# Getting a single feedback peak from a basic 4 stages phaser

I try to set a very basic 4 stages phaser (two 2-pole all-pass filters), with a particular feedback path where only one peak is raised at center frequency when the feedback is above 0.

The following amplitude responses describe what I try to get: in the first measure (light green) we get two notches (no feedback). In the second measure (purple) we get two notches and a single peak at center frequency (feedback). My chain is very simple: the input is sent into two 2-pole all-pass filters having the same center frequency, the output is then mixed 50/50 with the input. I've also added a feedback path, where the output of the last all-pass filter is feed back to the input of the first one (input + state * feedbackAmount).

Instead of getting a single peak I get two (as described in the following picture). I've also tested negative feedback it does not help. Any idea? Edit1: My signal flow leading to two peaks when the feedback is close to 1: Edit2: My implementation may be wrong indeed, to be sure I've also run a test in Ableton Live 8. Noise straight to the phaser.

First: no feedback: Second: full feedback, we get two peaks too. (Most phaser I've tested provide the same result): If I follow the OP’s description properly, the description is what would be a comb filter, given the all-pass filters operate as delay lines (over a certain frequency range of operation). The sum of a signal with a delayed copy of the signal would be a "comb" response, that would cyclically go between constructively summing and cancelling since the phase through a delay line is linearly proportional to frequency (so cylces between 0° and 180° as you sweep frequency). To eliminate the second peak, the bandwidth of the delay line or the feedforward path from the input that adds with the output would need to be limited.

To achieve what is wanted, consider a simple IIR filter with pole placement at the resonant frequency, given by :

$$H(z) = \frac{k}{(z-\alpha e^{j\omega})(z-\alpha e^{-j\omega})}= \frac{k}{z^2-z2\alpha\cos(\omega)+\alpha^2}$$

Where $$\omega$$ is the normalized radian frequency of the resonance and $$\alpha$$ sets the bandwidth of the resonance, and $$K$$ sets the gain. Use $$\alpha<1$$, with the closer it is to 1 the tighter the resonance.

For example, with $$\alpha = 0.99$$, $$k=0.1$$ and $$\omega = 1$$ the response is: • Thank you. I've added my signal flow in my original question. The center frequency of each 2-pole all-pass is always the same but may be altered. So how to know the frequency of the second peak generated by the feedback, 𝜔 in your formula and also other parameters? Can I use the same Q value as I used in the all-pass filters for 𝛼 and 𝐾 will be the amount of feedback? The comb filtering occurs indeed when I blend both dry and wet signal but I don't know where to put the filter you suggested. Is it an additional all-pass filter? – Luciano Micalizzi May 18 at 12:46
• @LucianoMicalizzi hmmm could you provide more details in your question as to the purpose and goals of this and what constrains you to that block diagram? Also “All-pass filters” do not have Q as to be a true “all-pass” the amplitude would be flat and only the phase would be modified. – Dan Boschen May 18 at 12:57
• I want to create a 4 stages phaser (two 2-pole all-pass filters), with a special behaviour for the feedback: only one peak is raised when the feedback amount is above 0, not two. The attached block diagram produces 2 peaks when the feedback amount is close to 1. Check the "first" spectrum (only one peak), this is what I would like. – Luciano Micalizzi May 18 at 13:26
• So instead of that, why can't you use what I provided, replacing the 2-pole all-pass filters and the feedback? (Please actually update in your question itself since it would be important for anyone know reading your question, as well as what is constraining you if you must use the block diagram as such ---and if not constrained what is this for? That may help get better answers). – Dan Boschen May 18 at 13:39
• I've done many tests and indeed by limiting the bandwidth of the delay line or the feedforward path as you suggested, I can get something close enough. I accept this answer. Thank you. – Luciano Micalizzi May 29 at 16:15

I've simulated your system, and since it's a four-stage phaser you do get two notches and one peak in between, if you don't count the more or less pronounced peaks at DC and at Nyquist.

The frequency response of your system is

$$G(e^{j\omega})=\frac{1+(1-f)H^2(e^{j\omega})}{1-fH^2(e^{j\omega})}\tag{1}$$

where $$f$$ is the feedback value, and $$H(e^{j\omega})$$ is the frequency response of the $$2^{nd}$$-order all-pass filter.

The plot below shows the total frequency response of your system with the feedback factor varying from $$0.1$$ to $$0.9$$. Lower feedback factors have sharper notches and a wider peak, whereas larger feedback values result in broader notches and a sharper peak: So there might be some bug in your implementation.

• Thank you. My implementation may be wrong indeed, so see "edit2" in my original question. The second peak at Nyquist on your graph looks like the second peak I want to avoid. – Luciano Micalizzi May 18 at 17:06
• @LucianoMicalizzi: What's your sampling frequency? – Matt L. May 18 at 17:23
• @LucianoMicalizzi: Also take a look at this answer. With that structure you'll always have peaks at DC and at Nyquist. You can also see this in the two plots in this wikipedia article. – Matt L. May 18 at 17:26