Cmpt 250 Unsigned Numbers January 11, 2008

Transcription

1 Cmt 25 Unsined Numbers Januar, 28 These notes serve two uroses in the contet of Cmt 25: as we develo the basic desin of an inteer ALU, we ll review the basics of number reresentation and combinational loic desin. Reresentation and Addition of Unsined Numbers First, let s review the reresentation of unsined inteers in binar. Consider an n-bit number. For n = 4, we have (2 3 ) + (2 2 ) + (2 ) + (2 ) = 5. (2 3 ) + (2 2 ) + (2 ) + (2 ) = 7. (2 3 ) + (2 2 ) + (2 ) + (2 ) = When we add two unsined inteers, we know that a carr-out from the most sinificant bit (msb) indicates an overflow. Overflow occurs when the result of a calculation is a number which is too lare to reresent with the number of bits available. For 4-bit numbers, (2 3 ) + (2 3 ) = (2 4 ), which is too lare to reresent. Reresentation and Addition/Subtraction of Sined Numbers Subtraction loic which uses unsined binar reresentation is clums. It s eas enouh to define the loic, but we ll still need a bit to kee track of the sin, and, as the tet oints out, we ll need additional loic to fi u the outut. All in all, it s a comlicated mess. Fortunatel, there s a much better alternative. We can reresent sined inteers with two s comlement reresentation. As we ll see, this will allow us to use the same FA loic for both addition and subtraction. Let s review two s-comlement notation. Consider an n-bit number. The msb the sin bit has a neative weiht, 2 n. For n = 4, we have

2 Cmt 25 Sined Numbers Januar, 28 ( 2 3 ) + (2 2 ) + (2 ) + (2 ) = 7. ( 2 3 ) + (2 2 ) + (2 ) + (2 ) = ( 2 3 ) + (2 2 ) + (2 ) + (2 ) =. ( 2 3 ) + (2 2 ) + (2 ) + (2 ) = 8 When addin numbers in two s-comlement reresentation, we have arithmetic overflow when There s a carr-in to the sin bit and no carr-out: The ositive comonent of the number is too lare to reresent. There s a carr-out from the sin bit and no carr-in: comonent of the number is too lare to reresent. The neative Now that we know where we re oin, let s connect to the underlin mathematics. The two s comlement of an n-bit number is formed b subtractin the number from 2 n. If we actuall had to do arithmetic to form the two scomlement reresentation of a number, there would be no advantae. But recall that with a bit of work it s ossible to show that = +, where reresents the loical comlement (NOT) of. How, ou ask? Well, consider 3, reresented with three bits. To form the two s comlement I subtract from 2 3 : = = ( ) + = ( ) + Now, recall that = and = (notice that there are no borrows since we re subtractin a value from ). This is loical neation of the value we re subtractin! We have = + = + = Let s have a look at an actual subtraction. Suose I want to calculate 7 3 =. = + ( ) 2

3 Cmt 25 Addition/Subtraction Loic Januar, 28 The art in arentheses is the mathematical two s comlement, which will be. Now, suose I etend all the values to four bits. = + We did sa that we were oin to ive the msb neative weiht in two s comlement reresentation, riht? So we can combine the and + as Continuin the addition, we have = + = () = = 7 3 = 4 A lot haened there at the ver end. Let s look closel. M four-bit addition hardware knows nothin about this neative weiht for the sin bit stuff. It siml follows the rules for addin binar bits. As far as the hardware is concerned, + =, and there will be a carr-out from the msb. It s u to us to interret the result correctl, usin the knowlede that the msb has weiht 2 3 and all other diit ositions have ositive weiht. Notice that I had a carr into the the sin osition (a ositive 2 3 ). The hardware doesn t care about diit weiht. It knows that + roduces a sum of and a carr-out of. But we know that addin +2 3 to the neative 2 3 alread resent should cancel, leavin. The zero is the correct result, iven our interretation of the sin bit. You can now see the derivation of one of the two s-comlement overflow rules: If there s a carr-in to the sin bit, and a carr-out, there s no overflow. The correct result is ( = ). Thus the advantae of two s-comlement reresentation: We can erform both addition and subtraction usin onl addition and loical comlement. Not onl that, but we can use the eact same addition loic for unsined and sined numbers. The onl difference lies in detectin overflow: We have different overflow rules for unsined and sined numbers. Combinational Loic for Two s-comlement Addition/Subtraction Let s remember how to desin the loic required for addition. It s easiest to do this b first desinin a comonent to add one bit osition. Once we have this comonent, we can use n of them to add two n-bit numbers. What are our basic buildin blocks? 3

