# Calibration for Sample-And-Hold Mismatches in \(M\)-Channel TIADCs Based on Statistics
## 5 Simulation and Comparison
### Comparisons
Figure 13: Output power spectra of the 2-channel TIADC (**a**) without calibration, (**b**) calibrated using the IFB-based method, (**c**) calibrated using the I/Q-based method, and (**d**) calibrated using the method in this paper. \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Order** & **PB Cut-Off Frequency** & **SB Cut-Off Frequency** & **PB Attenuation** & **SB Attenuation** \\ \hline \(N_{hp}=40\) & \(0.495f_{s}\) & \(0.4f_{s}\) & \(-1\) dB & \(-115\) dB \\ \hline \hline \end{tabular} Note: (1) “PB” is the abbreviation for pass-band. (2) “SB” is the abbreviation for stop-band.  
\end{table} 
Table 5: Simulation parameters for the high-pass filter in the input-free band (IFB) method. needs four multiplications of real numbers. \(\langle 3\rangle\) It consumes \(4(N_{cp}+1)\) multiplications to update the compensation filter's taps. Adding \(\langle 1\rangle\) to \(\langle 3\rangle\), the total number of multiplications needed in one calibration iteration is  
\[8N_{cp}+0.5N_{hb}+8. \tag{32}\]  
Here, \(N_{hb}=10\) and \(N_{cp}\) is set to be 0, which results in 13 multiplications in total. The IFB-based method and the proposed method use the common compensation technique and the consumption is calculated as follows. For simplicity, only the first Richardson iteration is considered for both the IFB method and ours. \(\langle 4\rangle\) It needs \(0.5N_{d}\) multiplications to get \(v^{\prime}\left[2n\right]\) because the differentiator's taps are anti-symmetric with a null center tap. \(\langle 5\rangle\) It requires 1 multiplication to scale \(v^{\prime}\left[2n\right]\) with estimated coefficients as (8). The consumption of the identification procedure for the IFB-based method is calculated as follows. \(\langle 6\rangle\) It consumes \((0.5N_{h}+1)\) multiplications to filter \(z[n]\) due to the anti-symmetric property of the high-pass filters' taps. \(\langle 7\rangle\) It requires 2 multiplications to update \(\hat{\beta}_{1}\). Adding \(\langle 4\rangle\) to \(\langle 7\rangle\), the total number of multiplications needed in one calibration loop is  
\[0.5\left(N_{d}+N_{hp}\right)+3. \tag{33}\]  
Here, \(N_{d}=20\) and \(N_{hp}=40\), which results in 33 multiplications in total. The consumption of the identification procedure for the proposed method is calculated as follows. \(\langle 8\rangle\) It takes \(2N\) multiplications to calculate \(D_{v_{1},v_{0}}\) and \(D_{v_{0},v_{1}}\) as (17). \(\langle 9\rangle\) It requires \((2N+1)\) multiplications to calculate \(\partial P/\partial\hat{\beta}_{1}\) as (25) to (28). \(\langle 10\rangle\) It requires 1 multiplication to update \(\hat{\beta}_{1}\). (All the constant coefficients used in (17), and (25) to (28) can be combined together into the step \(\mu\), so only one multiplication is needed to consider them all.) Adding \(\langle 4\rangle\) to \(\langle 5\rangle\) and \(\langle 8\rangle\) to \(\langle 10\rangle\), the total number of multiplications needed in one calibration loop is  
\[4N+0.5N_{d}+2. \tag{34}\]  
Here, \(N_{d}=20\) and \(N\) is set to be 32, which results in 140 multiplications in total. The comparison results of the above two cases are shown in Table 9. The proposed method in this paper is more complex compared with the other two methods when calibrating the same signals, but this method can apply to more types of signals and more channels whereas the other two methods cannot.