# What does this equation mean??

Discussion in 'General Electronics Chat' started by Engr1144, Oct 5, 2013.

1. ### Engr1144 Thread Starter New Member

Oct 5, 2013
8
0
Vac = √(Vrms^2 - Vdc^2)

Can any one help in understanding this equation?

2. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
It gives you the RMS amplitude of the AC component of a waveform that consists of an AC component riding on a DC component (i.e., a DC offset or a non-zero average value).

It comes from the more fundamental relationship that

Vrms^2 = Vdc^2 + Vac^2

3. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Note that this equation is only valid for AC waveforms with a 0 average value.

4. ### joeyd999 AAC Fanatic!

Jun 6, 2011
2,675
2,722
Ummm....any component without a zero average value is, by definition, not AC!

5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Not sure about the definition of AC but some waveforms which have both positive and negative peaks do not have zero average.

6. ### joeyd999 AAC Fanatic!

Jun 6, 2011
2,675
2,722
And therefore have one or more AC components with a single DC component with a magnitude of the average value.

7. ### wayneh Expert

Sep 9, 2010
12,093
3,033
The distinction between AC and rippling DC comes up often. Some - most I think - would say the ripple on a DC signal is the AC component on top of a DC voltage. Any time-varying element to voltage = AC.

I personally am resolved to reserve the term AC for circuits where the current actually alternates direction, not just magnitude. It's just a semantic difference but I hope to be a little more precise in my language. Unfortunately this distinction depends on the circuit, not the signal itself. If there's any time-varying element, you could always construct a circuit to produce an alternating current.

8. ### mik3 Senior Member

Feb 4, 2008
4,846
63
I agree that varying signals where the current does not change direction are not consider AC.

9. ### crutschow Expert

Mar 14, 2008
12,991
3,227
I favor the definition that any voltage variation riding on a DC voltage is AC, independent of the current direction. For example, if you have an AC signal going into and out of an amplifier with coupling capacitors that shift the DC level of the signal as it goes through the amp, the signal is still considered AC when it's inside the amp, even though the current may never technically reverse direction as it goes through.

Even though the term AC stands for alternating current, in common engineering use it refers to any time varying signal, independent of the actual current direction or DC level. Restricting the term to only signals where the current actually reverses is too narrow a definition. Why should a signal go from AC to not-AC just because you add a DC level shift? If you slowly increase the DC level shift of an AC signal does it suddenly change to not-AC when the last ns of the signal current no longer reverses direction?

Last edited: Oct 5, 2013
10. ### wayneh Expert

Sep 9, 2010
12,093
3,033
That's true. But it's also true that the term "DC" can be and often is applied to the same signal that would be called "AC" by that definition. Consider the output of a wall wart or even a half-wave bridge. Definitely labeled "DC" despite the same rippling that many would call "AC".

It's unfortunate that the language is sloppy, so that the term to describe the profile depends on the context. The output of a wall wart is only "DC" because it can be used to power something that needs DC. If you use it instead as an audio source, it sure likes like AC.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
One can propose "interesting" circuit behavior in which an AC voltage source driving an ideal inductor introduces a DC offset to the current without having a DC source or rectifier elements, etc.

File size:
21 KB
Views:
19
12. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
Barring specifying a suitable set of initial conditions, you would need some kind of non-linear element.

13. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
Ah, yes. A nonlinear element -- a.k.a. a switch.

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Good point. I guess I could remove the switch and apply Laplace transform solution as a "disguise".

15. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
I was thinking the same thing. Perhaps use it as a "cookie problem" for class.

16. ### GopherT AAC Fanatic!

Nov 23, 2012
5,991
3,742
Is "cookie problem" a common phrase? I've never heard of it except when talking about web browsers.

17. ### GopherT AAC Fanatic!

Nov 23, 2012
5,991
3,742
Wayne,
With your definition, how would you write it? How would you rename or redefine Vac^2?

Vrms^2 = Vdc^2 + Vac^2

18. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
Nope. It means that I throw out an optional problem to the class for them to go away and work on and submit as soon as they have an answer. The first person to submit a correct answer gets one of the large chocolate chip cookies that places in some shopping malls sell (they are typically 8" to 12" in diameter) that I buy on my own dime. It's something that one of my colleagues at the Air Force Academy did when I was teaching there.

Alas, sadly, I find that few people even attempt the problems, so it costs me very little money.

19. ### GopherT AAC Fanatic!

Nov 23, 2012
5,991
3,742
Ah, good idea but i am sorry to hear it is not costing you much.

20. ### Engr1144 Thread Starter New Member

Oct 5, 2013
8
0
why isnt this just Vrms= Vac+Vdc???
because for any ac signal with dc offset:
(vrms)^2=1/T∫((Vm)sinwt+Vdc)^2dwt
=(Vac+Vdc)^2