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I have a problem finding the following Vout for this circuit.

Is it possible to just use the following formulas that i have shown. I am just unsure due to the resistors.enter image description here

I assume that what i have done is wrong as i might need to take the other resistances for Req.

In short this question is asking:

1.How to find Vout and Vin

  1. How to find the transfer function for the whole band pass filter
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  • \$\begingroup\$ In your case it is reasonable to break the circuit at R3 since it is much greater than R1,R2. However in general you would write out a system of linear equations from KCL/KVL and solve for Vo in terms of Vi. \$\endgroup\$ – sstobbe May 14 '17 at 5:04
  • \$\begingroup\$ Oh true. I forgot that i could do that as a solution. Thanks \$\endgroup\$ – Student May 14 '17 at 5:17
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This type of passive circuit can be easily solved and expressed in a so-called low-entropy format using the fast analytical circuits techniques or FACTs. The principle is to apply the generalized transfer function formula for a second-order system. It is defined as:

\$H(s)=\frac{H_0+s(H_1\tau_1+H_2\tau_2)+s^2H_1H_{12}\tau_1\tau_{12}}{1+s(\tau_1+\tau_2)+s^2\tau_1\tau_{12}}\$

The \$\tau\$ are the natural time constants of the circuits determined when the excitation (the stimulus, \$V_1\$, is reduced to \$0\;V\$). Here, short the input source, implying that \$C_1\$ left terminal is grounded. Now, "look" at the resistance offered by the terminals of \$C_1\$ and \$C_2\$ in this condition: \$\tau_1=C_1(R_1+R_2)\$ and \$\tau_2=C_2(R_2+R_3)\$. Then, do the same but shorting \$C_1\$ and "looking" at the resistance offered by \$C_2\$. You should find \$\tau_{12}=C_2(R_3+R_2||R_1)\$. We have \$D(s)\$ now:

\$D(s)=1+s(C_1(R_1+R_2)+C_2(R_2+R_3))+s^2C_2(R_3+R_2||R_1)C_1(R_1+R_2)\$

The dc gain (\$s=0\$) is obtained by opening all caps and you have

\$H_0=0\$

The high-frequency gains \$H\$ are found by setting the corresponding energy-storing elements in their high-frequency states. For \$H_1\$ and \$H_2\$, respectively replace \$C_1\$ and \$C_2\$ by short circuits and find: \$H_1=\frac{R_2}{R_1+R_2}\$ while \$H_2=0\$. As \$H_{12}\$ implies that both caps are shorted, \$H_{12}=0\$. We have:

\$N(s)=sH_1\tau_1=s\frac{R_2}{R_2+R_1}C_1(R_1+R_2)=sR_2C_1\$

What you do are simple sketches from which you infer the above values. No equations!

enter image description here

The complete transfer function involving the zero at the origin is then:

\$H(s)=\frac{sR_2C_1}{1+s(C_1(R_1+R_2)+C_2(R_2+R_3))+s^2C_2(R_3+R_2||R_1)C_1(R_1+R_2)}=\frac{\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$

If I now factor the term \$\frac{s}{\omega_z}\$ in the numerator and \$\frac{s}{\omega_0Q}\$ in the denominator then rearrange, you obtain a true low-entropy transfer function defined as:

\$H(s)=H_{00}\frac{1}{1+Q(\frac{s}{\omega_0}+\frac{\omega_0}{s})}\$

in which (this is a raw result and you can rearrange and simplify):

\$Q=\frac{\sqrt{C_2(R_3+R_2||R_1)C_1(R_1+R_2)}}{C_1(R_1+R_2)+C_2(R_2+R_3)}\$

\$\omega_0=\frac{1}{\sqrt{C_2(R_3+R_2||R_1)C_1(R_1+R_2)}}\$

\$H_{00}=\frac{R_2C_1}{C_1(R_1+R_2)+C_2(R_2+R_3)}\$

\$H_{00}\$ is the gain in the flat region.