4 Cmt 25 Addition/Subtraction Loic Januar, 28 A half adder adds two bits to roduce a sum bit and a carr-out. A full adder adds two bits and a carr-in to roduce a sum bit and a carr-out. The desin of these comonents is straihtforward; ou learned this in Cmt 5. Let s desin a full adder that takes as inuts two addend bits and and a carr-in, and roduces as oututs a sum and a carr-out. The first ste is to sef a truth table: co From the truth table, we can construct a air of Karnauh mas (k-mas). Usin the rules for rouin terms and derivin loic eressions from k- mas, we can roduce boolean eressions for and co. = co = + + You should also reconise the alternatin attern in the k-ma for as the XOR function: =. 4

5 Cmt 25 Addition/Subtraction Loic Januar, 28 co From the boolean equations, we can construct a combinational loic rcuit to imlement our full adder. full adder We can then comose full adder (FA) rcuits in a rile-carr confiuration to create an adder that will handle n-bit unsined numbers. The adder shown here will handle 3-bit unsined numbers, because three bits fits nicel on the ae. 2 2 ovf co FA co FA co FA 2 The carr-in to the least sinificant bit (lsb) is zero. Overflow (ovf ) is detected as the carr-out from the msb. How lon will it take to add two n-bit numbers with a rile-carr adder? In the rile-carr confiuration each bit osition must wait for the carr from the revious bit osition before it can roduce its own co and bits. Insectin the standard full-adder rcuit, we can see that it requires one ate dela to roduce the and two ate delas to roduce the co once all inuts,, and are available. The total time to add n bits is 2n ate delas 5

6 Cmt 25 Carr Look-Ahead Januar, 28 Let s convert this 3-bit rile-carr adder to a 3-bit add/subtract module for two s-comlement numbers. Given three bits, the module will be caable of the addition and subtraction of binar numbers in the rane 4 to +3 ( to ). 2 2 add/sub co co FA co FA co FA 2 ovf Notice that we re usin XOR for its controlled loical neation function. When we re addin, add/sub =, and the value of i is assed throuh unchaned. When we re subtractin, add/sub = and the value i is resented at the inut of each full adder. Further, we use the same add/sub sinal to rovide the +, b forn a at the inut of the lsb. A sinle XOR ate sufifices to imlement the two s-comlement overflow rules. A rile-carr adder is simle to desin and easil etended to an number of bits, but it has a bi disadvantae: the roaation dela rows linearl with the number of bits, because we must allow sufifient time (two ate delas er bit osition) for a carr to roaate from the least sinificant bit (lsb) to the msb. As n becomes lare (ticall 32 or 64 bits in the current eneration of rocessors), the total addition time of 2n becomes unaccetabl lon. How can we reduce this? Carr Look-Ahead Connectin n full adders in a rile carr confiuration is straihtforward and uses the least amount of hardware for arallel addition of n-bit oerands. But the rile carr does not scale well: the time required for addition increases as O(n). 6

7 Cmt 25 Carr Look-Ahead Januar, 28 To reduce the time required for addition, we must break the rile carr chain and relace it with a faster method. Once such method is called carr look-ahead. To develo the necessar loic, let s look aain at a sinle bit osition. A full adder accets three inuts:,, and, and roduces two oututs, and co. When, eactl, will we have a carr-out? If both and are, we will enerate a carr-out. = If either of or are, we will roaate the carr-in throuh to the carr-out. = Now we can eress the value of the carr-out as co = + () A sliht otimisation is ossible here. We ve defined the roaate sinal as. In man imlementation technoloies, the rcuit which imlements an XOR ate requires more transistors than the rcuit that imlements an OR ate. Suose that we redefine to be +. Here are three aruments that () is still correct. Arument #: Carr out is a binar quantit, either or. The net effect of redefinin to be + is that both and will be when = and =. But + = + =, so there s no chane in the value of the carr-out. Arument #2: boolean alebra co = ( + ) + + ( + ) ( + ) + ( + ) + ( + ) + ( + ) co = + substitute, =, distributive one/zero, inverse relicate distributive inverse, one/zero, distributive rewrite with = and = + 7

8 Cmt 25 Carr Look-Ahead Januar, 28 Arument #3: Karnauh ma The k-ma to the left shows the carr-out function usin the smbols and instead of, to illustrate the contributions of = and = to the outut co = +. The minterm is alread covered b the function. When is converted from to + the result is that the minterm is covered three times: b,, and. Let s summarise: The fiure on the left shows the comonent we ve just desined: A sinle bit full adder which accets as inuts the addends and and a carr-in and roduces as oututs a sum and roaate and enerate sinals and. Ok, ou ask, what have I ained? Well, nothin et, but I ve laid the necessar roundwork. For concreteness, consider an adder that will add two four-bit values (3:) and (3:). In bit osition i, i = 3,..., we will have inuts (i), (i), and (i), and oututs (i) and co(i). Suose I want to know the value of the carr-in to the second bit osition, () = co(). The carr-in will be if a carr is enerated in the least sinificant bit osition, or if a carr-in roaates throuh. Eressed in terms of and, () = co() = () + () () (2) So far there s no imrovement over a rile carr I ve siml rewritten the boolean eression for the carr-out in terms of and. But what about the net bit osition? (2) = co() = () + () () = () + () (() + () ()) = () + () () + () () () (3) Similarl, (3) = co(2) = (2) + (2) () + (2) () () + (2) () () () (4) = co(3) = (3) + (3) (2) + (3) (2) () + (3) (2) () () + (3) (2) () () () (4) 8

