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This is an article about bilateral feedback analysis.

Could anyone explain the what is meant by this sentence?

The most important fact about this circuit is the symmetric bilateral transmission through the feedback impedance \$Z_{F}\$ albeit the forward amplifier is still unilateral.

What does "symmetric bilateral transmission" mean here?

Thank you.

enter image description here

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2 Answers 2

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What is the purpose of ZF? Answer: To provide a feedback path that produces a voltage across ZS||Zin. This feedback voltage is superimposed with a corresponding part of the "normal" signal input voltage Vi at the opamp input node.

However, this is an ideal case that happens only if there would be an output impedance Zo=0. If Zo has a finite value, a part of the input signal Vi arrives directly at the output (and NOT only via the forward amplifier Voc). This portion depends on the voltage divider Zo/(Zs+ZF+Zo). This neglects the large Zin.

That means: The feedback path works in both directions: (1) As wanted: From the output back to the input and (2) Not desired but unavoidable: Directly from the input to the output.

By the way: We face the same situation with a feedback resistor connected between the collector node and the base node of a transistor stage. This is the principle behind the so called "Miller-Effect".

It is a remaining question why this example circuit has positive feedback (perhaps just a drawing error ?).

EDIT: In equ. (2) we can introduce the following ideal values: Zin infinite, Zo=0 and the open-loop gain k as infinite. In this case, we arrive at A=-ZF/ZS . This is the classical inverting gain for an ideal opamp with negative feedback. Hence, in the shown circuit we have to exchange the non-inv. and the inv. input (drawing error).

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  • \$\begingroup\$ About feedback sign, maybe k<0... \$\endgroup\$ Jul 16, 2016 at 22:05
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    \$\begingroup\$ anhna, for a finite value of Zo a part of the input voltage arrives DIRECTLY across Zo (via ZF). Therefore we have two voltages at the output node: The amplified voltage (as always) and a small portion that is caused by the voltage divider Zo/(Zo+ZF). This last portion is not desired because ZF is intended to act only from the output back to the input - but this effect cannot be avoided for a finite Zo. Hence, we have the case of "bilateral feedback" (a large feedback signal and a smaller "forward" signal). That is the meaning of "bilateral feedback". \$\endgroup\$
    – LvW
    Jul 18, 2016 at 7:41
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    \$\begingroup\$ anhna, did you realize the sign error in the shown circuit? (see my EDIT). \$\endgroup\$
    – LvW
    Jul 18, 2016 at 7:51
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    \$\begingroup\$ anhnha, with "input voltage" I mean the applied signal voltage Vi as shown in the schematic. To be more exact, I have corrected the relevant voltage divider expression in my detailed answer. It must be Zo/(Zs+ZF+Zo). Regarding your last question: Of course, we have a current through Zs and ZF and, hence, corresponding voltages. However, the UNWANTED part at the ouput appears only for a finite value of Zo. For this, imagine that the opamp does not amplify. In this case you have a resistive chain Zs-ZF-ZO only, driven by Vi. Now, a voltage Vo does exist for a finite Zo only. \$\endgroup\$
    – LvW
    Jul 20, 2016 at 9:24
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    \$\begingroup\$ There are two sources: (1) Vin causes a (smaller) output voltage without phase inversion Vo1=+Vi*Zo/(Zs+ZF+Zo) (assuming the opamp output Voc=0) and (2) the opamp ouput voltage Voc causes another (larger) portion Vo2=-Voc*(Zs+ZF)/(Zo+ZF+Zs). The output voltage Voc can be found using the classical expression for an inverting opamp. \$\endgroup\$
    – LvW
    Jul 20, 2016 at 19:56
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In this context it basically means that Zf transmits signals in both directions.

Look at the amplifier model you are using - just the amplifier, take out all the surrounding components. If you force Vo from the outside, would this produce something at the amplifier input, i.e. across Zin? No.

Now take your Zf, make the same experiment... And you quickly understand that the same thing does not apply.

A unilateral system lets signals propagate in a unique direction, i.e. from one port to another, while a bilateral system allows the signal to go in both directions.

Symmetric here means that you can swap input and output and nothing changes: if you reverse Zf, nothing changes. Think of Zf as a generic two port system. For such a system you usually have two distinct transfer functions, one from port A to port B, the other from port B to port A. If the two transfer functions are identical the system is symmetrical, i.e. its ports are indistinguishable, and if you swap them nothing changes.

This is probably stressed in your textbook because the fact that Zf is not unilateral influences the closed loop gain transfer function in a non trivial way.

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  • \$\begingroup\$ Quote: "Symmetric here means that you can swap input and output and nothing changes: if you reverse Zf, nothing changes." I must admit that I do not understand this part of the answer (...nothing changes). Please, can you explain? \$\endgroup\$
    – LvW
    Jul 17, 2016 at 16:46
  • \$\begingroup\$ What I mean is that a two port system, such as an impedance, is symmetrical iff its ports are indistinguishable. I will clarifiy this in the answer. \$\endgroup\$ Jul 17, 2016 at 23:50
  • \$\begingroup\$ Thank you very much the answer Vladimir Cravero and LvW for the question. That is also what I was confused. \$\endgroup\$
    – emnha
    Jul 18, 2016 at 3:04
  • \$\begingroup\$ OK - and in the present case, the "two-port system" is simply a passive part Z ? But - is this an answer to a question regarding bilateral feedback? \$\endgroup\$
    – LvW
    Jul 18, 2016 at 7:33
  • \$\begingroup\$ If you feel my answer is incomplete or not clear please do edit it. I am not sure I understand your latest remark, unfortunately. \$\endgroup\$ Jul 18, 2016 at 7:35

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