Getting a single feedback peak from a basic 4 stages phaser

Question

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):

Answer 1

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:

Answer 2

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.