9 Cmt 25 Carr Look-Ahead Januar, 28 co(3) P G Carr Look-Ahead () (3) (3) (3) (2) (2) (2) () () () () () (3:) (3:) () (3) (2) () () sum(3:) Fiure : Four bit carr look-ahead adder (CLA Adder) Now we ve achieved somethin: All the carries co(3:) are available after a constant dela, as is the sum. The inuts (3:) and (3:) are all available simultaneousl. All oututs (3:) and (3:) are available after one ate dela. All oututs co(3:) are available after two more ate delas. All oututs (3:) are available after one more ate dela. (Two ate delas, if XOR is relaced with an equivalent AND/OR construction.) In total, we need = 4 ate delas to erform a four-bit addition. Suose that I bo u the loic of equations (2) (4) into a comonent and label it Carr Look-Ahead. Usin this carr look-ahead comonent and sinle bit full adders with roaate and enerate oututs, I can construct a four bit adder with carr look-ahead as shown in Fiure. The P and G oututs will be elained below. Usin the same carr look-ahead loic, we can create larer adders. We need to aument the carr look-ahead comonent to roduce two additional oututs, the rou roaate and enerate sinals G and P. Considerin bit ositions (3:) as a rou, a carr will be enerated if an individual bit osition enerates a carr and that carr is roaated to the 9

10 Cmt 25 Carr Look-Ahead Januar, 28 P(i+5)G(i+5) co(i+5) Carr Look-Ahead (i) (i+5) (i+5) (i+2) (i+) (i+) (i+8) (i+7) (i+7) (i+4) (i+3) (i+3) (3:) (3:) P G CLA Adder (5:2) co (3:) (3:) (3:) P G CLA Adder (:8) co (3:) (3:) (3:) P G CLA Adder (7:4) co (3:) (3:) (3:) P G CLA Adder (3:) co (3:) Fiure 2: Siteen Bit CLA Adder msb osition: G = (3) + (3) (2) + (3) (2) () + (3) (2) () () A carr will roaate throuh the rou if it can be roaated throuh all bit ositions: P = (3) (2) () () These are the P and G sinals shown on the carr look-ahead comonent above. We can use a second level of carr look-ahead loic to san four-bit rous, creatin a 6-bit adder. This second level of carr look-ahead loic will use the P and G values from the first level to enerate rou carr-in values. The schematic is shown in Fiure 2. The comonent labelled CLA Adder is the four bit CLA adder shown in Fiure. Notice that the first level of carr look-ahead loic is now hidden inside this comonent. Seficall, for each four bit rou, the first level carr look-ahead loic will roduce oututs P(5), P(), P(7), P(3), G(5), G(), G(7), and G(3). Usin these values as inuts, the second level will roduce oututs (2), (8), and (4), which will become inuts to the first level carr look-ahead loic. The total time required to roduce a sum will increase b four ate delas: It takes one ate dela to calculate all sinle bit roaate and enerate sinals (5:) and (5:). Usin (i) and (i), it takes two ate delas for the first laer of carr look-ahead loic to roduce the P and G sinals for each four-bit rou (P(5), P(),..., G(7), G(3)).

11 Cmt 25 Carr Look-Ahead Januar, 28 Usin P(), P(7), P(3), G(), G(7), G(3), and (), it takes two ate delas for the second laer of carr look-ahead loic to roduce rou carr-in values (2), (8), and (4). Usin (2), (8), (4), and (), it takes two ate delas for the first laer carr look-ahead loic to roduce the remainin carr-in values (5:3), (:9), (7:5), and (3:). Finall, it takes an additional ate dela to calculate the sum in each bit osition once the carr-in arrives. In total, the time required to add two 6-bit values is t = = 9 ate delas, as oosed to t = 6 2 = 32 ate delas for a 6-bit rile carr adder. The sinals P(5) and G(5) are not strictl necessar for a 6 bit addition, but are required if we want to build a 64 bit adder usin a third laer of look-ahead loic. Notice the tradeoff that we ve made: In return for a substantial increase in the amount of loic, we ve reduced the rowth rate of the time required for n-bit addition and subtraction from linear (2n) to loarithmic ( ( l n ) + ). 2