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

Calculate each output-signal sample as the sum of the input signal and the previous output signal with scaling. This defines a low pass single-pole filter:

  Yn = Xn.(1-T) + Yn-1.T

Where:

• X_n is the input
• Y_n is the output, Y_n–1 is the output from the previous loop.
• T is our frequency control: 0<T<1

Physically, T is the amount of decay between adjacent output samples for a 'step' input. You can directly specify the value of T or derive it from the 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. T=e^–1/d.

Instead of multiplying by 1–T, it is more convenient to divide by the F=1/(1–T):

  Yn = Yn–1 + (Xn–Yn–1)/F

You can determine the digital filter’s parameters using the following steps:

1. Choose a value for F. Suppose F==8, so you can do a right-shift instead of a divide.
2. Calculate T = 1–1/F = 1–1/8 = 0.875.
3. Calculate the time constant as d = –1/ln(T) = –1/ln(0.875) = 7.49 samples.
4. Multiply that into your sample rate: 22000/7.49 = 2.9kHz

Here are some sample values for a 22000Hz sample rate:

Freq. (Hz)	t	F		Freq. (Hz)	t	F		Freq. (Hz)	t	F		Freq. (Hz)	t	F
22000	0.367	1.58		21000	0.384	1.62		20000	0.402	1.67		19000	0.421	1.72
18000	0.441	1.78		17000	0.461	1.85		16000	0.483	1.93		15000	0.505	2.02
14000	0.529	2.12		13000	0.553	2.24		12000	0.579	2.37		11000	0.606	2.54
10000	0.634	2.73		9000	0.664	2.98		8000	0.695	3.28		7000	0.727	3.66
6000	0.761	4.18		5000	0.796	4.91		4000	0.833	6.01		3000	0.872	7.84
2000	0.913	11.5		1000	0.955	22.5		900	0.959	24.9		850	0.962	26.4
800	0.964	27.9		750	0.966	29.8		700	0.968	31.9		650	0.970	34.3
600	0.973	37.1		550	0.975	40.5		500	0.977	44.4		450	0.979	49.4
400	0.981	55.4		375	0.983	59.1		350	0.984	63.3		325	0.985	68.2
300	0.986	73.7		275	0.987	80.4		250	0.988	88.5		225	0.989	98.4
200	0.990	110.0		175	0.992	126.0		149	0.993	147.0		100	0.995	219.0
75	0.996	292.0		50	0.997	440.0

Yet another

Summarized from this paper.

First order filter

(Similar to Y_n = Y_n–1 + (X_n–Y_n–1)/F as described above)

Y = X.B0 + Z[0].B1 - Z[1].A1

Z[1] = Y

Z[0] = X

Where:

• X is the input
• Y is the output

Second order filter

Y = X.B0 + Z[0].B1 + Z[1].B2 + Z[2].A1 + Z[3].A2

Z[3] = Z[2]

Z[2] = Y

Z[1] = Z[0]

Z[0] = X

The first order version is just this with B2 = A2 = 0.

Input parameters:

• Fc = cutoff frequency (Hz)
• Fs = sample rate (Hz)
• W = tan(pi.Fc/Fs)
• Q = quality factor

Filter Parameters	H(S)	N	B0	B1	B2	A1	A2
Lowpass 1st order	1/(S + 1)	1/(W + 1)	W*N	B0	0.0	N*(W-1)	0.0
Highpass 1st order	S/(S + 1)	1/(W + 1)	N	-B0	0.0	N*(W-1)	0.0
Lowpass 2nd order	1/(S2 + S/Q + 1)	1/(W2 + W/Q + 1)	N*W2	2*B0	B0	2*N*(W2 + 1)	N*(W2 + W/Q + 1)
Highpass 2nd order	S2/(S2 + S/Q + 1)	1/(W2 + W/Q + 1)	N	-2*N	B0	2*N*(W2 + 1)	N*(W2 + W/Q + 1)
Bandpass 2nd order	(S/Q)/(S2 + S/Q + 1)	1/(W2 + W/Q + 1)	N*W/Q	0.0	-B0	2*N*(W2 + 1)	N*(W2 + W/Q + 1)

Multiple cascaded filters

The first order lowpass or highpass section has a slope of 6dB/Octave, and the second order section has a slope of 12dB/Octave. If the better slope is required, you have to cascade several sections.

Butterworth

All stages of Butterworth filter have the same cutoff frequency Fc. The parameter which is different for the stages is the filter Q. Here is the table for Q for Butterworth up to 6-th order:

1. ---
2. 0.71
3. --- 1.0
4. 0.54 1.31
5. --- 0.62 1.62
6. 0.52 0.71 1.93

If the value of Q is skipped, it means this is a first order section. Otherwise it is a second order section.

Linkwitz-Riley

LR consists of two identical Butterworth filters in series. Use the same table that is used for Butterworth filter design.

Bessel

Bessel design is a little bit more complicated, since not only Q but the Fc also differs from the stage to stage.

You should multiply the Fc for each stage by the following coefficients:

1. 1.00
2. 1.27
3. 1.32 1.45
4. 1.60 1.43
5. 1.50 1.76 1.56
6. 1.90 1.69 1.60

The corresponding Q values are:

1. ---
2. 0.58
3. --- 0.69
4. 0.81 0.52
5. ---- 0.92 0.56
6. 1.02 0.61 0.51

If the value of Q is skipped, it means this is a first order section. Otherwise it is a second order section.