Online Test 4 - Digital Logic and Circuit Design

 Since MSB of a is 1, 2's complement of A = 0000 0110. Therefore A = -6
B = 10.
A x B = -60
Since a negative number we have to find the 2's complement of -60
Represent -60 in 2's complement.
Binary of 60 = 0011 1100
2's complement = 1100 0100

 From circuit we see
T1=XQ1’+X’Q1—-(1)
AND
T2=(Q2 ⊕ Q1)’—-(2)
AND DESIRED OUTPUT IS 00->10->11->01->00
SO X SHOULD BE Q1Q2’+Q1’Q2 SATISFYING 1 AND 2.
SO ANS IS (4).

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The clock cycles are per block; if an instruction size is 2 then it requires twice no. of clock cycles.

So answer is (2). 

On solving above k-map we get $AB’D+C’D+BC’D’$
So Ans is (1).

2's complement of the number 10101 is 01010 + 1 = 01011.

XOR gate outputs 1 when an odd number of inputs are 1, and 0 when an even number of inputs are 1. This property makes it ideal for testing odd parity - if the XOR result is 1, the word has odd parity; if 0, it has even parity.

A flip-flop is a bistable multivibrator that can store one bit of binary information (0 or 1). It has two stable states and can maintain its state indefinitely until triggered by a clock or control signal. This makes it a fundamental 1-bit memory element in digital systems.

Semiconductor memory is a volatile memory.

FETs (Field Effect Transistors) in linear ICs serve two key purposes: they provide very large resistances in small chip area and significantly increase input impedance due to their high gate resistance. This makes them ideal for input stages and for creating large-value resistors that would be impractical with standard resistive components.

Q₀ will toggle in every cycle because Q₀‘ (Q0 complement) is fed as input to the D₀ flip flop.
For the D₁ flip flop, D₁ = Q₀ ⊕ Q₁‘ , i.e., Q₀ XOR Q₁‘.
So, the bit pattern Q₀ Q₁ will be :

Q₀ Q₁
0     0
1     1
0     1
1     0
0     0
.        .
.        .
.        .
Thus, the transition sequence will be

C = $\dfrac{1t}{V}$ = $\dfrac{0.1 \times 10^{-12} \times 20 \times 10^-3}{0.5}$ = 4 × 10^-15 F

This program multiplies BYTE1 by 10. Hence, content of A will be 46H.
07H = 07₁₀, 7 × 10 = 70₁₀ = 46H

For example, signed char is 8 bits, we can store from -128 to 127 using sign char.

A two's complement number system encodes positive and negative numbers in a binary number representation. The weight of each bit is a power of two.

The value $w$ of an $N$-bit integer $a_{N-1}a_{N-2}...a_{0}$ is given by the following formula.

$w = -a_{N-1}2^{N-1} + \sum_{i=0}^{N-2}a_i2^i$

The most significant bit determines the sign of the number and is sometimes called the sign bit $-(2^{N-1}$ shown above. Using $N$ bits, all integers from $-2^{N-1}$ to $2^{N-1}-1$ can be represented.