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SO i came across an exercise question in a book

Design a two-stage “bandpass” RC filter, in which the first stage is highpass with a breakpoint of 100 Hz, and the second stage is lowpass with a breakpoint of 10 kHz. Assume the input signal source has an impedance of 100Ω. What is the worst- case output impedance of your filter, and therefore what is the minimum recommended load impedance?

enter image description here I understand that the correct way to approach this is to break up the high pass filter and low pass filter separately and solve them by ensuring

  1. the load impedance of each circuit is 10x higher than the output impedance
  2. The cut-off frequency adheres to the equation 1/RC = 2pifreq.

With that the answer to worst-case output impedance of filter is the parallel addition of R1 = 1000 ohms and R2 = 100000 ohms, which is 909 ohms approximately. This means the load impedance should be at least 909 ohms. But i have an unusual approach to combine everything and resolve it. But i am not sure why this method is wrong. Can anyone advise me on this?

My approach:

  1. Combine a basic high pass and low pass filter enter image description here

  2. Account for input signal source of 100 ohms. Treat it as an ideal voltage source (voltage source has no impedance) with a 100 ohms resistor since question indicates that. enter image description here

  3. Calculate overall output impedance of the entire band-pass filter circuit by replacing voltage source with a short and rearranging

enter image description here

By taking the worst possible output impedance based on this analysis, we will use a low frequency signal and worst possible output impedance is R1+R2? Things look quite off and i am not sure where i went wrong

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    \$\begingroup\$ You cannot separately treat each stage. This is because the input impedance of the second stage loads the first one while the output impedance of the first drives the second one. You must determine the entire transfer function of this second-order network and apply the low-\$Q\$ approximation to highlight two poles in the denominator and one zero at the origin. Have a look at my APEC 2016 and you should be able to determine your transfer function with a few drawings (no equation). \$\endgroup\$
    – Verbal Kint
    Apr 19 at 5:39
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    \$\begingroup\$ Your approach in tail-ending these filter stages is OK, where R2>>R1 so that the low-pass filter doesn't load down the preceding high-pass filter. The same "rule" can be used for RL, so that RL>>R2. The calculation of \$RL_{min}=909\$ violates this "rule". Why do you think your combining method of these filter stages is unusual? It is commonly done. \$\endgroup\$
    – glen_geek
    Apr 19 at 12:17
  • \$\begingroup\$ @glen_geek Ahhhhh so the approach of combining is the correct one with R1+R2 being the load impedance? I think i was just confused as to why i am left with two different answers albeit i wasn't sure which is correct. \$\endgroup\$
    – Iberico
    Apr 19 at 14:30
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    \$\begingroup\$ @Iberico For example, I would find for the HPF: R=707 Ohm and C=2.25 uF, which presents 1k Zin load to the source and has a Zout of 500 Ohm. And for the LPF: R=3.5k Ohm and C=4.5 nF, which presents 5k Zin to the HPF and has a Zout of 2.5k Ohm. Like that? \$\endgroup\$
    – periblepsis
    Apr 19 at 22:31
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    \$\begingroup\$ @Iberico In the future, you don't need to be pasting CircuitLab schematics as images. Use the built-in schematic editor. It's on the "menu bar" on top of both the question and answer editor on this site. No need to use the CircuitLab website for that: it's all built in, and you get all features "for free". \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Apr 20 at 3:58

1 Answer 1

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The incorrect thinkning i had was

parallel addition of R1 = 1000 ohms and R2 = 100000 ohms, which is 909 ohms approximately.

this was the part that is incorrect because as glen_geek mentioned, the load impedance should not be less than the output impedance of the entire system. Hence, by treating the high pass and low pass filter separately, i can approximate the worst case output impedance of the combined band-pass filter to be R2 if i ensure that R2 is at least 10 times larger than R1.

Similarly, if i combined the entire circuit and analyse it at once, i will get that the worst case output impedance is R1+R2. This can be approximated to R2 and the tail-ending method will be consistent with the combined method.

Note that this is an approximation and not an exact solution which involves mathematical modelling of the circuit transfer function to solve

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    \$\begingroup\$ Since the corner frequency of an RC is always where the impedance of the capacitor equals the resistance of the resistor, or where the phase shift is halfway between its two extremes (at 45 degrees), this should drive your thinking process. Also, be aware that the impedance is frequency-dependent because of the capacitors. So when you are thinking about 10X you need to ask yourself, "At what frequency?" For example, the high-pass might be 10X at the corner frequency but if so then it will be less than 10X at higher frequencies. Just a note to consider. \$\endgroup\$
    – periblepsis
    Apr 20 at 20:09

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