1. because when overflow occurs the sign bit

1.     Overflow
in 2’s complement calculation occurs when the result exceeds the max number
possible in that form. The result of an overflow could be severe. When adding
two positive numbers, if overflow happens, the result would be a negative
number. This is because when overflow occurs the sign bit would be set. With
the sign bit set, the number would represent a negative quantity.

Given below is an example which showcases this for a 3-bit system where bit 2
represents the sign bit:

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A = 3 (011)
B = 2 (010)
The result for A+B as we know would be = 5
Using 2’s complement arithmetic:
011 + 010 = 101
The above number in 2’s complement form represents: -3

2.     The
two’s complement follows the following for an X-bit wide system:
The most significant bit (MSB) forms the sign-bit. Thus, the user knows for
every calculation that the MSB significant bit would give us the sign. This
makes it easy to get the sign of the number. It also solves the problem of
representing the negative numbers easily, with the MSB helping us to identify
whether the number is positive or negative.

Apart from the explicit sign bit, the magnitude can also be easily computed for
both positive (sign bit 0) and negative numbers (sign bit 1). The magnitude for
the positive number could be directly computed using the value the number
represent. For negative numbers, we would need to use the two’s complement
rules and then calculating the magnitude. In this way both negative and
positive numbers can be represented and calculated easily.

3.     Given
8-bit numbers:

a.      121
– Since it is a positive number and within the range of an 8-bit system. The
binary representation would itself give the 2’s complement:

121
= 0111 1001

b.     -53
– For negative number we would need to follow the 2’s complement arithmetic
rules:

i.     Represent
53 in binary: 0011 0101

ii.     Negate
the result discarding the sign bit: 100 1010

1 to the result and add the sign bit: 1100 1011

-53
= 1100 1011

4.     Given
8-bit numbers:

a.      1101
1001 – Since it is a negative number we would need to use 2’s complement arithmetic
to get the exact value:

i.     Negate
the number discarding the sign bit: 010 0110

1 to the result and calculate the magnitude: 010 0111 = -39
1101 1001 = -39

b.     0110
0010 – For positive number we could get the magnitude directly:

0110
0010 = 98

5.     The
IEEE-754 Single Precision Floating point representation is shown below:

The IEEE-754 single precision floating point format
uses 32-bits to represent the floating point number. The bits are divided as
shown in the figure above:
Bits 22:0         – Form the Mantissa (or the
Fraction part)
Bits 30:23       – Form the Exponent part
Bit 31              – Forms the sign bit

Using the above bit positions, we can represent a
floating-point number easily. Given below is the way, how a floating-point
quantity could be expressed. This is how the floating point number is
expressed:

(-1)S
x 1.F x 2(E-127)

Here:

S: Represents the Sign bit
F: Magnitude of the fraction part
E: Magnitude of the exponent part

Using the above information, we can express floating
point numbers easily. All the three different quantities can be calculated
separately and then the above equation can be used to get the actual value of
the floating-point number.

6.     The
following shows the calculation for 56.25 using single precision floating point
numbers. It follows the same rules discussed in the previous answers relating
to the floating-point numbers:

56.25:

The binary expansion of the above number can be
calculated using the equation mentioned above. The three quantities are shown
below:

Sign bit = 0
Fraction = 11000010000000000000000
Exponent = 10000100

7.     The
adder can be used to do both addition and subtraction using two’s complement
form. This is true because subtraction in 2’s complement form is nothing but
adding a number A with the 2’s complement version of the other number B.

The full-adder can be used as a
subtractor this depending on the following two inputs used with a controlled
inverter:
Carry in = 1 (The carry-in should be set)
B = ~B (Using the controlled inverter)

Here is the diagram which shows the
same:

Using the adder as a subtractor by
setting the Cin bit and negating B

8.     a)
Characteristic and excitation table for T flip-flop:

Q(next)
= TQ’ + T’Q

Excitation Table:

Q

Q(next)

T

0

0

0

0

1

1

1

0

1

1

1

0

b)  Characteristic and excitation table for J-K
flip-flop:

Q(next)
= JQ’ + K’Q

Q

Q(next)

J

K

0

0

0

X

0

1

1

X

1

0

X

1

1

1

X

0

9.     Output
equation is shown below:

Y = (QA
+ QB) X

State equations are shown below:

QA(next)
= XQA + XQB

QB(next)
= XQB’

10.  State
table and state diagram is shown below:

State Table:

Current State
(QAQB)

Input
(X)

Next State
(QA(next)QB(next))

00

0

00

00

1

01

01

0

00

01

1

10

10

0

00

10

1

11

11

0

00

11

1

10

State diagram is shown below:

State Mapping
(QAQB)

State Name

00

S0

01

S1

10

S2

11

S3