how to compute average power of the generated sinusoidal signal?

lim1 = 0;

lim2 = 2*pi;

t = lim1:0.01:lim2;

A = 1.5;

f = 0.1;

omega= 2*pi*f;

phase= 0;

x = A*cos(omega*t+phase);

plot(t,x);

the following is my signal graph ? hohw should i compute the average power of the signal? how do i write out the formula?? for the average power signal

Px= limN→∞(1/2N+1) ∑|x(n)|^2

Answer 1

[Abraham Boayue]

Try this code, just an extension of your code.

clear variables
close all;
lim1 = -10;% no need to have t in pi
           % units since omega is 
           % already measured in pi.
lim2 = 10;

M = 500;   % length of 
delta = (lim2-lim1)/(M-1);
t = lim1:delta:lim2;
A = 1.5;
f = 0.1;
omega= 2*pi*f;
phase= 0;
x = A*cos(omega*t+phase);

N = 10; %  let say you want to average 
        %  10 terms of x
px = zeros(1,M);
for n = 1:N
    px = px + (abs(A*cos(n*omega*t+phase))).^2;
end
px = (1/(2*N+1))*px;
%disp(px)

     figure
      plot(t,x,'linewidth',2);
      hold on
      plot(t,px,'linewidth',2)
      a = ylabel('Signal');
      set(a,'Fontsize',14);
      a = xlabel('x');
      set(a,'Fontsize',14);
      a=title('x(t) and its average px');
      legend('x(t)','px')
      set(a,'Fontsize',16);
      grid;

[Snow sin]

Hi, thanks for the code, i tried it and have some question... Why your px values have 2 answer? shouldn't the average power value is a single value??

Answer 2

[Abraham Boayue]

Are you saying that px = 1/(2×N+1)[(Acos(wt))^2 + (Acos (2wt))^2 +... (Acos(N*wt))^2] should equal a single number? How possible is that? No way! According to the formula that you provided, you are simply summing a series of cosine squares and then dividing the result by 1/(2*N+1). You are actually performing an ensamble averaging on a time varying function and the result show be a time varying function. If you choose N large enough, you should end up with a delta function, a single value function that occurs at the peak of the cosine wave.