Worksheets2nd yr week 1 stld
Total questions: 25
Worksheet time: 15mins
A signed integer has been stored in a byte using the 2’s complement format. We wish to store the same integer in a 16-bit word. We should
copy the original byte to the most significant byte of the word and fill the more significant byte with zeros.
copy the original byte to the more significant byte of the word and fill the less significant byte with zeros.
copy the original byte to the less significant byte of the word and make each bit of the more significant byte equal to the most significant bit of the original byte.
copy the original byte to the less significant bytes as well as the more significant byte of the word.
An equivalent 2’s complement of the 2’s complement number 1101 is
11010
001101
110111
111101
The 2’s complement representation of –17 is
101110
101111
111110
110001
4-bit 2’s complement representation of a decimal number is 1000. The number is
+8
0
–7
–8
The range of signed decimal numbers that can be represented by 6-bit 2’s complement number is
–31 to +31
–63 to +63
–64 to +63
–32 to +31
11001, 1001 and 11100 correspond to the 2’s complement representation of which one of the following sets of numbers?
25, 9 and 57 respectively
–6, –6 and –6 respectively
–7, –7 and –7 respectively
–25, –9 and –57 respectively
Decimal 43 in Hexadecimal and BCD system is respectively
2B, 0100 0011
2B, 0010 0011
2B, 0011 0100
B2, 0010 1011
A new Binary Coded Pentary (BCP) system is proposed in which every base-5 number is represented by a corresponding 3-bit binary code. For example, base-5 number 24 will be represented by code 010100. In this numbering system, code 10001001 corresponds to the decimal number in base-5 system
423
1324
2201
4321
X = 01110 and Y = 11001 are two 5-bit numbers represented in two’s complement format. The sum of X and Y represented in two’s complement format using 6 bits is
100111
001000
000111
101001
The two numbers represented in signed magnitude and 2’s complement form are P = 11101101 and Q = 11100110. If P is subtracted from Q, the value obtained in 2’s complement form is
10000111
00001111
11111001
11111101
The number of bytes required to represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is _____
P, Q, and R are the decimal numbers corresponding to the 4-bit binary numbers considered in signed magnitude, 1’s complement, and 2’s complement representations respectively. If P + Q + R is considered in 6-bit 2’s complement representation, the result is
111101
110101
110010
111001
If two 2’s complement numbers having sign bits x and y are added and the sign bit of the result is z, then the occurrence of overflow is indicated by the Boolean function
xy ˉz
x ˉyz
x ˉy ˉz+xyz ˉ
xy+yz+zx
The logical expression y=A+A ˉB is equivalent to
y=AB
y=A ˉB
y=A+B
y=A+B ˉ
The minimized form of the logical expression (A ˉBC ˉ+AB ˉC ˉ+A ˉBC+ABC) is
AC+BC+A ˉB
AC ˉ+B ˉC+A ˉB
AC ˉ+BC+A ˉB
AC ˉ+B ˉC+AB
The number of distinct Boolean expressions of 4 variables is
16
256
1024
65536
The functions W, X, Y, and Z are as follows: W&=R+PQ+R ˉS@X&=PQR ˉ+P ˉQR+PQ ˉR ˉS@Y&=RS+PR+PQ+P ˉQ@Z&=RS+PQ+Q ˉR+PQ ˉS Then
W=Z,X=Z ˉ
W=Z," " X=Y
W=Y
W=Y=Z
The Boolean expression AC+BC ˉ is equivalent to
AC+BC+AC ˉ
C ˉ+AC+BC ˉ+AC ˉ
AC+BC+BC+ABC
ABC+AB ˉC+ABC ˉ
The Boolean expression of the truth table shown is
B(A+C)(A ˉ+C)
B(A+C)(A ˉ+C)
B ˉ(A+C)(A ˉ+C)
B(A ˉ+C)(A ˉ+C ˉ)
The Boolean function Y=AB+CD is to be realized using only 2-input NAND gates. The minimum number of gates required is
2
3
4
5
The Boolean expression Y=A'B'C'D+A'BCD'+AB'C'D+ABC'(D') ˉ can be minimized to
Y=A ˉBCD+A ˉBC+ACD
Y=A ˉBCD+BCD+ABCD
Y=ABCD+BCD+ABCD ˉ
Y=A'BCD'+B'(C^' ) ˉD+ABC'D'
If X=1in the logic equation [X+Z]Y ˉ+(Z ˉ+XY)X ˉZ+Z(X+Y)=1 then
Y=Z
Y=2
Z=1
Z=0
In the sum of products function Σ(2,3,4,5) the prime implicants are
XY,XY ˉ
X ˉY,XY ˉZ
X ˉYZ,XYZ,XY ˉZ
X ˉYZ,XYZ,XY ˉZ,YZ
The Boolean expression (X+Y)(X+Y ˉ)+X simplifies to
X
Y
XY
X+Y
For an n-variable Boolean function, the maximum number of prime implicants is
2(n-1)
n/2
2^n
2^(n-2)
