计算机系统 I 作业 part1
Question
a)
\[E=\sum m(1,2,4,6)=\prod M(0,3,5,7)\]
\[F=\sum m(0,2,4,7)=\prod M(1,3,5,6)\]
b)
\[\bar E=\sum m(0,3,5,7)\]
\[\bar F=\sum m(1,3,5,6)\]
c)
\[E + F = \sum m(1,2,4,6) + \sum m(0,2,4,7) = \sum m(0,1,2,4,6,7)\]
\[E \cdot F = \sum m(1,2,4,6) \cdot \sum m(0,2,4,7) = \sum m(2,4)\]
d)
\[E=\sum m(1,2,4,6) = \bar X \bar Y Z + \bar X Y \bar Z + X \bar Y \bar Z + X Y \bar Z\]
\[F=\sum m(0,2,4,7) = \bar X \bar Y \bar Z + \bar X Y \bar Z + X \bar Y \bar Z + X Y Z\]
e)
\[E = \bar X \bar Y Z + Y \bar Z + X \bar Z\]
\[F = \bar X \bar Z + \bar Y \bar Z + X Y Z = (\bar X + \bar Y) \bar Z + X Y Z\]
也可以根据逻辑代数公式直接化简
Question
prime implicants : \(XZ, \bar X \bar Z, WX, W \bar Z\)
essential prime implicants : \(XZ, \bar X \bar Z\)
Question
\[\begin{aligned}
AB\bar C + B\bar C\bar D + BC + \bar CD &= AB\bar C + B\bar C\bar D + BC + \bar CD + B\bar CD \\
&= AB\bar C + B\bar C + BC + \bar CD \\
&= AB\bar C + B + \bar CD \\
&= B + \bar CD
\end{aligned}\]
Question
b)
prime implicants : \(XZ, \bar X \bar Z, \bar WX\bar Y, \bar W\bar Y\bar Z, WXY, WY\bar Z\)
essential prime implicants : \(XZ, \bar X \bar Z\)
\[F(W,X,Y,Z) = XZ + \bar X \bar Z + \bar WX\bar Y + WXY\]
Question
\(X = A\)
\[G_0 = G(0, B, C, D) = \bar BC + BD\]
\[G_1 = G(1, B, C, D) = \bar B \bar C + B\bar D\]
\(X = B\)
\[G_{00}=G_0(0, C, D) = C\]
\[G_{01}=G_0(1, C, D) = D\]
\[G_{10}=G_1(0, C, D) = \bar C\]
\[G_{11}=G_1(1, C, D) = \bar D\]
Question
\[F(A, B, C, D) = AB + B\bar C\bar D + \bar A\bar B D + ACD\]
\[F(0, 0, C, D) = D\]
\[F(0, 1, C, D) = \bar C\bar D\]
\[F(1, 0, C, D) = CD\]
\[F(1, 1, C, D) = 1\]
Question
Question
可以转化成判断 \(A+(2^4-1-B)+1\) 是否有进位，我们可以只需要 Adder 中的 Cout 来完成
\(A<B \to O = 0\)
\(A\geq B \to O = 1\)
Question
0x0C000000 = 0000 1100 0000 0000 0000 0000 0000 0000
two's complement interger = \(2^{26}\times 3\) = 201326592
unsigned interger = \(2^{26}\times 3\) = 201326592
不会 MIPS
0x0C000000 = 0000 1100 0000 0000 0000 0000 0000 0000
IEEE 754 single : 1 sign + 8 exponent + 23 fractions
\[\begin{matrix}\underbrace{0}\\\text{sign }\end{matrix}\begin{matrix}\underbrace{00011000}\\\text{exponent }\end{matrix} \begin{matrix}\underbrace{00000000000000000000000}\\\text{fractions}\end{matrix}\]
\((-1)^0 \times (1 + \text{0}) \times 2 ^ {\overline{00011000}\text{ - 127}} = 1 \times 2^{-103}\)
single : 1 sign + 8 exponent + 23 fractions, bias = 127
\[63.25 = (111111.01)_2 = 1.1111101 \times 2^{101}\]
\(5 + 127 = (10000100)_2\)
\[\begin{matrix}\underbrace{0}\\\text{sign }\end{matrix}\begin{matrix}\underbrace{10000100}\\\text{exponent }\end{matrix} \begin{matrix}\underbrace{11111010000000000000000}\\\text{fractions}\end{matrix}\]
double : 1 sign + 11 exponent + 52 fractions, bias = 1023
\(5 + 1023 = (10000000100)_2\)
\[\begin{matrix}\underbrace{0}\\\text{sign }\end{matrix}\begin{matrix}\underbrace{10000000100}\\\text{exponent }\end{matrix} \begin{matrix}\underbrace{1111101000000000000000000000000000000000000000000000}\\\text{fractions}\end{matrix}\]
half precision : 1 sign + 5 exponent + 10 fractions, bias = 15
\(-1.5625\times 10^{-1} = -0.15625 = (-1) \times (0.00101) = (-1) \times (1.01) \times (2^{-11})\)
\(-3 + 15 = (01100)_2\)
\[\begin{matrix}\underbrace{1}\\\text{sign }\end{matrix}\begin{matrix}\underbrace{01100}\\\text{exponent }\end{matrix} \begin{matrix}\underbrace{0100000000}\\\text{fractions}\end{matrix}\]
32-bit has larger range and presion than 16-bit.
Question
| D(t) |
X |
Y |
D(t+1) |
S |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
1 |
1 |
| 0 |
1 |
0 |
1 |
1 |
| 0 |
1 |
1 |
0 |
0 |
| 1 |
0 |
0 |
1 |
1 |
| 1 |
0 |
1 |
0 |
0 |
| 1 |
1 |
0 |
0 |
0 |
| 1 |
1 |
1 |
1 |
1 |
graph LR
0[0] --> |00/0, 11/0| 0[0]
0[0] --> |01/1, 10/1| 1[1]
1[1] --> |00/1, 11/1| 1[1]
1[1] --> |01/0, 10/0| 0[0]
Question
Question
a)
b) 待补充
Question
待补充
Question
Question
a) \(t_a=2t_{XOR-pd}=0.08\text{ns}\)
b) \(t_b=t_{XOR-pd}+t_{Inv-pd}+T_s=0.07\text{ns}\)
c) \(t_c=t_{FlipFlop-pd}+2t_{XOR-pd}=0.16\text{ns}\)
d) \(t_d=t_{FlipFlop-pd}+t_{XOR-pd}+t_{Inv-pd}+T_s=0.15\text{ns}\)
e) \(\text{Clock frequency} = \frac{1}{t_d} = 6667 \text{MHz}\)