「論理回路」 2009 年度定期試験解答例

2009

7

「論理回路」 2009 _{年度定期試験 解答例}

担当

:

1

(1)

− 16,

15 (2) 11001100

(3) (x + y + a)(y + z + b)(z + x + c)(x + a z)(y + b x)(z + c y)

= (x + (y + a))(x + a z)(y + (z + b))(y + b x)(z + (x + c))(z + c y)

= (x + (y + a)a z)(y + (z + b)b x)(z + (x + c)c y)

= (x + ayz)(y + bzx)(z + cxy)

= (xy + ayz)(z + cxy) = xyz (4) (a ⊕ b ⊕ c)(a ⊕ b c)

= abc ⊕ ab ⊕ ac ⊕ b c

= (a ⊕ 1)b c ⊕ ab ⊕ ac

= abc ⊕ ab ⊕ ac = ac(b ⊕ 1) ⊕ ab

= abc ⊕ ab = ab(c ⊕ 1) = abc

(5) f

(x, a, b) = f (x, a, b) = xa + xb + ab

= xa · xb · ab = (x + a)(x + b)(a + b)

= (x + a b)(a + b) = xa + xb + a b = f (a, b, c)

f (a, b, c)

(6) x ⊕ y = xy + xy

g(a, b, c) = abc + abc (7)

a b

c f

d

2

(1)

B/01 C/10

A/00

E/11 D/11

0 1

0

1

0 1

0 0 1

1

(2)

符号化された状態遷移表 現状態 次状態 出力

x = 0 x = 1 y z 0 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 1 1 0 0 1 0 1 1 1 1 0 1 0 0 1 0 1 1 0 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 1 1

d

= bx + cx

a

b c

x 1

X X 1 1

1 X X

X X

d

= abx + cx

a

b c

x 1 1 1

X X 1

X X X X

d

= abx + bcx

a

b c

x

1 1 1

X X X X 1 X X

y = a + b

a

b c

X 1 1 X 1 X

z = a + bc

a

b c

1 X

1 X 1 X

1

3

(最終的な状態遷移表だけでよい.)

現状態 次状態/出力

0 1

0 S1 S3/0 S5/1 0 0 S2 S6/0 S5/1 0 0 S3 S8/1 S6/1 0 0 S⁴ S¹/0 S⁶/0 0 0 S⁵ S⁷/1 S³/1 0 0 S⁶ S⁸/1 S³/1 0 0 S7 S2/0 S3/0 0 0 S8 S9/0 S6/0 0 0 S9 S1/0 S4/1 0 0

⇒

現状態 次状態/出力

0 1

0 S1 S3/0 S5/1 1 1 S² S⁶/0 S⁵/1 1 1 S⁹ S¹/0 S⁴/1 0 2 1 S³ S⁸/1 S⁶/1 2 1 S5 S7/1 S3/1 2 1 S6 S8/1 S3/1 2 1 2 S4 S1/0 S6/0 0 1 S7 S2/0 S3/0 0 1 S⁸ S⁹/0 S⁶/0 0 1

⇒

現状態 次状態/出力

0 1

0 S¹ S³/0 S⁵/1 1 1 S² S⁶/0 S⁵/1 1 1 3 S9 S1/0 S4/1 0 2 1 S3 S8/1 S6/1 2 1 S5 S7/1 S3/1 2 1 S6 S8/1 S3/1 2 1 2 S⁴ S¹/0 S⁶/0 0 1 S⁷ S²/0 S³/0 0 1 S8 S9/0 S6/0 3 1

⇒

現状態 次状態/出力

0 1

0 S¹ S³/0 S⁵/1 1 1 S2 S6/0 S5/1 1 1 3 S9 S1/0 S4/1 0 2 1 S3 S8/1 S6/1 4 1 S⁵ S⁷/1 S³/1 2 1 S⁶ S⁸/1 S³/1 4 1 2 S⁴ S¹/0 S⁶/0 0 1 S7 S2/0 S3/0 0 1 4 S8 S9/0 S6/0 3 1

⇒

現状態 次状態/出力

0 1

0 S1 S3/0 S5/1 1 5 S2 S6/0 S5/1 1 5 3 S⁹ S¹/0 S⁴/1 0 2 1 S³ S⁸/1 S⁶/1 4 1 S⁶ S⁸/1 S³/1 4 1 5 S5 S7/1 S3/1 2 1 2 S4 S1/0 S6/0 0 1 S7 S2/0 S3/0 0 1 4 S⁸ S⁹/0 S⁶/0 3 1

よって

現状態 次状態/出力

0 1

S

S

/0 S

/1 S

S

/0 S

/1 S

S

/1 S

/1 S

S

/1 S

/1 S

S

/0 S

/0 S

S

/0 S

/0

4

(1) s = a ⊕ b ⊕ c, c

= ab + bc + ca

(関数が正しければ,

(2)

x

FA s co

a b ci FA

s co

a b ci FA

s co

a b ci FA

s co a b ci

a

b

a

b

a

b

a

b

s

s

s

s

c

c

c

c

x

5

S⁰⁰⁰/0

S¹⁰⁰/0 S⁰⁰¹/0

S⁰¹⁰/0

S¹⁰¹/1

S¹¹⁰/1 S⁰¹¹/1

S¹¹¹/1 0

1 0

1

0

1 0

1

1 0

1 0

1 0

0

1

Nagisa ISHIURA

2