Calculating RMS Value of a 2V Amplitude, 2s Period, 50% Duty Cycle Square Wave

Question as above. Amplitude Um=2V, period T=2s, duty cycle 50%.

total from ) to T UM/T dt == 1V

Are you sure? The RMS formula for any periodic waveform looks a little different.

Hello
Colleague Filip calculated well ... but the average value of the voltage.
From the definition of the rms value, this voltage (a rectangle with a duty cycle of 50%) must give off the same amount of heat as a DC voltage in the same period (I messed up a bit, but I'll give you a picture below that will explain everything). So P=U*I, I=U/R that gives P=U^2/R, and the heat generated is energy, so we multiply the power by time, which in our case is 2 seconds. It was for direct current.

Now we calculate for a rectangle with a filling of 50%, i.e. for one second we can assume the waveform as a constant voltage of 2 V.
The thermal energy generated in both cases must be equal, so (U^2/R)*2seconds = (u^2/R)*1second (the second second is zero, so we don't take it into account). Now simple mathematical relations and we get the formula U=u/2^0.5, which means that the rms value of a square wave with a duty cycle of 50% and an amplitude of 2 V is 1.41 V.

or

U=((1/T)*integral from 0 to T zu^2 after dt)^0.5 (for more advanced).

We associate the average value of the current with the amount of flowing charge, i.e. with the surface area in a given period.
We associate the effective value with the heat generated in a given period.

Goodbye (it's nice to go back to school)

In my maths terms, the rms value is the average of the absolute value of the function** (I change all negative values to positive ones), i.e. in this case in the U[V]-t[s] axis system:

from t=0 to t=1s: U is constant and equals 2V (U=Um=2V),
from t=1s to t=2s: U=0, ***

change all negative values to positive ones (remains unchanged)

and now I calculate the average normally, so in this case the average will be 1V, because for the first second it was 2V, and for the second: 0V.

** this def. is equivalent to the formula given by Aleksander_01 (squared first, then square root - positive values remain);
*** I'm not sure if this is how the waveform looks like, but I'm giving the principle (isn't it about alternating current?);
-

And I always thought that the RMS value of a symmetrical square wave with 50% duty cycle equals its amplitude, which is 2V

Hello
Jony I read your previous posts and you seem like a sensible guy, and here is this post. You probably haven't thought about what you're writing, after all, it's not about a symmetrical waveform (then, yes, the rms value is equal to the max value, i.e. in our case 2 V). Look carefully at your attachment (last item - square wave).
I'd like to believe that you just made a mistake, which happens to everyone.
Your previous posts prove that you have proper knowledge (and here's a gaffe).
Regards

Yes Aleksander_01 you are totally right my mistake sorry. The effective value is 1.41

I didn't express myself very clearly, but my friend Aleksander_01 understood what the mileage was (unsymmetrical). Before I wrote this post, I counted from the formula (integral over the period under the root), which was also given by my colleague Aleksander_01, but somehow I get a different result. Could you show what and how?

I figured if it was symmetrical it wouldn't be a problem.
You know the formula, then we substitute: for T we substitute 2 because the period lasts 2 seconds, we write the integral from zero to T / 2 (that is one second, we only care about this part of the waveform), then it is u^2 so for u we put 2 because the amplitude is 2 V. And we count. And it comes out 2^0.5.

You can create the pattern yourself, you just need to know the principle that I described in the previous post. In your case, it will be: PT=pT, ((U^2)/R)T=((u^2)/R)T/2 and from this equation comes this integral.
Regards

Thanks a lot buddy Aleksander_01 . I'm not very good at math and this is where it showed itself...

In sum:

Usk=Root(dT/T)*Umax where dT/T is the duty cycle, Umax is the amplitude