TEST BANK FOR Artificial Neural Networks By B. Yegnanarayana (Instructors Solution Manual).

Solution to problem 1 (a):
s
-1
-1 q=0
+1
a1
a2
a3
Using the logic function,
f(x) = 1; x >
= 0; x
where x =
X
i
wiai, the truth table is obtained by giving all possible combinations of a1; a2; a3. The
results are shown in the following table.
Truth table:
a1 a2 a3 s
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 0
To derive the logic function, one can use the following Karnaugh map (K-map) representation of
the truth table. The logic function is obtained from the map by expressing the terms corresponding
to the entries with 1.
0 0
0
1 0 0
0 1 0 0
1 00 01 10
a2a3
11
a
Logic function s(a1; a2; a3) = a1a2a3
1
(b)
q=1 q=0
+1
-1
+1
+1 s
a2
1
a3
a
Truth table:
a1 a2 a3 s
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 0
K-map representation of the truth table is as follows:
0 1
0
1 0 0
0 1 1 0
00 01 11 10
a1
a2a3
Logic function s(a1; a2; a3) = a2a3 + a1a3
(c)
+1 q=0 q=-1 s
-1
+1
a -1
Truth table:
a s(t 􀀀 1) s(t)
0 0 1
0 1 1
1 0 0
1 1 1
2
K-map representation of the truth table is as follows:
1
1
1
0
a
1
1
0
s(t-1) 0
Logic function s(t) = a + s(t 􀀀 1)
Solution to problem 2:
Using the following basic logic networks
nor
q=1 s =
or
q=0
q=0 q= -1 s = =
s = a1+
+1
+1
+1
+1
+1
+1
-1
a1
a2
a1
a2
a1
a2
a1 a2
a2
a1+ a2 a1 a2
and
(a) The logic network for s(a1; a2; a3) = a1a3 + a2a3 + a1a3 is given by
q=0
q=0
q= -1
q=0
q=1
q=1
a3
s
+1
+1
+1
-1
a1 a1 a3
a2 a3
a1+ a3 a1 a3
a2
+1 +1
+1
+1
+1
+1
+1
3
(b) The logic network for s(a1; a2; a3) = a1a2a3 is given by (c)
q=1 +1 s
+1
+1
+1 q=1
a2
a1
a3
The logic network for s(a1; a2; a3) = a1a2a3 is given by
q=0 +1 +1
s
+1
a2
q=1
a3
a1
a1 a3
q= -1
a1+ a3
Solution to problem 3:
By applying the operations of an M-P neuron, the output of the network shown in Fig.P1.2 for the
input [1 1 1]T is s = 0.
Solution to problem 4:
For the output function: f(x) =
8><
>:
1; x > 0
0; x 0
Input vectors: i1 = [1 1 0 0]T ; i2 = [1 0 0 1]T ; i3 = [0 0 1 1]T ; i4 = [0 1 1 0]T
Choose the learning parameter = 1
Let the initial weight vector w(0) = [1 0 0 0]T
(a)
For the output function shown in Fig.P1.3b.
Output of the rst input vector s = f(wT (0)i1) = 1
Weight update w(1) = w(0) + si1 = [1 0 0 0]T + [1 1 0 0]T = [2 1 0 0]T
Output of the second input vector s = f(wT (1)i2) = 1
Weight update w(2) = w(1) + si2 = [2 1 0 0]T + [1 0 0 1]T = [3 1 0 1]T
4