![6
11
Refinement of the von Neumann model
Communication between components handled by the
system bus
The System Bus Model
12
Fetch Phase:
o RAM[ PC ] ==> IR
Execute Phase:
o decode IR
o take appropriate action
o update PC
The Fetch-Execute Cycle](https://image.slidesharecdn.com/chapter1-191007203301/85/Chapter-1-6-320.jpg)
![7
13
Assume each instruction is 4 bytes long
Assume PC: 00010700
Fetch phase:
o access RAM[ 00010700 ]
o copy 4 bytes (E0827003) to IR
IR now contains: E0827003
Example
14
Assume IR: E0827003
Execute phase:
o decode IR
ADD instruction on ARM
o take appropriate action
R[2] + R[3] ==> R[7]
o update PC
PC + 4 ==> PC
Example (continued)](https://image.slidesharecdn.com/chapter1-191007203301/85/Chapter-1-7-320.jpg)
![8
15
RAM
Address
Data
00000000
00000001
00000002
.
.
.
.
00010700 E0
. 82
. 70
. 03
.
.
FFFFFFFD
FFFFFFFE
FFFFFFFF
CPU
PC: 00010700
E0827003IR:
16
CPU
PC: 00010700
IR: E0827003
ADD
R[2]
R[3]
R[7]](https://image.slidesharecdn.com/chapter1-191007203301/85/Chapter-1-8-320.jpg)
Chapter 1
- 1. 1 1 CSE 320 Computer Organization and Architecture 2 Course organization Computing environment Overview of course topics Today’s Lecture
- 2. 2 3 Course website http://www.cse.msu.edu/~cse320/ Syllabus and calendar Enrollment Course Organization 4 Hardware: array of ARM processors o pi1.cse.msu.edu . . o pi22.cse.msu.edu Operating system: Linux Accounts Computing Environment
- 3. 3 5 Due Thursday, 9/5 (by 11:59 PM) Focuses on computing environment: o UNIX tutorial o Using the "vim" editor o Using the "handin" system Computer Project #1 6 The Information Revolution Computers have led to a third revolution for civilization, with the information revolution taking its place alongside the agricultural and industrial revolutions. This race to innovate has led to unprecedented progress since the inception of electronic computing in the late 1940s. Had the transportation industry kept pace with the computer industry, for example, today we could travel from New York to London in a second for a penny. (Patterson and Hennessy)
- 4. 4 7 The Information Revolution Driven by rapid innovation in technology Complex applications now feasible • World Wide Web • Cell phones • Computers in automobiles • Human genome project 8 Computer architecture focuses on the functional behavior of a computing system as viewed by the programmer (such as the size of an integer data object in bytes). Computer organization focuses on the structural relationships which are not visible to the programmer (such as the clock frequency or the total size of RAM). Architecture and Organization
- 5. 5 9 We can view a computing system at several levels, from the highest level (users running programs) to the lowest level (transistors operating according to the laws of physics). Levels of Abstraction 10 The von Neumann model consists of five major components: 1) input unit 2) output unit 3) arithmetic logic unit 4) memory unit 5) control unit The von Neumann Model
- 6. 6 11 Refinement of the von Neumann model Communication between components handled by the system bus The System Bus Model 12 Fetch Phase: o RAM[ PC ] ==> IR Execute Phase: o decode IR o take appropriate action o update PC The Fetch-Execute Cycle
