I have a very space and power constrained application, where I want to apply a twin-T notch filtering to an audio signal and add a little gain of about ~10x.

The following filter section has the desired frequency response (a widish deep notch at ~1 kHz):


simulate this circuit – Schematic created using CircuitLab

The problem is that my signal source has a rather high output impedance of ~20 kOhm, and that my signal sink has a rather low input impedance of 10 kOhm. So, I need buffering.


simulate this circuit

I am now trying to implement this functionality using a single op-amp like so:


simulate this circuit

The impedances Z1 and Z2 could be networks of resistors and capacitors to create an inverse notch filter function, i.e. a pole-like response. My intuition tells me that this should be possible, but I struggle to create it.

Is such a feedback network possible, that will lead to the same response as the dual op-amp version above? Maybe by using some inverting amplifier form or a hybrid of inverting and non-inverting (as long as input impedance permits).

  • \$\begingroup\$ Question: Is it your primary goal to use the high-impedance input of only one single opamp for such a notch? And why? Just to save one opamp? \$\endgroup\$
    – LvW
    Jun 27, 2022 at 13:29
  • \$\begingroup\$ @LvW I plan on using a dual opamp, but need the second op-amp for another purpose. Adding a third op-amp would either mean a much larger quad component or a second component. I'm not set on using the non-inverting opamp configuration, but my experiments with inverting were even worse because of the highly frequency dependent impedance of the filter. \$\endgroup\$
    – tobalt
    Jun 27, 2022 at 13:54
  • \$\begingroup\$ Are the source or load impedances known and stable? \$\endgroup\$ Jul 7, 2022 at 7:49
  • \$\begingroup\$ @TimWilliams yes, stable but frequency dependent. The load is a line-in, the source is a passive magnetic e-guitar pickup \$\endgroup\$
    – tobalt
    Jul 7, 2022 at 7:51
  • \$\begingroup\$ Ah. Oh, let me guess, 50Hz notch? Can you not just shield the thing better? :P \$\endgroup\$ Jul 7, 2022 at 9:46

2 Answers 2


Think of it this way: your final transfer function is of the form:

$$H(s)=\dfrac{s^2+1}{s^2+as+1} \tag{1}\label{1}$$

Then, you are using a non-inverting amplifier, so \$\eqref{1}\$ will have to be the result of:

$$\begin{align} H(s)&=1+G(s) \quad\Rightarrow \\ G(s)&=\dfrac{-as}{s^2+as+1} \tag{2}\label{2} \end{align}$$

Which is an inverting bandpass. If you consider a multiple feedback bandpass, for the same 1 Hz notch frequency you'll need to make the input minus that bandpass.

But you can't just use the MFB input as ground and connect the input signal to the non-inverting input, because that will not achieve \$H(s)\$. And you also can't add the input signal to both the MFB input and the non-inverting input, because then the input would go through the t.f., then the difference would follow using the same t.f..

But if you first subtract the offending extra t.f. from the input, using the same RC network to the non-inverting input, as it is used in the MFB topology, you might make it work:

gettin' crafty

It's costly in terms of passive elements, but it saves you an opamp. Best use low(er) tolerance RC, for best results (getting a notch right the analog way is an art).

  • \$\begingroup\$ The OP mentioned the notch was at 1kHz, not 1Hz. You could probably scale it, but you should note that in the answer. \$\endgroup\$
    – Aaron
    Jun 27, 2022 at 17:12
  • \$\begingroup\$ @Aaron That's true, but I started my answer with "[the] final transfer function is of the form", that's why I used the normalized H(s). I would think OP is able to deduce the necessary scaling, and the tuning of the second RC network to give a high impedance input. After all, all it takes is to add k to the resistors and transform m to n for capacitors (which is the reverse of what I did from the link in the answer). \$\endgroup\$ Jun 27, 2022 at 21:33
  • \$\begingroup\$ Thanks, and sorry for the late reply (other project got in the way). It sounds intriguing, but I am struggling to understand this circuit in terms of an op amp. Could you please tell me, which nodes correspond to the in+, in- and out nodes of the opamp in your schematic? \$\endgroup\$
    – tobalt
    Jul 6, 2022 at 19:12
  • \$\begingroup\$ @tobalt An opamp can be viewed as a VCVS: differential inputs and grounded output. WYSIWYG: + is the noninverting, - the inverting, ground can be ignored, the output is the + node. But, note that this "alternative" is really smoke and mirrors: if you intended to avoid an impedance seen by the signal's output, then this sort of breaks the deal. You may add high values to the second (bottom) network, but then, why not just make a straight bandpass with those high values, without using twice as much components? So, there's a grain of salt in there, somewhere. \$\endgroup\$ Jul 6, 2022 at 19:35
  • 1
    \$\begingroup\$ @tobalt Without an additional active element, no, since the correctional network is passive. LvW's explanation is quite on point. \$\endgroup\$ Jul 7, 2022 at 8:55

"My intuition tells me that this should be possible, but I struggle to create it."

I am afraid that your intuition cannot lead to a solution.

Let us assume that the feedback transfer function is Hr(s). Then, the overall transfer function for an ideal opampwill be

When you require a notch filter with H(s=wo)=0 the feedback function must be: Hr(s=wo)>>>infinite.

In words: The feedback path must contain a bandpass with infinite gain at w=wo

  • \$\begingroup\$ I see. But isn't this the scenario for a perfect notch filter with an exact zero in the transfer function ? Let's say the notch attenuation is finite at 0.03 (-30 dB). In that case the pole impedance doesn't need to be infinite. \$\endgroup\$
    – tobalt
    Jun 27, 2022 at 14:40
  • \$\begingroup\$ Right, but the bandpass gain must be very large (instead of infinite). That means: You need an active bandpass - and a second opamp. However, there are many other two-opamp-notch circuits with better properties.... \$\endgroup\$
    – LvW
    Jun 27, 2022 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.