Fundamentals of Computer Systems

Fundamentals of Computer Systems Review for the Midterm Stephen A. Edwards Columbia University Spring 22

The Midterm 75 minutes 4 5 problems Closed book Simple calculators are OK, but unnecessary One double-sided 8.5 sheet of your own notes Anything discussed in class is fair game Much like homework assignments Problems will range from easy to difficult; do the easy ones first. Historical developments & trivia will not be on the test.

The Axioms of (Any) Boolean Algebra A Boolean Algebra consists of A set of values A An and operator An or operator A not operator A false value A A true value A Axioms a b = b a a b = b a a (b c) = (a b) c a (b c) = (a b) c a (a b) = a a (a b) = a a (b c) = (a b) (a c) a (b c) = (a b) (a c) a a = a a = We will use the first non-trivial Boolean Algebra: A = {, }. This adds the law of excluded middle: if a = then a = and if a = then a =.

Simplifying a Boolean Expression You are a New Yorker if you were born in New York or were not born in New York and lived here ten years. Axioms x ( x) y a b = b a a b = b a a (b c) = (a b) c a (b c) = (a b) c a (a b) = a a (a b) = a a (b c) = (a b) (a c) a (b c) = (a b) (a c) a a = a a = Lemma: x = x (x x) = x (x y) if y = x = x

Simplifying a Boolean Expression You are a New Yorker if you were born in New York or were not born in New York and lived here ten years. Axioms x ( x) y = x ( x) (x y) a b = b a a b = b a a (b c) = (a b) c a (b c) = (a b) c a (a b) = a a (a b) = a a (b c) = (a b) (a c) a (b c) = (a b) (a c) a a = a a = Lemma: x = x (x x) = x (x y) if y = x = x

Simplifying a Boolean Expression You are a New Yorker if you were born in New York or were not born in New York and lived here ten years. Axioms x ( x) y = x ( x) (x y) = (x y) a b = b a a b = b a a (b c) = (a b) c a (b c) = (a b) c a (a b) = a a (a b) = a a (b c) = (a b) (a c) a (b c) = (a b) (a c) a a = a a = Lemma: x = x (x x) = x (x y) if y = x = x

Simplifying a Boolean Expression You are a New Yorker if you were born in New York or were not born in New York and lived here ten years. Axioms x ( x) y = x ( x) (x y) = (x y) = x y Lemma: a b = b a a b = b a a (b c) = (a b) c a (b c) = (a b) c a (a b) = a a (a b) = a a (b c) = (a b) (a c) a (b c) = (a b) (a c) a a = a a = x = x (x x) = x (x y) if y = x = x

Alternate Notations for Boolean Logic Operator Math Engineer Schematic Copy x X X or X X Complement x X X X AND x y XY or X Y X Y XY OR x y X + Y X Y X + Y

Definitions Literal: a Boolean variable or its complement E.g., X X Y Y Implicant: A product of literals E.g., X XY XYZ Minterm: An implicant with each variable once E.g., XYZ XYZ X YZ Maxterm: A sum of literals with each variable once E.g., X + Y + Z X + Y + Z X + Y + Z

Sum-of-minterms and Product-of-maxterms Two mechanical ways to translate a function s truth table into an expression: X Y Minterm Maxterm F X Y X + Y X Y X + Y X Y X + Y X Y X + Y The sum of the minterms where the function is : F = X Y + X Y The product of the maxterms where the function is : F = (X + Y)(X + Y)

Minterms and Maxterms: Another Example The minterm and maxterm representation of functions may look very different: X Y Minterm Maxterm F X Y X + Y X Y X + Y X Y X + Y X Y X + Y The sum of the minterms where the function is : F = X Y + X Y + X Y The product of the maxterms where the function is : F = X + Y

The Menagerie of Gates Buffer Inverter AND NAND OR NOR XOR XNOR + +

De Morgan s Theorem (a b) = ( a) ( b) (a b) = ( a) ( b) AB = A + B = A + B = A B =

Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X X X X Y W The Karnaugh Map Sum-of-Products Challenge Cover all the s and none of the s using as few literals (gate inputs) as possible. Few, large rectangles are good. Covering X s is optional.

Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X X X X Y W The minterm solution: cover each with a single implicant. a = W X Y Z + W X Y Z + W X Y Z + W X Y Z + W X Y Z + W X Y Z + W X Y Z + W X Y Z 8 4 = 32 literals 4 inv + 8 AND4 + OR8

Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X X X X Merging implicants helps Y W Recall the distributive law: AB + AC = A(B + C) a = W X Y Z + W Y + W X Z + W X Y 4 + 2 + 3 + 3 = 2 literals 4 inv + AND4 + 2 AND3 + AND2 + OR4

Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X X X X Y W Missed one: Remember this is actually a torus. a = X Y Z + W Y + W X Z + W X Y 3 + 2 + 3 + 3 = literals 4 inv + 3 AND3 + AND2 + OR4

Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X X X X Y W Taking don t-cares into account, we can enlarge two implicants: a = X Z + W Y + W X Z + W X 2 + 2 + 3 + 2 = 9 literals 3 inv + AND3 + 3 AND2 + OR4

Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X X X X Y W Can also compute the complement of the function and invert the result. Covering the s instead of the s: a = W X Y Z + X Y Z + W Y 4 + 3 + 2 = 9 literals 5 inv + AND4 + AND3 + AND2 + OR3

Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X X X X Y W To display the score, PONG used a TTL chip with this solution in it: (3) (2) OUTPUT a OUTPUT b

Decoders Input: n-bit binary number Output: -of-2 n one-hot code 2-to-4 in out 3-to-8 decoder in out in 4-to-6 decoder out

The 7438 3-to-8 Decoder 5 Y A 4 Y Select Inputs B 2 3 Y2 C 3 2 Y3 Y4 Data Outputs Y5 G 6 9 Y6 Enable Inputs G2A 4 7 Y7 G2B 5

General n-bit Decoders Every minterm I n I 2 I Implementing a function with a decoder: E.g., F = AC + BC I n. I 2 I I n I 2 I. I n I 2 I I n I 2 I I n I 2 I I n I 2 I C B A F C B A 7 6 5 4 3 2 F

The Two-Input Multiplexer B A S Y S B A Y B A S S A B Y S B A Y X X X X S Y A B

The Four-Input Mux D C B A 3 2 Y D C S S 2 B Y S 2 S Y A B C D A S 2 S

General 2 n -input muxes I 2 n. I I 2 n S n S 2 S Y I 2 n. I Y Y = I S n S 2 S + I S n S 2 S + I 2 S n S 2 S +. I 2 n 2S n S 2 S + I 2 n S n S 2 S I 2 n n-to-2 n decoder S n S 2 S

Using a Mux to Implement an Arbitrary Function F = AC + BC C B A F

Using a Mux to Implement an Arbitrary Function F = AC + BC Apply each value in the truth table: C B A F 7 6 5 4 3 2 C B A F

Using a Mux to Implement an Arbitrary Function F = AC + BC C B A F

Using a Mux to Implement an Arbitrary Function F = AC + BC Can always remove a select and feed in,, S, or S. C B A F C B F A A 3 2 C B Y

Using a Mux to Implement an Arbitrary Function F = AC + BC Can always remove a select and feed in,, S, or S. C B A F C B F A A A A 3 2 C B Y

Using a Mux to Implement an Arbitrary Function F = AC + BC Can always remove a select and feed in,, S, or S. C B A F C B F A A B B A A 3 2 C B Y In this case, the function just happens to be a mux: (not always the case!) B A Y C

The Simplest Timing Model In Out t p Each gate has its own propagation delay t p. When an input changes, any changing outputs do so after t p. Wire delay is zero.

A More Realistic Timing Model In Out t p(max) t p(min) It is difficult to manufacture two gates with the same delay; better to treat delay as a range. Each gate has a minimum and maximum propagation delay t p(min) and t p(max). Outputs may start changing after t p(min) and stablize no later than t p(min).

Critical Paths and Short Paths A B C D Y How slow can this be?

