design a degree 4 butterworth lowpass filter for a 3 dB frequency of 1 GHz and a system impedance of 100
Work on paper all the required calculations explaining every steps, and draw the circuit design assigning the above requirements.


PLEASE NOTE THAT I DO NOT NEED ANY MATLAB CODES FOR THIS LOWPASS FILTER BUT I NEED ALL ITS CALCULATIONS AND ITS FINAL CIRCUIT DRAWN ON PAPER. EXPLAIN EACH AND EVERY STEP CLEARLY.

The difference between the two classifications of PM motors is whether rotor and field are made of permanent magnets or coils.

The two classifications of PM motors are Brushless DC (BLDC) and Brushed DC (BDC) motors. The primary difference between the two classifications is that the rotor and field can be either permanent magnets or coils. BLDC motors have permanent magnet rotors and a stator that is a set of coils. Meanwhile, the rotor in BDC motors is a set of coils, and the stator has permanent magnets.

The speed of a wound armature PM motor is varied by changing the current applied to the stator. The speed of a wound armature PM motor is regulated by adjusting the current applied to the stator, not the armature. This current produces an electromagnetic field that interacts with the magnetic field produced by the permanent magnets on the rotor. The interaction between the magnetic fields produces a torque that drives the motor.

The faster the current changes, the greater the speed of the motor.3. Applications of a wound armature PM motor Wound armature PM motors are used in several applications, including: Motorcycles Scooters Electric Bicycles (E-Bikes)

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The written form of R in terms of unit vectors along the x- and y-axes shown are; R Cos θ and R sin θ respectively.

Given that the force R, which is the resultant of the unit vectors given has magnitude R;

It follows that the resolved form of unit vector, R along the x- and y- axes is;

Where, angle measure, θ = angle subtended by the resultant force with the horizontal.

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Answer:

byee byee bbbbbbbbbbbb

Answer:

The free body diagram of the system is, 558 368 368 508 O ?? O, Consider the equilibrium of horizontal forces. F

Explanation:

I hope this helps you but I think and hope this is the right answer sorry if it’s wrong.

We can say that the given function n^3/1000 - 100n^2 - 100n + 3 is in the Θ(n^3) notation, as it is bounded above and below by some multiple of n^3 for sufficiently large n.

To explain why the given function n^3/1000 - 100n^2 - 100n + 3 is in the Θ(n^3) notation, we need to show that it is bounded above and below by some multiple of n^3 for large enough n.

First, let's consider the highest-order term in the function, which is n^3/1000. As n grows larger, the influence of the other terms in the function diminishes relative to n^3/1000. Therefore, for sufficiently large n, the function is dominated by the term n^3/1000. We can drop the lower-order terms and coefficients and write the function as Θ(n^3/1000).

Next, we can simplify Θ(n^3/1000) to Θ(n^3) because Θ notation ignores constant factors. Specifically, any function that is bounded above and below by some multiple of n^3/1000 is also bounded above and below by some multiple of n^3. In other words, we can multiply n^3/1000 by any constant factor and still have a function that is Θ(n^3).

Therefore, we can say that the given function n^3/1000 - 100n^2 - 100n + 3 is in the Θ(n^3) notation, as it is bounded above and below by some multiple of n^3 for sufficiently large n.

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Answer:

look online

Explanation:

Answer:

The fundamental difference between effective and less effective matrix organizations is whether the tension between different perspectives is creative or destructive. While various processes, systems and tools can help, what matters most is what top leadership says and does and how that flows through the organization in shared targets, clear accountabilities, live team interactions and team-building transparency and behaviors.

Getting matrix management right is linked inextricably to an organization’s culture - the only sustainable competitive advantage. Key components of a culture can be grouped into behaviors, relationships, attitudes, values and the environment.

Environment and values: Each organization has its own environment, context and bedrock values. Everyone needs to know what matters and why. Don’t try to do anything else until you’ve got that set.

Attitude is about choices: An organization’s overall strategy drives choices about which of its parts will be best in class (superior), world class (parity), strong (above average), or simplified/outsourced to be good enough. These choices help determine the need for a matrix and how best to design it.

