Lecture 6. Notes. Notes. Notes. Representations Z A B and A B R. BTE Electronics Fundamentals August Bern University of Applied Sciences

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1 Lecture 6 epresenttions epresenttions TE52 - Electronics Fundmentls ugust 24 ern University of pplied ciences ev. c2d5c88 6. Integers () sign-nd-mgnitude representtion The set of integers contins the Nturl Numers N nd their negtives (,, 2,, 4,...) In the deciml system we use the sign-nd-mgnitude nottion, e.g. +4 nd -4 where the + is often omitted Importnt: the sign-nd-mgnitude nottion hs two s, nmely + N -! In the inry system, we do not hve + nd -, we only hve nd However, y the introduction of the M s sign-it we cn crete inry form of the sign-nd-mgnitude representtion y interpreting =+ nd =-: I I 2 I +2 x2! I I 2 I 2 x! epresenttions ev. c2d5c ddition nd ustrction with Integers () deciml sustrction =- we normlly rewrite to deciml ddition =+C y forming the negte of the sustrctor, e.g. C = -, this cn e esily ccomplished y chnging the sign in sign-nd-mgnitude system. For inry sign-nd-mgnitude interprettion this opertion would e n inversion of the sign-it. In sign-nd-mgnitude representtion we cn write this ddition s: s z = s + s c C = 4 + (+ ) The ddition is complex : epresenttions s s c Condition: s z Exmples: + + none + + C + 2 = none + C + ( 2) = + C + C 2 + () = < C C + ( 2) = + C C 2 + = < C + C + 2 = ev. c2d5c88 6.

2 Integers () one s-complement representtion ddition nd sustrction in the sign-nd-mgnitude representtion requires n dder nd sustrctor, cn t we find nother representtion tht requires less complexity? Wht if insted of inverting only the sign it to represent negtive numer, we invert (one s-complement) ll its: ec.: Inputs: sign-nd-mgnitude: s-complement: epresenttions + I 2 I +2 I 2 I + I 2 I + I 2 I I 2 I I 2 I 2 I 2 I I 2 I NOTE: one s-complement representtion lso hs 2 representtions for! ev. c2d5c Integers () one s-complement representtion ddition nd sutrcting one s-complement numers is esier thn doing the sme in sign-nd-mgnitude representtion: sustrction =- is gin trnsformed into n ddition =+C y forming the one s-complement C=- (y inverting [complementing] ll the its of ). The ddition =+C cn now e performed y inry ddition s for N N, with the difference tht crry c k hs to e dded t the L position; this is the so clled end-round crry ddition: epresenttions One s complement: ev. c2d5c Integers () one s-complement representtion One s complement: epresenttions One s-complement is gret, s for oth the ddition nd sustrction of integers we only require inverters nd n end-round crry dder However, n end-round crry dder is slow! Oservtion: The end crry c n is only when one of the vriles or C in =+C is negtive! Why not dd to ll negtive numers? ev. c2d5c88 6.6

3 Integers () two s-complement representtion Two s complement: One s complement: epresenttions The positive numers in two s-complement nottion hve sme it encodings s the other presented numer interprettion systems The + nd - mp on the sme it pttern in the two s-complement nottion resulting in single! The remining it pttern is mpped to -4 ev. c2d5c Integers () one s versus two s complement One s complement ddition/sustrction: I 2 I 2 I I +/ Full dder Full dder Full dder epresenttions O = I +/ I O 2 O O One s complement is slow due to the end-round crry Two s complement is fster, nd hs only one! ev. c2d5c Integers () Excess-N representtion The two s complement representtion is commonly used to represent integers in for exmple Processors, ignl Processors, nd others However, there exists nother importnt representtion: the Excess-N representtion, lso clled the ised representtion The Excess-N representtion uses pre-specified numer N s ising vlue. The deciml numer N is represented y ll s Exmple: The three it Excess- representtion The Excess-N representtion is minly used in nlog to converters nd the IEEE floting-point stndrd Performing clcultions in the Excess-N representtion is only in few cses simple nd out of the scope of this course! epresenttions ev. c2d5c88 6.9

4 Integers () Overflow nd Underflow id we solve the overflow nd underflow prolem? No, s we still hve only k-its to represent numers! We just chnged the rnges: epresenttion: Numer: eciml rnge: inry x N x 2 k ign-nd-mgn. x (2 k ) x 2 k One s-compl. x (2 k ) x 2 k Two s-compl. x (2 k ) x 2 k Excess-N x N x 2 k (N + ) NOTE: The two s complement nd excess-n representtions hve n symmetric rnge nd do not hve + nd ; they hve single! In cse the deciml clcultion results in vlue x outside the given rnges we phse n overflow or underflow prolem! The detection of overflow nd underflow is outside the scope of this course! epresenttions ev. c2d5c88 6. ummry (N nd ) ll numer representtions re noted s series of its from M...L: For numer representtion of k-its we cn represent the deciml equivlents like: epresenttion: Numer: eciml rnge: inry x N x 2 k ign-nd-mgn. x (2 k ) x 2 k One s-compl. x (2 k ) x 2 k Two s-compl. x (2 k ) x 2 k Excess-N x N x 2 k (N + ) To convert positive (negtive) integer into negtive (positive) one we hve to complement the representtion s: epresenttion: Complement: ign-nd-mgn. Invert sign it One s-compl. Invert ll its Two s-compl. Invert ll its nd dd Excess-N lookup in tle epresenttions ev. c2d5c88 6. Multipliction ememer: The clssic deciml pen nd pper multipliction: inry multipliction: l l x + + x epresenttions inry multipliction is performed similr, ut much simpler! s: = nd = (Note: This represent n N gte!) ev. c2d5c88 6.2

5 Full dder Full dder Full dder Full dder Full dder Full dder Multipliction I I 2 I I I 2 I epresenttions Hlf dder Full dder Full dder Hlf dder Hlf dder Full dder O = I x I Overflow O O 2 O O ev. c2d5c88 6. ivision ememer: The clssic deciml pen nd pper division: inry division: l epresenttions 6 8 eminder eminder inry division is performed identicl, ut simpler! s for ech resulting digit we only hve to sustrct once! ev. c2d5c ivision I O O I 2 O 2 epresenttions I O O = I % I O = I / I O I O I 2 O 2 I O How cn we represent the numer when eminder, e.g. 2 = 2? ev. c2d5c88 6.5

6 Fixed Point representtion () The set on el numers contins ll integers nd ll frctions m n el numers re normlly noted in the deciml system y plcing deciml point: 2.55 We cll this nottion the fixed point representtion of rel numer. ecll: Plce-nd-vlue nottion epresenttions i i Therefore: N = d p p ecomes = d p p p= p= j We cn do the sme for ll plce-nd-vlue notted numers of ny se! ev. c2d5c Floting Point representtion () Wht out very ig nd very smll numers? ig numer like we normlly write s or E5 nd smll numers line.2 we similrly write s or E 5 We cll this the floting point representtion, where numer is represented y mntiss nd exponent. The mntiss is in fixed-point representtion, whilst the exponent is in plce-nd-vlue representtion: = mntiss = where: < mntiss < exponent = 5 where: < exponent < epresenttions ev. c2d5c Floting Point representtion () We cn lso represent floting point numers using set of its (IEEE 754 stndrd) This is for exmple used in every computer to represent the types flot, doule, etc. For flot often single precision floting point representtion is used: epresenttions ign it Exponent < Mntiss < 2 Excess x 2 + Fixed Point ev. c2d5c88 6.8

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