My name is Andria and I am attending college at 46yrs old. I am majoring in electrical engineering and this is for study purposes.

Thank you for your time

  • \$\begingroup\$ It doesn't mean much. One can't even say that the voltage waveform is periodic or, even assuming for a moment that it were periodic, that the average was taken over an integer number of periods. Context always matters. A reasoned assumption may be that this is periodic and that the average was performed over a positive integer number of such periods. But then it still only tells you that the average isn't zero. What that means, again, depends upon context even with reasonable assumptions. What textbook are you working from? It may help somewhat to know. \$\endgroup\$ Jun 10, 2023 at 6:45
  • \$\begingroup\$ It means exactly what it says: it means the average is not zero. Sometimes this is a problem, sometimes this s normal. For example, some circuits use decoupling capacitors that force the average to be zero, and if you send this waveform through a circuit like that, it will remove the average voltage. \$\endgroup\$ Jun 10, 2023 at 6:51
  • \$\begingroup\$ A waveform can have any average value. If for some reason you expect the waveform to have an average of 0 but, measurement shows a different value, then, the waveform needs to be added to the question \$\endgroup\$
    – sai
    Jun 10, 2023 at 7:15
  • \$\begingroup\$ It means it has a DC component, nothing more. A non-zero avergae is a non-zero average. \$\endgroup\$
    – Neil_UK
    Jun 10, 2023 at 7:57
  • \$\begingroup\$ A non-symmetrical waveform has an average voltage that is not zero caused by even-harmonics distortion. The top or bottom of the waveform is squashed or is extended. \$\endgroup\$
    – Audioguru
    Jun 10, 2023 at 14:45

4 Answers 4


The question is very open, and there are a thousand ways in which a non-zero average may be significant in some way.

The most basic answer I can give you, in the context of a periodic waveform, one that repeats over and over in time, is that the waveform can have some offset that shifts it up or down on a graph. Below are three sinusoidal waveforms. The one in blue has no offset, and its average is zero. Intuitively you can see that this zero-mean results from the waveform's symmetry above and below its centre-line.

The other two, orange and tan, are the result of adding 1V and subtracting 2V respectively from the blue waveform. Therefore they have averages (means) of +1V and -2V:

enter image description here

Some waveforms do not have such symmetry. Take for example a periodic rectangular waveform that spends 10% of its time at +1V, and the remaining 90% at -1V, as shown here:

enter image description here

Even though this signal has extremes of +1V and -1V, and seems centered about zero, it spends much more time at -1V than at +1V, and its average will not be zero. Rather it will be much closer to -1V, precisely \$\frac{(+1V \times 10\%) + (-1V \times 90\%)}{100\%} = -0.8V\$.

Consider what we would have to do to that signal, if we wished it to have an average of exactly zero; we would have to "raise" it up, by adding 0.8V, which is what I have done to produce this:

enter image description here

That trace certainly isn't centered about zero, but I assure you its average is exactly zero.

I used CircuitLab to simulate these changing voltages, and produce those graphs. CircuitLab features are built in to this Stack Exchange site, and you can click on "simulate this circuit" below, to try this out for yourself. It's a fantastic, easy to use tool to play around with circuits and signals. Here are the circuits I used to create these graphs:


simulate this circuit – Schematic created using CircuitLab


Mathematically, it describes a whole class of waveforms, far more general than waveforms that average to zero. In fact, when we say that the average of a waveform is zero it's usually either the average taken over a number of integral cycles of a periodic waveform or an average taken over an extremely large number of cycles so that the exact beginning and ending times don't matter.

For example, if I take the average of a mains 50Hz or 60Hz sinusoidal-ish waveform (like comes out of the wall) over a 10 microsecond period it might be just about any value between almost +/- peak voltage and zero. On the other hand, over a period of seconds or minutes it will approach zero very closely. Or if I average it over some integral number of cycles (for example, exactly 0.1 second, which is 5 or 6 full cycles).

There are consequences to a non-zero average, for example an inductor (or a transformer primary) can conduct very high steady state current (essentially the average voltage - or "DC component" divided by the wire resistance), which can cause problems (for example, inductor saturation, loss of inductance and failure of switching elements as a result).

Sometimes it is desirable to eliminate the DC component in a signal (so it averages to zero), and a series "blocking" capacitor is used to couple the AC component while blocking the DC (average) component in steady state (after the capacitor charges to the magnitude of the DC value).


It means the waveform has a DC component which offset the amplitude from origin, therefore when you calculate the average value - it is not 0, it is the DC value. More accurate - DC value is the AC RMS value

  • 1
    \$\begingroup\$ "More accurate - DC value is the AC RMS value" No. For a mixed signal with a DC component and one or more AC components (each on a separate frequency) there's no requirement for the value of the DC component to equal the AC RMS value. I expect you're thinking of the case of a single frequency AC signal (a sine wave with no DC offset) which has an RMS voltage or current value equal to a DC signal which delivers the same power to a resistive load (averaged over a whole cycle, or over many cycles). To find the RMS value of a mixed signal you'll need to use Pythagorean addition on each component. \$\endgroup\$
    – Graham Nye
    Jun 10, 2023 at 9:38

What does it mean if a voltage waveform does not have average zero?

It means that there is a signal of some AC amplitude superimposed on a DC level.


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