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CHAPTER 7
Solutions for Exercises
E7.1
(a) For the whole part, we have:
Quotient
Remainders
23/2
11
1
11/2
5
1
5/2
2
1
2/2
1
0
1/2
0
1
Reading the remainders in reverse order, we obtain:
23
10
= 10111
2
For the fractional part we have
2
×
0.75 = 1 + 0.5
2
×
0.50 = 1 + 0
Thus we have
0.75
10
= 0.110000
2
Finally, the answer is 23.75
10
= 10111.11
2
(b)
For the whole part we have:
Quotient
Remainders
17/2
8
1
8/2
4
0
4/2
2
0
2/2
1
0
1/2
0
1
Reading the remainders in reverse order we obtain:
17
10
= 10001
2
For the fractional part we have
2
×
0.25 = 0 + 0.5
2
×
0.50 = 1 + 0
Thus we have
0.25
10
= 0.010000
2
1
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View Full Document Finally, the answer is 17.25
10
= 10001.01
2
(c)
For the whole part we have:
Quotient
Remainders
4/2
2
0
2/2
1
0
1/2
0
1
Reading the remainders in reverse order we obtain:
4
10
= 100
2
For the fractional part, we have
2
×
0.30 = 0 + 0.6
2
×
0.60 = 1 + 0.2
2
×
0.20 = 0 + 0.4
2
×
0.40 = 0 + 0.8
2
×
0.80 = 1 + 0.6
2
×
0.60 = 1 + 0.2
Thus we have
0.30
10
= 0.010011
2
Finally, the answer is 4.3
10
= 100.010011
2
E7.2
(a)
1101.111
2
= 1
×
2
3
+ 1
×
2
2
+0
×
2
1
+1
×
2
0
+1
×
2
1
+1
×
2
2
+1
×
2
3
= 13.875
10
(b)
100.001
2
= 1
×
2
2
+0
×
2
1
+0
×
2
0
+0
×
2
1
+0
×
2
2
+1
×
2
3
= 4.125
10
E7.3
(a)
Using the procedure of Exercise 7.1, we have
97
10
= 1100001
2
Then
adding two leading zeros and forming groups of three bits we have
001 100 001
2
= 141
8
Adding a leading zero and forming groups of four bits we obtain
0110 0001 = 61
16
(b)
Similarly
229
10
= 11100101
2
= 345
8
= E5
16
2
E7.4
(a)
72
8
=
111 010 = 111010
2
(b)
FA6
16
= 1111 1010 0110 = 111110100110
2
E7.5
197
10
= 0001 1001 0111 = 000110010111
BCD
E7.6
To represent a distance of 20 inches with a resolution of 0.01 inches, we
need 20/0.01 = 2000 code words.
The number of code words in a Gray
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This note was uploaded on 03/31/2008 for the course ENGINEERIN ES3 taught by Professor Zatet during the Fall '07 term at Tufts.
 Fall '07
 Zatet
 Electrical Engineering

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