Question

Use c++ language to solve the following question

In the following transfer function, find the value of s that makes the function goes to infinity
G(s)= (S+3)/(s^4+9s^3+24s^2+20s)
Using modified Newton Raphson Method (developed by Ralston-Rabinowitz).

Answer 1

the function ques to infinity when the denominator tends to O. Hence We need to find the roots of the denominator function find code to roots. #include < bits/st de + th> #include EPSILON 0001 using namespace std; 11 The function is a 4 + 9x^3 + 24x^2 + 20 x double fun (double x) Ş return fow (2,4) + q* pow (2, 3) + 24*pow (X12) +20*x; 3 11 Derivativate of the above function which is 4x^3 + 27 2^2 +48 x + 20 double derivEuno (double x) 3 return 4* pow (2, 3) + 27 pow (X, 2) + 48*x + 20; ។ " Double derivative which is 12x^2 +54x +48 double doub deriv Fune (double x) { return 12 pow (812) + 54*x +48; 2 11 Function to find the root Void mod Newton Raphson (doublex) double h (func (X) * derive func(X) / ((pow (deriv Func (X), 2)) - (fund (x) & doub veriv Eune (2))); while (abs (h) >= EPSILON) Yh =(func(x) deriv fune (x))/ CC pow (deriifune (X), 2)) - (fune (x) + doubpriv Fune (x))) 1x(i+1)=2(2)-f(x) = f'(x) / Cf (x))^2 = f(x) = f(x) x=2-4 ?

cout << "The value of the root is 1LLX; Il Driver porogram to test above int main() { double no = -8:// initial value of root assumed modNewtonRaphson (x0); returno; 4 11 The other roots en found by changing the initial value of assumed root

Screenshot of code

#include<bits/stdc++.h> #define EPSILON 0.001 using namespace std; //The function is x^4 + 9x^3 + 24x^2 + 20x double func(double x) { return pow(x,4) + 9*pow(x,3) + 24*pow(x, 2) + 20*X; } // Derivative of the above function which is 4x^3 + 27x^2 + 48x + 20 double derivFunc( double x) { return 4*pow(x,3) + 27*pow(x, 2) + 48*x + 20; } //Double derivative which is 12x^2 + 54x + 48 double doubDerivFunc(double x){ return 12*pow(x, 2) + 54 x + 48; } 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 49 41 42 // Function to find the root void modNewton Raphson(double x) { double h = (func(x) derivFunc(x))/((pow(derivFunc(x),2)) (func(x) doubDerivFunc(x))); while (abs (h) >= EPSILON) { h = (func(x) derivFunc(x))/((pow(derivFunc(x),2)) (func(x) doubDerivFunc(x))); // x(i+1) = x(i) - f(x).f'(x) / (f'(x))^2 - f(x).f''(x) x = x - h; } cout << "The value of the root is : << x; } // Driver program to test above int main() { double x0 = -8; // Initial value of root assumed modNewtonRaphson (x0); return 0; 44

Output:

The value of the root is : -2Program ended with exit code: 0

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