ENEE 244 (01**). Spring 2006
Homework 1 solutions
1. Convert the following binary numbers to decimal: (a) 1101; (b) 10111001.
(a) 1101 = 1.23 + 1.22 + 0.21 + 1.20 = 8 + 4 + 0 + 1 = (13)10.
(b) 10111001 = 1 + 8 + 16 + 32 + 128 (written in backwards order) = 185.
2. Convert the following decimal numbers to binary: (a) 23; (b) 498.
(a) By repeated division by radix 2:
23
11 1
5
1
2
1
1
0
0
1
Answer from above, bottom to top: (10111)2 .
(b) By repeated division, as above (steps not shown) = (111110010)2.
3. Convert the following numbers to decimal: (a) (34)7; (b) (76)8.
(a) (34)7= 3.71 + 4.70 = 21 + 4 = 25.
(b) (76)8 = 7.81 + 6.80 = 56 + 6 = 62.
4. Convert (a) (47)10 to radix 8; (b) (63)10 to base 16.
(a) By repeated division by radix 8:
47
5
7
0
5
Answer from above, bottom to top: (57)8.
(b) By repeated division by radix 16:
63
3
F
0
3
Answer from above, bottom to top: (3F)16 /* Also written as 0x3F */
5. Convert (a) (11010110111)2 to hexadecimal; (b) (4532)8 to binary.
(a) Grouping into groups of 4 bits from right: 0x6B7.
(b) Converting each digit separately: (100 101 011 010)2.
6. Convert the following decimal numbers to binary: (a) 0.75; (b) 5.3.
(a) 0.75 X 2 = 1 + 0.5
0.50 X 2 = 1 + 0.0
Answer from above, top to bottom: (0.11)2.
(b) 5 is 101.
0.3 X 2 = 0 + 0.6
0.6 X 2 = 1 +
0.2
0.2 X 2 = 0 +
0.4
0.4 X 2 = 0 +
0.8
0.8 X 2 = 1 +
0.6 /* We stop here since 0.6 repeats */
Answer:
(101.01001)2.
/* Last four digits of fraction repeat forever */
7. Add the binary numbers (a) 0101 and 1001; (b) 01101.01 and 1001011.101.
(a) 01101
+10 01
11 10
(b) 101111101.01
+10 0 1 0
11.101
10 1
1 0 00.111
8. Do the following subtractions: (a) 1100 - 1001; (b) 101101 - 11010.
(a) Answer: 0011. (See book for workings for similar problems).
(b) Answer: 010011.
9. Multiply: 10100×101.
10100
101
10100
00000
10100
1100100
/* Answer obtained by adding preceding three rows */
10. Find the diminished radix complement of (a) (76)8; (b) (10010)2.
(a) 7's complement = 77-76 = (01)8.
(b) 1's complement = (01101)2. /* Obtained by switching all the bits */
11. Find the radix complement of (445)10.
9's complement = 999-445 = 554.
Þ
10's complement = 554 + 1 = 555.
12. Represent the following binary numbers in an 8-bit signed-magnitude scheme: (a) +1001; (b) -1011.
(a) 00001001.
(b) 10001011. /* First bit is sign bit */
13. A 6-bit computer uses the signed 2's complement representation. It adds 10011 + 11101. State the answer. How is the overflow detected? /* Hint: In such a representation, negative numbers are stored as 2's complement of their absolute value. Positive numbers are represented just like in the sign-magnitude scheme. */
10111010111
+ 0 1 1 1 01
1 1 0 0 00
Since
carry into sign bit (1) ≠ carry out of sign bit (0) Þ
Overflow is detected.
14. An 8-bit computer uses the signed 2's complement representation. Show how this computer performs (a) 111010 - 1100; (b) 101 - 10110.
(a) 111010 - 1100 = 111010
+ (-1100)
-1100 = 2's complement of
00001100 = 11110011+ 1= 11110100
Expression =
00111010 + 11110100 = 100101110.
Discarding carry
out of sign bit (left most digit) = 00101110. /*Answer */
In doing this
calculation, carry into sign bit (1) = carry out of sign bit (1)
Þ No overflow; answer
correct.
(b) 00000101 +
(-00010110)
-00010110 =
2's complement of 00010110 = 11101001 + 1 = 11101010.
Expression =
00000101 + 11101010 = 11101111. /* Nothing to discard, so this is answer */
In doing this
calculation, carry into sign bit (0) = carry out of sign bit (0)
Þ No overflow; answer
correct.
15. A 7-bit computer uses the signed 2's complement representation. What is the range of integers that can be represented in this scheme?
Range is [-(26) to +26- 1] = [-64 to +63].