Engineering Does this (textbook) circuit diagram contain a mistake?

AI Thread Summary
The discussion centers on a circuit diagram purportedly used to solve differential equations, with concerns raised about the absence of a resistor for the input signal -f(t) at the inverting terminal of an op-amp. Participants argue that without this resistor, the circuit's functionality is compromised, as it would not allow for proper signal addition. Some suggest that the input may be functioning as a transimpedance amplifier, indicating it could be a current input rather than a voltage input. However, there is skepticism about the practicality of using a circuit designed for differential equations with such an input-output configuration. The consensus leans towards identifying a mistake in the diagram regarding the necessary resistor for proper operation.
kostoglotov
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Homework Statement


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This circuit solves some differential equation, the question is asking for the equation based on the circuit diagram

b1npQSK.png


imgur link: http://i.imgur.com/b1npQSK.png

Homework Equations

The Attempt at a Solution



I refer to the input terminal for -f(t)...the signal from it doesn't seem to pass through any resistor on its way into the inverting summer, so I've no way (I think) of finding how the signal from -f(t) is scaled in the summer.

Is this a mistake on the diagram? If not, what am I missing?
 
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Could be a current input rather than a voltage input... or a mistake.
 
Yes - I am sure, it is a mistake. The right most amplifier must add up three signals (voltages) - and for this purpose another series resistor is necessary.
 
LvW said:
The right most amplifier must add up three signals (voltages)...

Where does the problem statement say that? Did you leave something out?
 
Tom.G said:
Where does the problem statement say that? Did you leave something out?
It is the purpose of the circuit to "solve differential equations".
Hence, it needs an input. Such an input (time domain) is shown as "-f(t) directly at the inv. terminal of an opamp, which - at the same time - receives the sum of two other voltages. However, this addition works only if f(t) also is connected through another resistor (remember the virtual ground principle).
Thus, without such a resistor the whole circuit makes no sense.
 
LvW said:
...this addition works only if f(t) also is connected through another resistor...

That input is acting as a transimpedance amplifier. It is a current input. You can use KCL to help wrap your head around it.
https://en.wikipedia.org/wiki/Transimpedance_amplifier
 
Tom.G said:
That input is acting as a transimpedance amplifier. It is a current input. You can use KCL to help wrap your head around it.
https://en.wikipedia.org/wiki/Transimpedance_amplifier
A "current input" is nothing else than a voltage input with a corresponding large source resistance (see my answer).
On the other hand - do you really assume that such a circuit (analog computer circuitry for solving differential equations) is intended to work with current in and voltage out? This would make no sense at all.
 

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