Critical Paths and Short Paths A B C D Y How slow can this be? The critical path has the longest possible delay. t p(max) = t p(max, AND) + t p(max, OR) + t p(max, AND)

Critical Paths and Short Paths A B C D Y How fast can this be? The shortest path has the least possible delay. t p(min) = t p(min, AND)

Glitches A glitch is when a single input change can cause multiple output changes. B A B C A C B B A B BC A B + BC A B + BC C A

Glitches A glitch is when a single input change can cause multiple output changes. B A B C A B + BC + AC C A Adding such redundancy only works for single input changes; glitches may be unavoidable when multiple inputs change.

Arithmetic: Additiono Adding two one-bit numbers: A and B Produces a two-bit result: C S (carry and sum) A B Half Adder C S A B C S

Full Adder In general, you need to add three bits: C o + + = + + = + + = + + = + + = + + = C i A B C o S C i A B C i A B C o S S

A Four-Bit Ripple-Carry Adder A B A 3 B 3 A 2 B 2 A B A B C o FA C i FA FA FA FA S S 4 S 3 S 2 S S

A Two s Complement Adder/Subtractor To subtract B from A, add A and B. Neat trick: carry in takes care of the + operation. A 3 B 3 A 2 B 2 A B A B SUBTRACT/ADD FA FA FA FA S 4 S 3 S 2 S S

Overflow in Two s-complement Representation When is the result too positive or too negative? + 2 2 + + + + + + + + + +

Overflow in Two s-complement Representation When is the result too positive or too negative? + 2 2 + % + % + The result does not fit when the top two carry bits differ. A n Bn A n Bn + + + S n S n + + + + % Overflow

Ripple-Carry Adders are Slow A3 B3 A2 B2 A B A B C C4 S3 S2 S S The depth of a circuit is the number of gates on a critical path. This four-bit adder has a depth of 8. n-bit ripple-carry adders have a depth of 2n.

Carry Generate and Propagate The carry chain is the slow part of an adder; carry-lookahead adders reduce its depth using the following trick: C A B K-map for the carry-out function of a full adder For bit i, C i+ = A i B i + A i C i + B i C i = A i B i + C i (A i + B i ) = G i + C i P i Generate G i = A i B i sets carry-out regardless of carry-in. Propagate P i = A i + B i copies carry-in to carry-out.

Carry Lookahead Adder Expand the carry functions into sum-of-products form: C i+ = G i + C i P i C = G + C P C 2 = G + C P = G + (G + C P )P = G + G P + C P P C 3 = G 2 + C 2 P 2 = G 2 + (G + G P + C P P )P 2 = G 2 + G P 2 + G P P 2 + C P P P 2 C 4 = G 3 + C 3 P 3 = G 3 + (G 2 + G P 2 + G P P 2 + C P P P 2 )P 3 = G 3 + G 2 P 3 + G P 2 P 3 + G P P 2 P 3 + C P P P 2 P 3

Bistable Elements Equivalent circuits; right is more traditional. Two stable states:

SR Latch R S S R

SR Latch R S S R R S Set

SR Latch R S S R R S Hold, State

SR Latch R S S R R S Reset

SR Latch R S S R R S Hold, State

SR Latch R S S R R S Huh?

SR Latch R S S R R S Set

SR Latch R S S R R S Hold, State

SR Latch R S S R R S Huh?

SR Latch R X S X S R R S Undefined

D Latch D C D C inputs outputs C D X

Positive-Edge-Triggered D Flip-Flop D C M Master D C Slave D D D C S C C C D C M transparent D C S opaque

Positive-Edge-Triggered D Flip-Flop D C M Master D C Slave D D D C S C C C D C M D C S transparent opaque opaque transparent

Positive-Edge-Triggered D Flip-Flop D C M Master D C Slave D D D C S C C C D C M transparent opaque transparent D C S opaque transparent opaque

Positive-Edge-Triggered D Flip-Flop D C M Master D C Slave D D D C S C C C D C M transparent opaque transparent opaque D C S opaque transparent opaque transparent

D Flip-Flop with Enable D D E C C E D X X X X X D E C D What s wrong with this solution?

