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My teacher says that this signal is DC, but if the polarity varies from 0V to -8V , shouldn't it be AC?

He says that

DC + AC = DC

enter image description here

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    \$\begingroup\$ I doubt you'll get much agreement on this. If the waveform goes from +10 V to -8 V I'd call it AC with a +1 V DC offset. \$\endgroup\$ Mar 16 at 3:49
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    \$\begingroup\$ Note that some people call any signal such a sine-wave that has been completely offset to always be of one polarity (positive for example), DC. It's wrong and it's annoying and confusing as all hell. Your teacher seems to start veering into this category. Unipolar and bipolar are other descriptive terms that can be used as well. \$\endgroup\$
    – DKNguyen
    Mar 16 at 4:28
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    \$\begingroup\$ @jsotola Your reasoning is valid, but move your reference point to -10 kV and I’d call it DC with a tiny AC ripple superimposed on it. \$\endgroup\$
    – winny
    Mar 16 at 15:28
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    \$\begingroup\$ @jean-leonardo please clarify the question (is it +10V or +0V to -8V) and label the peak, trough and 0V points on the axes more legibly. At what voltage level is the x-axis intended to be? \$\endgroup\$
    – Ed Randall
    Mar 17 at 7:56
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    \$\begingroup\$ "AC or DC" is a question of semantics - meanings of words. It's AC if you think that's what AC is, and it's DC if you think that's what DC is. We can describe the signal more accurately, "a 18V p-p signal superimposed on a 1V DC offset" and then the problem disappears entirely. \$\endgroup\$
    – user253751
    Mar 17 at 13:14
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DC + AC = DC

Plainly put: No.

It's AC with a DC offset, if you will.

This is less about whether the sign changes (that sign is just relative to some arbitrarily defined reference), but about whether the energy transport happens mainly in the DC or AC component.

That's all that makes the difference between "AC with DC offset" and "DC with AC ripple".

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    \$\begingroup\$ @JasonGoemaat, how about e.g. a transistor amplifier where the input signal is coupled on top of a DC bias so that it stays positive all the time? E.g. the example amplifier from the Falstad simulator with a scope showing the transistor base voltage: tinyurl.com/yzmsghuh \$\endgroup\$
    – ilkkachu
    Mar 16 at 21:15
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    \$\begingroup\$ I would say the parts of the circuit that had current (the yellow dots) always flowing in one direction would be DC and the parts where they changed directions would be AC because the Current Alternates. \$\endgroup\$ Mar 16 at 21:25
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    \$\begingroup\$ Hmm, what about a 14.1VAC with 10VDC offset then? Isn't it more about the intent, which we can't tell without knowing what we're measuring? \$\endgroup\$ Mar 17 at 8:01
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    \$\begingroup\$ Applied to a resistive load, if the sign of the potential difference remains the same, the current is (as commented elsewhere) not alternating. But applied to a reactive load, the key point identified by @mmmm is whether the energy transfer is predominantly AC or DC. \$\endgroup\$
    – grahamj42
    Mar 17 at 11:23
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    \$\begingroup\$ I agree with this, but I would change "plainly put: no" to "it depends": all three are AC, but V(a) is pure AC, V(b) is DC+AC, but overall it's still AC (current changes sign, as you say), and V(c) is DC (borderline, but anything from then on is DC), since the current has one sign, only. \$\endgroup\$ Mar 17 at 15:06
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It has both AC and DC components. Which of these you regard as "the signal" depends on its intent (i.e. on the information content).

If it is say analogue audio with some DC in there for some reason, then it is an AC signal with a DC offset or bias.

But if it is say a relay control line picking up interference from somewhere, then it is a DC signal with AC noise present.

You can see this difference of viewpoint among other answers offered here.

But be wary of hitting your teacher over the head in such cases; check the answer expected for the exam, as teachers are more focused on getting you through that than they are on the ultimate truth.

