# Signal-to-Noise Ratio Analysis of Bandpass Sampling Time-Modulated Fourier Transform Spectroscopy
## 2 Modeling Analysis of SNR
### Signal-to-Noise Ratio Modeling of BPS-FTS  
For BPS-FTS, due to the use of the bandpass sampling theorem, compared with the traditional Nyquist sampling theorem in FTS, the number of interferogram sampling points is greatly reduced, and the integration time is synchronously increased, which makes its SNR different from that of FTS. Assuming that its OPD sampling interval is \(\Delta x_{B}\) and the working spectrum range is \([v_{\text{min}},v_{\text{max}}]\), according to the basic bandpass sampling theorem, in BPS-FTS, the interferogram sampling interval is [14; 15]:  
\[\frac{m-1}{2v_{\text{min}}}\leq\Delta x_{B}\leq\frac{m}{2v_{\text{max}}} \tag{9}\]  
where \(v_{\text{min}}\) and \(v_{\text{max}}\) are the minimum and maximum wave numbers of the working spectrum, respectively, and \(m\) is any positive integer equal to or less than \(v_{\text{max}}/(v_{\text{max}}-v_{\text{min}})\). When \(m=1\), bandpass sampling is equivalent to Nyquist sampling; at this time, the SNR of BPS-FTS is equal to the SNR of FTS. It can be seen from Equation (9) that \(\Delta x_{B}\) is limited to change within a certain range, and the integration time also changes accordingly. At this time, the integration time \(\Delta t_{B}\) is a function of \(\Delta x_{B}\), \(r_{B}\) is the ratio of the OPD change within the integration time \(\Delta t_{B}\) to the OPD sampling interval \(\Delta x_{B}\), then:  
\[2s\Delta t_{B}=r_{B}\Delta x_{B}(0\leq r_{B}\leq 1) \tag{10}\]  
In the sampling interval range to each \(m\), when the interferogram OPD sampling interval takes the maximum sampling interval corresponding to \(m\), the maximum integration time can be obtained under this sampling interval. At this time, the maximum SNR is obtained, and the sampling interval can be expressed as follows:  
\[\Delta x_{B}=\frac{m}{2\
u_{\max}} \tag{11}\]  
From this, the number of interferogram sampling points under the maximum sampling interval can be calculated as follows:  
\[iB=\frac{2L}{\Delta x_{B}}=\frac{4L\
u_{\max}}{m} \tag{12}\]  
where \(L\) is the max OPD. The sampling OPD sequence corresponding to the maximum sampling interval is as follows:  
\[x_{ib}=-L+(ib-1)\cdot\Delta x_{B} \tag{13}\]  
where \(ib\) is all integers in the range \([1,iB]\);  
From Equation (6), we can get that at any \(x_{ib}\), within \(\Delta t_{B}\), the number of electrons output within the pixel band of the detector is as follows:  
\[ \begin{array}{l}e_{B\_xxt}(x_{ib})=\frac{A_{d}\pi(1-\epsilon)}{4F^{2}}\cdot \tau\cdot\eta\cdot\frac{1}{\hbar c\upsilon}\cdot\frac{1}{2s}\cdot 2 RT\cdot\\ \int\limits_{v_{\min}}^{\infty}\{[B(v_{iv})\cdot 2s\Delta t_{B}+2s\Delta t_{B} \cdot B(v_{iv})\cdot\cos(2\pi\
u_{iv}x_{ib})\text{sinc}(2\pi\
u_{iv}s\Delta t _{B})]\}dv\end{array}  \tag{14}\]  
where  
\[\Delta t_{B}=\frac{r_{B}\Delta x_{B}}{2s}=\frac{r_{B}\cdot m}{4s\
u_{\max}} \tag{15}\]  
Substituting Equation (10) into Equation (14), we can get the number of electrons represented by \(\Delta x_{B}\) and \(r_{B}\) as follows:  
\[ \begin{array}{l}e_{B\_x\_x_{B}}(x_{ib})=\frac{A_{d}\pi(1-\epsilon)}{4F^{2}} \cdot\tau\cdot\eta\cdot\frac{1}{\hbar c\upsilon}\cdot\frac{1}{2s}\cdot 2 RT\cdot\\ \int\limits_{v_{\min}}^{v_{\max}}[B(v_{iv})\cdot r_{B}\Delta x_{B}+r_{B}\Delta x _{B}\cdot B(v_{iv})\cdot\cos(2\pi\
u_{iv}x_{ib})\text{sinc}(\pi\
u_{iv}r_{B} \Delta x_{B})]dv\end{array}  \tag{16}\]  
At this time, for BPS-FTS, substitute Equation (11) into Equation (16), and consider the discretization form of the subsequent simulation input spectrum, where \(iv\) is the wave number position after discretization and \(V\) is the number of bands. It can be obtained that at each \(x_{ib}\), the number of electrons output by the detector is approximately as follows:  
\[