Examination of SID triangle waveform
Data recorded on the triangle waveform ($10)
The data are recorded using a REU. The recording is done like this:
;* Program 1 *
;Turn off screen and sprites
...
;Set frequency of voice 3
LDA #<freq
LDX #>freq
STA $D40E
STX $D40F
;Reset waveform
LDA #$08 ;Set test bit
STA $D412
;Setup REU to sample register $D41B for $10000 cycles
...
;Start recording
LDA #$10 ;Waveform: Triangle
LDX #%1001000 ;REU-command: Start transfer
STA $D412 ;Start waveform
STX $DF01 ;Start transfer
...
Furthermore, a construct like this is used to get a value from the waveform
as fast as possible (via software) from the SID:
;* Program 2 *
...
LDA #$10
STA $D412 ;Start waveform
LDA $D41B ;Record early value
STA Haps ;Save value
...
The table below summarizes the recordings (all values in hex):
Frequency Haps REU
8000 04 05 06 07 ... FD FE FF FF FE ... 01 00 00 01 ...
4000 02 02 03 03 04 04 ...
2000 01 01 01 01 02 02 02 02 03 03 03 03 04 ...
1000 00 00 00 00 01 01 01 01 01 01 01 01 01 02 ...
FFFF 07 09 0B 0D 0F 11 13 15 17
DEAD 06 08 0A 0C 0D 0F 11 13 14 16 18 1A 1B 1D 1F 21 22 24 26
28 29 2B 2D 2E 30 32 34 35 37 39 3B 3C 3E 40 42 43 45
47 49 4A 4C 4E 50 51 53 55 56 58 5A 5C 5D 5F 61 63 64
Legend:
Frequency is the value stored to $D40E and $D40F.
Haps is the value of the Haps variable from program 2.
REU is the data collected with the REU after running program 1.
Conclusion: The triangle waveform starts with value $00 and goes
to $ff linearly. Then it goes from $ff to $00 linearly.
(As a sidenote, I'll mention that the data for the sawtooth
waveform at frequency $8000 is Haps= 02, REU= 02 03 03 04 04 ... FE FE FF
FF 00 00 01 01 ...)
The Haps variable is the value of the waveform on one cycle earlier than
the REU starts recording.
Further data used to investigate the frequency/waveform mapping:
Frequency
Haps+REU, Number of this*value
0001 7FFC*00 8000*01 8000*02 ...
0002 3FFC*00 4000*01 4000*02 ...
0003 2AA7*00 2AAB*01 2AAA*02 2AAB*03
0004 1FFC*00 2000*01 2000*02 ...
0005 1996*00 199A*01 1999*02 199A*03 1999*04
0006 1552*00 1555*01 1555*02 1556*03 1555*04 1555*05 1556*06
0007 1245*00 1249*01 1249*02 1249*03 1249*04 1249*05 1249*06 124A*07
1249*08
0008 0FFC*00 1000*01 1000*02 ...
0009 0E35*00 0E39*01 0E39*02
000A 0CC9*00 0CCD*01
000B 0B9F*00
Legend:
Frequency is the value stored to $D40E and $D40F.
Haps+REU is the contents of the Haps variable from program 2 and the contents
of the REU after running program 1.
The format is number of times of repeation * value. (i.e. 7FFC*00 means
that the value "00" repeated $7FFC times after each other in the
data. 0003*FE 0004*DE would mean this sequence of data was recorded: FE
FE FE DE DE DE DE. The first FE would come from the variable Haps, while
the rest would be obtained with the REU.)
Conclusion: We know that with the STA $D412; LDA $D41B construct,
A contains the value of the waveform after 4 cycles. This is consistence
with the observation that the number of times the value $00 appears generally
is 4 lower than the number of times the other values appear.
