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4 years ago

Why is it useful to have both digital and analog input devices available when solving a problem?

Engineering

1 answer:

nadya68 [22]4 years ago

5 0

It is useful to have both analog and digital devices while solving a problem.

<u>Explanation:</u>

It is due to the fact that, both devices have their unique properties.
In Analog devices, input and output are analog signals which helps us to manipulate the signal in its original form alone.
But in Digital devices we can convert the analog input signal to the digital signals. Here we can manipulate and rectify the error signal in it and we can convert it back to its original form using a digital to analog converter. Hence signals are not lost.

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scrapers are used to haul dirt from a borrow pit to the cap of a landfill. the estimated cycle time for the scrapers is 9.5 minu

ladessa [460]

Answer is: 12.8 because when you multiply it by the 2nd power of 8 then the scraper equals to be 12.8 in height and that’s how much each scraper requires to operate :)

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3 years ago

Material selection is one of the design considerations for electronic packages. Explain how each of the material properties belo

ValentinkaMS [17]

Answer:

The explanations are provided below.

Explanation:

a) The accelerated thermal cycling test is a test that is performed to evaluate the fatigue strength of electronic solder connections. The glass transition temperature given as $T_{g}$ is a frequently used parameter to test the degree of cure of an epoxy encapsulant.

b) The Coefficient of Thermal Expansion of Encapsulant

This is an important parameter to prevent crackling. Crackling occurs when a joint is subjcted to high temperatures during reflow soldering. Thus, the property is important in determining the limits of the joint.

7 0

3 years ago

The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without us

devlian [24]

Answer:

The phasor technique can't be applied directly in the following cases:

a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)

b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)

c) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)

d) 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)

Explanation:

For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.

This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.

This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function $e^{j(wt+\alpha)}$ ), so it is not possible to combine with functions expressed as a sine, even though both have the same frequency.

If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:

sin (ωt) = cos (ωt-90º)

This means that in order to be able to represent a sine function as a cosine, we need to rotate it 90º in the plane clockwise.

This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).

8 0

3 years ago

1. You use

lorasvet [3.4K]

4-ways tell me if I’m wrong

4 0

3 years ago

Given an N x M matrix and a dictionary containing K distinct words of length between 1 and 30 symbols, the Word Search should re

Ket [755]

Answer:

import java.util.*;

public class Main {

public static String[] wordSearch(char[][] matrix, String[] words) {

int N = matrix.length;

List<String> myList = new ArrayList<String>();

int pos = 0;

for(String word : words)

{

for (int i = 0; i < N; i++) {

for (int j = 0; j < N; j++) {

if (search(matrix, word, i, j, 0, N)) {

if(!myList.contains(word))

{

myList.add(word);

}

String[] newDict = myList.toArray(new String[myList.size()]);

return newDict;

}

public static boolean search(char[][] matrix, String word, int row, int col,

int index, int N) {

// check if current cell not already used or character in it is not not

if (word.charAt(index) != matrix[row][col]) {

return false;

}

if (index == word.length() - 1) {

// word is found, return true

return true;

}

// check if cell is already used

if (row + 1 < N && search(matrix, word, row + 1, col, index + 1, N)) { // go

// down

return true;

}

if (row - 1 >= 0 && search(matrix, word, row - 1, col, index + 1, N)) { // go

// up

return true;

}

if (col + 1 < N && search(matrix, word, row, col + 1, index + 1, N)) { // go

// right

return true;

}

if (col - 1 >= 0 && search(matrix, word, row, col - 1, index + 1, N)) { // go

// left

return true;

}

if (row - 1 >= 0 && col + 1 < N

&& search(matrix, word, row - 1, col + 1, index + 1, N)) {

// go diagonally up right

return true;

}

if (row - 1 >= 0 && col - 1 >= 0

&& search(matrix, word, row - 1, col - 1, index + 1, N)) {

// go diagonally up left

return true;

}

if (row + 1 < N && col - 1 >= 0

&& search(matrix, word, row + 1, col - 1, index + 1, N)) {

// go diagonally down left

return true;

}

if (row + 1 < N && col + 1 < N

&& search(matrix, word, row + 1, col + 1, index + 1, N)) {

// go diagonally down right

return true;

}

// if none of the option works out, BACKTRACK and return false

return false;

}

public static void main(String[] args) {

char[][] matrix = { { 'j', 'a', 's' },

{ 'a', 'v', 'o'},

{ 'h', 'a', 'n'} };

String[] arr_str = {"a", "java", "vaxn", "havos", "xsds", "man"};

arr_str = wordSearch(matrix, arr_str);

for(String str : arr_str)

{

System.out.println(str);

}

8 0

4 years ago