# Effect of the Transition Points Mismatch on Quanta Image Sensors
## 3 Imaging Models for the Effect of the Transition Points Mismatch  
### Single-bit QIS  
The imaging model for the single-bit QIS employs the method described in [10] as shown in Figure 8. The grayscale matrix \(C_{n\times n}\) of a standard 256 \(\times\) 256 image (\"lena\") as shown in Figure 9a is abstracted, and every element is converted to an \(m\times m\) two-dimensional matrix through bicubic interpolation. The grayscale values (0-255) cannot be used as the exposure levels because they do not match with the dynamic range of a single-bit QIS, so the enlarged matrix is multiplied by an illumination factor \(h_{0}\) that is smaller than 1, and each new element is the average exposure level \(H\) for a jet as well as its ideal output signal expressed as an electron number. Each exposure \(H\) is converted to a random number according to a Poisson distribution. Then, the integer is converted to a random number according to a normal distribution with a standard deviation \(U_{n}\). These two conversions simulate shot noise and read noise, respectively. In the quantization process, the output signal containing shot noise and read noise is compared to a transition point \(U_{th}\) and a binary jet bit  
Figure 7: The BER as a function of \(U_{thm}\) for different QIS bit depths \(n\) and their relevant worst exposure for read noise \(U_{n}=0.15\)e\({}^{-}\). is obtained. The variable \(U_{th}\) could be a constant value for the whole jet array for ideal 1-bit ADCs or vary from column to column for actual 1-bit ADCs when ADCs are column parallel. Repeat the operations on the exposure \(H\) for \(z\) times to generate \(z\) captured sequential fields. Thus, the original value of a pixel is converted to an \(m\times m\times z\) binary jet bit cubic. In the image reconstruction process, the bit data of a cubic are added together, and the result represents the gray level of a pixel in the final image. The sum of a cubic has to be multiplied by \(255/m^{2}z\) to match with the grayscale range of the original image. In this imaging model, the spatial oversampling factor is \(m^{2}\), the temporal oversampling factor is \(z\), and the equivalent FWC of the final image is \(m^{2}z\). The BER is not suitable for comparing reconstructed images with different spatial oversampling factors or bit depths. This is explained in later discussion. Instead, this paper employs an evaluation index, called the grayscale similarity index (GSI), to calculate the difference between a reconstructed image and a reference image. It can be expressed as:  
Figure 8: A block diagram of the single-bit QIS imaging model. Figure 9: **(a)** The standard \(256\times 256\) image “lena”; (b) a reconstructed image contaminated by shot noise and read noise (\(U_{n}=0.15\text{e}^{-}\)) without the mismatch of a 1-bit ADC transition point (\(U_{th}=0.5\)) for \(m=16\) and \(z=16\), and the grayscale similarity index (GSI) is 0 LSB; (c) a reconstructed image with a transition point mismatch, \(U_{thn}=0.15\text{e}\)- added on the basis of (b), and the GSI is 3.31 LSB; (d) a reconstructed image from the same condition as (c), except that \(m=64\), \(z=1\), and the GSI is 1.74 LSB. \[GSI=\sqrt{\frac{1}{256\times 256}\sum_{i=1}^{256}\sum_{j=1}^{256}\left[c_{rec}(i,j)-c _{ref}(i,j)\right]^{2}}, \tag{14}\]  
where \(c_{rec}(i,j)\) and \(c_{ref}(i,j)\) is the grayscale value of pixel \((i,j)\) in the reconstructed image and the reference image. The reference image is a reconstructed image with ideal ADCs in the models. In the following discussion, the equivalent FWC is \(m^{2}z=4096\), and the illumination factor \(h_{0}=1/400\) to guarantee that the average exposure level \(H\) of each jet is smaller than full exposure \(H=1\). In fact, \(h_{0}=1/255\) is enough and is compatible with that in the multi-bit case. However, in the simulation, it was found that the reconstructed image with a little smaller \(h_{0}\) has better contrast. The reference image is reconstructed with a constant transition point \(U_{th}=0.5\) for \(m=16\) as shown in Figure 9b.