Tak jsem si udelal mapu na tu tvoji verzi a ono to zas uplne jednoducha funkce neni. Tady je priklad pro RS 1:
Funkce S1
|T2| 0 | 0 | 1 | 1 ||
|T1| 0 | 1 | 1 | 0 ||T3 |T4 |
-- --- --- --- --- --- ---
| | 0 | 1 | 0 | 0 || 0 | 0 |
| | 0 | 0 | 0 | 0 || 1 | 0 |
| | 0 | 0 | 0 | 0 || 1 | 1 |
| | 0 | 0 | 0 | 0 || 0 | 1 |
S1 = T1 & !T2 & !T3 & !T4 = !(!T1 | T2 | T3 | T4)
Pro dalsi pozice to bude stejny, jen se vzdycky obrati stav postupne pro T2, T3 a T4. Nicmene 1 NOR se 4-ma vstupama to resi.
Funkce R1:
|T2| 0 | 0 | 1 | 1 ||
|T1| 0 | 1 | 1 | 0 ||T3 |T4 |
-- --- --- --- --- --- ---
| | 0 | 0 | 0 | 1 || 0 | 0 |
| | 1 | 0 | 0 | 0 || 1 | 0 |
| | 0 | 0 | 0 | 0 || 1 | 1 |
| | 1 | 0 | 0 | 0 || 0 | 1 |
Takto vyjde "cista" logicka funkce pro pripady kdy je nutne delat RESET. Vytknul jsem !T1 at se to vejde aspon na radek.
R1 = !T1 & [ T2 & !T3 & !T4 + T3 & !T2 & !T4 | T4 & !T2 & !T3 ]
Teoreticky potrebujeme mit RESET v 0 pouze pokud je SET = 1 a nebo vsechny tlacitka v 0 . Takze si muzu dovolit udelat treba toto:
|T2| 0 | 0 | 1 | 1 ||
|T1| 0 | 1 | 1 | 0 ||T3 |T4 |
-- --- --- --- --- --- ---
| |_0_|_0_| 0 |_1_|| 0 | 0 |
| |_1_| 1 | 1 | 1 || 1 | 0 |
| | 0 | 0 | 0 | 1 || 1 | 1 |
| |_1_| 1 | 1 | 1 || 0 | 1 |
R1 = T3 & !T4 | !T3 & T4 | !T1 & T2 = ![ !(T3 & !T4) & !(!T3 & T4) & !(!T1 & T2) ]
|T2| 0 | 0 | 1 | 1 ||
|T1| 0 | 1 | 1 | 0 ||T3 |T4 |
-- --- --- --- --- --- ---
| |_0_|_1_| 0 |_0_|| 0 | 0 |
| |_1_| 1 | 1 | 1 || 1 | 0 |
| | 0 | 1 | 0 | 0 || 1 | 1 |
| |_1_| 1 | 1 | 1 || 0 | 1 |
R2 = T3 & !T4 | !T3 & T4 | T1 & !T2
...analogicky zbytek. Vstupni NANDy pujde recyklovat ve vsech vyrazech, takze staci jen 4.