sunshine102030

Thanks , it was useful for me

Use cdma to construct the aggregate frame for two denders s1 frame 0 , s2 frame 1, the code for s1 is (+1,-1,+1,+1,+1,-1,+1,+1), the code for s2 is (-1,+1,-1,-1,-1,+1,-1,-1) I tried to solve it but I am not sure. I multiplied s1 code by -1 and s2 code by +1 …

If we have the following message bits > 11010001010001001010 we split this meassage into blocks of 5 bits, then we use two dimensional matrix for error correction to give the sent frame But what if we have a message with odd bits , for example : > 1010010 To how …

> I know that the max value for short data type is 32767 and d=2*a=2*(20000)=40000 that is greater than 32767 > therefore the result of d is in minus d= -25536 while a,b and c displayed as they are scince they are< 32767 ,but the question is from where we …

please can you help to get this answer :  I have a problem in the result of kilometer formula #include<iostream> #include<fstream> #include<iomanip> #include<string> using namespace std; int main() { ifstream infile; ofstream outfile; string outfilename; string filename; string Kilo; string dayfees; double f1, f2, f3, Kilometers, cost_km, …

please , can you help to find the formulas accourding to the results in the pictures ?  The first formula is based on number of kilometers you plan to go. The expense for this formula is computed as follows: - For the first 500 kms, the cost of …

#include <iostream> #include <cmath> #include <iomanip> using namespace std; const float pi=3.14; const float e=2.71828; int main() int r,h; double A,E1,E2; cout<< "please enter two positive integers for r and h: "<<endl; cin>>r>>h; A=(2*pi*r*r+2*r*r*h); E1=(pow(e,h))+(3*(log10(A))); E2=sqrt(abs(cos(pow(r,2.0)+sin(pow(r,2.0))); cout<<"Radius Height Area E1 E2"<<endl; cout<<fixed<<setprecision(2)<<showpoint; cout<<"====== ====== ==== == =="<<endl; cout<<setw(6)<<r<<setw(12)<<h<<setw(10)<<A<<setw(8)<<E1<<setw(8)<<E2<<endl; return 0; …