Describe how data is transmitted using half-duplex serial data transmission.

Answers

Answer 1
Answer:

Half-duplex serial data transmission allows for communication in both directions, but only one direction at a time. In applications such as walkie-talkies and traditional telephone systems.

In half-duplex serial data transmission, data is transmitted in one direction at a time

1. Sender initiates transmission: The sender, also known as the transmitter, starts the data transmission process. It prepares the data to be sent and waits for the appropriate time to start transmitting.

2. Sender sends data: The sender begins sending the data in a sequential manner. It breaks down the data into smaller units called frames or packets.

3. Receiver acknowledges receipt: After receiving each frame, the receiver acknowledges its successful reception to the sender.

4. Sender waits for acknowledgment: Upon sending a frame, the sender waits for the receiver's acknowledgment. If an acknowledgment is received, the sender proceeds to send the next frame.

5. Receiver processes the data: The receiver receives the frames and processes the data within each frame. It checks for errors using techniques like checksums or cyclic redundancy checks (CRC).

6. Roles switch for bidirectional communication: Once the sender finishes transmitting its data, the roles switch and the receiver becomes the sender, initiating its own transmission.

7. Communication continues: The process continues as the receiver sends its data and waits for acknowledgments, while the new sender processes the received data and sends acknowledgments.

Overall, half-duplex serial data transmission allows for communication in both directions, but only one direction at a time. This type of transmission is commonly used in applications such as walkie-talkies and traditional telephone systems.

Learn more about half-duplex serial here;

brainly.com/question/33829144

#SPJ3

Answer 2
Answer:

Answer:

In half duplex mode, the signal is sent in both directions, but one at a time. In full duplex mode, the signal is sent in both directions at the same time. In simplex mode, only one device can transmit the signal. In half duplex mode, both devices can transmit the signal, but one at a time.


Related Questions

In a clustered column chart, the names of each column are part of the ____ series. select one:a. category b. data c. label d. legend
How many generations of computer languages have there been since the middle of the 20th century?
A computer with a frequency 2 GHZ and its average cycle per instruction is 2. what is the MIPS of the computer?
The Internet is considered a WAN. *TrueFalse
What does I/O mean in computing​

Assume hosts A and B are each connected to a switch Svia 100-Mbps links. The propagation delay on each link is 25μs. The switch Sis a store-and-forward device and it requires a delay of 35μs to process a packet after is has received the last bit in the packet. Calculate the total time required to transmit 40,000 bits from Ato B in the following scenarios. (The total time is measured from the start of the transmission of the first bit at A, until the last bit is received at B. We always assume that links are bi-directional with the same transmission rate and propagation delay in each direction unless specifically instructed otherwise.)

Answers

Answer:

885 μs

Explanation:

Given that:

Switch via = 100 Mbps links

The propagation delay for each link = 25μs.

Retransmitting a received packet = 35μs

To determine:

The total time required to transmit 40,000 bits from A to B.

Considering the fact as a single packet:

Transmit Delay / link = size/bandwith

= 4×10⁴ bits / 100 × 10⁶ bits/sec

= 400 μs

The total transmission time = ( 2 × 400 + 2 × 25 + 35) μs

= (800 + 50 + 35) μs

= 885 μs

Given an array A of n + m elements. It is know that the first n elements in A are sorted and the last m elements in A are unsorted. Suggest an algorithm (only pseudo code) to sort A in O(mlogm +n) worst case running time complexity. Justify.

Answers

Answer:

Explanation:

We can divide array A into two arrays B , C

B would contain the sorted n elements.

C would contain the unsorted m elements.

Now we can sort C using merge sort with worst time complexity mlogm.

When you will have array C sorted you can concatenate with B and put it back in A.

Volume of Pyramid = A*h/3 where A is the area of the base of the pyramid and h is the height of the pyramid. Write a C++ program that asks the user to enter the necessary information about the pyramid (note that the user would not know the area of the base of the pyramid, you need to ask them for the length of one of the sides of the base and then calculate the area of the base). Using the information the user input, calculate the volume of the pyramid. Next display the results (rounded to two decimal places). Example Output (output will change depending on user input): The area of the base of the pyramid is: 25.00 The height of the pyramid is: 5.00 The volume of the pyramid is: 41.67 *Pseudocode IS required for this program and is worth 1 point. The program IS auto-graded.

