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Vadim26
2 months ago
12

An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side

frequencies. What is the total bandwidth of the AM signal
Engineering
1 answer:
Mrrafil [318]2 months ago
6 0

Answer:

Total bandwidth: 8 kHz

Explanation:

Data provided:

Transmitter frequency: 3.9 MHz

Modulation up to: 4 kHz

Solution:

For the upper side frequencies:

Upper side frequencies = 3.9 × 10^{6} + 4 × 10³

Upper side frequencies = 3.904 MHz

For the lower side frequencies:

Lower side frequencies = 3.9 × 10^{6} - 4 × 10³

Lower side frequencies = 3.896 MHz

Consequently, the total bandwidth is computed as:

Total bandwidth = upper side frequencies - lower side frequencies

Total bandwidth = 8 kHz

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Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or
Kisachek [356]

Answer:

Here is the JAVA program:

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed to 4 assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two arrays origList[] and offsetAmount[] get assigned their values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // variable for storing the product results

for(i = 0; i <= origList.length - 1; i++){

/* iterates from 0 to the end of origList */

/* multiplies each origList entry with the corresponding offsetAmount entry, stores results in product */

product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }

System.out.println(product); }}

Output:

80 180 80 400

Explanation:

If you wish to print the product of origList alongside offsetAmount values vertically, this can be done in this manner:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

origList[i] *= offsetAmount[i];

System.out.println(origList[i]);}

}}

Output:

80

180

80

400

The program is shown with the output as a screenshot along with the example's input.

8 0
2 months ago
Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe
iogann1982 [368]

An item of protective gear that shields individuals passing by from stray sparks or metal during the welding process performed by another worker is known as: E. Welding Screens.

An operator is a person tasked with joining two or more metals using a technique called wielding.

In the course of wielding, both sparks and tiny metallic fragments are released, which pose a danger to the operator and others working nearby.

As a result, the equipment outlined below should be worn or utilized directly by a worker actuating the wielding process:

  • Visors.
  • Goggles.
  • Protective clothing.
  • Dark walls.

Nonetheless, a type of protective gear that defends other workers nearby from stray sparks or metallic fragments while the operator (worker) is in the act of welding is called welding screens.

Find more information:

4 0
1 month ago
A thin-film gold conductor for an integrated circuit in a satellite application is deposited from a vapor, and the deposited gol
mote1985 [299]

Answer:

Explanation:

The equilibrium vacancy concentration can be described by:

nv/N = exp(-ΔHv/KT),

where T is the temperature at which vacancies form,

K = Boltzmann's constant,

and ΔHv = enthalpy of vacancy formation.

Rearranging this equation to express temperature allows you to calculate it using the provided values. A detailed breakdown of the process is included in the attached file.

5 0
2 months ago
You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
pantera1 [306]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

The half-life for the specified RC circuit can be expressed as

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

The circuit has a resistance of 40 ohms, and by adding a new resistor of 48 ohms, the total resistance becomes 40 + 48 = 88 ohms.

Thus, the new half-life is

t'_{1\2} =R'Cln2

Now, divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

After substituting all values, we can calculate the revised half-life

t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
2 months ago
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