## 2 Introduction to Bandpass Sampling Theory and BPS-FTS  
### Bandpass Sampling Theory  
In practice, the spectra measured by Fourier transform spectroscopy instruments are usually limited to a finite bandwidth [3; 5; 6; 7]. According to bandpass sampling theory, the bandpass signal sampling frequency is determined by both the signal bandwidth and maximum frequency, which can overcome the limitations of the Nyquist sampling theorem and recover the signal without aliasing [5; 6; 7; 8; 9]. We assume that the highest band signal frequency is \(f_{H}\), the lowest frequency is \(f_{L}\), and the signal bandwidth is \(D=f_{H}-f_{L}\). According to bandpass sampling theory, to prevent aliasing in the sampled signal frequency domain, it is necessary to ensure that the spectral curve obtained after \(m\) and \(m-1\) translations of the component in the negative frequency domain of the signal cannot overlap with the original positive frequency domain component of the signal in this spectral region. Therefore, the following equation must be satisfied:  
\[\left\{ \begin{array}{l}-f_{L}+(m-1)f_{s}\leq f_{L}\\ -f_{H}+mf_{s}\geq f_{H}\end{array} \right. \tag{1}\]  
where \(f_{s}=2D\). From Equation (1), it follows that:  
\[\frac{2f_{H}}{m}\leq f_{s}\leq\frac{2f_{L}}{m-1} \tag{2}\]where \(1\leq m\leq\left\lfloor f_{H}/D\right\rfloor_{\
u}\left\lfloor\bullet\right\rfloor\) are rounded down. For Fourier transform spectroscopy, the frequency of the interference signal corresponds to the wavenumber \(\
u=1/\lambda\) of the spectrum [6]. According to bandpass sampling theory, the interference signal frequency range can be determined by the wavenumber boundary of the spectrum. Assuming that the detected spectrum wavenumber range is \(\
u_{\min}\sim\
u_{\max}\), then the interferogram sampling step \(\Delta x\) can be determined by Equation (3):  
\[\frac{m-1}{2\
u_{\min}}\leq\Delta x\leq\frac{m}{2\
u_{\max}} \tag{3}\]  
where \(m\) is any positive integer equal to or less than \(\
u_{\max}/(\
u_{\max}-\
u_{\min})\). When \(m_{\max}=\left\lfloor\
u_{\max}/(\
u_{\max}-\
u_{\min})\right\rfloor\), \(\Delta x\) is the maximum sampling step, which is given by:  
\[\frac{m_{\max}-1}{2\
u_{\min}}\leq\Delta x_{\max}\leq\frac{m_{\max}}{2\
u_{ \max}} \tag{4}\]  
Substituting \(m_{\max}=\left\lfloor\
u_{\max}/(\
u_{\max}-\
u_{\min})\right\rfloor\) into Equation (4), we can obtain:  
\[\Delta x_{\max}\leq\frac{1}{2\
u_{\max}}\cdot\left\lfloor\frac{\
u_{\max}}{ \
u_{\max}-\
u_{\min}}\right\rfloor=\frac{1}{2\
u_{\max}}\cdot\left\lfloor \frac{\
u_{\max}}{D}\right\rfloor \tag{5}\]  
From Equation (5), it can be observed that the maximum sampling step is inversely proportional to the bandwidth, which means that the narrower the bandwidth, the larger the maximum sampling step \(\Delta x_{\max}\). As bandwidth \(D\) converges to \(\
u_{\max}\), bandpass sampling approaches the Nyquist sampling theorem. For Fourier transform spectrometers, although the detected spectrum is a band signal, its bandwidth typically remains relatively wide. Considering atmospheric sounding applications as an example, their detection bands typically cover the medium wave band [10]. If the wavenumber is in the spectral range \(2100\) cm\({}^{-1}\sim 3050\) cm\({}^{-1}\), according to Equation (5), we can obtain the maximum interferogram sampling step in this range as \(\Delta x_{\max}\leq[1/(2\
u_{\max})]\cdot[3050/(3050-2100)]=[1/(2\
u_{\max}) ]\cdot 3\). Compared to the Nyquist sampling theorem, its sampling step is only three times larger. However, if we divide the spectrum in the range \(2100\) cm\({}^{-1}\sim 3050\) cm\({}^{-1}\) into \(100\) narrow bands, at this point, the bandwidth \(D\) of each narrow spectral band is \(1\%\) of the original full spectral bandwidth, the corresponding maximum sampling step \(\Delta x\) will increase by approximately two orders of magnitude, and the sampling frequency will decrease by approximately two orders of magnitude.