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I'm studying the following filter:

enter image description here

So this is a RC second order high pass filter. I am trying to answer the following question:

Only by inspection and simple analysis of the circuit calculate the filter gain at low frequency (f -> 0) and at high frequency (f -> infinity). Write out the value in dB and write the symbolic equations you use.

So for infinite frequency it is pretty easy. The capacitors is so low they behave as short-circuit, therefore we can simplify the parallels C1//R1 and C3//R3 as short-circuits. Therefore, the output resistance R4 is now in parallel with the input source voltage therefore Vo=Vi and the filter gain is 1 (0 dB). I'm not sure what the question means with symbolic equation but I assume it's all this in math format

$$Vi=ZRC1*I1+ZRC3*I3+Vo$$

$$ZRC1=\frac{1}{j 2 \pi f C1 }//R1=0 // R1 = 0$$ $$ZRC3=\frac{1}{j 2 \pi f C3 }//R3=0 // R3 = 0$$ $$Vi=Vo$$

Simple. Now my question is when we go through zero frequencies. So now the impedance of the capacitors is infinitely large and they can be seen as open-circuits. We can simply remove them from the circuit. So now we have all the resistances.

My question is if now there is an easy way to calculate the low-frequency gain. I know it should be very small (close to zero). But I wonder what approximations I can make and what is a simple way to calculate it.

I thought about calculating the input and output resistance but I'm not sure how that would help me, as it doesn't make it simple to relate output and input voltage.

I just need some guidance on what path I should follow, I can then analyse the circuit by myself. Any help is much appreciated.

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  • \$\begingroup\$ In reading the above, I'm not sure you are thinking in the right way about the problem. So let me kind of reset things for you. At DC, you can just remove the capacitors and work out the output voltage. At sufficiently high frequency, you can just replace the capacitors with wires and then solve the circuit, trivially (out=in.) In between, there will be a gradual shift. But you don't need to analyze the "very small (close to zero)" frequency case. What you should care about most is when things are under rapid change from the DC condition to the infinite-AC condition. Yes? \$\endgroup\$ – jonk Aug 21 at 21:15
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The DC case is also quite straightforward. Just analyze the resistor network:

$$\frac{V_o}{V_i}=\frac{R_2||(R_3+R_4)}{R_1+R_2||(R_3+R_4)}\frac{R_4}{R_3+R_4}\approx 0.003974$$

which corresponds to an attenuation of about \$48\$ dB.

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  • \$\begingroup\$ Omg, of course it is! I did not try that approach because in my head I was thinking "oh but this way I'm considering the cascaded circuit as unidirectional. But that is not the case at all, that would be if I calculated V1/V2 and then V2/V3, not taking into account both circuits "giving current" to R2. Thank you after looking at your expression I immediately knew what to do and worked the equations to find it! \$\endgroup\$ – Granger Obliviate Aug 21 at 21:50

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