# How to find the equivalent resistances for each filter so that the cutoff frequencies can be found?

Hello i am having trouble for this following circuit. (Top circuit in diagram1) The problem i am having is finding how to calculate the cutoff frequencies for the highpass filter and lowpass filter. This is since i am not sure on what equivalent resistance to use for each formula for each filter (fc=1/(2piRC)). For the highpass i just did that R9 was in series with R8 and then in parallel with R7, then in series with R6 to find that eq resistance so:

Req_highpass=((R9+R8)||R7)+R6

However i am not sure if this is correct for one but also i am not sure i can just ignore C6

Although for the lowpass filter i am unsure as how to find the equivalent resistance for the cutoff frequency for the lowpass filter? Note: I understand i can just place a buffer between it so interaction does not occur between them but i want to know how to calculate it doing it this way

As you can see when V is probed on the bottom half of circuit diagram 1 when using my equivalent resistance for the high pass filter i get what the cutoff frequency is. Note i have tried different resistor values an i always get it matching (for the example below the circuit values used are below this diagram in the bottom circuit)  • Please, stop flooding EE with the same question over and over! Take your time to read and understand the plethora of answers you already scored. – carloc May 21 '17 at 10:46
• Im sorry but i was told to ask this question again from another post – Student May 21 '17 at 10:55
• I didn't know, anyway don't mind too much, that was just my opinion – carloc May 21 '17 at 10:58

1) Start with the hypothesis that those two poles, the low pass and high pass ones, are widely frequncy separated so they virtually do not interact at all. Note that so far this is only an hypothesis and MUST be confirmed later upon workout completed

2) In detail this means:

• the high cut capacitor C3 is virtually open circuit while (at low frequencies) C2 is doing its job.

• Conversely, C2 is virtually short circuit when C3 comes in (high frequencies).

in short $f_\text{L}\ll f_\text{H}$

3) So it's now easy to work out C2 and C3 seen resistances

$R_\text{C2}=R_2+(\;R_3 \| (R_4+R_\text{in LM386})\;)\approx 75\,\text{k}\Omega$

and

$R_\text{C3}=R_\text{in LM386}\|(\;R_4+(R_2\|R_3)\;)\approx 27\,\text{k}\Omega$

4) and associated pole frequencies

$f_\text{L}=\frac{1}{2\pi R_\text{C2}C2}\approx 21\,\text{Hz}$

and

$f_\text{H}=\frac{1}{2\pi R_\text{C3}C3}\approx 59\,\text{kHz}$

5) It's now time to verify the start off hypothesis $f_\text{L}\ll f_\text{H}$, clear enough it is widely fulfilled.

Note: This applies to first OP post. As he keeps on changing his question what I wrote may as well be not true any longer.

• Thankyou! So the reason that this can be done is due to the factthat frequencies aare so widely separated they dont interact oh ok. Also when you state fL this is the high pass filter frequency or did you just call it fl as it cuts off low frequencies? – Student May 21 '17 at 11:02
• Also sorry but if i were to measure R4 with a multimeter, how would i find this value with calculations? Would it be the RC3? This is since for this particular circuit when i measured it with a multimeter it wasnt 43k for R4? – Student May 21 '17 at 11:03
• Common bandpass systems (as your ones) tipically show an "in band gain" and a low cut frequncy and a high cut one. That's why I named them fL and fH, but you can call them as you like. – carloc May 21 '17 at 11:07
• Or is this measurement fine as it is measuring 43k, in parallel with 24k and 50k in series. I was just unsure if this would affect my circuit as i thought i needed 43k ohms? Like as in when i measured it was around 37kohms – Student May 21 '17 at 11:12
• I don't quite get what you mean about R4. Anyway you cannot measure R4 (nor any other res around) with a multimer, you must read its color code or mark. This because other circuits around will impair multimeter reading. – carloc May 21 '17 at 11:14