Relationships and behaviors: This is why organizations have matrices. The most effective of them best balance focus and collaboration. They allow leaders and teams to build differential strengths and then work together to make the best possible decisions and scale enterprises with a creative tension that they could not do on their own.

My colleague Joe Durrett has worked all sides of matrix organizations in marketing at Procter & Gamble, sales and general management at Kraft General Foods and CEO of Information Resources, Broderbund Software and Advo. He has seen matrices at their best and at their worst and offered his perspective for this article along with his partners John Lawler and Linda Hlavac. The 12 ways to make matrix organizations more effective were built on their ideas.

Explanation:

The best reason for the fact that pure bending produces no longitudinal shearing stress is that there is no lateral shearing stress in a section due to pure bending.

This can be attributed to the mechanics of pure bending and the equilibrium conditions that exist within a bending beam.

During pure bending, a beam is subjected to bending moments that cause it to deform. The top fibers of the beam are compressed, while the bottom fibers are stretched.

This deformation creates internal forces that resist the bending, resulting in compressive and tensile stresses along the longitudinal axis of the beam.

Since there are no external forces acting perpendicular to the longitudinal axis of the beam, there is no shear force (V) present. Shear stress is directly related to shear force, and in the absence of shear force, there will be no longitudinal shearing stress.

Therefore, the absence of lateral shearing stress and shear forces is the key reason why pure bending produces no longitudinal shearing stress. This fundamental understanding of the mechanics of pure bending is essential in structural analysis and design, helping engineers to accurately predict the behavior of beams and ensure their structural integrity.

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The best reason for the fact that pure bending produces no longitudinal shearing stress is that there is no shear force acting on the section, which means that there is no lateral shearing stress in a section due to pure bending.

During pure bending, the top and bottom fibers of a beam are subjected to tensile and compressive stresses, respectively, which results in the formation of a neutral axis where the longitudinal stresses are zero. This neutral axis passes through the centroid of the section, which means that the bending moment causes a uniform distribution of stresses throughout the cross-section.

Since there is no shear force acting on the section during pure bending, there is no lateral shearing stress that is produced. This is because lateral shearing stress is caused by shear forces that act on a section, which result in deformation of the section perpendicular to the direction of the applied force.

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Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water  so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of  x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Answer:

  C. 56.56 VAC

Explanation:

The meaning of "average" is unclear in this context.

If the waveform is sinusoidal, or any other symmetrical shape, its average value is zero.

If the waveform is a full-wave rectified sinusoid, its average value is 2/π times the peak, about 50.93 V.

If you are concerned with the RMS value (not the average), that is 1/√2 times the peak, about 56.57 VAC.

Answer:

  C. the object is moving forward

Explanation:

A positive slope means position is increasing when time is increasing. Generally, increasing position is "moving forward."

Answer:

C. The object is moving forward

Answer:

Get ready and comfortable.

List all of the tasks you need to accomplish over the next week. .

Next schedule everything.

Get a planner/calender.

Cut those tasks that do not fit into your

Answer: There is four elements of Scratch. The stage, the sprites, the script and the programming palette.

Explanation:

Using APA reference style, the references is attached below from Feldman, R. S. (2013) to American Psychology Association (2020)

APA reference style is a style of citing sources used by the American Psychological Association. It is primarily used in the social sciences and includes guidelines for citing sources both within the text of a document and in the reference list at the end of the document. The guidelines provide a consistent way of citing sources to make it easier for readers to identify and locate the sources cited in a document.

In the given problem, if we need to write the reference using APA style, it will take this format below

Reference Page:

Feldman, R. S. (2013). Psychology and your life (2nd ed.). New York, NY: McGraw Hill Companies.

Frontline. (n.d.). League of denial. Retrieved from https://www.pbs.org/wgbh/frontline/film/league-of-denial/

Friends. (2004). The one after Joey and Rachel kiss [Television series episode]. In D. Crane, & M. Kauffman (Executive producers), Friends. Burbank, CA: Warner Bros. Television.

Silver Linings Playbook. (2012). [Motion Picture]. United States: The Weinstein Company.

