Sprawozdanie z lab. 206
Treść zadania
Dla automatów podanych niżej wykonać syntezę automatu parametrycznego (stosując metodę opisaną w instrukcji). Przyjąć najprostszy sposób kodowania sygnałów (np. wejścia z0 i z1 mogą być zakodowane na jednej linii).
A1 | y0 | y1 | y0 | y1 | |
---|---|---|---|---|---|
q0 | q1 | q2 | q3 | ||
Z0 | q0 | q2 | q2 | q0 | |
Z1 | q1 | q0 | q3 | q2 |
A2 | y0 | y1 | y1 | |
---|---|---|---|---|
q0 | q1 | q2 | ||
Z3 | q0 | q0 | q0 | |
Z4 | q1 | q2 | q0 |
Grafy automatów oraz ich wyrażenia G+
Automat A1
q0 b0 , q1 b1 , q2 b2 , q3 b3
G1+ = 0(q01(z1q12(z0q23(z1q34(z0q0, z1q2)4, z0q2)3,z1q0)2,z0q0)1)0
G1* = 0(b01(b12(b23(b34(b0,b2)4,b2)3,b0)2,b0)1)0
Automat A2
q0 b0 , q1 b1 , q2 b2
G2+ = 0(q01(z3q12(z3q23(z2q0, z3q0)3,z2q0)2,z2q0)1)0
G2* = 0(b01(b12(b23(b0, b0)3,b0)2,b0)1)0
Grafy nałożone na siebie
$$G^{'*} =_{}^{0}{(b_{0}_{}^{1}\left( b_{1}{_{}^{2}\left( b_{2}_{}^{3}\left( b_{0},b_{0},b_{3}_{}^{4}\left( b_{0},b_{2} \right)^{4},b_{2} \right)^{3},b_{0} \right)}^{2},b_{0} \right)^{1})}^{0}$$
$$G^{'*} =_{}^{0}{(b_{0}_{}^{1}\left( s_{1}b_{1}{_{}^{2}\left( {s_{1}b}_{2}_{}^{3}\left( {s_{1}b}_{0},{s_{2}b}_{0},s_{3}b_{3}_{}^{4}\left( {s_{1}b}_{0},s_{2}b_{2} \right)^{4},s_{4}b_{2} \right)^{3},s_{2}b_{0} \right)}^{2},s_{2}b_{0} \right)^{1})}^{0}$$
Tabela przejść i siatki Karnaugha
Q1 | Q0 | |
---|---|---|
b0 | 0 | 0 |
b1 | 0 | 1 |
b2 | 1 | 0 |
b3 | 1 | 1 |
S1 | S0 | |
---|---|---|
S1 | 0 | 0 |
S2 | 0 | 1 |
S3 | 1 | 0 |
S4 | 1 | 1 |
00 | 01 | 11 | 10 | |
---|---|---|---|---|
00 | 01 | 10 | 00 | 00 |
01 | 00 | 00 | 10 | 00 |
11 | - | - | - | 10 |
10 | - | - | - | 11 |
00 | 01 | 11 | 10 | |
---|---|---|---|---|
00 | 1 | - | - | |
01 | - | - | ||
11 | - | - | - | - |
10 | - | - | - | - |
00 | 01 | 11 | 10 | |
---|---|---|---|---|
00 | - | - | 1 | 1 |
01 | - | - | - | |
11 | - | - | - | |
10 | - | - | - |
$$J_{1} = \overset{\overline{}}{S_{0}}Q_{0}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }K_{1} = \overset{\overline{}}{S_{1}}\overset{\overline{}}{Q_{0}} + \overset{\overline{}}{S_{1}}\overset{\overline{}}{S_{0}}$$
00 | 01 | 11 | 10 | |
---|---|---|---|---|
00 | - | 1 | 1 | - |
01 | - | 1 | 1 | - |
11 | - | - | - | - |
10 | - | - | - | - |
00 | 01 | 11 | 10 | |
---|---|---|---|---|
00 | 1 | - | - | |
01 | - | - | ||
11 | - | - | - | |
10 | - | - | - | 1 |
$$J_{0} = \overset{\overline{}}{S_{0}}\ \overset{\overline{}}{Q_{1}} + S_{1}\overset{\overline{}}{S_{0}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }K_{0} = 1$$
Przyporządkowanie sygnału S
Nr automatu | Sygnał | Stan pocz. | Stan końcowy | Odpowiadający sygnał s |
---|---|---|---|---|
p1 | z0 | b0 (q0) | b0 (q0) | S2 |
p1 | z0 | b1 (q1) | b2 (q2) | S1 |
p1 | z0 | b2 (q2) | b2 (q2) | S4 |
p1 | z0 | b3 (q3) | b0 (q0) | S1 |
p1 | z1 | b0 (q0) | b1 (q1) | S1 |
p1 | z1 | b1 (q1) | b0 (q0) | S2 |
p1 | z1 | b2 (q2) | b3 (q3) | S3 |
p1 | z1 | b3 (q3) | b2 (q2) | S2 |
p2 | z2 | b0 (q0) | b0 (q0) | S2 |
p2 | z2 | b1 (q1) | b0 (q0) | S2 |
p2 | z2 | b2 (q2) | b0 (q0) | S1, S2 |
p2 | z2 | b3 (*) | * | * |
p2 | z3 | b0 (q0) | b1 (q1) | S1 |
p2 | z3 | b1 (q1) | b2 (q2) | S1 |
p2 | z3 | b2 (q2) | b0 (q0) | S1, S2 |
p2 | z3 | b3 (*) | * | * |
„ * ” oznacza dowolną wartość
P | Z | Q1 | Q0 | S1 | S0 |
---|---|---|---|---|---|
