Midterm 2
Section I. IEEE-754
Question 1
Convert 13.5625F into IEEE-754 in hexdecimal.
13.5625F = 1101.1001 x 2^0
1101.1001 = 1.1011001 x 2^3
Sign = + (0)
Exponent = 3 + 127 = 130 = 1000_0010
Fraction = 1.1011001
0 1000_0010 1011001 0000000000000000
0100 001 0101 1001 0000 0000 0000 0000
4 1 5 9 0 0 0 0
0x4159_0000
Question 2
Convert the IEEE-754 value 0xbfd4_0000_0000_0000 into base 10.
0xbfd4_0000_0000_0000
1011 1111 1101 0100 0000 0000 0000 0000 ... 0000
[1] [01111111101] [0100 0000 0000 0000 0000 0000 ... 0000
Sign = 1 (-)
Exponent = 01111111101 = 1021 – 1023 = -2
Fraction = 1.01
-1.01 x 2^-2 = -0.0101 = -(0.25 + 0.0625) = -0.3125
Section II. Encoding Instructions
Question 1
Encode beq x7, x21, -16
rs1 = 0b00111, rs2 = 0b10101, -16 = 1_1111_1111_0000
Bits 12 = [1], 11 = [1], 10:5 = [111111], 4:1 = [1000]
imm[12|10:5] = 1_111111, imm[4:1|11] = 1000_1
1_111111 10101 00111 000 1000_1 1100011
1111 1111 0101 0011 1000 1000 1110 0011
F F 5 3 8 8 E 3
Section II. Digital Logic
Question 1
Draw the circuit diagram depicting the following circuit equation:
\[Q = RS + T^\prime (U \oplus V)\]
Question 2
Write the circuit equation given the follwing truth table:
| A | B | C | Q |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
$Q = A^\prime B C + A B^\prime C + A B C$
**OR**
$Q = (A + B + C)(A + B + C^\prime)(A^\prime + B + C)(A^\prime + B^\prime + C)$
Section III. Piplelining
Question 1
Draw the 5-stage, RISC pipeline. Document each stage, what it does, and the functional unit(s) each stage uses.
IF – ID – EXE – MEM – WB uses IM – RF – ALU/FPU – DM – RF IF – fetches an instruction from current PC.
ID – fetches source registers and/or sign extends immediates.
EXE – Calculates the requested operation, such as ADD/SUB/SLL/XOR/etc.
MEM – loads or stores values from/to data memory.
WB – writes the result into the destination register.
Section IV. Floating-point Assembly
Question 1
Write the assembly that performs the following.
1
2
3
4
double myfunc(char op, double a, int b) {
if (op) return a / b;
else return a * b;
}
.section .text .global myfunc myfunc: # a0 – char op # fa0 – double a # a1 - int b fcvt.d.w fa1, a1 beqz a0, 2f # If we get here, return a / b fdiv.d fa0, fa0, fa1 ret 2: # If we get here, return a * b fmul.d fa0, fa0, fa1 ret