The power of a signal is something different from the level of the signal. I'm not sure how to give a simple explanation of power, so here are a few key points:

• Power is not a linear function of the signal; when you double \, you don't double the power of \ - you quadruple it.
• Power does not depend on the polarity of the signal. A negative signal has the same amount of power as a positive signal.

There are two ways to talk about power:

• The instantaneous power of a signal is \. This is the power at time \ (ie: right now) and it doesn't depend on what happens to the signal before or after this moment. Notice that this formula fits both of the points from above.
• The average power of a signal is the average of the instantaneous power - if your signal has a power of \ half of the time and \ the other half, then the average power is \.

If you remember that the average of \ points is

then you can see that your formula is a calculation of the average power of \.

Use the buffer function to segment the signal into 1-second increments. Since you want 1-second samples, the vector length is simply ‘Fs’, so — y = randn(73113,1); % Create Signal To Test Code (One Channel) fs = 8192; % Sampling Frequency (Also Frame Length) N = size(y,1); % Signal Length time = linspace(0, N, N-1)/fs; % Time Vector y_b = bandpass(y,[300 2000],fs); % Filter Signal y_seg = buffer(y_b,fs); % Matrix Of Signal Segments y_pwr = sum(y_seg.^2); % Calculate Power figure for k = 1:numel(y_pwr)-1 tv = time((1:fs)+fs*(k-1)); subplot(5,2,k) plot(tv,y_seg(:,k)) grid title(sprintf('Segment %d Power %.1f',k,y_pwr(k))) end subplot(5,2,k+1) tv = time(fs*k:end); plot(tv,y_seg(1:numel(tv),k+1)) grid title(sprintf('Segment %d Power %.1f',k+1,y_pwr(k+1))) EDIT — (28 Apr 2021 at 12:40) Corrected typoographical error.

I suggest that instead of using f = str2func(['@(x)' vectorize(f)]); that you use f = matlabFunction(f); However, you could just comment out all of your handling of f, as you never use f after you make it into a function handle. You have z(t)=integral(y,-t,t); At this point your y is a symbolic variable. integral() cannot be applied to symbolic variables: you would need to use int(y, -t, t) -- which is a value you can easily predict will be 0, since the integral of y with respect to y over y = a to y = b is 1/2 b^2 - 1/2 a^2 and with a = -t and b = -t that is going to be 1/2 t^2 - 1/2 (-t)^2 which is going to be 0.

A signal either has finite energy, finite power or even infinite power. If it has finite energy, it will have zero average power, according to your definition $$P_x=\lim_{T_0\rightarrow\infty}\frac{1}{T_0}\int_{-\frac{T_0}{2}}^{\frac{T_0}{2}}|x(t)|^2dt.$$

Knowing that the sincs are orthogonal to each other, as well as the cos function, the average power is given by

$$P_x=P(2)+P(2\cos(2\pi f_0t))+P(\operatorname{sinc})$$

where $P(...)$ is the power of the components in brackets. The sinc function has finite energy, hence it has zero power on average. Hence, we have

$$P_x=P(2)+P(2\cos(2\pi f_0 t)$$ with $$P(2)=4$$ and $$P(2\cos(2\pi f_0 t)=2^2\cdot\frac{1}{2}=2.$$

Therefore, the overall power is given by

$$P_x=6.$$

For the calculation of the $\cos$, consider $$P_{\cos}=\frac{1}{T}\int_{0}^{T}\cos^2(2\pi t/T)dt=\frac{1}{2}$$

If the signal $r(t)$ is of finite length (and if integral of its square during its definition interval is also finite) then its average power given by $$ \bar{P_x} = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |r(t)|^2 dt = \lim_{T \to \infty} \frac{1}{T} \int_{0}^{L} |r(t)|^2 dt = \lim_{T \to \infty} \frac{[\text{constant}]}{T} = 0$$ will be zero...

Btw, you may compute an instantaneous power of the signal, simply as the square of it $|r(t)|^2$...