APOLLO181 DAC Sound Generator: the software

How to play sound with APOLLO181 (Part 2): the software

After having built a basic 4-bit Digital to Audio converter, in this section we compute algorithms that generate audible notes in the range of 350-3500 Hz, starting from sampled waveforms, digitized at 8000 samples per second and using 4 bits per sample, that is equal to generate a constant bit rate signal of 32 Kbit/s.

The first, simplest waveform is a ~1KHz pure unsigned sine wave, which I have plotted in MS Excel with -45 degrees phase shift, and digitized with 8 samples (that means 8kHz of sample rate), using 16 rounded integer levels (that means 4-bit accuracy). The following table returns the eight level values which I have calculated for each sample:

Sample number	Degrees	D/A	D/A binary
1	0	13	1101
2	45	15	1111
3	90	13	1101
4	135	8	1000
5	180	2	0010
6	225	0	0000
7	270	2	0010
8	315	8	1000

Graphs plotted in MS Excel

A sine wave, with -45° phase shift, digitized at 8x sample rate, with 4-bit level accuracy

The first algorithm in APOLLO181 is very simple: just output to "port 0" the eight 4-bit binary samples, one by one, as an infinite cycle. In order to get a ~1kHz frequency, I temporarily used the trick to slow down the clock rate on the board, without computing any delay routine. The audio tone was nice to my ears and sufficiently loud.

1) Load accumulator with binary 13

2) Output accumulator to Port 0

3) Goto 4)

4) Load accumulator with binary 15

5) Output accumulator to Port 0

6) Goto 7)

7) Load accumulator with binary 13

...

... etc etc, for binary 8, 2, 0, 2

...

22) Load accumulator with binary 8

23) Output accumulator to Port 0

24) Goto 1)

Captured by a FTT spectral analysis system

A 4-bit ~1000 Hz sine wave, sampled at 8x rate, as it is reconstructed by Apollo181 DAC

How to invoke subprograms without Call/Return statements

As usual with APOLLO181, it is a bit of a challenge to write effective software in only 256 available memory locations . The major limitation is the no provision at all of calling subroutine instructions. Since the delay between samples must be identical, it would be a waste of memory writing the same delay routine many times, across the whole program.

Since in the next exercise we are going to produce a cycle of 32 different samples to reconstruct a complex periodic waveform, by using 256 Bytes of RAM we would need to write 32 nearly identical subprograms, each occupying maximum eight locations of memory (256/32=8), that generate the sample and compute the delay between two adjacent samples. This is not only impractical, but also impossible to perform.

The solution (source: Choosing and using 4 bit microcontrollers, by Philip McDowell, 1993, pg. 116) is making use of a table of jumps, where the return statement would be. The 'calling' section of code loads a progressive number into a dedicated register prior to jump to the 'subroutine', and at the end of the routine, this number is used as an index to a table of jumps. This means that every use of the routine must have an entry in the table, and this restricts the number of 'calls' to a maximum of sixteen for a 4-bit processor; in our case the maximum number of 'calls' is only eight, because APOLLO181's jump statement requires two instructions to be performed: the first to load the accumulator with the low-order nibble of the RAM location, and the second for the jump instruction itself, which contains the high-order nibble.

Thus, in our exercise, we will need to repeat the pseudo-call routines four times inside the main algorithm (32/8=4), that is wasteful, but anyway more efficient than writing thirty-two identical delay routines.

Here below you may find the structure of the algorithm for calling a subprogram many times, , by using a table of jumps without the use of Call/Return statements:

0) Return_index = 0

1) Load accumulator with sample n.1

2) Output accumulator to Port 0

3) Goto Delay_subroutine

4) Load accumulator with sample n.2

5) Output accumulator to Port 0

6) Goto Delay_subroutine

7) Load accumulator with sample n.3

8) …

… …

Delay_subroutine: CalculateDelay

Increment Return_index

Goto Return_index

Return_index 1): Goto 4)

Return_index 2): Goto 7)

Return_index 3): Goto ...

… …

Sound recontruction by additive synthesis

Joseph Fourier, mathematician and physicist of the 18th century, realized that any signal, no matter what looks the shape of the waveform, can be expressed as the sum of sinusoidal waves at various amplitudes and frequencies.

