14:332:231 DIGITAL LOGIC DESIGN. Why Binary Number System?

Transcription

1 :33:3 DIGITAL LOGIC DESIGN Ivan Marsic, Rutgers University Electrical & Computer Engineering Fall 3 Lecture #: Binary Number System Complement Number Representation X Y Why Binary Number System? Because computers work with binary values &, LOW & HIGH, TRUE & FALSE Recall the basic building blocks -- AND, OR, NOT logic gates Z = X Y X Y Z = X Y X Z = X AND gate OR gate NOT gate or inverter We need to learn to work with the Binary Number System of 5

2 Number Systems A positional number system has a radix (or base of the number) any integer r d p d k d d. d d j d n Most Significant Digit radix point Least Significant Digit r k r j p D = Σ d i r i i = n d i (r) Example: radix The integer and the fractional part are processed separately. 3of 5 Powers of : n It will be convenient to remember these powers: n n n 3 n of 5

3 Integer Part p Σ d i r i = (( (d p r + d p )r + + d )r + d )r + d i = Divide by r, the remainder is d, d, d, from the least significant to the most significant digit. Will use r =, binary conversion: d i = {, } Example: 5 =? 5: = R : = 6R d = d = 6: = 3R d = 3: = R d 3 = : = R d = 5of 5 Integer Part p Σ d i r i = (( (d p r + d p )r + + d )r + d )r + d i = Divide by r, the remainder is d, d, d, from the least significant to the most significant digit. Will use r =, binary conversion: d i = {, } Example: 5 =? 5: = R : = 6R d = d = 6: = 3R d = 3: = R d 3 = : = R d = 5 = Verify the result: = = 5 Least Significant Bit LSB MSB Most Significant Bit 6of 5 3

4 Fractional Part Σ d i r i = r (d + r (d + ))) i = n Same like before, but now we multiply with the radix. Example:.375 =?.375 =.75 < d =.75 =.5 > d =.5 =. d 3 = 7of 5 Fractional Part Σ d i r i = r (d + r (d + ))) i = n Same like before, but now we multiply with the radix. Example:.375 =?.375 =.75 < d =.75 =.5 > d =.5 =. d 3 = MSB LSB.375 =. Verify the result: + 3 = =.375 8of 5

5 Important Radices Radices important to computer engineers are: r =, 8, 6 Binary Decimal Octal Bit String Hexadecimal A B C D E F -Bit String Example:. =. = = = Fourth digit was added to the fractional part. =. = E.6 6 E.6 6 = = of 5 Binary Addition x y c out s c in s = sum c in = carry in c out = carry out X, Y, C in s, C out Decimal: Binary: + = + = + = sum carry out X + Y + C in = s C out + + = + + = + + = of 5 5

6 Binary Addition x y c out c in c in = carry in c out = carry out Example addition: s X, Y, C in s, C out augend + addend = sum X + Y + carries sum = X + Y + carries sum = X + Y + carries sum = of 5 Subtraction Decimal: = 8 b o rr borrow from the next leftward digit o w 5 7 = 8 minuend subtrahend difference Binary: = borrow (= ) from the next leftward digit b o r r o w = b o r r o w = because = of 5 6

7 Binary Subtraction x y b out d b in b in = borrow in b out = borrow out X, Y, B in s, B out X Y B in = d B out = = = 3 of 5 Example Subtraction () Must borrow, yielding the new subtraction = After the first borrow, the new subtraction for the column is ( ), so we must borrow again. The borrow ripples through leftwards until there is a non-zero digit from which to borrow. minuend subtrahend X 9 Y 6 borrows difference X Y 83 of 5 7

8 Example Subtraction () Verify the result: + minuend X 77 subtrahend Y 3 borrows difference X Y 5 5 of 5 Signed-Magnitude Representation Use the MSB for the sign: n bits: d n d n d sign magnitude d n = positive number d n = negative number n bits represent n numbers largest positive number: n bits n Σ i = n i = smallest negative number: two representations for zero: ( n ) and 6 of 5 8

9 Signed-Magnitude Arithmetic Arithmetic operations must process the sign separately. For example, subtraction: A B?. Compare the magnitudes A B. Subtract smaller magnitude from larger magnitude 3. If B > A, then change the sign of the result too complicated will NOT use it for computations. Instead, we use two s complement representation 7 of 5 Radix-Complement Representation Assumptions: fixed number of digits, D = d n d k d d, n radix r radix-complement representation of D: [D] r = r n D The involution property: [[D] r ] r = r n (r n D) = D How to compute it? (would like to avoid subtraction) 8 of 5 9

10 Radix-Complement Computation [D] r = r n D Rewrite r n = (r n ) + Then, [D] r = r n D = ((r n ) D) + Observe that (r n ) has the form mm mm where m = r n For example, for r = and n =, (r n ) = 9999 for r = and n = 5, (r n ) = Define the complement of a digit d to be d r = r d For example, for r =, the complements of 3, 5, and 8 are 3 = 3 = 6 5 = 5 = 8 = 8 = Then, the complement of D is obtained by complementing individual digits of D and adding 9 of 5 s-complement Representation n-bit s-complement representation of D: [D] = n D Compute two s complement as: [D] = ( n D ) + n : D: n bits d n d n d d n d n d + [D] d i = d i = = of 5

11 s-complement Computation. Complement the digits. Add to the Least Significant Bit 3. Discard carry out from Most Significant Bit [D] = ( n D ) + n : D: n bits d n d n d d n d n d + [D] d i = d i = = of 5 Two s Complement Number System of 5

12 s-complement Representation () Range of n-bit s complement: n A n Example: n = 5 represent 3 in s complement: 3 = = 3 What decimal number is represented in 5-bit s complement:? 3 of 5 s-complement Representation () Range of n-bit s complement: n A n Example: n = 5 represent 3 in s complement: 3 = = 3 What decimal number is represented in 5-bit s complement: negative number for magnitude: complement the digits and add + that is 6 so the number is: 6 of 5

13 So, what is this number? =? Answer: depends on the representation! Unsigned: = 89 Signed-magnitude: = 5 Two s complement: = 39 5 of 5 3