Instantaneous & Effective Values: Oscilloscope Waveform Definitions Explained (Max 90)

Exactly ... A completely lame problem, looking through the prism of forum members' knowledge, so I'm ashamed to ask a little about something like that, but ... I would like to ask for a definition of these two terms:
instantaneous and effective value of oscilloscope waveforms.
Thanks in advance for your answer.

Check e.g. in this encyclopedia.

Do you want integrals or peasant reason?

The instantaneous value (as the name suggests) is the value of the voltage, current at a given measuring moment (e.g. in a given second) and it changes very often (e.g. a sine wave). The rms value applies to current, including alternating voltage, and corresponds to the effect that this voltage, current will cause in the system compared to the direct V, A.

$$I=\sqrt{\frac{1}{T}\int{i^2dt}}$$ integral in the interval [0, T]

$$U=\sqrt{\frac{1}{T}\int{u^2dt}}$$ integral in the range [0, T], T = 2pi

$$U=\sqrt{\frac{1}{T}\int{u^2dt}}=\sqrt{\frac{1}{T}\int{(U^M\sin(\omega t))^2dt}}$$

The integral of the expression is equal to:

$$\sqrt{\frac{1}{T}\int{(U^M\sin(\omega t))^2dt}}$$

unfortunately I can not write it here too much work, the point is that for sine it should come out that Um = 1.41 U -> U = Um / first of (2)

I'm gonna do it with some decent equation editor.

here is the solution - how to calculate it (I had to do it in 2 minutes during the exam) for the current is the same, but it is different for non-solar waveforms.

Thanks guys, both simple definitions and a derivation will come in handy. Regards

The instantaneous value - at a given moment - that is the Y value of a given point on the graph (oscillogram).
Effective value - you will not see it on the oscilloscope. It is the equivalent value of the direct voltage (or current - when it comes to current), causing the separation in the resistor e.g. 1 kOhm of the same power as our voltage (current) waveform, for which we calculate the effective value. For example, you have a rectangular waveform of one second 2V, one second 0V. Hence, in the first sec: 2V * 2V / 1kiloom = 4mW (power = U square / R), energy in a second - 4 millijoules. In the next second, zero voltage, and therefore zero power and energy. On average, we have 2 millijoules in a second and we calculate the equivalent voltage: Ueff squared / 1 kilohm is supposed to be 2 mW. So the effective is 1.41V (square root of 2).
In general, you need an integral.
There is also the average value of the voltage (current) - normally we "cover the holes with hills" (this time without squares) and see what comes out. In our rectangle, 1V will come out (much easier, right?).
Different meters measure different values. The most popular measure the average value (and in the AC ranges the rectified average), but the AC ranges are scaled in the effective value for the sine wave to make it easier (i.e. without recalculation, when measuring the voltage in the socket, we get 230V). However, with a different shape of the waveform, the meter actually shows stupidity - it is not an effective value (because it is only for a sine wave), not an average value (because it is scaled for a sine wave). It is worth remembering and having limited confidence in the readings.
There is also a peak value - it probably does not need to be explained, on the oscilloscope screen you can see (one of the instantaneous values).

I am not talking here about deformed non-sinusoidal waveforms where the formulas resemble more the Pythagorean theorem

And it looks worse:

U = ? ((U0) ? + (U1) ? + (U2) ? + ... + (Un) ? where U0 - constant component - U1 - harmonics

And now the patterns with integrals that my colleagues presented, translate into peasant reason as follows:

- the instantaneous value is the point displayed by the oscilloscope at the moment
e.g. a (t) = Am sin (w * t) - for a specific moment t

- effective value is the area under the graph / duration of this part of the graph (usually the full period)
For sine: Am / sqrt (2)

- the average value is similar to the value, but we treat the field under 0 with the sign "-".
For sine = 0 - because the area under the x axis is always equal to the area above the axis (for one full period)

A little note
- RMS value is the root of the square (square means up to the power of 2) of the graph / duration of this fragment of the waveform (usually the full period).

According to the pattern as given by Tremolo

and what is given above as the effective value is also the average value.

