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I have a signal flow diagram like the one below for a biquad NHK. I'm trying to understand how to derive the equations \$\frac{v_1}{v_i}, \frac{v_2}{v_i}, \frac{v_3}{v_i}\$ from it.

I don't quite get the picture on how should I convert the diagram configuration to an equation. For example, what is supposed to be the \$\sum\$ signal in an equation? Is it the sum of the following transfer functions? What exactly is the \$\frac{-1}{sT}\$? How about the \$\frac{1}{Q}\$ - how is it interpreted?

enter image description here

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2 Answers 2

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Symbol Sigma is the sum of what "enter" in the block ...
i.e. \$ V1 = Vi * K + (-1) * V3 + V2 * (1/Q) \$

-1/( s * T) is an inverting integrator ...

Factors are an "amplifier" block. Just define the amplification factor.

The Q factor is a characteristic of the entire function (second-order equation). Defined as a parameter in simulation.

Here is the simulated function. Made with microcap v12, interactive.
Only point out "define" line text and change value with up or down arrows.
Here, "stepping" is used.

enter image description here

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  • \$\begingroup\$ I see! I understand where the expression for \$v_1\$ comes from - the block algebraic. But, as you mentioned, for \$v_2\$ I have the inverting integrator. So would it be written like \$V_1 (\frac{-1}{sT})\$? \$\endgroup\$
    – ludicrous
    Commented Sep 27, 2022 at 14:43
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    \$\begingroup\$ Yes. I will add simulated function in answer. \$\endgroup\$
    – Antonio51
    Commented Sep 27, 2022 at 15:03
  • \$\begingroup\$ You said Q factor is the solution to the second order equation? How can I find it? Would it be the cut frequency over the bandwidth? \$\endgroup\$
    – ludicrous
    Commented Sep 27, 2022 at 15:36
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    \$\begingroup\$ No. Not the solution, sorry. Q is an important factor in the equation. As you can see, it is the parameter that fixes the "maxima" of the function. The function is a bandpass filter and the center frequency is not affected by Q. \$\endgroup\$
    – Antonio51
    Commented Sep 27, 2022 at 17:24
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Hint to get you started: -

$$V_1 = KV_i + \dfrac{V_2}{Q} - V_3$$

And you can probably see that \$V_2 = \dfrac{-V_1}{sT}\$ and, \$V_3 = \dfrac{-V_2}{sT} = \dfrac{V_1}{s^2T^2}\$

You should be able to figure it out from there.

Regarding interpretation of sT and Q, there is nothing in your question that permits a further analysis.

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