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RS-2 32 P ro to co l f o r N AD P ro ducts v 1 .0 3 Page 2 o f 3
S in ce t h e c o ntr o l c h ara cte rs a re s im ply s e n t a s s in gle b yte v alu es, t h ere n eed s t o b e a w ay t o d if fe re n tia te
b etw ee n a c o ntr o l c h ara cte r a n d r e g ula r d ata . A s ta n d ard w ay o f d oin g t h is , w hic h t h is p ro to co l u se s, i s t o
u se a fla g c h ara cte r ( w hic h i s a c o ntr o l c h ara cte r) t o e n co de t h e r e g ula r d ata w hic h h av e t h e s a m e v alu e
a s a c o ntr o l c h ara cte r.
B efo re d esc rib in g h o w t h e d ata i s e n co ded , t h e c o ntr o l c h ara cte rs w ill b e l is te d s o t h ey c an b e u se d i n
e x am ple s. T he f o llo w in g t a b le s h o w s a ll t h e c o ntr o l c h ara cte rs a n d t h eir c o rre sp ond in g v alu es:
Contr o l C hara cte r Valu e
< sta rt> 1
< ch eck su m > 2
R ese rv ed f o r F utu re E xp an sio n 0, 3 -“ 1 9
< fla g > 94
C ontr o l c h ara cte rs 0 a n d 4 t h ro ug h 1 9 h av e b een r e se rv ed f o r f u tu re e x p an sio n. F or e x am ple , s ta n d ard
X ON/X OFF f lo w c o ntr o l u se s c h ara cte rs 1 7 a n d 1 9.
A ny d ata v alu es w hic h a re t o b e s e n t t h at c o rre sp ond t o a c o ntr o l c h ara cte r m ust b e e n co ded w it h t h e
f o llo w in g m eth o d:
1 . Sen d t h e f la g c o ntr o l c h ara cte r.
2 . Sen d t h e d ata v alu e b it w is e O Red w ith t h e v alu e 6 4.
A ll o th er d ata v alu es w hic h d o n o t c o rre sp ond t o a c o ntr o l c h ara cte r c an b e s e n t a s i s .
S om e e x am ple s:
Data V alu e t o S en d Enco ded b yte s e q uen ce
3 94, 6 7
5 94, 6 9
2 0 20
9 4 94, 9 4
O n t h e r e ceiv in g s id e, a n yti m e a v alu e o f 9 4 i s r e ceiv ed t h e f o llo w in g c h ara cte r m ust b e d eco ded . T he
c h ara cte r i s d eco ded u sin g t h e f o llo w in g a lg o rith m :
1 . If t h e v alu e r e ceiv ed i s 9 4, t h en t h e a ctu al v alu e i s 9 4.
2 . Els e , t h e r e ceiv ed v alu e s h o uld b e b itw is e A N Ded w it h t h e v alu e 1 91.
F or e x am ple , i f t h e s e q uen ce 9 4, 6 9 i s r e ceiv ed , i t s h o uld b e d eco ded t o t h e v alu e 5 s in ce 6 9 A N D 1 91
e q uals 5 .
C hecksu m C alc u la tio ns
The c h eck su m i s s im ply t h e i n vers e ( o nes c o m ple m en t) o f t h e s u m o f t h e C om man d a n d D ata b yte s o f t h e
p ack et. F or e x am ple , i f t h e c o m man d b yte w as 2 0, a n d t h e d ata b yte w as 2 0, t h en t h e c h ec k su m w ould b e
2 15. ( 2 0 + 2 0 = 4 0, c o m ple m en t o f 4 0 = 2 15). W hen t h e d ata v alu es a re e n co ded b ecau se t h ey o verla p
w it h a c o ntr o l c h ara cte r, t h e a ctu al v alu e s h o uld b e u se d i n t h e s u m a n d n o t t h e e n co ded v alu e.
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RS-2 32 P ro to co l f o r N AD P ro ducts v 1 .0 3 Page 3 o f 3
C om man d S et
T he c o m man d s e t i s d ocu m en te d i n s e v era l t a b le s i n a s e p ara te d ocu m en t. I t d esc rib es w hat a ll t h e
c o m man d s a re a n d w hic h p ro ducts s u p port w hic h c o m man d s.
A dditio ns t o t h e c o m man d s e t w il l b e d one o n a n o ngo in g b asis a s n ew f u nctio ns a re n eed ed f o r n ew
p ro ducts .
E xam ple 1
This e x am ple w ill d esc rib e e x actl y w hat i s s e n t w hen t h e I d en ti f ic atio n o f t h e u nit i s Q uerie d . T his i s d one
b y s e n d in g t h e Q uery F unctio n c o m man d w hic h i s c o m man d 2 0. S in ce i t i s t h e I d en ti f ic atio n t h at i s b ein g
q uerie d , t h e d ata v alu e 2 0 i s s e n t w it h t h e c o m man d .
T he a b ove i s d one b y s e n d in g t h e f o llo w in g b yte s e q uen ce:
< sta rt> Com man d Data Chec k su m
1 20 20 2 215
If t h e i d en ti f ic atio n s tr in g o f t h e u nit w as T 772 t h en t h e r e sp onse t o t h e a b ove c o m man d w ould b e:
< sta rt> Com man d Data Chec k su m
1 20 20,8 4,5 5,5 5,5 0 2 227
E xam ple 2
This e x am ple w ill d esc rib e h o w t o s e t t h e v o lu m e t o a l e v el o f 3 d B . T his i s d one b y s e n d in g t h e S et
F unctio n c o m man d w hic h i s c o m man d 2 1. T he d ata s e n t w ould b e t h e v alu e 2 3 t o s e t t h e v o lu m e, f o llo w ed
b y 3 w hic h i s t h e v alu e f o r t h e v o lu m e. N ote t h at s in ce t h e v alu e 3 i s a r e se rv ed v alu e f o r a c o ntr o l
c h ara cte r, i t m ust b e e n co ded .
T he a b ove i s d one b y s e n d in g t h e f o llo w in g b yte s e q uen ce:
< sta rt> Com man d Data Chec k su m
1 21 23, 9 4 , 6 7 2 208
If t h e v o lu m e i s s u cce ss fu ll y s e t t o + 3 d B , t h en t h e r e sp onse s e n t b ack w ould b e:
< sta rt> Com man d Data Chec k su m
1 20 23, 9 4 , 6 7 2 209
In t h e a b ove e x am ple s, n o te t h at t h e n o n-e n co ded v alu e o f 3 f o r t h e d ata i s u se d f o r t h e c alc u la tio n o f t h e
c h ec k su m .
