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Homework Help: Trouble with a Passive RC Twin-T Notch Filter

  1. Nov 16, 2013 #1


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    Gold Member

    1. The problem statement, all variables and given/known data

    I'm trying to understand how Twin-T notch filters work. I get that they are made of two passive RC filters, one low-pass and one high-pass in parallel. The trouble I'm having is how to determine the corner frequencies (where Vout = 0.707Vin). The gain of the corner frequencies is 0.707... or so I thought... If the low-pass and high-pass filters are both 2nd-order then the corner frequencies would reside at a gain of 0.5 but I can't seem to see how the low-pass and high-pass parts of the circuit are 2nd-order.

    2. Relevant equations

    Here is the circuit:


    Page 7 of "www.niu.edu/~mfortner/labelec/lect/p575_07a.pdf‎" [Broken] gives the gain of the low-pass filter as [itex]V_{out(LP)}=\left ( \frac{1/j\omega C}{R+1/j\omega C} \right )^{2}V_{in}[/itex].

    3. The attempt at a solution

    I redrew the low-pass part of the circuit as:


    which does not give the same output voltage as the 2nd-order low-pass filter in this diagram:


    I get a low-pass filter gain of [itex]\frac{2X_{c}}{R+2X_{c}}[/itex] due to the parallel capacitors being the same as one capacitor with value 2C and the fact that the top-left resistor and the capacitor(s) form a voltage divider network. The output voltage is the voltage across the capacitor(s).

    Why do I get a different gain equation for the "2nd order" low-pass filter part of the Twin-T circuit than the one listed above?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 16, 2013 #2


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    Staff: Mentor

    Hi JJBladester. You are asking why two quite different circuits have different response equations? :wink: We'd be surprised if they did not! They are different circuits, one having a second capacitor that further filters the output.

    A low pass filter has its low-pass gain measured at DC. If you omit all capacitors (since we are talking about DC, right?) then these circuits you are querying both have a gain of x1. The output voltage (when unloaded) is undiminished from the input (the capacitors are equivalent to an infinite resistance). Capacitances do not appear in the equation for low-frequency gain of a low-pass filter.

    Any time you need to consider the "potential divider" effect, then you must take into account the phase angle introduced by the "j" co-efficient in capacitive reactances, e.g., jXc. You can't add real resistance and imaginary reactance except by treating them like vectors (termed phasors).

    By attaching one identical circuit to the output of another you cannot simply shrug and throw in a ½ factor (you seem to have guessed it would be x2) and hope that it's how things work out. Because it isn't! The second circuit imposes a load on the first, so it changes the response of the first. Circuits connected in series always interact unless you interpose a buffer amplifier between each stage to eliminate those loading effects. To find the result of connecting two identical circuits in series you have to analyse the whole circuit, not just half and then "double it" or whatever shortcut you would like. :smile:

    Now, back to the twin-T. It, likewise, is not the response of a 3 element low-pass superimposed over the separate response of the 3 element high-pass. At the output where the junction of the two filters is joined together, that is not how we add voltages! Twisting two wires together is not electronic voltage addition! The outputs are joined, certainly, and this causes them to interact so that the low-pass section changes how it behaves, and the high-pass section also changes. The only way to see the notch in the twin-T is to analyse the full network as a complete entity.

    Looks like you are about to embark on thorough revision of your circuit theory here! http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]

    Good luck!
    Last edited by a moderator: May 6, 2017
  4. Nov 17, 2013 #3


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    Gold Member


    I feel like a stooge after posting that original post in frustration. I sucked it up and did a mesh analysis on the circuit and found the exact solution that explains what I found on the oscilloscope during last week's lab... to within 1% error.

    I've been taking dumbed-down courses for so long and we hardly ever have to do real analysis. This was a fun one to work out. Of course, as soon as you start to load a passive twin-T, loading effects make it garbage (unless you hook the output up to a voltage follower op-amp).

    Here is my work if you are interested.

    You would be proud. I did remember the "j" phase shift term in my capacitive reactances.

    I went to my professor about this problem and she told me to use superposition. I told her this is a single-source network so what is she talking about?... Not sure but it sounds wrong.
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