csci 136 computer architecture ii – constructing an arithmetic logic unit
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Csci 136 Computer Architecture II – Constructing An Arithmetic Logic Unit. Xiuzhen Cheng [email protected]. Announcement. Homework assignment # 4: Due on Feb 22, before class Readings: Sections 3.2, 3.3, B.5, B.6 Problems: 3.7, 3.9, 3.10, 3.12, 3.18-3.22 Project #1 is due on 11:59PM, Feb 13. - PowerPoint PPT PresentationTRANSCRIPT
Csci 136 Computer Architecture IICsci 136 Computer Architecture II –– Constructing An Arithmetic Logic UnitConstructing An Arithmetic Logic Unit
Xiuzhen [email protected]
Announcement
Homework assignment #4: Due on Feb 22, before class
Readings: Sections 3.2, 3.3, B.5, B.6
Problems: 3.7, 3.9, 3.10, 3.12, 3.18-3.22
Project #1 is due on 11:59PM, Feb 13.
Project #2 is due on 11:59PM, March 10.
What Are We Going To Do?
Implementing the MIPS ISA architecture!Must support the arithmetic/logic operations
Tradeoffs of cost and speed based on frequency of occurrence, hardware budget.
32
32
32
operation
result
a
b
ALU
Negating a two's complement number: invert all
bits and add 1
remember: “negate” and “invert” are quite different!
Converting n bit numbers into numbers with more
than n bits:
MIPS 16 bit immediate gets converted to 32 bits for arithmetic
copy the most significant bit (the sign bit) into the other bits
0010 -> 0000 0010
1010 -> 1111 1010
"sign extension" (lbu vs. lb)
Review on Two's Complement Operations
Just like in grade school (carry/borrow 1s) 0111 0111 0110+ 0110 - 0110 - 0101
Two's complement operations easysubtraction using addition of negative numbers
0111+ 1010
Overflow (result too large or too small for finite computer word):
Example: -8 <= 4-bit binary number <=7
Review on Addition & Subtraction
No overflow when adding a positive and a negative number
No overflow when signs are the same for subtraction
Overflow occurs when the value affects the sign:overflow when adding two positives yields a negative
or, adding two negatives gives a positive
or, subtract a negative from a positive and get a negative
or, subtract a positive from a negative and get a positive
Detecting Overflow
An exception (interrupt) occursControl jumps to predefined address for exception
Interrupted address is saved for possible resumption
Details based on software system / language
Don't always want to detect overflow— new MIPS instructions: addu, addiu,
subu
note: addiu still sign-extends!note: sltu, sltiu for unsigned comparisons
Effects of Overflow
Not easy to decide the “best” way to build somethingDon't want too many inputs to a single gate
Don’t want to have to go through too many gates
For our purposes, ease of comprehension is important
Let's look at a 1-bit ALU for addition:
Building blocks: AND, OR, Inverter, MUX, etc.
How could we build a 1-bit ALU for add, and, and or?
How could we build a 32-bit ALU?
Different Implementations for ALU
cout = a b + a cin + b cin
sum = a xor b xor cinSum
CarryIn
CarryOut
a
b
A 1-bit ALU
AND and ORA logic unit performing logic AND and OR.
Full AdderA 1-bit Full Adder ((3,2) adder).
Implementation of a 1-bit adder
A 1-bit ALU that performs AND, OR, and addition
Figure B.5.6
A 32-bit ALU, Ripple Carry Adder
A 32-bit ALU for AND,OR and ADD operation:connecting 32 1-bit ALUs
What About Subtraction
Invert each bit (by inverter) of b and add 1
How do we implement?
A very clever solution: a + (-b) = a + (b’ +1)
What About NOR Operation?
