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chris kennedy said:
1) What is an AC wave form with DC offset?

2) What are dBm and dBv measurements?

Thanks
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1) I am not sure in what context you are asking this question but this site will let you add
an offset to the AC wave and see the results http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Signal/Signal2.htm.
A DC offset will occur during a fault and is added to the resulting AC current but you probably know this.
2) This site has some good information on your question http://w7zoi.net/dbdbm.pdf
 
chris kennedy said:
1) What is an AC wave form with DC offset?

2) What are dBm and dBv measurements?

Thanks
Click to expand...

dBm is a dimensionless measure of power measured relative to 1mw.

P(dBm) = 10*log[P(mw)/1mw]

1mW = 0 dBm
2 mW = 3 dBm
etc.,

dBm is commonly used in telephony.

dBv is a dimensionless unit of power measured relative to 1V.

P(dBv) = 20*log[V(volts)/1 volt] where voltage is measured across a standard load say 600 Ohms.

.
 
A dc offset is normal in many situations. Just plot out:

v(t) = Vdc + Vpeak(sin(wt))

Choose any values you wish.
 
Cris
I tried to find some information that was not too technical. There are several site on symmetrical, asymmetrical fault currents and DC offset.
The basic idea is the reactance tries to resist any change in the current flow at the time the fault occurs. This site is fairly good explaining the concept.

http://www.skm.com/newsletters/winter99.pdf
 
bob said:
Cris
I tried to find some information that was not too technical. There are several site on symmetrical, asymmetrical fault currents and DC offset.
The basic idea is the reactance tries to resist any change in the current flow at the time the fault occurs. This site is fairly good explaining the concept.

http://www.skm.com/newsletters/winter99.pdf
Click to expand...
The basic principles of reactive devices are that "current cannot change instantaneously in an inductor" and "voltage cannot change instantaneously across a capacitor." However, that does not mean that the fault current "becomes" DC. It means that the higher the frequency of the signal, the more it is attenuated. A fault does not create a DC bias. However, the inductance of the loads will help to reduce the initial spike of a fault by preventing the instantaneous change in current.
 
bob said:
Cris
I tried to find some information that was not too technical. There are several site on symmetrical, asymmetrical fault currents and DC offset.
The basic idea is the reactance tries to resist any change in the current flow at the time the fault occurs. This site is fairly good explaining the concept.

http://www.skm.com/newsletters/winter99.pdf
Click to expand...

Another fine link, page 2 is very informative. Yet it does not explain how the DC offset was generated.
 
If the source has inductance, the fault current has to start at zero because you can't change current instantaneously in an inductor. If the fault voltage is > 0, however, at the instant on the sine wave when the fault starts, the current starts out with a high rate of change (V = L?di/dt) and it overshoots the normal peak, then eventually settles down to the steady state (symmetric) value. The current averages more than zero until it settles down and can be represented by a sine wave plus a decaying dc component. It is not symmetrical about the zero axis (the positive part is different from the negative part), so is called asymmetrical fault current.
 
chris kennedy said:
1) What is an AC wave form with DC offset?
Thanks
Click to expand...

In its simplest form it is an AC waveform like you are used to seeing, except that it is shifted up or down on the Y (voltage) axis so that it is symmetrical about a voltage other than 0V. That voltage is the DC offset.
 
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