I have captured these equations in a Mathcad sheet to show how the reference equation (read raw expression) compares with the low-entropy format.

enter image description here

enter image description here

They perfectly match. The difference is that you now have a transfer function letting you calculate the values for all components depending on how you want to tune this filter and what attenuation you want at the peak. What truly matters is the low-entropy well-ordered form which tells you what terms contribute gains (attenuation), poles and zeros. Without this arrangement, there is no way you can design your circuit to meet a certain goal. To my opinion, the FACTs are unbeatable to obtain these results in one clean shot (you would need to rework the raw reference function to obtain the form I gave). If you are designing circuits (passive or active) and need to determine transfer functions, I encourage you to acquire that skill because once you have it, you won't go back to the classical approach. If you start slowly step by step, it is quite simple actually. Complicate expressions when you master 1st-order circuits.

You can discover FACTs further here

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

and also through examples published in the introductory book

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf

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  • \$\begingroup\$ Great answer. But next time try to use a PNG instead of JPG. \$\endgroup\$ – G36 May 14 '17 at 10:04
  • \$\begingroup\$ Thanks, I was just realising for the bode plot does it matter if it starts at say -7db? \$\endgroup\$ – Student May 14 '17 at 10:10
  • \$\begingroup\$ Wow that's what is called OVERKILLING! Good low entropy design and analysis instead calls for spotting those two poles are both very weakly coupled (one cap seen resistance virtually do not change upon open/shorting the other one) and widely frequency separated (\$100\text{ nF}\times 86\text{ k}\Omega =8.6\text{ ms}\$ and \$ 9\text{ pF}\times 2.2\text{M}\Omega=20\;\mu\text{s}\$ are over two decades apart). In short transfer function is just DC gain, zero in the origin and the two poles already worked out. \$\endgroup\$ – carloc May 14 '17 at 10:50
  • \$\begingroup\$ @G36, thanks, no problem but just for my reference, why PNG rather than JPG? \$\endgroup\$ – Verbal Kint May 14 '17 at 11:33
  • \$\begingroup\$ @user6186979, what do you mean "start at -7 dB"? Because of the zero at the origin, the attenuation increases as you compute lower frequencies magnitudes. In a Log scale, since you can't have 0 Hz in the x-axis, as you go down the frequency axis and adopt a low starting frequency, let's say 100 mHz, 10 mHz or 1 mHz, the dB magnitude will dive to higher negative values. Does it answer the question? \$\endgroup\$ – Verbal Kint May 14 '17 at 12:36
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Left hand side is ok,

Right hand side equation is not.

$$R_{eq}= R_2//(R_3+\frac{1}{SC_2})$$

Then

$$V_{IN}=V_{IN1}\frac{R_{eq}}{R_{eq} + R_{1} + \frac{1}{SC_{1}} }$$

Note: $$X//Y = \frac{1}{X^{-1} + Y^{-1}}$$ and $$s = wj$$

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  • \$\begingroup\$ Oh ok thanks. So why is it possible to leave the LHS equation? As in is it since Vin is known at that point and so that is the only voltage going through R3 and C2? (My thought process anyway) \$\endgroup\$ – Student May 14 '17 at 5:15
  • \$\begingroup\$ "since Vin is known at that point and so that is the only voltage going through R3 and C2?" -> Yes: Vin has the value Vin =VIN1*Req/(Req + R1 + (SC1)^-1). So you could think of it as a voltage source with that specified value on that node. \$\endgroup\$ – Tomás Arturo Herrera Castro May 14 '17 at 5:42
  • \$\begingroup\$ Thanks, i was also wondering if you knew how i would find the transfer function for this band pass filter? \$\endgroup\$ – Student May 14 '17 at 5:51
  • \$\begingroup\$ Yes, just replace Vin from my equation into your LHS equation (Vin is just an auxiliary variable after all). I would recommend "s" notation as it is easier to work with and to find the zeroes and poles of the circuit. Note: The TF is Output/Input so leave it like Vout/Vin1. \$\endgroup\$ – Tomás Arturo Herrera Castro May 14 '17 at 6:00

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