I know that the transfer function of an inverting summing amplifier gives this, by using the Superposition principle: (for example with 2 input voltages)

\$U_{ out }=-(\frac { R_{ 1 } }{ R_{ f } } U_{ in1 }+\frac { R_{ 2 } }{ R_{ f } } U_{ in2 })\$

However, I don't exactly know how to derive this from the transfer function definition:

\$H = U_{out}/U_{in} = G(U_{+}-U_{-})/U_{in}\$

Inverting Summing Op-Amp

This is not explained in any book or on the Web. Can anyone explain this?

Thanks a lot!

  • \$\begingroup\$ U+ is zero and U- can be found using again superposition. That`s all. But you have to realize that there are TWO transfer functions because you have two input voltages, unless you define Uin as the sum of both input signals. \$\endgroup\$
    – LvW
    Commented Oct 25, 2015 at 16:35
  • \$\begingroup\$ I know that U+ is 0. \$\endgroup\$
    – ant0nisk
    Commented Oct 25, 2015 at 16:37
  • \$\begingroup\$ So, I need to transfer functions and then sum them? Like: H1 = Uout/Uin1 and H2 = Uout/Uin2 ? \$\endgroup\$
    – ant0nisk
    Commented Oct 25, 2015 at 16:38
  • \$\begingroup\$ Ok! I got it! I did it by having 2 transfer functions! Would you like to post it as an answer so that I accept it? \$\endgroup\$
    – ant0nisk
    Commented Oct 25, 2015 at 16:52
  • \$\begingroup\$ I think it is OK so. Congratulations. \$\endgroup\$
    – LvW
    Commented Oct 25, 2015 at 19:25

1 Answer 1


For anyone interested how I derived the transfer function, here is the solution:

(This is the derivation for the schematic as shown in the question. If you have more inputs, just do the same procedure for each one.)

Transfer function derivation

So, the solution was to have as many transfer functions as your inputs, and then the derivation is the same as an Inverting Amplifier.


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