Y = (A + B' + A'B)C'
Y = AC' + B'C' + A'BC'
Y = (A + A'B)C' + B'C'
Y = (A + B)C' + B'C'
Y = AC' + BC' + B'C'
Y = AC' + C'(B + B') → B + B' = 1
Y = AC' + C'
Y = C'

The given function is equivalent to the following expression:
Y = (((A + B)' + C)' + ((D + E)')')'
Y = ((A + B)' + C)')' . (D + E)'
Y = ((A + B)' + C) (D'E')

The above diagram is full subtractor. The equation is D=X⊕Y⊕Bin and B_{out=X'Bin+X'Y+YBin}

Subtract 0011 if there is no carry otherwise add 0011.
Example:
x+3
y+3
-------
(x+y+6)
Here, sum is excess-6. Hence, subtract 0011 to make it excess-3.

The characteristic table of an SR flip-flop is:

So, by simplifying using k-maps:
characteristic equation of an SR flip-flop = Q_n+1 = S + R^’Q_n

Consider a graph with two nodes(n1&n2) and number of edges are 1, There may be chance self edge with node n1 then graph is disconnected.
Consider the graph with three nodes(n1,n2&n3) and has 2 edges.
n1-->n2 and n2--->n1 then the graph is disconnected.

The chromatic number of a graph is the smallest number of colors needed to color the vertices of so that no two adjacent vertices share the same color and if a graph requires k different colours for its proper colouring, then k is the chromatic number of the graph.

The read and write functionality depends on the type of microcontroller peripheral. Generally, the status bits have a read only status and can be modified by peripherals only. So, read bit can be read by CPU and written by the peripheral.

Throughput means total number of tasks executed per unit time i.e. sum of waiting time and burst time.
Shortest job first scheduling is a scheduling policy that selects the waiting process with the smallest execution time to execute next.
Thus, in shortest job first scheduling, shortest jobs are executed first. This means CPU utilization is maximum. So, maximum number of tasks are completed.

→ Feasible schedule is completing all the jobs within deadline.
→ From the dead line, we can deduce that Job J2 & J4 will complete by time “2” whereas remaining two requires time “4”.
→ So the order of completion of Jobs are Either J2 or J4 and followed by either J1 or J3.
From the given options , Option A,C & D gives the solution because after completion of Jobs J2 and J4 then only jobs J1 and J3 is going to complete.
→ But option B , order of completing jobs is J4,J1,J2 ,J3 which is not possible and it is not feasible schedule

An optical encoder is an electromechanical device which has an electrical output in digital form proportional to the angular position of the input shaft.
Optical encoders enable an angular displacement to be converted directly into a digital form.
Encoder resolution is often referred to in bits, which are binary units: a 16 bit resolution rotary encoder will have 65,536 (216) increments per turn, or PPR.
In the given question, 8-bit optical encoder will have 2⁸ increments Resolution = 360/2ⁿ = 360/2⁸ = 1.4^o

Spatial Locality of reference – this says that there is chance that element will be present in the close proximity to the reference point and next time if again searched then more close proximity to the point of reference.
Temporal Locality of reference – In this Least recently used algorithm will be used. Whenever there is page fault occurs within word will not only load word in main memory but complete page fault will be loaded because spatial locality of reference rule says that if you are referring any word next word will be referred in
its register that’s why we load complete page table so complete block will be loaded.
Principle of locality of reference justifies the use of cache.

Initialisation: x = 1, i = 1;
Loop: x i
21 2
22 3
24 4
216 5
After this condition becomes false.

When five items: A, B, C, D, and E are pushed in a stack: Order of stack becomes: A, B, C, D, and E (A at the bottom and E at the top.) stack is popped four items and each element is inserted in a queue: Order of queue: B, C, D, E (B at rear and E at the front) Order of stack after pop operations = A. Two elements deleted from the queue and pushed back stack: New order of stack = A, E, D(A at the bottom, D at the top) As D is on the top so when pop operation occurs D will be popped out. So, correct option is (D).

FIFO is when implemented in preemptive version, it acts like round-robin algorithm.

3 × 512 + 7 × 64 + 5 × 8 + 3
= (2 + 1)× 512 + (4 + 2 + 1)× 64 + (4 + 1)× 8 + 2 + 1
= 1024 + 512 + 64 x 4 + 64 x 2 + 64 + 32 + 8 + 2 + 1
= 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 2 + 1
As 1024 has ten 0’s followed by 1, 512 has nine 0’s followed by 1 and so on..
So, the expression will contain total nine 1’s and will be be represented as 11111101011.

Synchronous communication requires that the clocks in the transmitting and receiving devices are synchronized – running at the same rate – so the receiver can sample the signal at the same time intervals used by the transmitter. No start or stop bits are required. For this reason “synchronous communication permits more information to be passed over a circuit per unit time.
2400 baud means that the serial port is capable of transferring a maximum of 2400 bits per second.
Number of 8-bit characters that can be transmitted per second = 2400/8 = 300

Bandwidth = Highest frequency - Lowest frequency.
Highest frequency = Bandwidth+ Lowest Frequency= 5kHz + 52kHz = 57kHz.

A Hub is a network hardware device for connecting multiple Ethernet devices together and making them act as a single network segment.
A repeater is an electronic device that receives a signal and re-transmits it. Repeaters are used to extend transmissions so that the signal can cover longer distances or be received on the other side of an obstruction.
An Ethernet Hub or a Repeater Hub also acts as a repeater and regenerates the signal to avoid its loss.