The set of residuals for the RO system are as follows.

$r_{RO1_1}: \\ \dot{\hat{q}}_{fp}=\frac{1}{I_{fp}}(-R_{fp}\hat{q}_{fp}- y_1+ u_1) \\ r_{RO1_1} = y_1 - \hat{q}_{fp}$

$r_{RO1_2}: \\ \dot{\hat{q}}_{rp} = \frac{1}{I_{rp}} (- R_{rp} - R_{forward}) \hat{q}_{rp}- y_2 + u_2 \\ \dot{\hat{e}}_{Ck}= \frac{\hat{q}_{rp}}{C_k}(6 y_4 +0.1)/(1.667*10^{-8}) \\ r_{RO1_2} = y_5 - \hat{e}_{Ck}$

$r_{RO1_3}: \\ \hat{R}_{memb} =0.202 (4.137*10^{11}(\frac{y_5-12000}{165}+29)) \\ \hat{q}_{rp} = \frac{1.667*10^{-8} C_k \dot{y}_{5}}{6 y_4 +0.1} \\ \dot{\hat{p}}_{memb} = \frac{1}{{C_{memb}}} (\hat{q}_{rp} - \frac{\hat{p}_{memb}}{ \hat{R}_{memb}} - \frac{\hat{p}_{memb} - y_1}{R_{return_l}} ) \\ r_{RO1_3} = y_2 - \hat{p}_{memb}$

$r_{RO2_1}: \\ \dot{\hat{q}}_{fp}=\frac{1}{I_{fp}}(-R_{fp}\hat{q}_{fp}- y_1+ u_1) \\ r_{RO2_1} = y_1 - \hat{q}_{fp}$

$r_{RO2_2}: \\ \dot{\hat{q}}_{rp} = \frac{1}{I_{rp}} (- R_{rp} - R_{forward}) \hat{q}_{rp}- y_2 + u_2 \\ \dot{\hat{e}}_{Ck}= \frac{\hat{q}_{rp}}{C_k}(6 y_4 +0.1)/(1.667*10^{-8}) \\ r_{RO2_2} = y_5 - \hat{e}_{Ck}$

$r_{RO2_3}: \\ \hat{R}_{memb} =0.202 (4.137*10^{11}(\frac{y_5-12000}{165}+29)) \\ \hat{q}_{rp} = \frac{1.667*10^{-8} C_k \dot{y}_{5}}{6 y_4 +0.1} \\ \dot{\hat{p}}_{memb} = \frac{1}{{C_{memb}}} (\hat{q}_{rp} - \frac{\hat{p}_{memb}}{ \hat{R}_{memb}} - \frac{\hat{p}_{memb} - y_1}{R_{return_s}} ) \\ r_{RO2_3} = y_2 - \hat{p}_{memb}$

$r_{RO3_1}: \\ \dot{\hat{q}}_{fp}=\frac{1}{I_{fp}}(-R_{fp}\hat{q}_{fp}- y_1+ u_1) \\ r_{RO3_1} = y_1 - \hat{q}_{fp}$

$r_{RO3_2}: \\ \hat{R}_{memb} =0.202 (4.137*10^{11}(\frac{y_5-12000}{165}+29)) \\ \hat{q}_{rp} = \frac{y_1-\hat{p}_{memb}}{R_{forward}} \\ \dot{\hat{p}}_{memb} = \frac{1}{{C_{memb}}} (\hat{q}_{rp} - \frac{\hat{p}_{memb}}{ \hat{R}_{memb}} - \frac{\hat{p}_{memb} }{R_{return_{ASE}}} ) \\ r_{RO3_2} = y_2 - \hat{p}_{memb}$