Thus, theoretically, it is possible to synthesize and reproduce any complex sound from the combination of pure sinusoidal tones. This is the concept behind additive synthesis, which involves constructing a wide range of sounds by adding up multiple sine waves at different frequencies and amplitude.

Unfortunately, a person’s voice or natural sounds are so complicated that require a nearly infinite amount of data processing to be recreated accurately. In practice, of course, the sum of sinusoidal waves is truncated to a finite number of terms, in order to considerably reduce the computational load.

C Major chord made up of three notes sounding simultaneously, transponed from C4 to C6 (at ~1kHz)

In this exercise we will compute an algorithm to play a major triad, that is a chord composed of three notes alone, played simultaneously.

In particular, we are going to synthesize a C Major chord, that is one of the most common chord, used in many songs. This chord is basic because it comprises only three notes: C, E, and G. These, physically, have frequencies in the ratio of 4:5:6 and when they are played together, the three notes blend very well and are pleasant to the human ear.

To simplify all calculations, from now on we will adopt the scientific scale, based on 'Middle C' having a frequency of 256 Hz. This is also known as the diatonic scale, that is an eight-note musical scale composed of seven pitches and a repeated octave.

The frequencies of the three notes in the C Major chord (starting at 'Middle C') are: C4= 256 Hz, E4= 320 Hz and G4= 384 Hz.

The ratio of E to C is 5/4 (or 1,25). The ratio of G to C is 6/4= 3/2 (or 1,5) and the ratio of G to E is 6/5 (or 1,2). Since the ratio between every note can be expressed as the ratio of two small whole numbers, and the frequencies of the notes are equally distant from each other, they all sound pleasant together; hence the entire major triad is consonant.

The resulting waveform of C+E+G, sum of sinusoids, is like:

C + E + G = sin(x) + sin(5/4 x) + sin(3/2 x)

where:

C4 = sin (2πf t)

E4 = sin (5/4 * 2πf t)

G4 = sin (3/2 * 2πf t)

and f = 256 Hz.

Since the period of C is 2π, the period of E is 4/3π and the period of G is 8/5π, the resulting period of the sum C+E+G is 8π (that is the lowest common multiple of 2π, 4/3π and 8/5π). This means that the period of the C Major chord is 4 times (8π / 2π = 4) the period of the lowest note in the chord (i.e. four times the period of C4 note).

Since frequency is equal to the reciprocal of the period, the fundamental frequency for the C Major chord is one fourth the frequency of the lowest note in the chord, that's one fourth of C4: thus, in the C Major chord the fundamental frequency is missing from the tone and there is no spectral energy at that frequency.

The fundamental missing frequency for the C Major chord results to be 256/4= 64 Hz: of course if we need to digitize such a chord with 8kHz sample rate, we would need 8000/64= 125 different samples to reconstruct it.

Since we have only 256 bytes of RAM available for the program, we have to transpose the C Major chord two octaves up: by transposing sound two octaves up (that means four times the frequency) we need only one fourth of the number of samples while maintaining constant the sampling rate.

The frequencies of the three notes in the transposed C Major chord (starting now at 'Top C') are: C6= 1024 Hz, E6= 1280 Hz and G6= 1536 Hz, and the resulting chord has a missing fundamental frequency of 1024/4= 256 Hz.

Since in the C Major chord, the fundamental frequency has no spectral energy, the chord can be anyway reproduced by APOLLO181, even if the small speaker cannot reproduce sounds lower than 350 Hz.

In order to digitize such a chord using a sampling rate of 8000 Hz, we need now to generate only 8000/256 ~ 32 different samples useful to reconstruct the whole tone.

Graphs plotted in MS Excel

C Major Chord is synthetized by adding together bit-by-bit the samples of the 3 notes

Due to the lack of hardware resources, all the cumbersome calculations and the digital additive synthesis were made outside APOLLO181, under MS Excel. The algorithm is limited only to read the samples already stored in the program memory and issue them cyclically to the DAC, with the proper delay.

