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VashaNatasha [74]
3 years ago
10

What are the desired characteristics or values for the following parameters of an ideal amplifier?o Phase change as a function o

f the log of the frequency o Common Mode Rejection Ratioo Input Resistanceo Output Resistance
Engineering
1 answer:
kvv77 [185]3 years ago
4 0

Answer and Explanation:

- Phase change as a function of the log of the frequency

The phase shift should be independent of the frequency of the input signal. It should have an Infinite bandwidth with zero phase shift

- Common Mode Rejection Ratio

An ideal amplifier has Infinite common-mode rejection ratio (CMRR). A perfect operational amplifier amplifies only the voltage difference between its two inputs, completely rejecting all voltages that are common to both.

- Input Resistance

Infinite input impedance and so, zero input current & Zero input offset voltage.

The differential input impedance of the operational amplifier is defined as the impedance between its two inputs; the common-mode input impedance is the impedance from each input to ground.

And the impedance between its inputs should be infinite.

- Output Resistance

Zero output impedance and Infinite output voltage range.

Output Resistance leads to voltage drop. Thee voltage drop across the output impedance effectively reduces the open loop gain. So, the optimal value for this should be zero.

N.B - For AC circuits such as this one being discussed, resistance is known as impedance.

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The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt
Step2247 [10]

Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

Gs = 759.99W/m² ------- Approximated

3 0
3 years ago
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m
zvonat [6]

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3,  c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient  = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

\frac{T_{0}-T_{x}  }{T_{1}-T_{x}  } = C_{1} e^{(-0.4888^{2}*Fo )}

= 0.4167 = 1.0396e^{-0.4888*Fo}

therefore Fo =  3.8264

Now determine how long the plate should be left in the furnace

Fo = (\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )

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p = 7830

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8 0
3 years ago
Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of
rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

                     if ( values[month] > largest ){

                 largest = values[month];

return largest;

          }

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

                     if ( values[month] < smallest ){

                 smallest = values[month];

return smallest;

          }

 

int main()

{

double rainFall[NUM_MONTHS];

 for (int month = 0; month < NUM_MONTHS; month++)

  {

     cout << "Enter the rainfall (in inches) for month #";

     cout << (month + 1) << ": ";

     cin >> rainFall[month];

 

     while (rainFall[month] < 0)

     {

      cout << "Rainfall must be 0 or more.\n"

             << "Please re-enter: ";

      cin >> rainFall[month];

     }

  }

  cout << fixed << showpoint << setprecision(2) << endl;

  cout << "The total rainfall for the year is ";

  cout << getTotal(rainFall, NUM_MONTHS)

      << " inches." << endl;

   cout << "The average rainfall for the year is ";

  cout << getAverage(rainFall, NUM_MONTHS)

      << " inches." << endl;

   int subScript;

cout << "The largest amount of rainfall was ";

  cout << getLargest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl;

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      << " inches in month ";

  cout << (subScript + 1) << "." << endl << endl;

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}

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3 years ago
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Answer:

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Explanation:

6 0
3 years ago
Initialize the tuple team_names with the strings 'Rockets', 'Raptors', 'Warriors', and 'Celtics' (The top-4 2018 NBA teams at th
Drupady [299]

Answer:

#Initialise a tuple

team_names = ('Rockets','Raptors','Warriors','Celtics')

print(team_names[0])

print(team_names[1])

print(team_names[2])

print(team_names[3])

Explanation:

The Python code illustrates or printed out the tuple team names at the end of a season.

The code displayed is a function that will display these teams as an output from the program.

4 0
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