12
\$\begingroup\$

We had a big argument last night with vague conclusions. Is the current with a frequency less than 1 Hz considered DC?

It would still resemble a wave...

\$\endgroup\$
19
  • 16
    \$\begingroup\$ What's special about 1Hz? \$\endgroup\$
    – OJFord
    Dec 20, 2014 at 17:56
  • 2
    \$\begingroup\$ (The guy you talked to is wrong, at least at a theoretical level. Though it might (in the right context) be reasonable to say that it's "for all intents and purposes" DC.) \$\endgroup\$
    – Hot Licks
    Dec 20, 2014 at 19:18
  • 1
    \$\begingroup\$ See also: en.wikipedia.org/wiki/Pitch_drop_experiment Just because its solid doesn't mean its not a liquid. \$\endgroup\$
    – Passerby
    Dec 20, 2014 at 20:11
  • 1
    \$\begingroup\$ @Alnitak it just have to be "periodic" over whatever timespan you choose (not even necessarily continous), infinite's got nothing to do with it. \$\endgroup\$
    – Anton
    Dec 20, 2014 at 22:34
  • 1
    \$\begingroup\$ xkcd.com/594 \$\endgroup\$
    – Phil Frost
    Jan 18, 2015 at 11:41

9 Answers 9

34
\$\begingroup\$

AC and DC are relative terms. If you're looking at a 10kHz waveform for 100ns, you will think it is DC. It works the other way around too: if you forget about what's providing you with "DC", who knows if this waveform is not going to change in the next seconds, minutes, days, years? Think the voltage of a capacitor for example during slow discharge. If you monitor the voltage on an oscilloscope, you'll see a flatline. DC you say? Wait longer, and the flatline will decrease in voltage towards zero, which means there is some AC in there as well.

Besides, no signal is actually pure DC, you always have AC components as well due to noise and all sorts of causes. It is only "DC-enough" or "AC-enough" for the application you're intending to use it with/for.

Fourier transforms are a good way to picture what DC and AC components are in a waveform. The transform is constant for periodic signals and depends on time for any non-periodic signals like the capacitor example. For the square wave: (source: wikipedia) enter image description here

\$\endgroup\$
17
  • 4
    \$\begingroup\$ Some people call things that oscillate but don't cross 0 DC. Some devices can only take current in one direction even though they can tolerate wild voltage swings. \$\endgroup\$
    – Joshua
    Dec 20, 2014 at 22:00
  • 4
    \$\begingroup\$ Joshua: "things" that oscillate but don't cross 0V would typically be the sum of a DC component (average of the signal) and of an AC component (in turn perhaps the sum of different frequencies, c.f. Fourier transform of periodic signal). Transients are more difficult to categorize, but again it's a matter of timeframe. The average on the time window would give DC, and the rest AC. A Fourier transform is more strict, defining DC as 0Hz. Theoretically Fourier transforms are only for periodic signals, but one can assume any signal capture repeats itself and proceed. \$\endgroup\$ Dec 21, 2014 at 0:13
  • 4
    \$\begingroup\$ I disagree with your example of a capacitor during slow discharge being implied that it is AC. Throughout the entire discharge it is Direct Current (DC). At no time during the discharge is it Alternating Current (AC). Something being AC implies that the direction of the current changes. You can have a fluctuating DC voltage, but unless the current direction actually changes, it is not AC. Something being DC does not imply that the voltage must be constant, only that the direction of current flow does not reverse. \$\endgroup\$
    – Makyen
    Dec 21, 2014 at 2:03
  • 5
    \$\begingroup\$ And I'm saying you're incorrect in that characterization. If it is changing, it contains an AC component. Period. End of story. A fluctuating DC signal is a oxymoron. \$\endgroup\$ Dec 21, 2014 at 13:49
  • 3
    \$\begingroup\$ "Alternating Current" MUST mean that the current changes direction. Otherwise it is not "alternating", just "fluctuating". \$\endgroup\$
    – Floris
    Dec 22, 2014 at 3:56
19
\$\begingroup\$

Yes, you can have AC with a frequency less than 1Hz, in the same way you can have numbers between 0 and 1.