In particular I plotted for several periods each of the three sinusoids in MS Excel with +30 degrees phase shift, and I digitized, with an Excel formula, each period of the slowest wave (~1KHz sine wave) into eight samples (that means 8kHz of sample rate), by using sixteen rounded integer levels (that means 4-bit accuracy). I digitized the other two waves with the same sample rate, so, at the end, all the samples of the three waves last the same time and result synchronous. I then added together bit-by-bit the level of the corresponding samples, till covering a full period of the waveform. In fact, we have seen that the resulting wave is still periodic, with period equal to four periods of the slowest wave. For each sample I then divided the result of the sum by 3 and then I rounded it to an integer, in order to normalize the level to a 4-bit precision (that means that the peak amplitude of the level of any sample, so of the entire resulting waveform, cannot exceed binary 15).

The following table shows all the 32 samples (=8 samples*4 periods) which I have calculated:

Sample number

Degrees

D/A

D/A binary

Sample number

Degrees

D/A

D/A binary

1

0

11

1011

17

720

9

1001

2

45

15

1111

18

765

10

1010

3

90

12

1100

19

810

9

1001

4

135

5

0101

20

855

7

0111

5

180

2

0010

21

900

6

0110

6

225

4

0100

22

945

7

0111

7

270

8

1000

23

990

7

0111

8

315

11

1011

24

1035

6

0110

9

360

10

1010

25

1080

5

0101

10

405

8

1000

26

1125

7

0111

11

450

7

0111

27

1170

11

1011

12

495

8

1000

28

1215

12

1100

13

540

8

1000

29

1260

9

1001

14

585

6

0110

30

1305

3

0011

15

630

5

0101

31

1350

0

0000

16

675

6

0110

32

1395

4

0100

Graphs plotted in MS Excel

The resulting C6+E6+G6 Chord is still a periodic wave, that can be digitized with 32 samples

Captured by a FTT spectral analysis system

C6+E6+G6 Chord, sampled at 8x rate, 4-bit accuracy, as it is reconstructed by Apollo181 DAC

Here you can listen to APOLLO181 C Major chord outcome and below the page you may check the full program.

PLAY APOLLO181 reconstructed chord

Sound reconstruction: a pencil-paper method

The last exercise is easy and practical. The waveform of a single tone of a real violin was recorded and printed on a paper. After that, with the use of a special grid 32x16, it was sampled with a red pencil. The 32 samples, having each 4-bit accuracy, were loaded into APOLLO181 RAM and played by using the same algorithm as was used before.

Here you can listen to the true tone of the original violin first and, after a brief pause, to APOLLO181 playing the same but reconstructed tone.

PLAY VIOLIN tone comparison

A part the natural vibrato timbre, characteristic envelope modulation of the real violin’s waveform, the two tones sound very similar to the human ear.

A real violin was recorded and digitized by hand using 32 samples with 4-bit precision

