JEE 160/EE 260 Exam 1 Solution
Convert (1434.015625) base 10 to binary, octal, and hexadecimal (10 points).
(10110011010.000001) base 2
(2632.01) base 8
(59A.04) base 16
Multiply (514) base 8 and (26) base 8 without converting to decimal
(15 points).
514
26
======
3710
1230
======
16210
(16210) base 8
Provide the minimum sum-of-products for the following (10 points).
W(A,B,C,D) = SUM m(1,2,3,7,8,9,11,13)
X(A,B,C,D) = PRODUCT M(1,3,4,5,10,11,12,14)
\ CD W
AB \
\ 00 01 11 10
+----+----+----+----+
00| | 1 | 1 | 1 |
+----+----+----+----+
01| | | 1 | |
+----+----+----+----+
11| | 1 | | |
+----+----+----+----+
10| 1 | 1 | 1 | |
+----+----+----+----+
W = B'D + A'CD + A'B'C + AC'D + A B'C'
\ CD X
AB \
\ 00 01 11 10
+----+----+----+----+
00| | 0 | 0 | |
+----+----+----+----+
01| 0 | 0 | | |
+----+----+----+----+
11| 0 | | | 0 |
+----+----+----+----+
10| | | 0 | 0 |
+----+----+----+----+
X = A'B'D' + A'BC + ABD + AB'C'
X = BCD' + AC'D + BCD + A'CD'
Provide the minimum product-of-sums for the following (10 points).
Y(A,B,C,D) = SUM m(1,2,3,5,7,9,10,11)
Z(A,B,C,D) = PRODUCT M(0,2,3,7,8,10,11,15)
\ CD Y
AB \
\ 00 01 11 10
+----+----+----+----+
00| | 1 | 1 | 1 |
+----+----+----+----+
01| | 1 | 1 | |
+----+----+----+----+
11| | | | |
+----+----+----+----+
10| | 1 | 1 | 1 |
+----+----+----+----+
Y = (C + D)(A' + B')(B' + D)
\ CD Z
AB \
\ 00 01 11 10
+----+----+----+----+
00| 0 | | 0 | 0 |
+----+----+----+----+
01| | | 0 | |
+----+----+----+----+
11| | | 0 | |
+----+----+----+----+
10| 0 | | 0 | 0 |
+----+----+----+----+
Z = (C' + D')(B + D)
Gray code is a code where only one bit changes as it counts. The count
sequence is 0000, 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101,
1111, 1110, 1010, 1011, 1001, 1000. Follow the design procedure to create
a circuit to convert binary to gray code (truth table, maps, simplified
equations). Do not draw the circuit. Only work the problem to
sum-of-products or product-of-sums (25 points).
A B C D | W X Y Z
============+============
0 0 0 0 | 0 0 0 0
0 0 0 1 | 0 0 0 1
0 0 1 0 | 0 0 1 1
0 0 1 1 | 0 0 1 0
0 1 0 0 | 0 1 1 0
0 1 0 1 | 0 1 1 1
0 1 1 0 | 0 1 0 1
0 1 1 1 | 0 1 0 0
1 0 0 0 | 1 1 0 0
1 0 0 1 | 1 1 0 1
1 0 1 0 | 1 1 1 1
1 0 1 1 | 1 1 1 0
1 1 0 0 | 1 0 1 0
1 1 0 1 | 1 0 1 1
1 1 1 0 | 1 0 0 1
1 1 1 1 | 1 0 0 0
\ CD W \ CD X
AB \ AB \
\ 00 01 11 10 \ 00 01 11 10
+----+----+----+----+ +----+----+----+----+
00| | | | | 00| | | | |
+----+----+----+----+ +----+----+----+----+
01| | | | | 01| 1 | 1 | 1 | 1 |
+----+----+----+----+ +----+----+----+----+
11| 1 | 1 | 1 | 1 | 11| | | | |
+----+----+----+----+ +----+----+----+----+
10| 1 | 1 | 1 | 1 | 10| 1 | 1 | 1 | 1 |
+----+----+----+----+ +----+----+----+----+
\ CD Y \ CD Z
AB \ AB \
\ 00 01 11 10 \ 00 01 11 10
+----+----+----+----+ +----+----+----+----+
00| | | 1 | 1 | 00| | 1 | | 1 |
+----+----+----+----+ +----+----+----+----+
01| 1 | 1 | | | 01| | 1 | | 1 |
+----+----+----+----+ +----+----+----+----+
11| 1 | 1 | | | 11| | 1 | | 1 |
+----+----+----+----+ +----+----+----+----+
10| | | 1 | 1 | 10| | 1 | | 1 |
+----+----+----+----+ +----+----+----+----+
W = A
X = A'B + AB'
Y = BC' + B'C
Z = C'D + CD'
Construct a 8-line-to-3 priority encoder using 4-line-to-2 priority encoders,
2-to-1 multiplexers, and the minimum amount of additional combinational logic
(20 points).
