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This simple 3rd order low-pass circuit

schematic

simulate this circuit – Schematic created using CircuitLab

can be described with the following 3 equations:

$$-V_1 = \displaystyle \frac{-1}{sC_1} \left( \displaystyle \frac{V_{in} - V_1}{R_S} - I_2 \right)$$

$$-I_2 = \displaystyle \frac{-1}{sL_2} \left( \displaystyle V_1 - V_3 \right)$$

$$V_3 = \displaystyle \frac{-1}{sC_3} \left( -I_2 + \displaystyle \frac{V_3}{R_L} \right)$$

It can be realized through an active-RC filter using integrators and so opamps. The resulting signal flow graph should be

schematic

simulate this circuit

Note that \$ I_2 \$ is a voltage value and not a current. The values or numbers alongside of the branches are their transfer functions.

Suppose that it is possible to realize the negative resistances corresponding to the right-hand side branches. The active RC realization is as follows:

schematic

simulate this circuit

Each integrator is realized with an opamp circuit; it is assumed that the input signals of each integrator are currents and the outputs are voltages.

It is explicitly specified in the book that the second stage is a non-inverting integrator (realized through the negative resistances of the right-hand side branches, so that the minus sign in its transfer function is cancelled by the minus sign of these resistances).

So, why is it used the equation

$$-I_2 = \displaystyle \frac{-1}{sL_2} \left( \displaystyle V_1 - V_3 \right)$$

instead of

$$I_2 = \displaystyle \frac{1}{sL_2} \left( \displaystyle V_1 - V_3 \right)$$

?

As stated in the comments, I think there is no correspondance between the equations, the signal flow graph and the realization with opamps:

  • if the second block is a non-inverting integrator, in the signal flow graph its transfer function should be \$ 1 / sL_2 \$ instead of \$ - 1 / sL_2 \$; the output should be \$ I_2 \$ instead of \$ -I_2 \$ and the resistances on the left-hand braces should both be \$ -1 \$ (so, even if the signal flow graph can perform only summation, it is possible to subtract \$ I_2 \$ in the first and in the third stage);

  • the active RC circuit doesn't exactly realize the signal flow graph: its second stage is a non-inverting integrator (while in the signal flow graph it appears as inverting) but it realizes the positive left-hand braces resistances.

Did I make some mistakes or is the circuit built in a wrong manner?

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    \$\begingroup\$ That's because I2 has to be subtracted from Vin and the flow graph only makes use of summing blocks. That's why the transfer function is -1/sL2. Not because both equations are mathematically equivalent. \$\endgroup\$ – Ambiorix Jan 28 '15 at 2:16
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    \$\begingroup\$ The two equations are identical as far as I can see. \$\endgroup\$ – Andy aka Jan 28 '15 at 8:37
  • \$\begingroup\$ @Ambiorix You are right about the summing nodes. But there is no correspondace between the signal flow graph and the realization: that is, the output of the second opamp is actually \$ [ 1/(sL_2) ] (V_1 - V_3) \$ due to the negative resistors. So, it is \$ I_2 \$ and not \$ - I_2 \$. My question is: why? \$\endgroup\$ – BowPark Jan 28 '15 at 10:30
  • \$\begingroup\$ @Andyaka Yes, the equations are identical, but they define two different circuits. The former equation, with both minus sign, states that there is an inverting integrator; the latter states that there is a non-inverting integrator. The former agrees with the signal flow graph but (in my opinion) not with the realization with opamps; the latter agrees with the realization with opamps but not with the signal flow graph. What's wrong? \$\endgroup\$ – BowPark Jan 28 '15 at 10:34

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