# Research on Performance Evaluation and Optimization Theory for Thermal Microscope Imaging Systems
## 2 Materials and Methods
### SNR of the TMISs Output Image  
The SNR of the target image received by the imaging system can be expressed as  
\[SNR=\frac{S}{N}=\frac{\Delta\Phi(\lambda,T)R(\lambda)}{\sqrt{\int_{0}^{\infty} s^{\prime}(f)MTF_{e}^{2}(f)df}} \tag{7}\]  
where \(R(\lambda)\) is the responsivity of the detector; \(s^{\prime}(f)\) is the noise power spectrum of the system; \(MTF_{e}(f)\) is the transfer function of the electronic filter; \(\Delta\Phi(\lambda)\) is the radiant flux difference of the target and the background:  
\[\Delta\Phi(\lambda)=ab[\varepsilon_{T}(\lambda,T)M_{0}(\lambda,T_{T})- \varepsilon_{B}M_{0}(\lambda,T_{B})]\tau_{0}(\lambda)\left(\frac{d_{0}}{2} \right)^{2}\left/[\left(\frac{d_{0}}{2}\right)^{2}+l^{\prime 2}]\right. \tag{8}\]  
Figure 5: Optical path diagram of the TMISs.  
where \(T_{T}\) and \(T_{B}\) are the target and background temperatures. When \(T_{T}\) and \(T_{B}\) are not significantly different, calculus can be used instead of the difference. Then, the radiant flux difference of the target and the background is  
\[\Delta\Phi(\lambda)\approx ab[\Delta\varepsilon M_{0}(\lambda,T_{B})+ \varepsilon_{B}\frac{\partial M_{0}(\lambda,T_{B})}{\partial T}\Delta T]\tau_{0 }(\lambda)\left(\frac{d_{0}}{2}\right)^{2}\Bigg{/}[(\frac{d_{0}}{2})^{2}+l^{ \prime 2}] \tag{9}\]  
where \(\Delta\varepsilon=\varepsilon_{T}-\varepsilon_{B}\) is the spectral radiation difference between the target and background; \(\varepsilon_{T}\) and \(\varepsilon_{B}\) represents the spectral emissivity of the target and background, respectively; \(\Delta T=T_{T}-T_{B}\) is the target and background temperature difference; \(\partial M_{0}(\lambda,T_{B})/\partial T\) is the rate of change of the blackbody spectral emittance at the background temperature \(T_{B}\) with respect to the temperature difference.  
Assuming that the measurement normalized detection rate is \(D^{*}\); \(V_{n}\) is the noise voltage corresponding to the unit bandwidth at the measurement point \(f_{0}\); \(\Delta f\) is the measurement bandwidth; \(A_{d}=a\times b\) is the area of the detector. Then, the expression relating \(D^{*}\) and \(R\) is  
\[R(\lambda)=\frac{D^{*}(\lambda)V_{n}}{\sqrt{A_{d}\Delta f}}=D^{*}(\lambda) \sqrt{\frac{s^{\prime}(f_{0})}{A_{d}}} \tag{10}\]  
substituting Equation (10) into Equation (7) yields:  
\[SNR=\frac{\Delta\Phi(\lambda,T)D^{*}(\lambda)}{\sqrt{A_{d}}\sqrt{\int_{0}^{ \infty}s(f)MTF_{e}^{2}(f)df}}=\frac{\Delta\Phi(\lambda,T)D^{*}(\lambda)}{\sqrt {A_{d}\Delta f_{n}}} \tag{11}\]  
where \(s(f)=s^{\prime}(f)/s^{\prime}(f_{0})\) is the normalized noise power spectrum; \(\Delta f_{n}\) is the noise equivalent bandwidth:  
\[\Delta f_{n}=\int_{0}^{\infty}s(f)MTF_{e}^{2}(f)df \tag{12}\]  
Combining Equations (9) and (11), we obtain the image SNR of the TMISs with working wavelength in the range of \([\lambda_{1},\lambda_{2}]\):  
\[\begin{split}& SNR=\frac{\Delta\Phi(\lambda,T)D^{*}(\lambda)}{ \sqrt{A_{d}\Delta f_{n}}}\\ &=\frac{\int_{\lambda_{1}}^{\lambda_{2}}ab[\Delta cM_{0}(\lambda, T_{B})+\varepsilon_{B}\frac{\partial M_{0}(\lambda,T_{B})}{\partial T} \Delta T]\tau_{0}(\lambda)\left(\frac{d_{0}}{2}\right)^{2}\Big{/}[(\frac{d_{0 }}{2})^{2}+l^{\prime 2}]D^{*}(\lambda)d\lambda}{\sqrt{A_{d}A_{n}}}\\ &=\int_{\lambda_{1}}^{\lambda_{2}}\big{[}\Delta cM_{0}(\lambda,T _{B})+\varepsilon_{B}\frac{\partial M_{0}(\lambda,T_{B})}{\partial T}\Delta T ]D^{*}(\lambda)d\lambda\times\frac{a\eta_{0}(\lambda)(d_{0}/2)^{2}}{\sqrt{A_ {d}\Delta f_{n}[(d_{0}/2)^{2}+l^{\prime 2}]}}\end{split} \tag{13}\]  
assuming that \(\varepsilon_{T}=\varepsilon_{B}=1\) and \(\tau_{0}(\lambda)=\tau_{0}\). Then, the signal-noise ratio of the target image received by the system is  
\[SNR=\int_{\lambda_{1}}^{\lambda_{2}}\frac{\partial M_{0}(\lambda,T_{B})}{ \partial T}\Delta TD^{*}(\lambda)d\lambda\times\frac{\sqrt{A_{d}}\tau_{0}{D_ {NA}}^{2}}{\sqrt{\Delta f_{n}}} \tag{14}\]  
where \(D_{NA}=sinu_{1}=\left.(\frac{d_{0}}{2})\right/\sqrt{\left[(\frac{d_{0}}{2})^{ 2}+l^{\prime 2}\right]}\).