Captured by a FTT spectral analysis system

The violin waveform, sampled at 8x rate, as it is reconstructed by Apollo181 DAC

Address Description APOLLO181 Code

00		49	SET ALU to 9#
01		1E	ACC = E#
02		20	REG [0] = ACC
03		21	REG [1] = ACC
04		22	REG [2] = ACC
05		23	REG [3] = ACC
06	Output sample 1	1B	ACC = B#
07		F0	OUT PORT [0] = ACC
08		16	ACC = 6#
09		08	GOTO 86#
0A	Output sample 2	1F	ACC = F#
0B		F0	OUT PORT [0] = ACC
0C		16	ACC = 6#
0D		08	GOTO 86#
0E	Output sample 3	1C	ACC = C#
0F		F0	OUT PORT [0] = ACC
10		16	ACC = 6#
11		08	GOTO 86#
12	Output sample 4	15	ACC = 5#
13		F0	OUT PORT [0] = ACC
14		16	ACC = 6#
15		08	GOTO 86#
16	Output sample 5	12	ACC = 2#
17		F0	OUT PORT [0] = ACC
18		16	ACC = 6#
19		08	GOTO 86#
1A	Output sample 6	14	ACC = 4#
1B		F0	OUT PORT [0] = ACC
1C		16	ACC = 6#
1D		08	GOTO 86#
1E	Output sample 7	18	ACC = 8#
1F		F0	OUT PORT [0] = ACC
20		16	ACC = 6#
21		08	GOTO 86#
22	Output sample 8	1B	ACC = B#
23		F0	OUT PORT [0] = ACC
24		16	ACC = 6#
25		08	GOTO 86#
26	Output sample 9	1A	ACC = A#
27		F0	OUT PORT [0] = ACC
28		14	ACC = 4#
29		09	GOTO 94#
2A	Output sample 10	18	ACC = 8#
2B		F0	OUT PORT [0] = ACC
2C		14	ACC = 4#
2D		09	GOTO 94#
2E	Output sample 11	17	ACC = 7#
2F		F0	OUT PORT [0] = ACC
30		14	ACC = 4#
31		09	GOTO 94#
32	Output sample 12	18	ACC = 8#
33		F0	OUT PORT [0] = ACC
34		14	ACC = 4#
35		09	GOTO 94#
36	Output sample 13	18	ACC = 8#
37		F0	OUT PORT [0] = ACC
38		14	ACC = 4#
39		09	GOTO 94#
3A	Output sample 14	16	ACC = 6#
3B		F0	OUT PORT [0] = ACC
3C		14	ACC = 4#
3D		09	GOTO 94#
3E	Output sample 15	15	ACC = 5#
3F		F0	OUT PORT [0] = ACC
40		14	ACC = 4#
41		09	GOTO 94#
42	Output sample 16	16	ACC = 6#
43		F0	OUT PORT [0] = ACC
44		14	ACC = 4#
45		09	GOTO 94#
46	Output sample 17	19	ACC = 9#
47		F0	OUT PORT [0] = ACC
48		12	ACC = 2#
49		0A	GOTO A2#
4A	Output sample 18	1A	ACC = A#
4B		F0	OUT PORT [0] = ACC
4C		12	ACC = 2#
4D		0A	GOTO A2#
4E	Output sample 19	19	ACC = 9#
4F		F0	OUT PORT [0] = ACC
50		12	ACC = 2#
51		0A	GOTO A2#
52	Output sample 20	17	ACC = 7#
53		F0	OUT PORT [0] = ACC
54		12	ACC = 2#
55		0A	GOTO A2#
56	Output sample 21	16	ACC = 6#
57		F0	OUT PORT [0] = ACC
58		12	ACC = 2#
59		0A	GOTO A2#
5A	Output sample 22	17	ACC = 7#
5B		F0	OUT PORT [0] = ACC
5C		12	ACC = 2#
5D		0A	GOTO A2#
5E	Output sample 23	17	ACC = 7#
5F		F0	OUT PORT [0] = ACC
60		12	ACC = 2#
61		0A	GOTO A2#
62	Output sample 24	16	ACC = 6#
63		F0	OUT PORT [0] = ACC
64		12	ACC = 2#
65		0A	GOTO A2#
66	Output sample 25	15	ACC = 5#
67		F0	OUT PORT [0] = ACC
68		10	ACC = 0#
69		0B	GOTO B0#
6A	Output sample 26	17	ACC = 7#
6B		F0	OUT PORT [0] = ACC
6C		10	ACC = 0#
6D		0B	GOTO B0#
6E	Output sample 27	1B	ACC = B#
6F		F0	OUT PORT [0] = ACC
70		10	ACC = 0#
71		0B	GOTO B0#
72	Output sample 28	1C	ACC = C#
73		F0	OUT PORT [0] = ACC
74		10	ACC = 0#
75		0B	GOTO B0#
76	Output sample 29	19	ACC = 9#
77		F0	OUT PORT [0] = ACC
