What is the RMS value of a 5 volt peak AC signal if it is referenced at 5V DC?

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gauthamtechie

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A 5 volt peak signal will have 5/1.414= 3.53 V RMS.
What is the RMS value for the case described in the question: A 5 V peak referred from 5 V DC, which will have a positive peak of +10V and lowest point at 0 V ?

Does it make sense to go with peak/1.414?
 
If you get a true RMS meter and measure the signal relative to the 5V line, it will still be 3.53V.

Maybe you should post the actual question.

Mike.
 
Pommie said:
If you get a true RMS meter and measure the signal relative to the 5V line, it will still be 3.53V.

Maybe you should post the actual question.

Mike.
Click to expand...

Let me put it this way: What is the effective voltage across a terminal that contains two sources connected: A 5 V DC + a sinusoidal Voltage source of 5 V peak ? If I measure it in LTspice, I get a sine waveform that starts at 5 volt, goes upto 10 volt positive peak, all the way down to 0 volt negative peak. Now what do I call it? A 6.123 V DC as suggested by mister T in #3? So is this value that is equivalent to a 6.123 V constant ripple free DC; which is what RMS is all about?
 
Use this equation to find RMS value:

[LATEX]V_{RMS}=\sqrt{Vdc^2 + \frac{V_{ACpeak}^2}{2}} = \sqrt{5V^2 + \frac{5V^2}{2}} = \sqrt{ 25V + 12.5V} = \sqrt{37.5V} = 6.123V [/LATEX]
 
Yes, that voltage is the equivalent in heating power of 6.123V DC (only) through a resistive load.
 
crutschow said:
Yes, that voltage is the equivalent in heating power of 6.123V DC (only) through a resistive load.
Click to expand...

True if referenced from ground. However the OP stated that it should be referenced from 5V. Hence my request that the full question be posted.

Mike.
 
Be aware that if you measure such a waveform with a DVM, you will have to use a so-called "True RMS AC+DC" go get a value of 6.123 volts. If the meter doesn't include the DC part in the measurement, the the reading will be 3.536 volts.
 
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