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For my first problem, I am trying to find the input resistance of the circuit itself, is it just simply the input voltage V1 minus the calculated voltage at the negative terminal of the upper left op-amp divided by 1k?

For my second problem, I'm trying to construct an op-amp circuit that produces an output corresponding to this formula with one summing amplifier and one differentiating amplifier, but the input impedance seen by both V1 and V2 must be 5k:

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Any tips on how to proceed? I am a bit stuck on how to force the input impedance seen by the two voltages to be 5k.

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Here is what I have come up with so far for the two separate op-amps. How can I combine them while forcing both input impedances seen by V1 and V2 to be 5k?

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  • \$\begingroup\$ Your two questions are big enough to post individually. That will help readers searching the site know if the questions they see are helpful to them. \$\endgroup\$
    – The Photon
    Commented Apr 3, 2016 at 15:28

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I am trying to find the input resistance of the circuit itself, is it just simply the input voltage V1 minus the calculated voltage at the negative terminal of the upper left op-amp divided by 1k?

Check yourself with dimensional analysis. If you divide a voltage by a resistance, do you get a resistance, or something else?

The input resistance of a circuit is how much the voltage has to change to cause a change in input current:

$$R_{\mathrm{in}}=\frac{\mathrm{d}V_{\mathrm{in}}}{\mathrm{d}I_{\mathrm{in}}}$$

If you can't figure out this value by inspection, you can calculate the input current for two different input voltages and use the differences to find the derivative (since this is a linear circuit, it doesn't matter what two points you choose).

On your second question,

I am a bit stuck on how to force the input impedance seen by the two voltages to be 5k.

Think about what op-amp circuit gives a high input impedance and doesn't change the voltage signal. How could you modify it to have whatever input impedance you want?

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  • \$\begingroup\$ Sorry, for the first question I meant to say that that was the input current, and then I would divide V1 by that input current to get the input resistance. Is that the right concept for input resistance? \$\endgroup\$
    – David M
    Commented Apr 3, 2016 at 15:31
  • \$\begingroup\$ @DavidM, Honestly I think I'd have to do a full analysis to do the first one. \$\endgroup\$
    – The Photon
    Commented Apr 3, 2016 at 15:34
  • \$\begingroup\$ Well I was saying if you had all the node voltages worked out in the first circuit, is finding the input current like that and then the input resistance by dividing V1 by the input current the way to solve for the input resistance? \$\endgroup\$
    – David M
    Commented Apr 3, 2016 at 15:36
  • \$\begingroup\$ Also, I'm still a bit confused on the second portion. For this problem I'm trying to construct a circuit to that equation but I can only use one summing amplifier and one differentiating amplifier. I know how to make them individually, but I'm not sure how to make sure the input impedance is 5k when they are connected together \$\endgroup\$
    – David M
    Commented Apr 3, 2016 at 15:48
  • \$\begingroup\$ Does the text of the question ask you to make the input impedance 5 kOhms, or the input resistance, or the magnitude of the input impedance? If it actually asks for the input impedance to be 5 kOhms, as you stated it in the post, I don't think it's possible without an additional op-amp. \$\endgroup\$
    – The Photon
    Commented Apr 3, 2016 at 15:52

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