Because for some waveforms the mean value is equal to the effective value.

for the sake of distinction, it is said to be effective because mean = average (different nomenclature) and also called expected in another way, which is more related to the probability. We would not like sine to be zero.

sawitar wrote:

Because for some waveforms the mean value is equal to the effective value.

And somehow no examples of this come to my mind, unless a square wave in half a period.

there is one case - when f = 0 (valid for all hihi shapes)

When f = 0, the waveform is constant over time. No example.
After all, no one will write f (t) = Amsin (0 * t) etc., only f (t) = Am.

For any periodic waveform, the mean value cannot be equal to the RMS value, because the mean value is always zero. There is one exception - when we have a constant component. So even this rectangle with 0.5 padding does not satisfy the above relation.

Behind half a period yes it does.

And the statement quotation "There is one exception - when we have a constant component."
It is untrue.

Examples are just multiplied:
- a (t) = Am * | sinwt | - is a periodic waveform, and its mean value is equal to the rms value Usk = Uśr = 1 / sqrt (2),
- b (t) = 1 (t-kT) -1 (t- (0.5 + kT)) - for T = 1, nl. natural, Usk = Uśr = 0.5

and already in the first point, the error Uśr = 2Am / ? for a (t) = Am * | sinwt |. It is enough to calculate the integral by definition.
and for a unipolar square wave with a duty cycle of 0.5 we have Usk = amplitude / sqr (2) here usk = 1 / ?2, aver = 0.5

for permanent nap the value of sk is equal to the average.
for other waveforms the effective value is always greater than the average

it is known that for the sum of sinusoidal waveforms and the constant component, the effective value is
sqrt (A0 * A0 + A1 * A1 / 2 + A2 * A2 / 2 + .... An * An / 2).
A0 - constant component
A1 ... An - harmonic amplitudes
and you know what the average value is

continuing the above thread.
Uoer is equal to the constant component.
And everything is correct.

Actually. I counted everything by definition and I have to admit the herb is right. The examples I have given are not correct. Sorry.

Behind half a period yes it does.
If you take a square wave for half a period - then you have a constant value - which is what you defended so fiercely (i.e. f = 0). Therefore, there is a concept of "period" for floating signals at all. After all, no one writes a half-time waveform ... or a double-frequency waveform ...


And here was my incomplete quote, so I complete - because it could be misunderstood:
"Because the mean value is always zero. There is one exception - when we have a constant component. "
Quote

ziolek wrote:

"It's not true.

then take the integral of some waveform with a constant component - you will not get zero.

plesniak wrote:

And here was my incomplete quote, so I complete - because it could be misunderstood:
"Because the mean value is always zero. There is one exception - when we have a constant component. "

Obviously, the quote is read the other way around. It sounded like the constant component usk = aver, not that usr is different from 0.

Apart from that, if we were to stick to all of them, there is no word for periodic waveforms in the supplement.

And the square wave was a special example, because only for the constant waveform usk = aver. And some said that for periodic waveforms as well.

I can say the same for your quote:

plesniak wrote:

there is one case - when f = 0 (valid for all hihi shapes)

We also deal with a constant value waveform.
And it is not said, for example, that it is a sinusoidal (or other periodic) waveform with the frequency f = 0.

But I suppose it was also written a bit for a joke, as with my rectangle. Although mathematically, everything is ok.

it would be possible to devise periodic waveforms in which it is possible to calculate the usr if the period changes are regular. There is a whole differential-integral-border calculus for this.

Imagine a sine wave in which, after a full period, the next period is twice as long, followed by a period of 3 times as long ... so nothing can change in terms of the average.

Just as we calculate the average of the entire sinusoidal waveform, we calculate only one particle from 0 to 2pi (in the generalized case 0 to T)

The real waveform from the contact may differ by, inter alia, harmonics of higher odd orders 150,250 may be frayed so our calculations are not correct, however, close enough to the real ones as to be good. As if someone persistently wanted to exactly, one would have to cut the band below 140Hz, count what is happening with the current at 150Hz, then add up the average value of this waveform. Then square this value and the current value for 50Hz, add and take the square root.