# cutoff frequency of low pass filter with two resistors (differential signal)

How do I calculate the cutoff frequency for this design? simulate this circuit – Schematic created using CircuitLab Or simply, • Hello rumman, welcome to EE.SX! Yes the resistances just add. The formulas look good, except the designators are a component off (R2+R3). $f$ will be the -3dB point, where V_out is about half of VIN. – rdtsc Feb 24 '16 at 23:19

Whenever you have a symmetrical circuit you can cut it in half to make the analysis easier. The equivalent half-circuit will show the same behavior. Please note that I am assuming that R2=R3.

In this case the symmetry might not be apparent, but by replacing the capacitor C1 by two series connected capacitors, each one having twice the value, the circuit becomes symmetrical. And the half circuit is a low-pass with R2 and 2*C that determine the corner frequency. Hence, $$f=\frac 1{2\pi\cdot R_2 \cdot 2C}$$ which is equivalent to your first equation, for R2=R3.

Why not simply applying the basic rules of circuit analysis?

The current I through the circuit is I=Vsin/(R2+R3+1/jwC1) and the voltage across C1 is

Vc=I*1/jwC1=Vsin/[jwC1(R2+R3+1/jwC1)].

Therefore:

Vc/Vsin=1/[1+jwC1(R2+R3)]

This is the classical first-order lowpass function of the form 1/(1+jw/wc) with the 3dB angular (cut-off) frequency

wc=1/C1(R2+R3).