78		10	ACC = 0#
79		0B	GOTO B0#
7A	Output sample 30	13	ACC = 3#
7B		F0	OUT PORT [0] = ACC
7C		10	ACC = 0#
7D		0B	GOTO B0#
7E	Output sample 31	10	ACC = 0#
7F		F0	OUT PORT [0] = ACC
80		10	ACC = 0#
81		0B	GOTO B0#
82	Output sample 32	14	ACC = 4#
83		F0	OUT PORT [0] = ACC
84		10	ACC = 0#
85		0B	GOTO B0#
86	1st Delay routine	90	SET NO CARRY
87		30	ACC = REG [0]
88		52	ACC = Arithm f (ACC, 2#, CARRY)
89		20	REG [0] = ACC
8A		90	SET NO CARRY
8B		E7	ACC = IN PORT [7]
8C		2F	REG [F] = ACC
8D		3F	ACC = REG [F]
8E		51	ACC = Arithm f (ACC, 1#, CARRY)
8F		2F	REG [F] = ACC
90		30	ACC = REG [0]
91		DC	IF CARRY GOTO Cx#
92		1D	ACC = D#
93		08	GOTO 8D#
94	2nd Delay routine	90	SET NO CARRY
95		31	ACC = REG [1]
96		52	ACC = Arithm f (ACC, 2#, CARRY)
97		21	REG [1] = ACC
98		90	SET NO CARRY
99		E7	ACC = IN PORT [7]
9A		2F	REG [F] = ACC
9B		3F	ACC = REG [F]
9C		51	ACC = Arithm f (ACC, 1#, CARRY)
9D		2F	REG [F] = ACC
9E		31	ACC = REG [1]
9F		DD	IF CARRY GOTO Dx#
A0		1B	ACC = B#
A1		09	GOTO 9B#
A2	3rd Delay routine	90	SET NO CARRY
A3		32	ACC = REG [2]
A4		52	ACC = Arithm f (ACC, 2#, CARRY)
A5		22	REG [2] = ACC
A6		90	SET NO CARRY
A7		E7	ACC = IN PORT [7]
A8		2F	REG [F] = ACC
A9		3F	ACC = REG [F]
AA		51	ACC = Arithm f (ACC, 1#, CARRY)
AB		2F	REG [F] = ACC
AC		32	ACC = REG [2]
AD		DE	IF CARRY GOTO Ex#
AE		19	ACC = 9#
AF		0A	GOTO A9#
B0	4th Delay routine	90	SET NO CARRY
B1		33	ACC = REG [3]
B2		52	ACC = Arithm f (ACC, 2#, CARRY)
B3		23	REG [3] = ACC
B4		90	SET NO CARRY
B5		E7	ACC = IN PORT [7]
B6		2F	REG [F] = ACC
B7		3F	ACC = REG [F]
B8		51	ACC = Arithm f (ACC, 1#, CARRY)
B9		2F	REG [F] = ACC
BA		33	ACC = REG [3]
BB		DF	IF CARRY GOTO Fx#
BC		17	ACC = 7#
BD		0B	GOTO B7#
BE	NOP	10	ACC = 0#
BF	NOP	10	ACC = 0#
C0	1st Jump Table	1A	ACC = A#
C1		00	GOTO 0A#
C2		1E	ACC = E#
C3		00	GOTO 0E#
C4		12	ACC = 2#
C5		01	GOTO 12#
C6		16	ACC = 6#
C7		01	GOTO 16#
C8		1A	ACC = A#
C9		01	GOTO 1A#
CA		1E	ACC = E#
CB		01	GOTO 1E#
CC		12	ACC = 2#
CD		02	GOTO 22#
CE		16	ACC = 6#
CF		02	GOTO 26#
D0	2nd Jump Table	1A	ACC = A#
D1		02	GOTO 2A#
D2		1E	ACC = E#
D3		02	GOTO 2E#
D4		12	ACC = 2#
D5		03	GOTO 32#
D6		16	ACC = 6#
D7		03	GOTO 36#
D8		1A	ACC = A#
D9		03	GOTO 3A#
DA		1E	ACC = E#
DB		03	GOTO 3E#
DC		12	ACC = 2#
DD		04	GOTO 42#
DE		16	ACC = 6#
DF		04	GOTO 46#
E0	3rd Jump Table	1A	ACC = A#
E1		04	GOTO 4A#
E2		1E	ACC = E#
E3		04	GOTO 4E#
E4		12	ACC = 2#
E5		05	GOTO 52#
E6		16	ACC = 6#
E7		05	GOTO 56#
E8		1A	ACC = A#
E9		05	GOTO 5A#
EA		1E	ACC = E#
EB		05	GOTO 5E#
EC		12	ACC = 2#
ED		06	GOTO 62#
EE		16	ACC = 6#
EF		06	GOTO 66#
F0	4th Jump Table	1A	ACC = A#
F1		06	GOTO 6A#
F2		1E	ACC = E#
F3		06	GOTO 6E#
F4		12	ACC = 2#
F5		07	GOTO 72#
F6		16	ACC = 6#
F7		07	GOTO 76#
F8		1A	ACC = A#
F9		07	GOTO 7A#
FA		1E	ACC = E#
FB		07	GOTO 7E#
FC		12	ACC = 2#
FD		08	GOTO 82#
FE		16	ACC = 6#
FF		00	GOTO 06#

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