1. Converting from Base b to Base c
Start with base b
representation
Initialize base c
value x to 0
Get next xi
Going from
left to right
x m-1…x1x0
Repeat for
All digits
Convert (xi)b
number (Di)c
by using table
Initialize base c
value x to 0
Update the base c
value by
x=xb+Di

2. Review

3. Fixed Point Numbers

4. CS501 Advanced Computer Architecture
Lecture 34
Dr.Noor Muhammad Sheikh

5. Example: Convert the Hexadecimal number B316 to base 10
Solution:
According to the above algorithm,
X=0
X=x+B(=11)=11
X=16*11+3= 179
Hence B316=17910

6. Converting from calculator’s Base c to Base b
Start with bace c
integer
Initialize i=0
and v=x
Set Di=v mod b
And v=v/b 
Convert (Di )cto (xi )b
Set i=i+1
If v!=0, repeat

7. Example: convert 39010 to base 16
c=10, b=16
Solution:
According to the above algorithm
390/16 =24( rem=6), x0=6
24/16= 1(rem=8), x1=8, x2=1
Thus 39010=18616

8. Converting a Base b fraction to calculator’s Base c
Start with base b
representation
Convert to Base c
number (Di)c
by using table
Initialize f=0.0 and
Set i = -m
f-1f-2…f-m
xn-1xn-2…x1x0.x-1x-2…x-m
Repeat for
all digits
until i=0
Here the digits are
treated from right to left,
with division implemented
at each step
Update f=(f+D)/b
And i=i+1

9. Example: Convert (.4cd)16 to Base 10
Solution:
F=0
F=(0+13)/16=0.8125
F=( )/16=
F=( )/16=( )10

10. Converting a fraction from calculator’s Base c to Base b
Start with fraction
f in Base c
Initialize i=1
and v=f
Set D-i=bv 
and v=bv - Di
Convert Di to f-i
Set i=i+1
If v!=0 repeat
Until enough digits
are generated
No division is required
and the process should be
terminated when enough
digits are generated

11. Representation of Negative Numbers

12. Example: Convert 0.2410 to base 2
Solution:
0.24*2=0.48, f-1=0
0.48*2=0.96, f-2=0
0.96*2=1.92, f-3=1
0.92*2=1.84, f-4=1
0.84*2=1.68, f-5=1,…
Thus =( )2

13. Sign-magnitude form
This is the simplest form for representing a signed number
A symbol representing the sign of the number is appended to the left of the number
This representation complicates the arithmetic operations

14. Radix Complement This is the most common representation
Given an m-digit base b number x, the radix complement of x is
xc= ( bm – x) mod bm
This representation makes the arithmetic operations much easier

15. Diminished Radix Complement
The diminished radix complement of an m-digit number x is
xc’=bm-1-x
This complement is easier to compute than the radix complement
The two complement operations are interconvertible, as
xc= ( xc’+1)mod bm

16. Complement representation of negative numbers
Diminished
Radix Complement
Radix Complement
Number
Representation
0 or bm - 1
0<x<bm/2 x
-bm/2<=x<0
|x|c=(bm-|x|
-bm/2+1<=x<0
|x’|c=bm-1-|x|

17. Example Base 2 Complement representation
8 bit 2’s complement
8 bit 1’s complement
Number
Representation
0 or 255
0<x<128 x
-128<=x<0
256-|x|
-127<=x<0
255 -|x|

18. Examples of number representations
Decimal
2’s
complement
1’s
Sign-
magnitude
16’s
Unsigned 27
011011 1B
11011
.17
0.2B
-26
100110
100101
111010 E6 -
-0.57
F.6E

19. Example: Multiplication and division using shift operation
Overflow would occur if we use four bits here
6x4
001102x410=110002=2410
60/16
/1610= =310
The fractional portion of the result is lost

20. Example: Multiplication and division of negative numbers
-6=(11010)2
-6x4=(01000)2=8 which is wrong!
using less no. of bits might change sign
So, -6=(111010)2
-6x4=(101000)2 = -24

21. Example: Multiplication and division of negative numbers
-24=(101000)2
-24x2=(010100)2=20
-24x2=(110100)2 = -12
Changing the size of the number,
24= (n=6) to (n=8)
-24= (n=6) to (n=8)

22. Unsigned addition operation
The digitwise procedure for adding m-digit
base b numbers x and y is as shown
Initialize digit j=0 and c-in c0=0
Compute j sum sj=(xj+yj+cj)mod b
And carry cj+1=(xj+yj+cj)/b
Increment j=j+1
Repeat if j<m

23. ALSU

24. Example: unsigned addition in Base 2 and 16
Base 16 addition
Base 2 addition
A B 4 216
+ 3 1 C 116
0 1 0
D D 0 316
00011

25. Addition Hardware Base b unsigned digit adder
(xj+yj+cj)/b
(xj+yj+cj)mod b xj yj
0<=cj+1<=1
0<=cj<=1
0<=sj<=b

26. 1-bit half adder

27. 1-bit full adder

28. Addition Hardware Base b unsigned m-digit adderx0 y0 co so x1 y1 s1
xm-1
ym-1
Cm-1
sm-1 c1 Cm … c2
The hardware can be implemented
using two levels of AND/OR logic