RMS and amplitude of a pulse wave

Thread Starter

Henry4321

Joined May 6, 2024
9
I am stuck at this issue with square waves.
The signal illustrated on the picture is measured:
- 4,0 V for a digital multimeter set to DC mode -- (so DC offset, I presume)
- 8,0 V for the TRMS
I must find the Vpeak.
Now, I am quite lost. With square waves, you normally expect Vpeak to be equal = RMS/sqrt(2) for the pulse duty factor 50%, but it's clearly not 50% here. Also, the only values given are DC offset and the TRMS (AC + DC offset). The answer given is 16 Volts, but I don't know where it could come from. Am I to understand the peak value is twice as high as TRMS? Then how do I explain it? and why the DC offset is given then?
Please help.
Thank you

pulse-signal.png
 

WBahn

Joined Mar 31, 2012
30,303
Don't try to rely on formulas that only apply to special cases, or try to infer general conclusions from this special case.

In terms of a voltage waveform, v(t), how do you calculate the DC and the RMS values of that waveform?

You have two parameters at play -- Vp and t1 -- and you have two equations -- the DC value of a waveform and the RMS value of a waveform.

Two equations, two unknowns.

BTW, the given answer of 16 V is correct.
 

Thread Starter

Henry4321

Joined May 6, 2024
9
Don't try to rely on formulas that only apply to special cases, or try to infer general conclusions from this special case.

In terms of a voltage waveform, v(t), how do you calculate the DC and the RMS values of that waveform?

You have two parameters at play -- Vp and t1 -- and you have two equations -- the DC value of a waveform and the RMS value of a waveform.

Two equations, two unknowns.

BTW, the given answer of 16 V is correct.
I understand how to come to the following: TRMS = Vp * sqrt(t1/T) through integration of the RMS formula (where this special case for 1/2 ratio comes from). I also know that RMS of AC signal is sqrt(TRMS^2 - DC^2). But as far as how these two are related in this case and if they are related at all, well... the answer eludes me
 

WBahn

Joined Mar 31, 2012
30,303
Forget relying on stuff like the RMS of an AC signal is sqrt(something or other).

If you understand how to get the RMS value of a signal through integration, you should understand how to get the DC value of a signal through integration, since it's much easier.
 

Thread Starter

Henry4321

Joined May 6, 2024
9
Forget relying on stuff like the RMS of an AC signal is sqrt(something or other).

If you understand how to get the RMS value of a signal through integration, you should understand how to get the DC value of a signal through integration, since it's much easier.
Well, if I understand what you are saying correctly, it should be something like:
TRMS^2 = Vp^2*t1^2 + DC^2
This is the only way to get DC offset that I can think of in this case
 

WBahn

Joined Mar 31, 2012
30,303
Again, what IS the definition of the DC value of a waveform?

It's nothing more than the average value. How do you compute the average value of a function?
 

Ian0

Joined Aug 7, 2020
10,277
Don't think in terms of voltage. rms is all about power. Work out how much power the waveform would deliver to a resistive load. Then work out V from P and R.
 

WBahn

Joined Mar 31, 2012
30,303
Don't think in terms of voltage. rms is all about power. Work out how much power the waveform would deliver to a resistive load. Then work out V from P and R.
He got the correct expression for the RMS voltage. He is struggling with finding the DC value.
 

Thread Starter

Henry4321

Joined May 6, 2024
9
Again, what IS the definition of the DC value of a waveform?

It's nothing more than the average value. How do you compute the average value of a function?
You see, that’s where I’m struggling exactly. Since DC is the mean value, the Vp should be twice as high. I mean, it’s a rectangular-shaped wave, so if one was to slice in half, Vp should be 8V. So I discarded that idea as obviously incorrect
 

BobTPH

Joined Jun 5, 2013
9,299
For a true square wave (50% duty cycle) you would get the average voltage like this:

Vavg = Vp+ * 1/2 + Vp- * 1/2

If Vp+ = 8V and Vp- = 0V

Vavg = 8 * 1/2 + 0 * 1/2 = 4
 

WBahn

Joined Mar 31, 2012
30,303
You see, that’s where I’m struggling exactly. Since DC is the mean value, the Vp should be twice as high. I mean, it’s a rectangular-shaped wave, so if one was to slice in half, Vp should be 8V. So I discarded that idea as obviously incorrect
What is the average value of the following list of numbers:

[100, 100, 0, 0, 0, 0, 0, 0, 0, 0, 100, 100, 0, 0, 0, 0, 0, 0, 0, 0]

Is the average equal to one-half of the peak value?

Again, how do you calculate the average value of ANY waveform?

Say you have:

x(t) = 10 - 2t + 3t²

and you want to determine the mean of x(t) from t = 1 to t = 7, how would you do it.
 

Thread Starter

Henry4321

Joined May 6, 2024
9
What is the average value of the following list of numbers:

[100, 100, 0, 0, 0, 0, 0, 0, 0, 0, 100, 100, 0, 0, 0, 0, 0, 0, 0, 0]

Is the average equal to one-half of the peak value?

Again, how do you calculate the average value of ANY waveform?

Say you have:

x(t) = 10 - 2t + 3t²

and you want to determine the mean of x(t) from t = 1 to t = 7, how would you do it.
Oh God, I think I got it:
DC (equal to mean value) = 1/T*integral(u(t) dt) from 0 to T can be represented as 1/T*integral of Vpeak from 0 to t1 (as the rest is zero)
Solving it gives
Vpeak*t1/T = 4
And the other equation is
Vpeak^2*(t1/T) = TRMS^2 = 64
Vpeak will be 64*T/t1 divided by 4*T/t1 = 16 volts

Thank you so much!!
 
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