U.S. Energy Information Administration. (2013). Annual energy outlook 2013 with projections to 2040. Retrieved from http://www.eia.gov/forecasts/aeo/er/pdf/0383er(2013).pdf

Snickers. (2013, April 2). You’re not you when you’re hungry [Video file]. Retrieved from https://youtu.be/2rF_FRCd_LA

American Psychological Association. (2020). Publication manual of the American Psychological Association (7th ed.). doi:10.1037/0000165-000

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Answer:

I think its A

Explanation:

The values of h at the two internal boundaries are :

Given data :

Z₁ = Z₂ = Z₃ = 25 m

h top = 120 m

h bottom = 100 m

K₁ = 0.0001 m/s

K₂ = 0.0005 m/s

K₃ = 0.0010 m/s

we will apply the formula below since flow is perpendicular to the bedding plane

Keq = \(\frac{Z1 + Z2 + Z3 }{\frac{Z1}{K1}+\frac{Z2}{K2} + \frac{Z3}{K3} }\)  ----- ( 1 )

Insert values given above into equation 1

Therefore ; Keq = 2.307 * 10⁻⁴ m/s

Hydraulic gradient ( Ieq ) = head loss / length

                                          = ( 120 - 100 ) / 3 * 25

                                   Ieq  = 0.266

Given that the flow is perpendicular to bedding plane

q1 = q2 = q3

V₁ = V₂ = V₃ = V

K₁i₁ = K₂i₂ = K₃i₃ = Keq * ieq

Hence :

V = Keq* Ieq

   = 2.307 * 10⁻⁴ * 0.266

   = 6.15 * 10⁻⁵ m/s .

Also;

K₁i₁ =  Keq * ieq = K₂i₂ = K₃i₃

therefore :

  i₁ = 0.615

  i₂ =  0.123

  i₃ = 0.0615

Pressure at point 1 ( i.e. pressure between first two formations )

h₁ = h top - i₁L₁

   = 120 - 0.615 * 25

   = 104.625 m

Pressure at point 2 ( i.e. pressure between the 2nd and 3rd formation )

h₂ = h₁ - i₂L₂

    = 104.625 - 0.123 * 25

    = 101.55 m

Therefore we can conclude that The values of h at the two internal boundaries are :  h₁ = 104.625 m , h₂ = 101.55 m

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The solution obtained from OPL is, X = 1, Y = 4, and the optimal value of Z is 10. She estimates that she is going to earn grade points depending on the number of periods spent on each course.

Tyler Galpin, a friend of Wednesday Addams, arrives in town and calls Wednesday Addams for a date. Assessing her situation, Wednesday decides that all she really needs is a total of 16 grade points gained from any of the courses to graduate.

She wants to allocate her time so that she spends the fewest number of study periods necessary to guarantee her receiving at least 16 grade points. We need to formulate this decision problem as an integer programming model and solve using OPL.

So she gets 4 grade points if she spends one period studying an easy course and 2 grade points if she spends one period studying a difficult course. Number of periods studied Grade points from Easy course Difficult course04                 0             0        14                 4             2        24                 4             2        37                 7             4        48                 8             6        58                 8             9        

 So, the given problem can be formulated as follows: Minimize \(Z = x11 + x12 + x13 + x14\)

Subject to\(4 x11 + 4 x12 + 7 x13 + 8 x14 ≥ 16\)

(Easy courses)\(2 x11 + 2 x12 + 4 x13 + 6 x14 ≥ 16\)

(Difficult courses\()Y ≥ 1x11, x12, x13, x14,\)

Y are integers\(xij ≥ 0 (i = 1,2,3,4; j = 1, 2, …, 10)\)

Below is the OPL code:

int easy\([1..4]=[4,4,7,8];\)

int hard\([1..4]=[2,2,4,6];\)

dvar int \(x[1..4][1..10] in 0..10\);

dvar int y in 1..4;

minimize sum\((i in 1..4, j in 1..10)\)

The solution obtained from OPL is, X = 1, Y = 4, and the optimal value of Z is 10.

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To write a program called "NumAverage" that inputs numbers (non-zero positive integers) from a user, averages those numbers, and then displays the result, check the code given below.

