} \tag{10}\]  
where \(N_{2}\) is the number of all of the erroneous sampling points in the auxiliary interferometer output, and \(P_{1}\) is the abscissa value of the peak point in the frequency spectrum of the measurement signal in Equation (9).  
Figure 2: (**a**) Extreme points in the experiment when there are noises in the auxiliary interferometer. (**b**) Extreme points in the experiment when there is the mode-hopping signal in the auxiliary interferometer.  
If \(N_{2}\) is much smaller than \(N\), the second item on the left side of Equation (10) can be neglected. So, the measurement error caused by the erroneous sampling points of the auxiliary interferometer output can be shown as:  
\[\Delta L_{m1} = L_{m1}-\ L_{m}\ =\ \frac{n_{f}}{n_{air}}L_{a}(P_{1}/N_{1}\ -\ P/N) \tag{11}\]  
where \(\Delta L_{m1}\) is the measurement error caused by the erroneous sampling points of the auxiliary interferometer, \(L_{m1}\) is the measured distance when there are erroneous sampling points, and \(L_{m}\) is the ideal measured distance when there are no erroneous sampling points. Figure 3a shows the comparison of the simulated result when there are 10 erroneous sampling points (the red line) and the ideal distance (the blue line). In our experiments, \(n_{f}\) is 1.4682, \(n_{air}\) is 1.0003, and \(L_{a}\) is 20.05832 m. When the ideal value of \(P\) is 9000, \(N\) is 60,000, and \(N_{2}\) is 10, the measured distance error is 758.9 \(\upmu\)m, which is agreement with the theoretical measurement error of 760 \(\upmu\)m, which is calculated by Equation (11). Obviously, the accuracy of the measurement distance is greatly affected. Besides, the main lobe energy of the spectrum decreases, and the side lobe's energy increase. Hence, it is essential to eliminate the influence of the erroneous sampling points in order to ensure the accuracy.  
As is shown in Figure 3b, if \(N_{2}\) is 100, the second item on the left-hand side of Equation (11) can't be neglected. In this situation, we cannot obtain the accurate measured distance.  
Figure 3: (**a**) The simulation result of 10 erroneous sampling points on accuracy when the ideal value of \(P\) is 9000 and \(N\) is 60,000. (**b**) The simulation result of 100 erroneous sampling points on accuracy when the ideal value of \(P\) is 9000 and \(N\) is 60,000.