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.
Answer:
#include<iostream>
int main() {
float num_1, num_2, num_3, num_4, average;
//Taking input for four numbers
std::cout << "Enter first number(integer or floating-point)";
std::cin >> num_1;
std::cout << "Enter second number(integer or floating-point)";
std::cin >> num_2;
std::cout << "Enter third number(integer or floating-point)";
std::cin >> num_3;
std::cout << "Enter fourth number(integer or floating-point)";
std::cin >> num_4;
average= (num_1+num_2+num_3)/3;
// Comparing average with fourth number
if (average==num_4)
{
std::cout << "Equal";
}
return 0;
}
Answer:
the required time for the specimen is 1109.4 min
Explanation:
Given that;
diameter of metal alloy d₀ = 1.7 × 10² mm
Temperature of heat treatment T = 450°C = 450 + 273 = 723 K
Time period of heat treatment t = 250 min
Increased grain diameter 4.5 × 10² mm
grain diameter exponent n = 2.1
First we calculate the time independent constant K
dⁿ - d₀ⁿ = Kt
K = (dⁿ - d₀ⁿ) / t
we substitute
K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250
K = (373032.163378 - 48299.511117) / 250
K = 1298.9306 mm²/min
Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )
dⁿ - d₀ⁿ = Kt
t = (dⁿ - d₀ⁿ) / K
t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306
t = ( 1489328.26061158 - 48299.511117) / 1298.9306
t = 1441028.74949458 / 1298.9306
t = 1109.4 min
Therefore, the required time for the specimen is 1109.4 min