0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 1 |
0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 1 |
0 | 1 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | * |
1 | 0 | 1 | 1 | * | * |
1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | * |
1 | 1 | 1 | 1 | * | * |
P | Z | |||
---|---|---|---|---|
p1 | 0 | z0 (z2) | 0 | |
p2 | 1 | z1 (z3) | 1 |
00 | 01 | 11 | 10 | 00 | 01 | 11 | 10 | ||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
00 | 1 | 00 | 1 | 1 | |||||||||||
01 | 1 | 01 | 1 | 1 | |||||||||||
11 | - | 11 | - | - | |||||||||||
10 | - | 10 | 1 | 1 | - | - |
$$\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }S_{1} = \overset{\overline{}}{P}Q_{1}\overset{\overline{}}{Q_{0}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }S_{0} = \overset{\overline{}}{P}ZQ_{0} + P\overset{\overline{}}{Z} + \overset{\overline{}}{\text{Z\ }}\overset{\overline{}}{Q_{0}}$$
Funkcja wyjść
00 | 01 | 11 | 10 | |
---|---|---|---|---|
0 | 1 | 1 | ||
1 | 1 | - | 1 |
Y | |
---|---|
y0 | 0 |
y1 | 1 |
Y = Q0 + PQ1
Zamiana bramek na NAND i NOR
$$J_{1} = \overset{\overline{}}{S_{0}}Q_{0} = \ \overset{}{\overset{\overline{}}{\mathbf{S}_{\mathbf{0}}}\mathbf{Q}_{\mathbf{0}}}\text{\ \ }$$
$$K_{1} = \overset{\overline{}}{S_{1}}\overset{\overline{}}{Q_{0}} + \overset{\overline{}}{S_{1}}\overset{\overline{}}{S_{0}} = \overset{\overline{}}{S_{1} + Q_{0}} + \overset{\overline{}}{S_{1} + S_{0}} = \overset{\overline{}}{\overset{\overline{}}{\overset{\overline{}}{\mathbf{S}_{\mathbf{1}}\mathbf{+}\mathbf{Q}_{\mathbf{0}}}}\mathbf{*}\overset{\overline{}}{\overset{\overline{}}{\mathbf{S}_{\mathbf{1}}\mathbf{+}\mathbf{S}_{\mathbf{0}}}}}$$
$$J_{0} = \overset{\overline{}}{S_{0}}\ \overset{\overline{}}{Q_{1}} + S_{1}\overset{\overline{}}{S_{0}} = \overset{}{\overset{\overline{}}{S_{0} + Q_{1}} + S_{1}\overset{\overline{}}{S_{0}}} = \overset{\overline{}}{\overset{\overline{}}{\overset{\overline{}}{\mathbf{S}_{\mathbf{0}}\mathbf{+}\mathbf{Q}_{\mathbf{1}}}}\mathbf{*}\overset{\overline{}}{\mathbf{S}_{\mathbf{1}}\overset{\overline{}}{\mathbf{S}_{\mathbf{0}}}}}$$
K0 = 1
$$S_{1} = \overset{\overline{}}{P}Q_{1}\overset{\overline{}}{Q_{0}} = \ \overset{}{\overset{\overline{}}{\mathbf{P}}\mathbf{Q}_{\mathbf{1}}\overset{\overline{}}{\mathbf{Q}_{\mathbf{0}}}}$$
$$S_{0} = \overset{\overline{}}{P}ZQ_{0} + P\overset{\overline{}}{Z} + \overset{\overline{}}{\text{Z\ }}\overset{\overline{}}{Q_{0}} = \overset{}{\overset{}{\overset{\overline{}}{P}ZQ_{0}} + \overset{}{P\overset{\overline{}}{Z}}} + \overset{\overline{}}{Z + Q_{0}} = \overset{}{\overset{\overline{}}{\overset{\overline{}}{\overset{\overline{}}{P}ZQ_{0}}*\overset{\overline{}}{P\overset{\overline{}}{Z}}} + \overset{\overline{}}{Z + Q_{0}}} = \overset{\overline{}}{\overset{\overline{}}{\overset{\overline{}}{\overset{\overline{}}{\overset{\overline{}}{\mathbf{P}}\mathbf{Z}\mathbf{Q}_{\mathbf{0}}}\mathbf{*}\overset{\overline{}}{\mathbf{P}\overset{\overline{}}{\mathbf{Z}}}}}\mathbf{+}\overset{\overline{}}{\overset{\overline{}}{\mathbf{Z +}\mathbf{Q}_{\mathbf{0}}}}\ }$$
$$Y = Q_{0} + PQ_{1} = \overset{\overline{}}{\overset{\overline{}}{\mathbf{Q}_{\mathbf{0}}}\mathbf{*}\overset{\overline{}}{\mathbf{P}\mathbf{Q}_{\mathbf{1}}}}$$
Schemat układu
Wnioski
Układ został podłączony i sprawdzony. Działał poprawnie.