User manual

24
Prvotni vzorčni program se še nahaja v pomnilniku od naslova 34h naprej, saj se prepišejo le
spodnji naslovi. Napišite kratek program, ki vsebuje le en skok na ta naslov. Tukaj je najprej
potrebno določiti stran 3. Saj potem absolutni skok s podanim naslovom 4, ki sledi, cilja na
dejanski naslov 34.
Naslov
Ukaz
Podatki
Komentar
00 8 3 Stran 3
01 9 4 Naslov = 34
83 94
Izpis programa 16: Absolutni skok na program časomera
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Prvotni vzorčni program se tako ponovno prikliče. To testirajte tudi za druge primere. Celoten
pregled vseh uporabljenih programov najdete v prilogi.
Pregled ukazov
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Vsi ukazi na enem mestu – to delo s krmilnikom poenostavi. Naslednja tabela vsebuje vse
ukaze v strnjeni obliki.
1
2
3
4
5
6
7
8
9
A
B
C
D
E
Port
=
Wait Jump
-
A= … = A A = … A = … Page Jump C* D* Skip
if …
Call R
et
0 0 1 ms 0 0 0 0 0 0 0
1 1 2 ms 1 1 B = A A = B A =
A+1
1 1 1 1 A>B 1
2 2 5 ms 2 2 C = A A = C A = A-
1
2 2 2 2 A<B 2
3 3 10
ms
3 3 D = A A = D A =
A+B
3 3 3 3 A =
B
3
4 4 20
ms
4 4 Dout
= A
A =
Din
A = A-
B
4 4 4 4 Din.
0 = 1
4
5 5 50
ms
5 5 Dout.
0 =
A.0
A =
Din.0
A =
A*B
5 5 5 5 Din.
1 = 1
5
6 6 100
ms
6 6 Dout.
1 =
A.0
A =
Din.1
A =
A/B
6 6 6 6 Din.
2 = 1
6
7 7 200
ms
7 7 Dout.
2 =
A.0
A =
Din.2
A = A
And B
7 7 7 7 Din.
3 = 1
7
8 8 500
ms
8 8 Dout.
3 =
A.0
A =
Din.3
A = A
Or B
8 8 8 Din.
0 = 0
8
9 9 1 s 9 9 PWM
= A
A =
AD1
A = A
Xor B
9 9 9 Din.
1 = 0
9
A 10 2 s 10 10 A =
AD2
A =
Not A
A A A Din.
2 = 0
A
B 11 5 s 11 11 B B B Din.
3 = 0
B
C 12 10 s 12 12 C C C S1 =
0
C
D 13 20 s 13 13 D D D S2 =
0
D