Assuming ideal op-amplifiers, I did the following steps:
\$v_{out1}=v_1\frac{R_2}{R_1}=2v_1\$
\$v_{out2}=v_o=\left(\frac{v_{out1}}{R_3}+\frac{v_2}{R_4}\right)R_5=4v_1+2v_2\$
My textbook claims \$4v_1-2v_2\$
\$\begingroup\$
\$\endgroup\$
3
-
\$\begingroup\$ There are 2 things you need to know here. The correct transfer function for an inverting opamp, and how to apply superposition. \$\endgroup\$– efox29Oct 9, 2015 at 0:03
-
\$\begingroup\$ I know the transfer function, and I am aware of the superposition theorem, but I'm not sure if I've done that step correctly, considering my answer comes out right except for a minus sign on the second term. \$\endgroup\$– B. LeeOct 9, 2015 at 0:05
-
\$\begingroup\$ Vout1 is not 2Vi. Transfer function is wrong. \$\endgroup\$– efox29Oct 9, 2015 at 0:07
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
These stages are inverting. Your equations are both missing a - in the transfer function.
V1 gets inverted twice, that's why it has a positive coefficient. V2 gets inverted once, that's why it has a negative coefficient.
answered Oct 9, 2015 at 0:14