Benha University
Benha High Institute of Technology
Jan 2011 – Fall semester
Exam (Regular),
Time
: 3 Hrs
Department : Electrical Engineering
Course name : E211 Digital Logic Circuit
Year
: 2nd
Time Allowed: Hours
Attempt all questions
Q1)
(12 points)
a) If the state of a 12 bit register is 1010 1101 1100. What is its content if it
represents
(1)-Three decimal digits in BCD
(2)- Three decimal digits in excess-3
code.
(3)-Three decimal digits in 84-2-1 (4)- A decimal number
*Answer*
1. XXX
2. 7X9
3. 6XX
4. 2780 whereX is invalid Code For Decimal Digits
b) List the numbers from 10 to 26 in base 12.
*Answer*
A, B, 10,11,12,13,,14,15,`16,17,18,19,1A,1B,20,21,22
c) Assume a 3- input AND gate with output F and a 3- input OR gate with
output G. The inputs are A, B, and C. show the signals by means of timing
diagram of the outputs F and G as function of the 3 inputs ABC (use all 8
possible ABC combinations).
A
B
C
F
G
Q2)
(12 points)
a) List the truth table of the function
b) *Answer*
F= xy + xy\ + y\z =Σ(1,4,5,6,7)
X
Y
Z
F
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
0
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
1
1
1
1
1
c) Express the Complement of the following functions in sum of minterms.
F1(A,B,C,D) = Σ(0,2,6,11,13,14) ,
F2(x,y,z)= П(0,3,6,7)
*Answer*
F1\ = Σ(1,3,4,5,7,8,9,10,12,15)
F2\ = Σ( 0,3,6,7)
d) show that the dual of a Two input XOR is equal to its complement
*Answer*
XOR= XY\ + X\Y
Complement = (XY\ + X\Y)\= (XY\)\ (X\Y)\ = (X\ +Y)( X + Y\)= X\ X + Y Y\ + xy +
x\y\= xy+x\y\=XNOR------ 1
Dual = (x+y\)(x\+y)=xx\+xy+x\y\+yy\=xy+x\y\=XNOR----------2
1=2
Q3)
(12 points)
For the function F (A,B,C,D) = Σ (0,1,2,3,4,8,9,12)
a) Simplify the function with the suitable map.
b) Implement the function with two level AND- OR gates
c) Implement the function with two level NAND gates.
d) Implement the function with two level NOR- OR.
*Answer*
F= A\B\ + C\D\ + B\C\
1
1
1
1
1
0
0
0
1
0
0
0
1
1
0
0
Q4)
(16 points)
a) For circuit that operate as a Full Adder, construct a truth table, simplify the
function obtained by the truth table, and implement the simplified function
with suitable gates.
*Answer*
X
Y
Z
S
C
0
0
0
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
0
1
1
0
0
1
0
1
0
1
0
1
1
1
0
0
1
1
1
1
1
1
b) Repeat the above item for code conversion circuit that converts from BCD
code into excess-3 code, taking the unused combinations as Don’t Care
conditions.
Q5)
(8 points)
Analyze the given circuit below to obtain the logic function for F1 & F2.
*Answer*
F1= T2 + T3
T1 = (A+B +C)\ = A\B\C\
T2 = ABC
F2 = AB + AC + (BC)\ = AB + AC + B\ + C\ ------------------------1
F2\ = (AB)\(AC)\(BC)
T3 = T1.F2\ = (A\B\C\)(AB)\(AC)\(BC)
F1 = T2 + T3 = ABC + (A\B\C\)(AB)\(AC)\(BC)--------------------2