+----------+ +--------+
| Priority | | MUX |
| Encoder | +--|S |
D0------|D0 | | | |
D1------|D1 A0|-----------------------|D0 Y|--------A0
D2------|D2 A1|-----+ +------------|D1 |
D3------|D3 | | | | | |
| V|--+ | | | +--------+
| | | | | |
+----------+ | | | |
| | | |
| | | |
+----------+ | +---------+ | +--------+
| Priority | | | | | | MUX |
| Encoder | | | | +--|S |
D4------|D0 | | | | | | |
D5------|D1 A0|----------+ +-------|D0 Y|--------A1
D6------|D2 A1|-----------------------|D1 |
D7------|D3 | | | | |
| V|--------------------+ +--------+
| | | |
+----------+ | |
| +--------------------A2
| |
| | +-----+
| +--| OR |
| | |-----------V
+--------------------| |
+-----+
Implement the follwoing Boolean function with a 4-to-1 multiplexer and a
single inverter (10 points).
Z(A,B,C) = PRODUCT M(1,3,4,5)
A B C | Z
================== +--------+
0 0 0 | 1 C' | MUX |
0 0 1 | 0 B-----|S0 |
---------+-------- A-----|S1 |
0 1 0 | 1 C' |\ | |
0 1 1 | 0 C--| >o---+--|D0 |
---------+-------- |/ +--|D1 Y|------Z
1 0 0 | 0 0 0----|D2 |
1 0 1 | 0 1----|D3 |
---------+-------- | |
1 1 0 | 1 1 +--------+
1 1 1 | 1
---------+--------
Draw the circuit to implement the mathematical function P = W*X + Y*Z,
where W, X, Y, and Z are unsigned and 2 bits long, using AND gates and
Full Adders (25 point).
W1 W0 Y1 Y0
* X1 X0 * X1 Z0
=================== ===================
X0W1 X0W0 Z0Y1 Z0Y0
+ X1W1 X1W0 + Z1Y1 Z1Y0
======================= =======================
B3 B2 B1 B0 A3 A2 A1 A0
A3 A2 A1 A0
+ B3 B2 B1 B0
=====================
P4 P3 P2 P1 P0
+------+
+-----+ | FA |
Z0---| AND | A0 0--|Cin |
| |--------------------------------------|A S|-------P0
Y0---| | +-------------|B Cout|---+
+-----+ | | | |
| +------+ |
| |
| +--------------+
+-----+ | |
Z0---| AND | +------+ | | +------+
| |--+ | FA | | | | FA |
Y1---| | | 0---|Cin | A1 | +---|Cin |
+-----+ +------|A S|---------------------|A S|-------P1
+------|B Cout|---+ | +----------|B Cout|---+
+-----+ | | | | | | | | |
Z1---| AND | | +------+ | | | +------+ |
| |--+ | | | |
Y0---| | +--------------+ | | +--------------+
+-----+ | | | |
| +------+ | | | +------+
+-----+ | | FA | | | | | FA |
Z1---| AND | +---|Cin | A2 | | +---|Cin |
| |---------|A S|---------------------|A S|-------P2
Y1---| | |B Cout|--------------+ +---|B Cout|---+
+-----+ | | A3 | | | | | | |
+------+ | | +----+ +------+ |
| | | | |
+-----+ | | | | +--------------+
X0---| AND | B0 | | | | |
| |------------------------+ | | | |
W0---| | | | | |
+-----+ | | | |
+-----+ | | | |
X0---| AND | +------+ | | | | +------+
| |--+ | FA | | | | | | FA |
W1---| | | 0---|Cin | B1 | | | +---|Cin |
+-----+ +------|A S|----------+ | +------|A S|-------P3
+------|B Cout|---+ | +---|B Cout|-------P4
+-----+ | | | | | | | |
X1---| AND | | +------+ | | | +------+
| |--+ | | |
W0---| | +--------------+ | |
+-----+ | | |
| +------+ | |
+-----+ | | FA | | |
X1---| AND | +---|Cin | B2 | |
| |---------|A S|------------+ |
W1---| | |B Cout|-----------------+
+-----+ | | B3
+------+