Asynchronous Preset/Clear PRE D CLR CLK D PRE CLR

The Synchronous Digital Logic Paradigm Gates and D flip-flops only INPUTS OUTPUTS Each flip-flop driven by the same clock STATE C L Every cyclic path contains at least one flip-flop CLOCK NEXT STATE

Flip-Flop Timing Setup Time: Time before the clock edge after which the data may not change CLK D t su

Flip-Flop Timing Setup Time: Time before the clock edge after which the data may not change Hold Time: Time after the clock edge after which the data may change t su t h CLK D

Flip-Flop Timing Setup Time: Time before the clock edge after which the data may not change Hold Time: Time after the clock edge after which the data may change t su t h CLK D Minimum Propagation Delay: Time from clock edge to when might start changing t p(min)

Timing in Synchronous Circuits D C L CLK t c CLK D t c : Clock period. E.g., ns for a MHz clock

Timing in Synchronous Circuits D C L CLK Sufficient Hold Time? t p(min,ff) t p(min,cl) CLK D Hold time constraint: how soon after the clock edge can D start changing? Min. FF delay + min. logic delay

Timing in Synchronous Circuits D C L CLK t p(max,ff) Sufficient Setup Time? t p(max,cl) CLK D Setup time constraint: when before the clock edge is D guaranteed stable? Max. FF delay + max. logic delay

Clock Skew: What Really Happens C L D CLK CLK 2 CLK Sufficient Hold Time? CLK CLK 2 t skew D t p(min,ff) t p(min,cl) CLK 2 arrives late: clock skew reduces hold time

Clock Skew: What Really Happens C L D CLK CLK 2 CLK Sufficient Setup Time? CLK CLK 2 t skew D t p(max,ff) t p(max,cl) CLK arrives early: clock skew reduces setup time

Cool Sequential Circuits: Shift Registers A 2 3 CLK A 2 3 X X X X X X X X X X

Universal Shift Register L 3 2 D 3 2 D 3 2 2 D 2 3 2 3 D 3 R S S CLK S S 3 2 R 3 2 D 3 D 2 D D 3 2 2 L S S Operation Shift right Load Hold Shift left

Cool Sequential Circuits: Counters Cycle through sequences of numbers, e.g.,

The 74LS63 Synchronous Binary Counter

Moore and Mealy Machines Inputs Next State Logic Next State CLK Current State Output Logic Outputs The Moore Form: Outputs are a function of only the current state.

Moore and Mealy Machines Inputs Next State Logic Next State CLK Current State Output Logic Outputs The Mealy Form: Outputs may be a function of both the current state and the inputs. A mnemonic: Moore machines often have more states.

Mealy Machines are the Most General Inputs Outputs Current State C L CLK Next State Another, equivalent way of drawing Mealy Machines This is exactly the synchronous digital logic paradigm

Moore vs. Mealy FSMs Alyssa P. Hacker has a snail that crawls down a paper tape with s and s on it. The snail smiles whenever the last four digits it has crawled over are. Design Moore and Mealy FSMs of the snail s brain.

State Transition Diagrams: Looking for S S S2 S3 S4 Moore Machine: States indicate output / / / / S / S / S2 / S3 / Mealy Machine: Arcs indicate input/output

Moore Machine Next State S A S S S S S S S S S2 S2 S3 S2 S2 S3 S S3 S4 S4 S S4 S2 Output S Y S S S2 S3 S4

Moore Machine A Next State S A S Output S Y S 2 CLK S 2 Y S CLK S CLK S S

Mealy Machine S A S Y S S S S S S S S2 S2 S3 S2 S2 S3 S S3 S

Mealy Machine A S A S Y S CLK S CLK Y S S

State Transition Diagram for the TLC C + L/T S/T S/T HG H : G F : R FY H : R F : Y Inputs: C: Car sensor S: Short Timeout L: Long Timeout CL/T C + L/T HY H : Y F : R FG H : R F : G S/T S/T C L/T Outputs: T: Timer Reset H: Highway color F: Farm road color S C S L T S HG X X HG HG X X HG HG X HY HY X X HY HY X X FG FG X FG FG X X FY FG X X FY FY X X FY FY X X HG S H F HG G R HY Y R FG R G FY R Y

