Dual Tone Digital Twist for the PIC

Scott Dattalo says

Speaking of mixing tones, during my hiatus I wrote a cute little twist routine. While primarily intended for adjusting the twist in DTMF generated signals, it has applications elsewhere too. It would work here to mix two arbitrary waves. You could cascade calls to this routine for mixing more than two tones. For N tones you'd need N-1 calls. That's not terribly efficient, but at least you'd have a way of combining fixed waveforms with non 2^k weighted coefficients.
The purpose of this routine is to evaluate:
    A*x + B*y
  ------------
     A  +  B
Such that A+B = 2^N
This formula is useful for several applications. For example, DTMF generartors sometimes need to weight the individual sine waves differently. The term used to describe the difference in amplitudes is 'twist'. It's the ratio of the amplitudes in dB.
Another application of this formulation is for low-pass filtering. If 'x' is the current filtered value, and 'y' is the latest sample, then this equation will be a recursive low-pass filter. z-transforms may be used to calculate the frequency response, but that's beyond the scope of this simple description. An intui- tive way to look at it is as weighted averages. If A is made large relative to B, then more emphasis is placed on the average. If B is large then the latest samples are given more weight.
Algorithm:
This algorithm exploits the relationship: A + B = 2^N. For example, consider N=4. Then the (A,B) pairs are:
 (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9), (8,8)
 i.e. B = 2^N - A , and , A = 2^N - B
Upon closer examination, it's observed that the (A,B) pairs are two's complements of one another (modulo 2^N). E.g., the 2's complement of 1 mod 16 is 15. In general:
     B = (~A + 1) % 2^N
Substituting this into the twist formula yields:
   A*x  + ((~A + 1) % 2^N)*y
  ---------------------------
         2^N
A modulo N operation means that only the lower N bits are significant. So dropping the notation and keeping this observation in mind we can simplify this equation:
   A*x + (~A + 1)*y     A*x + ~A*y + y
  ------------------ = ----------------
        2^N                  2^N
Which can be evaluated in psuedo-C:
 int A;     /* initialized somewhere's else */

 int twist ( int x, int y)
 {
   int num = y;

   num += (A & 0x01) ? x : y;
   num >>= 1

   num += (A & 0x02) ? x : y;
   num >>= 1

   num += (A & 0x04) ? x : y;
   num >>= 1

   num += (A & 0x08) ? x : y;
   num >>= 1

  /* etc. for N>4 */

   return num;
 }
Excution time: 1 + 5*N cycles (e.g. 21cycles for A+B=16=2^4)
twist:
        movf    y,w
        movwf   num

        btfsc   A,0
         movf   x,w
        addwf   num,f
        rrf     num,f

        movf    y,w     ;A macro would work well here...
        btfsc   A,1     ;All we're doing is an un-rolled
         movf   x,w     ;multiplication. If a bit in 'A' is
        addwf   num,f   ;set, we add 'y' to the running sum
        rrf     num,f   ;otherwise we add 'x'. In either case
                        ;we divide by 2.

        movf    y,w
        btfsc   A,2
         movf   x,w
        addwf   num,f
        rrf     num,f

        movf    y,w
        btfsc   A,3
         movf   x,w
        addwf   num,f
        rrf     num,f

        return