Last updated: 10/26/2006
ECEn/CS 224
Chapter 13 Homework Solutions
13.1
If one desires to count to
M
, what is the minimum number of bits required in the resulting
counter?
The maximum number,
M
, that can be represented by
K
bits is 2
K
1. In other words, a
minimum of
K
bits are required, where
K
is the smallest integer such that 2
K
1 is greater
than or equal to
M
. Solving for
K
gives us
(
1
log
2
+
=
M
K
However, this answer allows for fractions of bits. We must therefore limit our answer to
the next highest integer (i.e., we take the ceiling). Thus we have,
(
1
log
2
+
=
M
K
13.2
Design a 3bit counter that counts in multiples of 3. That is, the count sequence is: 000
–
011
–
110
–
000
–
· · ·
.
Use D flip flops.
Q
2
Q
1
Q
0
N
2
N
1
N
0
0
0
0
0
1
1
0
0
1
X
X
X
0
1
0
X
X
X
0
1
1
1
1
0
1
0
0
X
X
X
1
0
1
X
X
X
1
1
0
0
0
0
1
1
1
X
X
X
N
2
= Q
0
N
1
= Q
2
’
N
0
= Q
1
’
Q
2
Q
1
Q
0
0
1
00
0
X
N
2
01
X
X
11
1
X
10
X
0
Q
2
Q
1
Q
0
0
1
00
1
X
N
1
01
X
X
11
1
X
10
X
0
Q
2
Q
1
Q
0
0
1
00
1
X
N
0
01
X
X
11
0
X
10
X
0
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Last updated: 10/26/2006
D
Q
D
Q
D
Q
Q
2
Q
1
Q
0
CLK
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 Spring '10
 Schultz
 Logic gate, Q1 Q0 N2, N2 N1 N0, Q0 N2 N1

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