State and Output Encoding S C S L T S HG X X HG HG X X HG HG X HY HY X X HY HY X X FG FG X FG FG X X FY FG X X FY FY X X FY FY X X HG S H F HG G R HY Y R FG R G FY R Y A one-hot encoding: HG HY FG FY G Y R

State and Output Encoding S C S L T S X X X X X X X X X X X X X X X X X X S H F T = S CL + S S + S 2 (C + L) + S 3 S S 3 = S 2(C + L) + S 3 S S 2 = S S + S 2 (C + L) S = S CL + S S S = S (CL) + S 3 S H R = S 2 + S 3 H Y = S H G = S F R = S + S F Y = S 3 F G = S 2

State and Output Encoding C L S3 S2 FY HR FG T = S CL + S S + S 2 (C + L) + S 3 S S 3 = S 2(C + L) + S 3 S S 2 = S S + S 2 (C + L) S = S CL + S S S = S (CL) + S 3 S H R = S 2 + S 3 S HY H Y = S FR H G = S F R = S + S S S HG F Y = S 3 F G = S 2

The CMOS Inverter A Y A 3V p-fet Y n-fet An inverter is built from two MOSFETs: An n-fet connected to ground A p-fet connected to the power supply V

The CMOS Inverter A 3V A Y 3V Off V Y On V When the input is near the power supply voltage ( ), the p-fet is turned off; the n-fet is turned on, connecting the output to ground ( ). n-fets are only good at passing s

The CMOS Inverter A V A Y 3V On 3V Y Off V When the input is near ground ( ), the p-fet is turned on, connecting the output to the power supply ( ); the n-fet is turned off. p-fets are only good at passing s

The CMOS NAND Gate A B Y A Y Two-input NAND gate: two n-fets in series; two p-fets in parallel B

The CMOS NAND Gate A B Y A B Y Both inputs : Both p-fets turned on Output pulled high

The CMOS NAND Gate A B A B Y Y One input, the other : One p-fet turned on Output pulled high One n-fet turned on, but does not control output

The CMOS NAND Gate A B A B Y Y Both inputs : Both n-fets turned on Output pulled low Both p-fets turned off

The CMOS NOR Gate A B A B Y Y Two-input NOR gate: two n-fets in parallel; two p-fets in series. Not as fast as the NAND gate because n-fets are faster than p-fets

A CMOS AND-OR-INVERT Gate A C Y A B C D Y B D

Static CMOS Gate Structure Inputs p-fet pull-up network Y Pull-up and Pull-down networks must be complementary; exactly one should be connected for each input combination. n-fet pull-down network Series connection in one should be parallel in the other

Read-Only Memories: Combinational Functions A k. A 2 A A 2 k n ROM D n. D D row column A 6 A 5 A 4 A 3 A 2 A A 28 ROM D General ROM: 2 k words n bits per word Example: Space Race ROM

Implementing ROMs / Z: not connected Bitline 2 Bitline Bitline Wordline Wordline A A 2-to-4 Decoder 2 Wordline 2 Add. Data 3 Wordline 3 D 2 D D

Implementing ROMs / Z: not connected Bitline 2 Bitline Bitline Wordline A A 2-to-4 Decoder 2 Wordline Wordline 2 Add. Data 3 Wordline 3 D 2 D D

Implementing ROMs / Z: not connected A A 2-to-4 Decoder 2 Add. Data 3 D 2 D D

CMOS Mask-Programmed ROMs Add. Data ROM programmed by selectively connecting drain wires Active-high wordlines

EPROMs and FLASH use Floating-Gate MOSFETs

Static Random-Access Memory Cell Bit line Bit line Word line

Dynamic RAM Cell Bit line Word line

Our Old Pal, the Space Race ROM 2 3 4 5 means 6 7 8 9 and 2 means 3 4 5 A 3 A 2 A A D D D 2 D 3 D 4 D 5 D 6 D 7

Our Old Pal, the Space Race ROM 2 3 4 5 6 7 8 9 2 3 The decoder or AND plane In a RAM or ROM, computes every minterm Pattern is not programmable 4 5 A 3 A 2 A A D D D 2 D 3 D 4 D 5 D 6 D 7

