Hello everyone,

If the average value of a square wave is just the same as its peak vaue, then, how about if you just had a half cycle of the square wave?

I mean if, say, you had to work out the average value of the positive half of the square wave cycle?

I've tried to work this out myself logically, but I can't quite grasp what the answer would be and would appreciate any advice

 

The average value of a square wave is related to the duty cycle. I have in mind a square wave that has 0 volts as one extreme, and, say, 10 volts as the peak. At a 50% duty cycle, the average voltage is 50% of peak, or 5 volts.

If your 1/2 cycle was simply the positive excursion, the the average is the same as the peak.

 

Hello Been There,

Thanks very much for your quick reply.

The square wave that you described, with 0 volts at one extreme and 10 volts at the other extreme, would be all Direct Current wouldn't it? Is that right?

The wave that I am looking at, which I printed off the A.About.Circuits textbook, (the part that tells you how to work out the rms and average value of sine, square & triangular waves) looks like an alternating square wave to me, as there's a horizontal line going through it in the middle.

So, if it's an alternating square wave, would it be the same answer?

I am pleased to learn new terms like "duty cycle" (does that mean a whole cycle or what? Or does it mean the time during which a wave has an effect?) and "excursion".

I would guess, as a relative beginner, that you would probably only usually need to know the average value of the whole cycle of the square wave anyway, and that there would probably rarely be any need to work out what the average of 1/2 a cycle of the square wave would be? Is that right?

I apologize if my beginner's confusion is confusing and difficult to understand, so to speak!

 

Please visit the following internet web page. It explains how to calculate the average value of a periodic square wave. I hope it assists you in clarifying your question. Have a great day!

http://campus.murraystate.edu/tsm/tsm118/Ch6/Ch6_8/ch6_8.htm

 

hello Martinez,

Thanks very much for your help. The web site you recommended is very enjoyable to read (I've read about 1/3 of it so far, I started 10 minutes ago) and extremely interesting. I love it when people explain difficult complicated things to me and I can ACTUALLY UNDERSTAND IT! That is so rewarding.

best wishes

milly molly mandy

 

Hello Ed,
Thanks for your help, also for the information about RMS, which I am also trying to learn as thoroughly as I can. As I am doing all my learning independently at home (no teacher to ask!) it’s useful when people on this forum give you tips and titbits of info “by word of mouth”, so to speak.
May I ask then, if the average of a square wave with a 50 per cent duty cycle would be 0, then what would be the average of the full cycle of a square wave of -5V to + 5V? Would it just be 5V?

 

The average value of any function of time is just the integral from 0 to t of V(t)dt. Now if V(t) is a constant, then the result is V*t which is only equal to V if t = 1 unit of time. By extension, if one period of a square wave is one unit of time then the average over a half cycle is 0.5*V

 

One half cycle of a squarewave is simply a DC voltage level (either positive or negative, depending on which half). With DC, peak = RMS = voltage.

For a full cycle squarewave, +5V and -5V, the average is zero, as there are two 5VDC sources cancelling each other out. A Fully rectified perfect squarewave would simply be 5VDC (ignoring any voltage drop from rectifier, instant switching diodes, and 0 risetime on squarewave).