Implement the following function using two op-amps and the required number of resistors
Vo = V1 - 2V2 + 3V3 - 4V4
Applying the KCL at node 1
0−V11KΩ−0−2V21KΩ+0−V011KΩ=0\frac{0-V_1}{1K\Omega}-\frac{0-2V_2}{1K\Omega}+\frac{0-V_{01}}{1K\Omega}=01KΩ0−V1−1KΩ0−2V2+1KΩ0−V01=0
V01=−V1−2V2V_{01}=-V_1-2V_2V01=−V1−2V2
Applying the KCL at node 2
0−V11KΩ+0−3V31KΩ+0−4V41KΩ+0−V01KΩ=0\frac{0-V_1}{1K\Omega}+\frac{0-3V_3}{1K\Omega}+\frac{0-4V_{4}}{1K\Omega}+\frac{0-V_{0}}{1K\Omega}=01KΩ0−V1+1KΩ0−3V3+1KΩ0−4V4+1KΩ0−V0=0
−V01−3V3−4V4=V0-V_{01}-3V_3-4V_4=V_0−V01−3V3−4V4=V0
VO=−[−V1−2V2]−3V3−4V4V_O=-[-V_1-2V_2]-3V_3-4V_4VO=−[−V1−2V2]−3V3−4V4
VO=V1+2V2−3V3−4V4V_O=V_1+2V_2-3V_3-4V_4VO=V1+2V2−3V3−4V4
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