Music notes

Miscellaneous notes about computer music

Ode To Joy

This is a handy quicky test:

 int odeToJoy[] = { F2S,QUARTER, F2S,QUARTER, G2 ,QUARTER, A3 ,QUARTER,
                    A3 ,QUARTER, G2 ,QUARTER, F2S,QUARTER, E2 ,QUARTER,
                    D2 ,QUARTER, D2 ,QUARTER, E2 ,QUARTER, F2S,QUARTER,
                    F2S,DOTTED_QUARTER, E2,EIGHTH, E2,HALF,
                    F2S,QUARTER, F2S,QUARTER, G2 ,QUARTER, A3 ,QUARTER,
                    A3 ,QUARTER, G2 ,QUARTER, F2S,QUARTER, E2 ,QUARTER,
                    D2 ,QUARTER, D2 ,QUARTER, E2 ,QUARTER, F2S,QUARTER,
                    E2 ,DOTTED_QUARTER, D2,EIGHTH, D2,HALF, 
                    E2 ,QUARTER, E2 ,QUARTER, F2S,QUARTER, D2 ,QUARTER,
                    E2 ,QUARTER, F2S,EIGHTH,  G2,EIGHTH, F2S,QUARTER, D2 ,QUARTER,
                    E2 ,QUARTER, F2S,EIGHTH,  G2,EIGHTH, F2S,QUARTER, E2 ,QUARTER,
                    D2 ,QUARTER, E2 ,QUARTER, A,QUARTER, REST,ETERNITY } ;

Digital filter

 filter ( input, tunefactor, damping )
 {
   lowpass  += tunefactor * bandpass ;
   highpass  = input - lowwpass  - damping * bandpass ;
   bandpass += tunefactor * highpass ;
   notch     = lowpass + highpass ;  // Optional!
 }

Another Low-pass filter

Leaving aside the sophisticated design methods based on Z transformation with its extensive math, this idea uses another approach based on a recursive equation. You calculate each output-signal sample as the sum of the input signal and the previous output signal with corresponding coefficients. A recursive equation defines a single-pole lowpass filter as: Y[n]=X[n]×a0+Y[n–1]×b1, where X[n] and Y[n] are input and output values of sample [n], Y[n–1] is an output value of the previous sample [n–1], and a0 and b1 are weight coefficients that decrement δ controls. The coefficients have the value of 0<δ<1, a0=1–δ, and b1=δ. Physically, δ is the amount of decay between adjacent output samples when the input signal drops from a high level to a low level. You can directly specify the value of δ or find it from the desired time constant of the filter, d, which is the number of samples it takes the output to rise to 63.2% of the steady-state level for a lowpass filter. A fixed relationship exists between d and δ: δ=e–1/d, where e is the base of natural logarithms. The preceding equations yield Y[n]=Y[n–1]+(1–δ)×(X[n]–Y[n–1]).

Instead of multiplying a decimal-point number, 1–δ, it is more convenient for assembler programming to divide by the reciprocal integer, F=1/(1–δ): Y[n]=Y[n–1]+(X[n]–Y[n–1])/F. Thus, you can determine the digital filter’s parameters using the following steps:

  1. Choose the parameter F. For assembler, it is convenient to perform division as right shifts. For right shifts, the value of F should be 2S, where S is the number of shifts. Let F equal 8, which you reach after three right shifts.
  2. Calculate the decrement: δ=1–1/F=1–1/8=0.875.
  3. Calculate the time constant as d=–1/lnδ=–1/ln0.875=7.49 samples.

The equation Y[n]=Y[n–1]+(X[n]–Y[n–1])/F determines the design of the microcontroller’s algorithm for the filter