Answer:
The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then
and considering that 
then
For ramp disturbance d(t)=at
therefore, the steady state error is given by
![e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]](https://tex.z-dn.net/?f=e%28%5Cinfty%29%3D%20%5Clim_%7Bs%20%5Cto%200%7D%20s%20%5B%5Cfrac%20%7BD%28s%29%7D%7B1%2BG_c%28s%29G%28s%29%7D%5D)
![e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]](https://tex.z-dn.net/?f=e%28%5Cinfty%29%3D%20%5Clim_%7Bs%20%5Cto%200%7D%20s%20%5B%5Cfrac%20%7Ba%7D%7Bs%5E%7B2%7D%2Bs%5E%7B2%7DG_c%28s%29G%28s%29%7D%5D)
![e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]](https://tex.z-dn.net/?f=e%28%5Cinfty%29%3D%20%5Clim_%7Bs%20%5Cto%200%7D%20s%20%5B%5Cfrac%20%7Ba%7D%7Bs%2BsG_c%28s%29G%28s%29%7D%5D)
![e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]](https://tex.z-dn.net/?f=e%28%5Cinfty%29%3D%20%5Clim_%7Bs%20%5Cto%200%7D%20s%20%5B%5Cfrac%20%7Ba%7D%7BsG_c%28s%29G%28s%29%7D%5D)
Whenever 
has a double intergrator, the error 
becomes zero
Answer:
#include <iostream>//including iostream library to use functions such as cout and cin
using namespace std;
int main() {
int userInput = 0;
do
{
cout << "Enter a number < 100: " ;
cin >> userInput;
if (userInput < 100)//condition if number is less than 100
{
cout << "Your number < 100 is: " << userInput << endl;
}
} while (userInput > 100);//do while loop condition
return 0;
}
Explanation:
A do-while loop executes regardless in the first iteration. Once it has run through the first iteration, it checks if the condition is being met. If, the condition is TRUE, the loop begins the second iteration. If FALSE, the loop exits. In this case, the condition is the userInput. after the first iteration, lets say the userInput is 120, the condition userInput > 100 is true.Therefore, the loop will run again until eventually the number is less than hundred, lets say 25. In that case the condition would be 25 > 100, which would be false, so the loops will break.
Answer:
factor of safety = 0.8
Explanation:
given data
maximum gauge pressure in steel p = 10 MPa
outer diameter do = 200 mm
wall thickness t = 2 mm
ultimate stress of the steel σU = 400 MPa
solution
We check the type of shell that is
so we can see that that is 
so that it mean vessel is a thin shell now we can we get maximum pressure by the maximum tensile stress is
σ = 
.......................1
put here value and we get
σ = 
σ = 500 MPa
so factor of safety will be express as
factor of safety = 
..........2
factor of safety = 
factor of safety = 0.8
We are given a CSP with only binary can concentrate assume we run backtrackingSearch with ARC