Frequency isn't an integer number, but a "real" number. You can quite happily have a waveform of \$1 \times 10^{-100}Hz\$ if you wanted. You'd have to be quite patient to see it change, but it will change, and given time it would trace an AC waveform.

\$\endgroup\$
5
  • 12
    \$\begingroup\$ "Quite patient" is quite an understatement \$\endgroup\$
    – user253751
    Dec 21, 2014 at 5:41
  • 1
    \$\begingroup\$ "and given time it would trace an AC waveform." Indeed it wouldn't, since the protons in your oscilloscope don't survive that long. (Probably; realistically of course there'll be some other interruption much earlier.) \$\endgroup\$ Dec 21, 2014 at 16:30
  • \$\begingroup\$ 1: You don't have to watch an entire cycle to see it change. 2: You're assuming time is linear. \$\endgroup\$
    – Majenko
    Dec 21, 2014 at 22:09
  • 4
    \$\begingroup\$ Also: How do we know the Big Bang isn't just the zero-crossing point of the universe? \$\endgroup\$
    – Majenko
    Dec 22, 2014 at 14:14
  • 1
    \$\begingroup\$ Oh come on, a signal who's oscillation frequency is so slow that the present age of the universe is a measurement error is DC. \$\endgroup\$ May 23, 2017 at 17:12
5
\$\begingroup\$

As with any AC voltage, frequency is the inverse of the period in seconds, and vice versa:

$$f = 1 / T$$ $$T = 1 / f$$

As f gets asymptotically close to 0, T correspondingly becomes very large.

As a practical example, I have a function generator that generates any frequency up to 5MHz in 0.01 Hz steps. So at its lowest setting (0.01 Hz), it can generate a sine wave with a period of 100 seconds.

\$\endgroup\$
5
\$\begingroup\$

If you want to be strict, all real current is AC. I'll explain why.

Looking at it from a thermodynamic point of view, a direct current (which never changes magnitude) would require two terminal points of fixed charge; that is, one relative positive, one relative negative. (I'm using charge here instead of voltage or current in order to stick to my thermodynamic approach, and keep things simple.) The relative positive would dispense into the relative negative, without ever changing magnitude itself; thus, an infinite source of charge, dispensing into an infinite well. This is of course an ideal.

Since such black boxes do not exist in the real world, it is safer to say that "direct current" is simply a model. The rules that apply to it have been calculated and can be applied to a slowly varying voltage source, such as a gradually draining AA battery; but all sources of current will ultimately reach zero, and thus have a frequency.

So, in a broad sense, there are cases in which /any/ current frequency can be described as DC; and the AC laws can be derived from the DC laws. As to whether 1 Hz looks like DC, it depends on how short a time frame you are using it over, and how close it appears to be to level during that time. It's really up to you.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ This confuses me. I thought that alternating current changes flow of current for each half period. DC battery just discharges in same direction which ultimately makes it unstable DC current \$\endgroup\$
    – Brlja
    Dec 20, 2014 at 20:12
  • \$\begingroup\$ You need to remember that current flow is relative in direction; zero is wherever you want to put it. Thus, battery alternating current can be thought of as a low-frequency sinusoid, plus a constant; which qualifies as AC. \$\endgroup\$ Dec 21, 2014 at 2:09
  • \$\begingroup\$ Your argument doesn't quite hold if we consider superconductors, but in essence you're right of course: DC is just a model. \$\endgroup\$ Dec 21, 2014 at 16:32
  • \$\begingroup\$ Well, they're both models, really. DC versus AC is like arguing quantum mechanics versus general relativity; they're both correct, but the equations only really apply under certain circumstances, and are part of an overarching whole. (Also, superconductors still link a finite source to a finite drain, so I don't think I follow how they wouldn't change with time.) \$\endgroup\$ Dec 21, 2014 at 17:10
  • \$\begingroup\$ "Alternating current" has a fairly strong implication of being a periodic waveform. A lot of the non-DC situations being discussed here are more about the real term of the complex exponential than they are about the periodic imaginary one. \$\endgroup\$ Jan 18, 2015 at 17:20
2
\$\begingroup\$

As others have already pointed out, you can have AC of as low a frequency as you wish.