Answers

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

int main()

{

   double side, height;

   

   cout<<"Enter the length of one of the sides of the base: ";

   cin>>side;

   cout<<"Enter the height: ";

   cin>>height;

   

   double area = side * side;

   double volume = area * height / 3;

   

   cout<<"The area of the base of the pyramid is: "<<area<<endl;

   cout<<"The height of the pyramid is: "<<height<<endl;

   cout<<"The volume of the pyramid is: "<<fixed<<setprecision(2)<<volume<<endl;

   return 0;

}

Pseudocode:

Declare side, height

Get side

Get height

Set area = side * side

Set volume = area * height / 3

Print area

Print height

Print volume

Explanation:

Include <iomanip> to have two decimal places

Declare the side and height

Ask the user to enter side and height

Calculate the base area, multiply side by side

Calculate the volume using the given formula

Print area, height and volume (use fixed and setprecision(2) to have two decimal places)

What is not recyclable in a
hybrid car
hydrogen car
petrol car

Answers

There are different aspect of cars that cannot be recyclable. In petrol cars, The aspect that cannot be recycled is  used gear oil, windshield wiper solution, brake fluid, power steering fluid, etc.

This is because they are very toxic substances, as they have lead and poisonous ethylene glycol in them.

  • A lot of electric and hybrid cars often uses different kinds of lithium-ion batteries and nickel metal hydride batteries that are used and not all part are recyclable.

  • In hydrogen car, their fuel cells is not to have some measure of recyclable.  The Fuel cells has some recyclable materials but not all.

Learn more about recyclable from

brainly.com/question/376227

The computer output for integer programs in the textbook does not include reduced costs, dual values, or sensitivity ranges because these variables are not meaningful for integer programs. TrueFalse

Answers

Answer:

The answer is True

Jenae helps maintain her school web site and needs to create a web site poll for students. Which tool will she use?

Answers

She needs to use a web browser

Answer:

web browser

Explanation:

the other three will not let her excess her page at all

Other Questions
Language: JavaYour task is to complete the logic to manage a twenty-four-hour clock (no dates, just time) that tracksthe hours, minutes, and sections, and various operations to adjust the time. The framework for yourclock is in the Time class with the four methods you must complete.1.) advanceOneSecondâ A user calls this method to advance the clock by one second.a. When the seconds value reaches 60, it rolls over to zero.b. When the seconds roll over to zero, the minutes advance.So 00:00:59 rolls over to 00:01:00.c. When the minutes reach 60, they roll over and the hours advance.So 00:59:59 rolls over to 01:00:00.d. When the hours reach 24, they roll over to zero.So 23:59:59 rolls over to 00:00:00.2.) compareTo â The user wants to know if a second time is greater than or less than the timeobject called, assuming both are on the same date.a. Returns -1 if this Time is before otherTime.b. Returns 0 if this Time is the same as otherTime.c. Returns 1 if this Time is after otherTime.3.) add â Adds the hours, minutes, and seconds of another time (the offset) to the current timeobject. The time should ârolloverâ if it exceeds the end of a 24 hour period.4.) subtract â Subtracts the hours, minutes, and seconds of another time (the offset) fromthe current time object. The time should âroll backâ if it precedes the beginning of a 24 hourperiod.________________________________________________________________________package clock;/*** Objects of the Time class hold a time value for a* European-style 24 hour clock.* The value consists of hours, minutes and seconds.* The range of the value is 00:00:00 (midnight) to 23:59:59 (one* second before midnight).*/public class Time {private int hours;private int minutes;private int seconds;/*** Add one second to the current time.* When the seconds value reaches 60, it rolls over to zero.* When the seconds roll over to zero, the minutes advance.* So 00:00:59 rolls over to 00:01:00.* When the minutes reach 60, they roll over and the hours advance.* So 00:59:59 rolls over to 01:00:00.* When the hours reach 24, they roll over to zero.* So 23:59:59 rolls over to 00:00:00.*/public void advanceOneSecond(){}/*** Compare this time to otherTime.* Assumes that both times are in the same day.* Returns -1 if this Time is before otherTime.* Returns 0 if this Time is the same as otherTime.* Returns 1 if this Time is after otherTime.*/public int compareTo(Time otherTime){return 0;}/*** Add an offset to this Time.* Rolls over the hours, minutes and seconds fields when needed.*/public void add(Time offset){}/*** Subtract an offset from this Time.* Rolls over (under?) the hours, minutes and seconds fields when needed.*/public void subtract(Time offset){}