A specific type of data made up of characters, numbers, and strings must be processed by programmers in order for the results to be useful, so programming languages are created and designed to assist in this process. The term "programme" refers to a collection of instructions that process data.

.MODEL SMALL

.DATA

       VAL1    DB      ?

       NL1     DB      0AH,0DH,'ENTER HOW MANY NO U WANT:','$'

       NL2     DB      0AH,0DH,'ENTER NO:','$'

       cmp    0AH,Q

       jmp

       NL3     DB      0AH,0DH,'AVEARGE:','$'

.CODE

MAIN    PROC

       MOV AX,(atsymbol)DATA

       MOV DS,AX

       LEA DX,NL1

       MOV AH,09H

       INT 21H

       MOV AH,01H

       INT 21H

       SUB AL,30H

       MOV CL,AL

       MOV BL,AL

       MOV AL,00

       MOV VAL1,AL

LBL1:

       LEA DX,NL2

       MOV AH,09H

       INT 21H

       MOV AH,01H

       INT 21H

       SUB AL,30H

       ADD AL,VAL1

       MOV VAL1,AL

       LOOP LBL1

LBL2:

       LEA DX,NL3

       MOV AH,09H

       INT 21H

       MOV AX,00

       MOV AL,VAL1

       DIV BL

       ADD AX,3030H

       MOV DX,AX

       MOV AH,02H

       INT 21H

       MOV AH,4CH

       INT 21H

MAIN    ENDP

       END     MAIN

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Answer:

Routine

Explanation:

Loop Structures — The Method Of Repeating Routines In Statements. Repetition of code are called loops, and they are defined as statements that execute lines of code (or routines) repeatedly according to conditions or iterations. ... Take for example a routine that must write as output the string “Hello” 40 times

Explanation:

Dora is shopping!!! yayayyayaysyayyayayayayayyayayayayysya

Answer:

Both Tech A and B are correct

Explanation:

One the three method of turning off the MIL is when Vehicle drive on 3 consecutive trips with warm-up cycle without detecting fault. So, with this statement, Tech A is correct.

The MIL will be acrivated when DTC stored in memory and so, the Tech B is also correct.

Answer:

Both A and B

Explanation:

I did the test and got it right

Answer:

a)

normal circular pitch = 7.8539 mm

transverse circular pitch = 9.0689 mm

axial circular pitches = 15.7077

b)

transverse diametral pitch is 0.3464 teeth/mm

transverse pressure angle is 22.8°

c)

Addendum = 2.5 mm

dedendum = 3.125 mm

pinion diameter = 54.8482 mm and Gear diameter = 164.5448 mm

Explanation:

Given that;

module m = 2.5 mm

Number of teeth on Gear  nG = 57 TEETH

Number of teeth on Pinion  nP = 19 TEETH

Helix angle W = 30°

Normal Pressure angle β = 20°

finding the circular pitch

Pc = πm

we substitute

Pc = π * 2.5 mm = 7.8539 mm

now the diametral pitch p = π / Pc

= π / 7.8539

= 0.4 teeth/mm

a)

So the normal circular pitch

Pn = π / P

Pn = π / 0.4

Pn = 7.8539 mm

the transverse circular pitch

Pt = Pn / cosW

Pt = 7.8539 / cos30°

Pt = 9.0689 mm

for axial circular pitches

Px = Pt / tanW

Px = 9.0689 / tan30°

Px = 15.7077

b)

The transverse diametral pitch and the transverse pressure angle.

The transverse diametral pitch Pt = PcosW

= 0.4 * cos30°

= 0.3464 teeth/mm

transverse diametral pitch is 0.3464 teeth/mm

transverse pressure angle  β1 = tan^-1 ( tan βn / cos W)

=  tan^-1 ( tan20° / cos 30°)

= tan^-1 ( 0.42027 )

β1 = 22.8°

transverse pressure angle is 22.8°

c)

The addendum, dedendum, and pitch diameter of each gear

Now from table standard Tooth proportions for Helical Gears;

Addendum a = 1/p

= 1 / 0.4

= 2.5 mm

dedendum b = 1.25 / p

= 1.25 / 0.4

= 3.125 mm

now  pinion diameter dP = Np / PcosW

= 19 / 0.4 (cos30°)