Our Old Pal, the Space Race ROM 2 3 4 5 6 7 8 9 2 3 4 The decoder or OR plane One term for every output Pattern is programmable = the contents of the ROM 5 A 3 A 2 A A D D D 2 D 3 D 4 D 5 D 6 D 7

Simplifying the Space Race ROM A A 2 A 3 A Essential minterms mean don t expand these

Our New PAL, the Space Race ROM A 3 A 2 A A D D D 2 D 3 D 4 D 5 D 6 D 7 2 3 4 5 6 7 8 9 2 32 2 3 2 3 3 2 4 5 32 32 32 32 3 32 3 2 32 3 2 D = 32 D = 32 D 2 = 32 + 32 D 3 = 32 + 3+ 32 + 32 D 4 = 3 2 + 32 + 32 + 32 D 5 = 3 + 2 + 2+ 32 + 32 D 6 = 3 2 + 32 D 7 = 3 2 + 32 Saved two ANDs

I I I I A (t I/ I/ I/ I/ I/ A 22V PAL: Programmable AND/Fixed OR CLK/I First Fuse Numbers 2 3 4 5 396 44 88 924 452 496 22 256 286 Increments 4 8 2 6 2 24 28 32 36 4 P = 588 R = 589 P = 58 R = 58 P = 582 R = 583 P = 584 R = 585 Macrocell Macrocell Macrocell Macrocell Macrocell P = 586 R = 587 23 22 2 2 9

Field-Programmable Gate Arrays (FPGAs) 6 RAM programmable switch LUT LE LE LE Switch Block SB SB LE LE LE Switch Box: 6 programmable switches

Datapath for dx, dy, right, and down dx = x - x; right = dx > ; if (!right) dx = -dx; dy = y - y; down = dy > ; if (down) dy = -dy; void line(uint6 x, Uint6 y, Uint6 x, Uint6 y) { Sint6 dx; // Width of bounding box Sint6 dy; // Height of BB (neg) Uint6 x, y; // Current point Sint8 sx, sy;// - or Sint6 err; // Loop-carried value Sint6 e2; // Temporary variable int right; // Boolean int down; // Boolean err = dx + dy; x = x; y = y; for (;;) { plot(x, y); if (x == x && y == y) break; e2 = err << ; if (e2 > dy) { err += dy; if (right) x++; else x--; } if (e2 < dx) { err += dx; if (down) y++; else y--; } }

Datapath for dx, dy, right, and down x x y y x x subtract >? negate y y subtract >? negate right dx down dy dx = x - x; right = dx > ; if (!right) dx = -dx; dy = y - y; down = dy > ; if (down) dy = -dy; err = dx + dy; x = x; y = y; for (;;) { plot(x, y); if (x == x && y == y) break; e2 = err << ; if (e2 > dy) { err += dy; if (right) x++; else x--; } if (e2 < dx) { err += dx; if (down) y++; else y--; } }

Datapath for err loop dx = x - x; right = dx > ; if (!right) dx = -dx; dy = y - y; down = dy > ; if (down) dy = -dy; dx dy add err = dx + dy; x = x; y = y; dy add dy << < e2 dx dx add e2_gt_dy > e2_lt_dx err for (;;) { plot(x, y); if (x == x && y == y) break; e2 = err << ; if (e2 > dy) { err += dy; if (right) x++; else x--; } if (e2 < dx) { err += dx; if (down) y++; else y--; } }

Datapath for x and y + right e2_gt_dy x loop x dx = x - x; right = dx > ; if (!right) dx = -dx; dy = y - y; down = dy > ; if (down) dy = -dy; err = dx + dy; x = x; y = y; x y = = + down e2_lt_dx y break loop y for (;;) { plot(x, y); if (x == x && y == y) break; e2 = err << ; if (e2 > dy) { err += dy; if (right) x++; else x--; } if (e2 < dx) { err += dx; if (down) y++; else y--; } }

Timing clk (x,y) (,) (5,3) (x,y) (7,4) (6,4) start done loop break x 2 3 4 5 6 7 5 6 y 2 3 4 3 4 err 3 6 2 5 4 3 dx, dy 7, 4,

Control FSM start break IDLE start start RUN loop = start DONE done = break