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Can we take the transfer function of the below 3rd-order high-pass filter as

$$\left (\frac{R}{R+\frac{1}{j\omega C}}\right )^3$$ ?

This circuit is a part of RC oscillator. To calculate the β of the oscillator we have to derive an equation for Vo/Vin. In the tutorial they derived it using KCL and KVL, but in some of the tutorials I watched they derive the transfer function for higher-order filters as above. Why can't we apply it here?

schematic

simulate this circuit – Schematic created using CircuitLab

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4 Answers 4

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Perhaps it helps to see it stacked up this way:

schematic

simulate this circuit – Schematic created using CircuitLab

The transfer function for \$\frac{V_{out}}{V_{b}}=\frac{R_3}{\frac1{s\,C_3}+R_3}\$. So that works. But what's the transfer function for \$\frac{V_{b}}{V_{a}}\$? Isn't it \$\frac{R_2 \,\mid\mid\,\left(\frac1{s\,C_3}+R_3\right)}{\frac1{s\,C_2}+\left[R_2 \,\mid\mid\,\left(\frac1{s\,C_3}+R_3\right)\right]}\$? And what about the one for \$\frac{V_{a}}{V_{in}}\$? That's even more complicated. No?

The final transfer function will be \$\frac{V_{out}}{V_{in}}=\frac{V_{out}}{V_{b}}\cdot\frac{V_{b}}{V_{a}}\cdot\frac{V_{a}}{V_{in}}\$. But that's not going to be the simple cubed expression as you show.

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  • \$\begingroup\$ What you propose is the brute-force approach. Nothing wrong here but you can imagine how easy it is to make mistakes when expanding and rearranging the final expression. Have a look at the solution using FACTs and see how you can get to the final result just by inspecting a series of simple sketches. \$\endgroup\$ Mar 11, 2023 at 15:20
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The fastest path is to use the fast analytical circuits techniques or FACTs. If we consider all resistors and capacitors of equal values (\$R\$ and \$C\$) then the expression is given below without writing a single line of algebra:

enter image description here

If now simulate this circuit with an adequately-tuned amplifier, you have your oscillator delivering a 650-Hz sinusoidal waveform:

enter image description here

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A passive filter assumes a low impedance source and a high impedance load. If this is not true, they must be included in the analysis.

This is not true for the output of the 1st and 2nd RC filter, and the input to the 2nd and 3rd RC filter.

You either need to add buffers between the stages, or use progressively higher impedance stages to approximate low/high impedances.

schematic

simulate this circuit – Schematic created using CircuitLab

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Your oscillator has nothing (AGC) to control the output level then the output will produce clipping distortion. Your filter is highpass which passes the clipping distortion. You should use a lowpass filter to reduce the clipping distortion like this: phase-shift osc

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