= 54.8482 mm

Gear diameter dG = nG / pcosW

= 57 / 0.4 (cos30°)

= 164.5448 mm

Answer:

class TriangleNumbers

{

public static void main (String[] args)

{

 for (int number = 1; number <= 10; ++number) {

  int sum = 1;

  System.out.print("1");

  for (int summed = 2; summed <= number; ++summed) {

   sum += summed;

   System.out.print(" + " + Integer.toString(summed));

  }

  System.out.print(" = " + Integer.toString(sum) + '\n');

 }

}

}

Explanation:

We need to run the code for each of the 10 lines. Each time we sum  numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.

\(f = force\)

\(m = mass \: of \: an \: object\)

\(a = acceleration\)

Newton's second law is one of the most important in all of physics. For a body whose mass m is constant, it can be written in the form F = ma, where F (force) and a (acceleration) are both vector quantities. If a body has a net force acting on it, it is accelerated in accordance with the equation

An engineer troubleshooting a network connection problem identifies a port flapping problem.

The engineer should investigate the following possible causes:

Therefore, the engineer should investigate all of the above options as possible causes of the port flapping problem.

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Health and safety standards in the wood processing industry are crucial for the safety and protection of workers in the industry. They must follow strict procedures to ensure that they are not exposed to harmful elements, and that their health and well-being are not compromised.

Following are the two points related to the importance of Health and Safety Standards in the wood processing industry.

In conclusion, health and safety standards in the wood processing industry are essential for the protection of workers and the success of companies in the industry. By prioritizing employee safety and compliance with regulations, employers can minimize the risks associated with wood processing and maintain a safe and healthy work environment.

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Answer:

Use GitHub or stackoverflow for this answer

Explanation:

It helps with programming a lot

This program prompts the user to input 10 test grades using dialog boxes. It then stores the grades in an array. The `displayGrades` method prints the grades in the console, while the `calculateAverage`, `findHighestGrade`, and `findLowestGrade` methods perform the corresponding calculations and display the results.

Here is a Java program that allows users to input test grades, calculates the average, highest, and lowest grade, and displays the results in the console:

```java

import javax.swing.JOptionPane;

public class TestScoresProgram {

   public static void main(String[] args) {

       int[] testGrades = new int[10];

       inputGrades(testGrades);

       displayGrades(testGrades);

       calculateAverage(testGrades);

       findHighestGrade(testGrades);

       findLowestGrade(testGrades);

   }

   public static void inputGrades(int[] grades) {

       for (int i = 0; i < grades.length; i++) {

           String input = JOptionPane.showInputDialog("Enter test grade #" + (i + 1));

           grades[i] = Integer.parseInt(input);

       }

   }

   public static void displayGrades(int[] grades) {

       System.out.print("These are all the test grades: ");

       for (int i = 0; i < grades.length; i++) {

           System.out.print(grades[i] + " ");

       }

       System.out.println();

   }

   public static void calculateAverage(int[] grades) {

       double sum = 0;

       for (int i = 0; i < grades.length; i++) {

           sum += grades[i];

       }

       double average = sum / grades.length;

       System.out.println("The total average of grades: " + average);

   }

   public static void findHighestGrade(int[] grades) {

       int highest = grades[0];

       for (int i = 1; i < grades.length; i++) {

           if (grades[i] > highest) {

               highest = grades[i];

           }

       }

       System.out.println("The Highest-Grade Number is: " + highest);

   }

   public static void findLowestGrade(int[] grades) {

       int lowest = grades[0];

       for (int i = 1; i < grades.length; i++) {

           if (grades[i] < lowest) {

               lowest = grades[i];

           }

       }

       System.out.println("The Lowest-Grade Number is: " + lowest);

   }

}

```

This program prompts the user to input 10 test grades using dialog boxes. It then stores the grades in an array. The `displayGrades` method prints the grades in the console, while the `calculateAverage`, `findHighestGrade`, and `findLowestGrade` methods perform the corresponding calculations and display the results.

Note that this program uses the `JOptionPane` class from the `javax.swing` package to display dialog boxes for user input.

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