Data transmission from node to a gateway

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Syed Fahad Hassan
Syed Fahad Hassan 2017 年 11 月 30 日
コメント済み: Walter Roberson 2020 年 12 月 31 日
Hi,
I am creating a wireless sensor network in Matlab in which nodes transmit information to the central gateway which then processes this information for analysis. I want to know
  1. How can I code this transmission in which a node transmits some amount of data to the gateway (say 1000kB at 250kBps) and gateway receives that transmission?
  2. The transmission should also be timed such that the delay experienced by rest of the nodes while in queue is calculated.
Thanks in advance for any suggestions and help.

回答 (1 件)

Deepanshu Udhwani
Deepanshu Udhwani 2017 年 11 月 30 日
編集済み: Walter Roberson 2017 年 11 月 30 日
e:\Bisection.txt
************************************************************************
clc;
clear all;
a=input('Initial point ');
b=input('Final Point ');
c=(a+b)/2;
f=@(x) x^3+4*(x^2)-10;
E=0.001;
ceil(n=(log(b-a)-log(E))/log(2))
while(abs(a-b)>E)
if f(a)*f(c)<0
b=c;
else
a=c;
end;
c=(a+b)/2;
end;
c
e:\Fixed_Point_Iteration.txt
************************************************************************
clc;
clear all;
a=input("a");
b=input('b');
%x0=(a+b)/2;
x0=1;
E=0.001;
N=ceil((log(b-a)-log(E))/(log(2)));
i=1;
g=@(x) x-((2*sin(3.14*x)+x)/(1+(2*3.14*cos(3.14*x)))); %Newton Method to find g(x);
while i<=N
xx=g(x0);
if(abs(xx-x0)<E)
break;
end;
x0=xx;i=i+1;
end;
xx
e:\Guass_Elimination.txt
************************************************************************
clc;
clear all;
n=input('N: ');
for i=1:n
for j=1:n+1
A(i,j)=input('Enter element');
end;
end;
for i=1:n-1
for j=i+1:n
m=A(j,i)/A(i,i);
for k=1:n+1
A(j,k)=A(j,k)-m*A(i,k);
end;
end;
end;
X(n)=A(n,n+1)/A(n,n);
for i=(n-1):-1:1
sum=0;
for j=i+1:n
sum=sum+(A(i,j)*X(j));
end;
X(i)=(A(i,n+1)-sum)/A(i,i);
end;
A
X
e:\GuassSeidal.txt
************************************************************************
clc;
clear all;
n=input('N ');
E=0.0001;
for i=1:n
for j=1:n+1
A(i,j)=input('Enter element');
end;
end;
for i=1:n
X(i)=0;
Xold(i)=0;
end;
sum=0;
normval=2;
while abs(normval)>E
Xold=X;
for i=1:n
for j=1:n
if j!=i
sum=sum+A(i,j)*X(j);
end;
end;
X(i)=(A(i,n+1)-sum)/A(i,i);
sum=0;
end;
normval=norm(Xold-X,inf);
end;
X
e:\Lagrange_Interpolation.txt
************************************************************************
% Q3
clc;
clear all;
n=input('N: ');
f=@(x) exp(2*x)*cos(3*x);
for i=1:n
X(i)=input('Enter X: ');
%Y(i)=input('Enter Y: ');
Y(i)=f(X(i));
l(i)=1;
end;
p=input('Enter the value of X where value is to be calculated');
for i=1:n
for j=1:n
if i!=j
l(i)=((p-X(j))/(X(i)-X(j)))*l(i);
end;
end;
end;
sum=0;
for i=1:n
sum=sum+l(i)*Y(i);
end;
sum
e:\Newton.txt
************************************************************************
clc;
clear all;
a=input('a ');
E=0.0001;
f=@(x) cos(x)-(x*exp(x));
g=@(x) -sin(x)-(x*exp(x)+exp(x));
while(true)
b=a-(f(a)/g(a));
if(abs(b-a)<=E)
break;
end;
a=b;
end;
b
e:\Newton_Divided_Difference_Interpolation.txt
************************************************************************
clc;
clear all;
n=input('N ');
p=input('p ');
for i=1:n
X(i)=input('X');
Y(i)=input('Y');
end;
for i=1:n
F(1,i)=Y(i);
end;
for i=2:n
for j=i:n
F(i,j)=(F(i-1,j)-F(i-1,j-1))/(X(j)-X(j-i+1));
end;
end;
for i=1:n
product(i)=1;
for j=1:(i-1)
product(i)=product(i)*(p-X(j));
end;
end;
F=transpose(F);
sum=0;
for i=1:n
sum=sum+F(i,i)*product(i);
end;
sum
e:\Power_Method.txt
************************************************************************
clc;
clear all;
n=input('N: ');
E=0.001;
for i=1:n
for j=1:n
A(i,j)=input('Enter Element');
end;
end;
for i=1:n
X(i)=input('Element');
end;
X=transpose(X);
Y=A*X;
K=norm(Y,inf);
Kdash=100;
X=(1/K)*Y;
while abs(K-Kdash) >E
Y=A*X;
Kdash=K;
K=norm(Y,inf);
X=(1/K)*Y;
end;
K
X
e:\RungeKutta.txt
************************************************************************
clc;
clear all;
a=input('a ');
b=input('b ');
h=input('h ');
n=(b-a)/h;
f=@(t,y) -y+(2*cos(t));
t0=0;y0=1;
for i=1:n
k1=h*f(t0,y0);
k2=h*f((t0+(h/2)),(y0+(k1/2)));
k3=h*(f((t0+(h/2)),(y0+(k2/2))));
k4=h*(f(t0+h,y0+k3));
y1=y0+(1/6)*(k1+2*k2+2*k3+k4);
y0=y1;
t0=t0+h;
end;
y1
e:\Secant.txt
************************************************************************
clc;
clear all;
a=input('a ');
b=input('b ');
c=0;
E=0.0001;
f=@(x) x^2-17;
while(abs(b-a)>E)
c=b-((b-a)/(f(b)-f(a)))*f(b);
if(f(c)==0)
break;
end;
a=b;
b=c;
end;
b
e:\Simpson.txt
************************************************************************
clc;
clear all;
a=input('a ');
b=input('b ');
N=input('N ');
h=(b-a)/N;
f=@(x) 1/(x*log(x));
sum=0;
for i=1:N-1
x=a+(i*h);
if i%2==0
sum=sum+(2*f(x));
else
sum=sum+(4*f(x));
end;
sum=(sum+f(a)+f(b));
end;
sum*(h/3)
e:\Trepozoidal.txt
************************************************************************
clc;
clear all;
a=input('a ');
b=input('b ');
N=input('N ');
h=(b-a)/N;
f=@(x) cos(x)*cos(x);
sum=0;
I=0;
for i=1:N-1
x=a+(i*h);
sum=sum+(2*f(x));
% Sum starts as Sigma i=1 to N-1 because f(a) and f(b) are adjusted later
end;
sum=sum+(f(a)+f(b));
sum*(h/2)
% I= h/2 * ( f(a) +f(b) ) + h*( sigma i=1 to N-1 f(a+h*i) )
  4 件のコメント
M fa M fa
M fa M fa 2020 年 12 月 28 日
please code for newton central difference interpolation
Walter Roberson
Walter Roberson 2020 年 12 月 31 日
newton central difference has nothing to do with data transmission between nodes; you should be asking in your own Question. http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

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