The overall behaviour can thus be simulated with this C-program:
void Triangle(long Frequency) {
long Cycle; /* Count of Phi2 cycles */
long Delay= 0x8000;
long Counter= 0; /* Initial value is 0 */
for (Cycle=0; ;Cycle= Cycle+1) {
/* Build waveform: Triangle */
if (Counter 256)
Waveform[cycle]= Counter;
else
Waveform[cycle]= 511- Counter;
/*
For the sawtooth waveform, use:
Waveform[Cycle]= Counter / 2;
*/
/* Handle frequency-function */
Delay = Delay- Frequency;
/* Frequency overflow? */
while (Delay =0 ) do {
/* For frequencies above $8000, this will be executed twice */
Delay = Delay + 0x8000;
/* Calculate new value for waveform */
Counter= (Counter+1) % 512; /* Keep in range 0..511 */
}
}
}
The above program (when checked) is only capable of reproducing the output
of register $1B. We can't be sure that the true output of pin 27 of the
SID is the same. In this respect, there are two issues:
Is the output digital and 8 bit?
Is the output aligned to the ø2 clock?
Is the output digital and 8 bit?
People have used oscilloscopes and concluded that this is the output is
digital and probably 8 bit. This can also be checked with this program,
if you have decent audio-equipment:
jsr $e544 ;Clear screen
ldx #0
stx $d021 ;Background color black
stx $d020 ;Border black
;(The black color reduces noise with my equipment)
lda #11 ;As of request from Andreas with kernal R2 :-)
col sta $d800,x ;Character color dark grey
inx
bne col
lda #$01 ;Frequency 1
sta $d40e
lda #$00
sta $d40f
lda #$00 ;ADSR:00F0
sta $d412
lda #$f0
sta $d413
lda #$81 ;Waveform: Random, changes rarely.
;Change for other waveforms
sta $d412
lda #$0f ;Max volume: $f
sta $d418
ldx $d41b
loop1 stx temp
loop2 ldx $d41b
cpx temp
bne loop2
lda $0400,x ;Display when value changes
eor #$80
sta $0400,x
jmp loop1
temp .byte $00
Turn up the volume of the TV-set and you can hear the "ticks"
when the waveform changes every few seconds. If the ticks happen in alignment
with the changes on the screen, the output is digital and 8 bit.
You can also try other waveforms (substitute the lda #$81 with lda #$11
for triangle or lda #$21 for sawtooth. Pulse ($41) requires definition of
the pulse width in $d410 and $d411), but these are all much faster and/or
the changes in the waveform are small, so it is not as easy to hear the
distinct ticks.
Is the output aligned to the ø2 clock?
This is an open question as of now (23 Mar 1995 :-). My feeling is that
the output is aligned, but this is only a feeling based on the fact that,
all, the SID has available to synchronize with, is the ø2 feed.
I believe examination of this requires an oscilloscope, which I have no
access to.
What should be done is to display the ø2 clock alongwith the pin
27 output with a triangle waveform at frequency $ffff (Just modify the above
program to use frequency $ffff and triangle waveform). If the waveform change
only once every ø2 cycle, the output is ø2 aligned and thus
"skipping values". If not, the waveform would change twice every
ø2 cycle and thus display all 256 values. If this is the case, I
have further studies to be done to clarify whether the output is happening
in a discrete partition of the ø2 cycle.
Observation
If you record the data of the waveform with a structure like this:
...
lda #$08 ;Reset waveform: Test bit is 1.
sta $d412
lda #$00 ;Frequency $8000
sta $d40e
lda #$80
sta $d40f
...
ldy #$00 ;Clear test bit
lda #$10 ;Waveform triangle
ldx #$90 ;REU command
sty $d412 ;The difference!
sta $d412
stx $df01
the recorded data are all 4 cycles behind! I.e. the first value in the REU
data is not $05, but $09. The similar thing happens if you use the capture
in program 2: Instead of getting $04, you get $08. The 4 cycle delay is
the time, the sta $d412 takes.
If you change the ldy #$00 (test bit=0) to ldy #$08 (test bit=1), the effect
disappears, so the conclusion is:
The triangle counter counts all the time, whether or not the waveform is
activated. The counter is held reset to 0 when the test bit is 1 (supposedly),
and as soon as the test bit is 0, the counter starts counting.
Another explanation, with the same result, is that the counter is reset
on falling edge of the test bit.
I have not examined whether this is the case for the sawtooth and the pulse
waveforms, but my bet is, this is the case.