I think it's worth adding, however, that at such low frequencies it mostly won't act much like most of us usually think of AC acting.

Just for an obvious example, you can typically think of a capacitor as allowing AC to flow through it, but as stopping DC. At extremely low frequencies like you're considering, you're probably not going to see any significant current flow, even though it is technically AC.

In particular, a capacitor basically acts like a (very gentle) high-pass filter. To pass such a low frequency well, you'd need a tremendously huge capacitor. By far the most common type of large capacitor is an electrolytic capacitor. An electrolytic capacitor is a little like a specialized battery--that is, part of how it works is chemical, not purely electrical. Like batteries, electrolytic capacitors can self-discharge over time. I've never tested to figure out an exact rate of self-discharge, but it wouldn't surprise me a lot if it were to self-discharge faster than (for example) a 0.01 Hz signal was charging it--if so, the net result would be that the capacitor never charged, and it would basically act like there was no capacitor there at all.1

The bottom line is that most AC circuits are designed for much higher frequencies, so even though there's no sharp cutoff below which a signal is no longer AC, quite a bit of typical thinking about AC circuit design may easily start to sort of fall apart as you reach such...subterranean frequencies.

Just for reference, the lowest frequency of AC in really common/wide use is probably in audio circuits. Although (again) it's not a hard cuttoff, the typical number used as the bottom-end of the audio range is 20 Hz.

There has been some work done in Extremely Low Frequency radio, but the lowest frequency of which I'm aware was around 50 Hz or so. For a 1 Hz signal, a half-wave dipole antenna would be substantially larger than planet earth.


1. In fairness, most electrolytic capacitors are polarized, so you normally use them for things like filters on DC power supplies. Here I'm assuming an (admittedly, less common) non-polarized electrolytic capacitor.

\$\endgroup\$
0
\$\begingroup\$

Of course. 1 Hz is once per second, and a second is a fairly arbitrary amount of time. If we had settled on 100 seconds per minute, 60 times per minute would have been 0.6 Hz.

\$\endgroup\$
1
  • \$\begingroup\$ Of note is that a "second" is (historically) a "second minute" -- an even more minute (my-newt) fraction of an hour than is a "minute". Things started with the hour and got smaller as clocks got better. Nothing at all special about the second. \$\endgroup\$
    – Hot Licks
    Dec 20, 2014 at 19:24
0
\$\begingroup\$

Yes, you can have alternating current (AC) that alternates with a frequency less than 1 cycle per second (a period longer than 1 second). If you connect a battery and a resistor using a properly wired DPDT switch, you would be able to reverse the voltage across the resistor, at will. So if you manually throw the switch once per second, or once every 2 seconds, or once every 100 seconds, etc. you would have "alternating current" with a frequency less than 1 cycle per second.

\$\endgroup\$
-2
\$\begingroup\$

Whether a voltage is AC or DC has nothing to do with frequency, but more to do with whether the voltage is alternating or not. If it's not alternating it's DC.

If a voltage always stays above zero (ie; positive) it is 'DC', although it may have a small 'AC' component. Such voltages have a mean value above zero (the DC level).

On the other hand, if the voltage alternates from positive to negative (no matter how slowly) it is 'AC'. Such voltages have a zero mean value.

\$\endgroup\$
1
  • 4
    \$\begingroup\$ Such signals are said to have "a DC component", not to be DC per se. \$\endgroup\$
    – glglgl
    Dec 20, 2014 at 16:46
-2
\$\begingroup\$

Yes. Hertz is a measure of how many cycles happen in a given time frame (1 second).

Since time is subjective, and a second is a unit defined by humans, you could (for example) have a "Zecond" that lasts 0.4 seconds.

Hence the definition of Hertz could be different but retain its meaning.

\$\endgroup\$
2
  • \$\begingroup\$ No, Hertz is a unit of measurement. Frequency is a measure. \$\endgroup\$
    – OrangeDog
    Dec 22, 2014 at 0:26
  • 1
    \$\begingroup\$ Time is anything but subjective, and it's defined by caesium 133, not humans. Humor is subjective and defined by humans. \$\endgroup\$
    – Phil Frost
    Jan 18, 2015 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.