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    \$\begingroup\$ +1 for this. You've got the rest of your career to reflect on the wrong things you learnt in school! \$\endgroup\$
    – CCTO
    Mar 16 at 15:09
  • \$\begingroup\$ +1 for use of BIAS and not OFFSET, suggesting the sinusoidal pattern is a signal. Not as much magnetic audio tape around these days. \$\endgroup\$
    – mckenzm
    Mar 17 at 21:01
  • \$\begingroup\$ @mckenzm Probably not so many quadrupole mass spectrometers either. \$\endgroup\$ Mar 18 at 9:06
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The signal is changing direction with time (alternating), as well as continuously changing amplitude with time.

I would call that an AC signal of 18 Vpp with a DC offset of +1 V.

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    \$\begingroup\$ It's not clear in the chart, but in the question text it goes from 0V to -8V, so only 8Vpp with a DC offset of -4. By my understandng this would be DC because the direction doesn't change since it spends ~100% of it's time in negative V territory? \$\endgroup\$
    – coagmano
    Mar 16 at 23:03
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    \$\begingroup\$ @FredStark maybe that example fits the definition of Pulsed DC? \$\endgroup\$
    – TylerW
    Mar 17 at 2:29
  • \$\begingroup\$ Definitely looks like it. I'm guessing the teacher deliberately used an example that sits on the borderline of definitions \$\endgroup\$
    – coagmano
    Mar 17 at 3:04
  • \$\begingroup\$ Agree with Tyler on that. That's right on the borderline as well. Because DC amplitude is now the same as AC amplitude for that signal. \$\endgroup\$
    – Mitu Raj
    Mar 17 at 4:09
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This is really a terminology question. For instance, Wikipedia has a classification under which your signal would be called either "Alternating" or "Pulsating":

enter image description here

Clarify with your teacher what is his definition of AC and DC, and then apply those definitions to the signals you need to classify.

Understand that such classification for arbitrary signals is descriptive rather than strict. There is no special Ohm's law for pulsating current. There are formulas/models which only work in special cases, such as constant voltage / current (which you would colloquially call DC) or for a periodic sinusoidal signal with zero offset (which you'd call AC). Obviously, the models which assume constant voltage won't apply to your signal, even if your teacher calls it DC.

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EVERY signal contains components of both AC and DC. The issue is one of dominance. Which of these components dominates the signal content.

Examples:

A "DC" power source is likely to contain a small amount of noise or "ripple". The noise/ripple will be AC however the dominant component will be the DC offset.

An "Audio signal" might contain a small DC offset. The signal will be primarily AC, but will contain a small amount of DC.

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I wonder what the time scale is... If it's hours, days or years you could describe the voltage as 'slowly varying DC'.

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    \$\begingroup\$ Interesting thought. \$\endgroup\$ Mar 18 at 9:47
  • \$\begingroup\$ This is what lock in amplifiers are for. You lock onto a DC signal that changes over time slowly b/c of a natural phenomenon. \$\endgroup\$
    – wbg
    Mar 19 at 6:34
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It is varying with time and also goes below zero, for all the definition is AC.

For some (few) people if a signal is always positive or always negative it could be defined DC (since the direction stays the same).

For all the practical uses DC is fixed, constant and only with some ripple or noise (if acceptable)

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  • \$\begingroup\$ It's not clear in the chart, but from the question text it goes from 0V to -8V, so it's always negative. I think that's why it's being called DC by the teacher \$\endgroup\$
    – coagmano
    Mar 16 at 23:05
  • \$\begingroup\$ @FredStark Are you sure? The picture reads 10V to -8V. \$\endgroup\$ Mar 17 at 9:51
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It's AC with a DC offset, or it's DC with an AC ripple. Your choice, depending on what you're going to use it for.

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Your question is a contradiction:

If the polarity varies from 0V to -8V , shouldn't it be AC?

If the VOLTAGE ranges between 0V and -8V then the polarity doesn't vary.

Furthermore your drawing says 10V not 0V, so your green graphline should be under the blue line.

VDC + VAC = VDC if only |VDC| >= VAC

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