All-Pass Filters, a good explanation

Type : information
References : Posted by Olli Niemitalo
Linked file : filters002.txt
(this linked file is included below)
Linked files
All-Pass Filters

A true allpass is not just:

|    | 
|    |    .       
|    |    |    |    .    

Instead, the first peak is negative and has a carefully chosen height.

     |    |    .

This can be achieved by adding scaled in(T) to the scaled output of: 

out(T) = in(T) + out(T-delay)*gain

Choosing the scalings right... Looking at the critical frequencies, 
those that go A) a number of full cycle per peak, and those that go B) a
number of full cycles and a half-cycle per peak, we can set the 
constraints to eliminate cancellation:

A) The sum of all peaks must be 1.

B) The sum of odd peaks minus the sum of even peaks (including the
negative peak which we give number zero) must be 1.

Let's get into business now that we know what we want. We call the
amplitude of the negative peak "a" and the amplitude of the first postive
peak "b". The ratio between adjacent positive peaks, the feedback gain, is
denoted by "g".

A) The sum of positive peaks is a geometric series and simplifies to
b/(1-g). Our first constraint becomes: 

a + b/(1-g) = 1.

B) Using similar math... The sum of odd peaks is b/(1-g^2). The sum of
even peaks is a + b*g/(1-g^2). So the second constraint is formed:

b/(1-g^2) - (a + b*g/(1-g^2)) = 1.

Solving the two equations we get:

a = -g
b = 1-g^2

Here i had a GREAT phewwww feeling of relief looking at

Choosing g is up to you. You can even make it negative, just remember to
keep |g| < 1. Gosh, perhaps complex g could be used to create a
pulsating response (forget i said that)!

We can write the allpass routine in a programmer-wise pleasant way, 
still preserving ease of mathematical analysis:

mid(T) = in(T) + g*mid(T-delay);
out(T) = (1/g-g)*mid(T) - (1/g)*in(T);

That can be thought of as two filters put in serial and a third one in
parallel with them. The first of the two serial ones is the delay with
feedback. The other "filters" are just different gains. 

For the first code line, the frequency response is the usual set peaks:

1 - g e^(-i w delay)

Adding the mixing on the second code line gives the flat-magnitude 
frequency response of the whole allpass system:

|       1/g - g              |
| -------------------- - 1/g | = 1
| 1 - g e^(-i w delay)       |

The phase response is a wavy one: (formula not double-checked!)
        (g^2-1) sin(delay w)
atan -------------------------
     (g^2+1) cos(delay w) - 2g

I hope this cleared things out - and that there aren't fatal errors! :)

-- Olli Niemitalo <>

Reprise des calculs (T = delay) :

       1/g - g        1
H = --------------- - -
    1 - g*exp(-iwT)   g

    1 - g^2 - (1 - g*exp(-iwT))
H = ----------------------------
       (1 - g*exp(-iwT)) * g

    -g + exp(-iwT)
H = ----------------
    1 - g*exp(-iwT)

                1 - g*exp(iwT)
H = exp(-iwT) * ---------------
                1 - g*exp(-iwT)

Le numerateur et le denominateur sont conjugues. d'ou :

|H| = 1


                             g * sin(wT)
arg (H) = -wT - 2 * arctan ---------------
                           1 - g * cos(wT)

Ce dephasage est le meme que celui trouve par Olli mais a l'avantage de
separer le retard global et le dephasage.

-- Laurent de Soras <>

More generally (in fact, maximally generally) you will get an all-pass
response with any transfer function of the form :

           z^-n * sum (a(i) * z^i)
H(z) = b * ------------------------
           sum (conj (a(i)) * z^-i)

where |b| = 1 and of course n should be large enough that your filter is

-- Frederick Umminger <>