- 7. 7 13 Assume each instruction is 4 bytes long Assume PC: 00010700 Fetch phase: o access RAM[ 00010700 ] o copy 4 bytes (E0827003) to IR IR now contains: E0827003 Example 14 Assume IR: E0827003 Execute phase: o decode IR ADD instruction on ARM o take appropriate action R[2] + R[3] ==> R[7] o update PC PC + 4 ==> PC Example (continued)
- 8. 8 15 RAM Address Data 00000000 00000001 00000002 . . . . 00010700 E0 . 82 . 70 . 03 . . FFFFFFFD FFFFFFFE FFFFFFFF CPU PC: 00010700 E0827003IR: 16 CPU PC: 00010700 IR: E0827003 ADD R[2] R[3] R[7]
- 9. 9 17 Must be able to convert between bases: Machines use base 2 (binary) Humans use base 10 (decimal) Humans abbreviate base 2 using base 16 (hexadecimal) or base 8 (octal) Number Systems 18 Decimal Hexadecimal Octal Binary 0 0 0 0000 1 1 1 0001 2 2 2 0010 3 3 3 0011 4 4 4 0100 5 5 5 0101 6 6 6 0110 7 7 7 0111 8 8 10 1000 9 9 11 1001 10 A 12 1010 11 B 13 1011 12 C 14 1100 13 D 15 1101 14 E 16 1110 15 F 17 1111
- 10. 10 19 Powers of Two • 20 = 1 • 21 = 2 • 22 = 4 • 23 = 8 • 24 = 16 • 25 = 32 • 26 = 64 • 27 = 128 • 28 = 256 • 29 = 512 • 210 = 1024 • 211 = 2048 • 212 = 4096 • 213 = 8192 • 214 = 16384 • 215 = 32768 20 Base 16 often used instead of base 2 Example: 0100101101111100 = 4b7c Groups of four bits (from right): 0100 1011 0111 1100 = 4b7c Shorthand for Binary
- 11. 11 21 Base 8 sometimes used instead of base 2 Example: 010010110111110 = 22676 Groups of three bits (from right): 010 010 110 111 110 = 22676 Shorthand for Binary 22 Example: ASCII characters A 1000001 100 0001 41 B 1000010 100 0010 42 C 1000011 100 0011 43 . . . X 1011000 101 1000 58 Y 1011001 101 1001 59 Z 1011010 101 1010 5a
- 12. 12 23 Example: UNIX file permissions Permissions for each file: rwx rwx rwx (owner, group, world) Make directory public: chmod 755 my_directory Make file private: chmod 600 my_file 24 Example: 2756 base 8 ==> base 10 2756 base 8 = 2 * 83 + 7 * 82 + 5 * 81 + 6 * 80 = 1518 base 10 Nested form: 2756 base 8 = (((((((2) * 8) + 7) * 8) + 5) * 8) + 6) = 1518 base 10 Convert Other Base to Decimal
- 13. 13 25 Algorithm: answer = 0 iterate over digits in original number answer = answer * base + current digit Convert Other Base to Decimal 26 2756 base 8 ==> base 10 answer = 0 = (0 * 8) + 2 = 2 = (2 * 8) + 7 = 23 = (23 * 8) + 5 = 189 = (189 * 8) + 6 = 1518 2756 base 8 = 1518 base 10 Convert Other Base to Decimal
- 14. 14 27 Example: 44 base 10 ==> base 2 44 / 2 = 22 R 0 22 / 2 = 11 R 0 11 / 2 = 5 R 1 5 / 2 = 2 R 1 2 / 2 = 1 R 0 1 / 2 = 0 R 1 44 base 10 = 101100 base 2 Convert Decimal to Other Base 28 Example: 44 base 10 ==> base 8 44 / 8 = 5 R 4 5 / 8 = 0 R 5 44 base 10 = 54 base 8 Convert Decimal to Other Base
- 15. 15 29 Algorithm: value = original number loop until value == 0 current digit = value % base value = value / base Convert Decimal to Other Base 30 Base 2 ==> Base 16: group digits ex: 1100011 base 2 ==> 63 base 16 Base 16 ==> Base 2: decompose digits ex: 5C base 16 ==> 1011100 base 2 Other Base ==> Base 10: multiply and add Base 10 ==> Other Base: repeated division Summary: base conversions
