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Can someone please explain to me in depth as to how this twin notch circuit is working. Also is what is the formula for finding the notch frequency because some sites give the answer as (1/2×pi×RC) while the others as (1/4×pi×RC). Twin notch filter circuit image

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  • \$\begingroup\$ does my answer work for you? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 22 '19 at 5:45
  • \$\begingroup\$ Notice there are some assumptions about the source resistance (to left) and the load resistance (to the right). \$\endgroup\$ – analogsystemsrf Jan 29 '19 at 5:35
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The correct answer is \$ T=RC/2 \$ only if using: {R,R/2 and C,2C}

Of course, using: {R,2R and C,2C} then it is \$T=RC/4\$ so both are true.

Can someone please explain to me in depth as to how this twin notch circuit is working?

It works by superposition of attenuated LPF and HPF causing exact same reduced amplitude and opposite phase cancelling the combined signal. The output impedance of the HPF also loads the output of LPF and visa-versa and contributes to the response of each.

The notch phase response starts at 0 deg at DC rising to +135' neaR notch then 0 at centre then dropping to -135' ( =-90'-45') with +/-45' shift for each midpoint.

The depth in dB of the notch depends on the external load to be high and the worst case tolerance error of each part. A worst case random part tolerance of 1% might reduce the notch to -40 dB and shift the frequency according to T=RC/2 tolerance stackup.

e.g. 60Hz The depth is limited by the resolution of the step frequency plot with an offset of 0.01 Hz at 60.01 Hz for a 60Hz notch.

enter image description here Using a 50 Ohm generator would work except above the notch attenuation occurs based on R ratios.

enter image description here

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  • \$\begingroup\$ How do you get -40dB out of RC filters? \$\endgroup\$ – Sparky256 Jan 22 '19 at 4:40
  • \$\begingroup\$ In theory with ideal parts the Notch filter is - ∞ from perfect amplitude and phase cancellation \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 22 '19 at 4:42
  • \$\begingroup\$ That would be with no inductors or op-amps? \$\endgroup\$ – Sparky256 Jan 22 '19 at 4:44
  • \$\begingroup\$ That is correct \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 22 '19 at 4:45
  • \$\begingroup\$ Thank you for the explanation..I have got a little hang of it but understood a lot.. \$\endgroup\$ – user183710 Jan 22 '19 at 6:56

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