# Suitable Integral Sampling for Bandpass-Sampling Time-Modulated Fourier Transform Spectroscopy
## 3 Analyzing Suitable Sampling Conditions for BPS-FTS  
In the context of BPS-FTS, we consider that the spectrum corresponding to each pixel (denoted by \(j\)) is monochromatic light. In an ideal scenario, the interferogram detected by this pixel, which continuously evolves over time, takes the form of a series of cosine curves. Sampling this interferogram at perfectly equidistant intervals across the entire OPD range, assuming that the OPD corresponding to a sampling point (denoted by \(i\)) is \(x_{i}^{j}\), yields an ideal sampling function comprising a series of impulse functions. However, practical sampling necessitates maintaining each sampling point for a certain integration time (but the OPD is still changing at this time), in order to accumulate sufficient energy and meet the signal-to-noise ratio requirements. This leads to the utilization of a rectangular sampling function. Figure 3 provides a schematic representation of monochromatic light sampling. Assuming that the moving mirror moves at a constant speed \(s\) during the sampling process and the integration time for each sampling is \(\Delta t\), the corresponding change in OPD is \(2s\Delta t\). The interference intensity obtained at sampling point \(x_{i}^{j}\) is represented in Equation (3):  
\[ \begin{split}& I\left(x_{i}^{j}\right)=\frac{1}{2}\int\limits_{x_{i}^ {j}\to s\Delta t}^{x_{i}^{j}+s\Delta t}B(\
u)\cos(2\pi\
u x)dx\\ &=\frac{1}{2}B(\
u)\cdot\frac{1}{\pi\
u}\cos\left(2\pi\
u x_{i}^{ j}\right)\sin(2\pi\
u s\Delta t)\\ &=B(\
u)\cdot s\Delta t\cdot\cos\left(2\pi\
u x_{i}^{j}\right) \sin(2\pi\
u s\Delta t)\end{split} . \tag{3}\]  
Following the definition of SMTF for FTS, the SMTF corresponding to Equation (3) forms a sinc function. According to the SMTF criterion, it can be deduced that when the value of the SMTF decreases to within 90% of its maximum value [28; 29; 30], the error tolerance of the reconstructed spectrum derived from the interferogram is deemed to be acceptable. In other words, the accuracy of the reconstructed spectrum can be considered precise. In this scenario,  
\[\text{SMTF}=\text{sinc}(2\pi\
u s\Delta t)\geq 0.9. \tag{4}\]  
In practical operations, each pixel \(j\) in BPS-FTS typically corresponds to a range of the spectrum, denoted by \([\
u_{\text{min}},\
u_{\text{max}}]\). According to Equation (2), in this context, for any valid \(m\), the corresponding range of sampling interval \(\Delta x\) is \(\left[\frac{m-1}{2\
u_{\text{min}}},\frac{m}{2\
u_{\text{max}}}\right]\). We define \(d\_\Delta x\) as the range covered by this interval, which is defined as  
\[d\_\Delta x=\frac{m}{2\
u_{\text{max}}}-\frac{m-1}{2\
u_{\text{min}}}. \tag{5}\]  
In this case, for any \(m\) that satisfies this condition, the corresponding sampling interval can be expressed as  
\[\Delta x=\frac{m}{2\
u_{\text{max}}}-k\cdot d\_\Delta x=\frac{(1-k)m}{2\
u_{ \text{max}}}+\frac{k(m-1)}{2\
u_{\text{min}}}, \tag{6}\]  
where \(0\leq k\leq 1\). If the integration sampling time is \(\Delta t\), then the sampling width of the rectangular function is \(2s\Delta t=r\Delta x(0\leq r\leq 1)\). For any wavenumber \(\
u\) within \([\
u_{\text{min}},\
u_{\text{max}}]\), the  
Figure 3: Schematic diagram of monochromatic light interferogram sampling. The black curve represents the ideal interference signal collected at pixel \(j\). The red rays represent the ideal sampling function under ideal conditions. The green curve represents the sampling function under integral sampling conditions. integrated sampling interferogram corresponding to Equation (3) can be updated using Equation (7):  
\[ \begin{split}& I\Big{(}x_{i}^{j}\Big{)}=B(\
u)\cdot\cos\Big{(}2\pi \
u x_{i}^{j}\Big{)}\frac{\sin\!\left\{\pi\
u r\left[\frac{(1-k)m}{2i_{\max}}+ \frac{k(m-1)}{2i_{\min}}\right]\right\}}{2\pi\
u}\\ &=\frac{1}{2}B(\
u)\cdot\cos\Big{(}2\pi\
u x_{i}^{j}\Big{)}\text {sinc}\Big{\{}\pi\
u r\left[\frac{(1-k)m}{2i_{\max}}+\frac{k(m-1)}{2i_{\min}} \right]\Big{\}}\cdot r\Big{[}\frac{(1-k)m}{2i_{\max}}+\frac{k(m-1)}{2i_{\min}} \Big{]}\end{split} .