- 16. 1 1 Today: Combinational Circuits (H&H 2.1-2.9) Next: continued Handouts Syllabus (old) Lecture Topics 2 Self-study module #1 (this week) Consulting hours posted Project #1 (due no later than 9/5) Reminder: check account password Reminder: use Pi array Announcements
- 17. 2 3 Due Thursday, 9/5 (by 11:59 PM) Focuses on computing environment: o UNIX tutorial o Using the "vim" editor o Using the "handin" system Computer Project #1 4 Circuit design based on Boolean algebra Three equivalent representations o algebraic expressions o truth tables o circuit diagram Combinational Circuits
- 18. 3 5 Expression in Boolean algebra: F(A,B) = A'B + AB' Truth table: Example: Exclusive OR A B F(A,B) 0 0 0 0 1 1 1 0 1 1 1 0 6 Circuit diagram: A B F(A,B)
- 19. 4 7 Boolean algebra 8 All possible functions on two inputs
- 20. 5 9 Basic Logic Gates 10 Other Logic Gates
- 21. 6 11 Any circuit can be defined using only: { NOT, AND, OR } Other complete gate sets: { NAND } { NOR } Complete Gate Sets 12 Useful to define gates which have more than 2 inputs. AND: output is 1 if all inputs are 1 OR: output is 1 if any input is 1 Not meaningful for NOT More Than Two Inputs
- 22. 7 13 Can be implemented using cascading: Can also be implemented directly (more efficient) More Than Two Inputs 14 Canonical Sum-of-Products Form: F(A,B) = A'B + AB' The expression is the sum of a series of products, where each product is a minterm A minterm is a product where each variable is present (complemented or uncomplemented) Standard Forms
- 23. 8 15 For a function with two inputs, there are four possible minterms: m0: A'B' m1: A'B m2: AB' m3: AB Canonical SOP form has a subset of all possible minterms Minterms 16 For a function with three inputs, there are eight possible minterms: m0: A'B'C' m4: AB'C' m1: A'B'C m5: AB'C m2: A'BC' m6: ABC' m3: A'BC m7: ABC For a function with four inputs, there are sixteen possible minterms Minterms
- 24. 9 17 The following are in canonical SOP form: F(A,B) = A'B' + A'B + AB' G(A,B,C) = A'BC + AB'C + ABC' + ABC H(A,B,C,D) = A'B'C'D' + A'B'CD + ABCD' Examples 18 The following are equivalent: F(A,B) = A'B' + A'B + AB' F(A,B) = m0 + m1 + m2 F(A,B) = minterms( 0, 1, 2 ) Alternate notation (minterm lists)
- 25. 10 19 G(A,B,C) = A'BC + AB'C + ABC' + ABC = minterms( 3, 5, 6, 7 ) H(A,B,C,D) = A'B'C'D' + A'B'CD + ABCD' = minterms( 0, 3, 14 ) Additional Examples: 20 Canonical Sum-of-Products Form: F(A,B) = A'B + AB' = minterms( 1, 2 ) Truth table: Truth tables and minterms A B F(A,B) 0 0 0 m0 0 1 1 m1 1 0 1 m2 1 1 0 m3
- 26. 11 21 Canonical Sum-of-Products form makes it easy to convert between representations: Given: G(A,B,C) = minterms( 3, 5, 6, 7 ) G(A,B,C) = A'BC + AB'C + ABC' + ABC truth table has 1's in rows m3, m5, m6, m7 circuit diagram has four AND gates (one for each minterm) and one OR gate 22 Ideally, a Boolean expression will be as simple as possible and still generate the correct values Note that there are an infinite number of Boolean expressions that represent the same function: F(A,B) = A'B + AB' = A'B + AB' + AB' = A'B + AB' + AB' + AB' Minimization
- 27. 12 23 The minimized (optimal, simplified) Boolean expression is the one which has: the fewest number of gates the fewest number of inputs to gates Reminder: we’re working with the complete gate set { NOT, AND, OR } Minimization Criteria 24 Three techniques: Algebraic manipulation Karnaugh map Quine-McCluskey algorithm Minimization Techniques