Explore existing hardware in the ALU
NOR (a,b) = not (a or b) = not(a) and not(b)
Only need to add an inverter for input a
32-Bit ALU for AND, OR, ADD, SUB, NOR
BinvertAinvert
In-Class Question
Prove that you can detect overflow by CarryIn31 xor CarryOut31
that is, an overflow occurs if the CarryIN to the most significant bit is not the same as the CarryOut of the most significant bit
Set on Less Than Operation
Idea: For slt $t0, $s1, $s2
Check $s1-$s2. If negative, then set the LSB of $t0 to 1, set all other bits of $t0 to 0; otherwise, set all bits of $t0 to 0.
How to set these bits? – Less input line
Implementation: Connect Result31 to Less
Overflow detection
Set on Less Than Operation
Question: In figure B.5.11, Set connects
directly to Less0. Can you find out any problem in this implementation?
How to fix it?
Conditional Branch
MIPS Instruction:beq $S1, $s2, label
Idea:Test $s1-$s2. Use an OR gate to test whether the result is 0 or not. It $s1=$s2, set a zero detector.
A Final 32-bit ALU
Operations supported: and, or, nor, add, sub, slt, beq/bnq
ALU Control lines: 2 operation control signal for and, or, add, and slt, 2 control line for sub, nor, and slt
ALU Control Lines Function
0000 And
0001 Or
0010 Add
0110 Sub
0111
1100
Slt
NOR
SPEED of the 32-bit Ripple Carry Adder
CarryOut, Result
The critical pathPath traversed by CARRY: contains 32 and gates, 32 or gates.We must sequentially evaluate all 1-bit adders. Ripple-carry is too low to be used in time-critical hardwareSpeedup: anticipate the carry! needs more hardware for parallel operation
An extreme case – sum of products“Infinite” hardware through two-level logicAn example implementation – too expensive! the number of gates grows exponentially! How many gates are needed to compute
c1: c2: c3: c4:
Carry-Lookahead Adder
The concept of propagate and generatec(i+1) = (ai . bi) + (ai . ci) + (bi . ci) = (ai . bi) + ((ai+bi).ci)
pi = ai + bi; gi = ai . bi
Write down c1, c2, c3, c4.Why fast?
First Level of abstraction: pi, gi Still too expensive! Why? still large number of gates How many gates are needed to compute
c1: c2: c3: c4:
Carry-Lookahead Adder
CarryOut is 1 if some earlier adder generates a carry and all intermediary adders propagate the carry.
Build Bigger Adders
Can’t build a 16-bit adder with carry lookahead!Could use ripple carry of 4-bit adder
Use carry lookahead at higher levels
“Super” Propagate Pi vs. “Super” Generate GiGroup concept, 4-bit adder as a building block
“Super” Propagate/Generate definitionc4 = g3 + (p3.g2) + (p3.p2.g1) + (p3.p2.p1.g0) + (p3.p2.p1.p0.c0) = C1
Super Propagate and Generate
A “super” propagate is true only if all propagates in the same group is trueP0= P1= P2= P3=
A “super” generate is true only if at least one generate in its group is true and all the propagates downstream from that generate are true.G0= G1= G2= G3=
A 16-Bit Adder
Give the equations for C1, C2, C3, C4?
Better: use the CLA principle again!
Second-level of abstraction!
An Example
Determine gi, pi, Gi, Pi, and C1, C2, C3, C4 for the following two 16-bit numbers:
a: 0010 1001 0011 0010b: 1101 0101 1110 1011
Do it yourself
Speed of Ripple Carry vs. Carry Lookahead
Example: Assume each AND and OR gate take the same time. Gate Delay is defined to be the number of gates along the critical path. Compare the gate delays of three 16-bit adder, one using ripple carry, one using first-level carry lookahead, and one using two-level carry lookahead.
Summary
Traditional ALU can be built from a multiplexor plus a few gates that are replicated 32 times
To tailor to MIPS ISA, we expand the traditional ALU with hardware for slt, beq, and overflow detection
Carry lookahead is much faster than ripple carry!CLA principle can be applied multiple times!
Questions?