- 28. 13 25 Apply the postulates and theorems of Boolean algebra: AB' + AB = A(B' + B) (distributive law) = A(1) (complement law) = A (identity law) Algebraic Manipulation 26 F(A,B,C) = A'BC' + A'BC + ABC' + ABC = A'B(C' + C) + ABC' + ABC = A'B(1) + ABC' + ABC = A'B + ABC' + ABC = A'B + AB(C' + C) = A'B + AB(1) = A'B + AB = (A' + A)B = (1)B = B
- 29. 14 27 Fill in the entries in a K-map, then inspect it to identify the optimal expression Karnaugh map is rectangular and has one entry for each minterm – adjacent minterms can be combined (same rules as algebraic manipulation) Karnaugh Map 28 K-map for function with 2 inputs A' A B' m0 m2 B m1 m3 A 0 1 B 0 m0 m2 1 m1 m3
- 30. 15 29 Minimized function: F(A,B) = A Example: F(A,B) = AB' + AB A' A B' 0 1 B 0 1 A 0 1 B 0 0 1 1 0 1 30 Minimized function: F(A,B) = AB' + A'B Example: F(A,B) = AB' + A'B A' A B' 0 1 B 1 0 A 0 1 B 0 0 1 1 1 0
- 31. 16 31 K-map for function with 3 inputs A'B' A'B AB AB' C' m0 m2 m6 m4 C m1 m3 m7 m5 32 K-map for function with 3 inputs AB 00 01 11 10 C 0 m0 m2 m6 m4 1 m1 m3 m7 m5
- 32. 17 33 F(A,B,C) = minterms( 2, 3, 6, 7 ) = A'BC' + A'BC + ABC' + ABC = A'B(C' + C) + ABC' + ABC = A'B + ABC' + ABC = A'B + AB(C' + C) = A'B + AB = (A' + A)B = B Ex: Algebraic Manipulation 34 F(A,B,C) = B Ex: Karnaugh Map A'B' A'B AB AB' C' 0 1 1 0 C 0 1 1 0
- 33. 18 35 F(A,B,C) = B Ex: Karnaugh Map AB 00 01 11 10 C 0 0 1 1 0 1 0 1 1 0 36 The majority function is true when more than half of the inputs are true. Majority function on 3 inputs: Application: Majority Function
- 34. 19 37 F(A,B,C) = minterms( 3, 5, 6, 7 ) = A'BC + AB'C + ABC' + ABC Gates: 8 Inputs: 19 38 F(A,B,C) = minterms( 3, 5, 6, 7 ) = A'BC + AB'C + ABC' + ABC = A'BC + AB'C + ABC' + ABC + ABC + ABC = (A'+A)BC + AC(B'+B) + AB(C'+C) = BC + AC + AB = AB + AC + BC
- 35. 20 39 F(A,B,C) = minterms( 3, 5, 6, 7 ) AB 00 01 11 10 C 0 0 0 1 0 1 0 1 1 1 40 F(A,B,C) = minterms( 3, 5, 6, 7 ) Minimized function: F(A,B,C) = AB + AC + BC AB 00 01 11 10 C 0 0 0 1 0 1 0 1 1 1
- 36. 21 41 Minimized function: F(A,B,C) = AB + AC + BC Gates: 4 Inputs to gates: 9 42 K-map for function with 4 inputs A'B' A'B AB AB' C'D' m0 m4 m12 m8 C'D m1 m5 m13 m9 CD m3 m7 m15 m11 CD' m2 m6 m14 m10
- 37. 22 43 K-map for function with 4 inputs AB 00 01 11 10 CD 00 m0 m4 m12 m8 01 m1 m5 m13 m9 11 m3 m7 m15 m11 10 m2 m6 m14 m10 44 F(A,B,C,D) = AB' + AC'D' + B'CD' + A'BC'D Ex: F(A,B,C,D) = minterms( 2, 5, 8, 9, 10, 11, 12 ) AB 00 01 11 10 CD 00 0 0 1 1 01 0 1 0 1 11 0 0 0 1 10 1 0 0 1
- 38. 23 45 F(A,B,C,D) = BD + A'BC' + AC'D + ABC + A'CD Ex: F(A,B,C,D) = minterms( 3,4,5,7,9,13,14,15 ) AB 00 01 11 10 CD 00 0 1 0 0 01 0 1 1 1 11 1 1 1 0 10 0 0 1 0 46 F(A,B,C,D) = A'BC' + AC'D + ABC + A'CD Ex: F(A,B,C,D) = minterms( 3,4,5,7,9,13,14,15 ) AB 00 01 11 10 CD 00 0 1 0 0 01 0 1 1 1 11 1 1 1 0 10 0 0 1 0
- 39. 24 47 Start with 1’s which are isolated Find 1’s that can only be included in 2-cover Find 1’s that can only be included in 4-cover Find 1’s that can only be included in 8-cover Continue until all 1’s covered at least once Order is important! 48 K-maps are "circular":