True-RMS, DC and AC?

Thread Starter

lippyf

Joined Sep 26, 2015
15
A voltage composed of a DC voltage and two sinusoidal with different frequencies:
u(t)= -3+3sin(100πt)-4sin(400πt)
What will a True-RMS-voltmeter show, if set to "DC" and "AC"?

I suppose in DC it will show -3, am I right?
How do I calculate the AC?
 

crutschow

Joined Mar 14, 2008
25,269
Depends upon the meter but, generally a true RMS voltmeter will show the sum of the ACrms and DC voltages when set to the DC range and only the ACrms value when set to the AC range.
 

MrAl

Joined Jun 17, 2014
7,761
A voltage composed of a DC voltage and two sinusoidal with different frequencies:
u(t)= -3+3sin(100πt)-4sin(400πt)
What will a True-RMS-voltmeter show, if set to "DC" and "AC"?

I suppose in DC it will show -3, am I right?
How do I calculate the AC?
Hi there,

You do not need to calculate each part individually, you can do it all at once.

RMS is defined as the root of the mean of the square of the signal.
Thus you would first square the entire signal, then find the mean, then take the square root, that gives you a result which should always be positive. For a complex signal you often have to solve for the integration times first. Do you know how to do all this?
A hint is the RMS value lies between 4 and 5 volts RMS.

If you dont like doing it that way, you can do each component individually then combine them to form one value. Do you know how to do it this way? This way avoids integration as long as you know the basic RMS values of each component first.
 
Last edited:

Thread Starter

lippyf

Joined Sep 26, 2015
15
Hi there,

You do not need to calculate each part individually, you can do it all at once.

RMS is defined as the root of the mean of the square of the signal.
Thus you would first square the entire signal, then find the mean, then take the square root, that gives you a result which should always be positive. For a complex signal you often have to solve for the integration times first. Do you know how to do all this?
A hint is the RMS value lies between 4 and 5 volts RMS.
U_rms=sqrt((-3)^2+(3^2)/2+(-4^2)/2) = 4.6368
is this the value?

so if I set DC range: U_dc_rms=sqrt((-3^2))=3
and if AC: U_ac_rms=sqrt((3^2)/2+(-4^2)/2)=3.5355

is this right?
 

MrAl

Joined Jun 17, 2014
7,761
Hi,

WOW, a resounding *YES*, very good work there :) for the 4.6 result.
That was the second way to do it.

But why did you divide by 2 in the second part?
Note that you must get the same result (4.6) if you do:
sqrt(DC^2+AC^2)
 

Thread Starter

lippyf

Joined Sep 26, 2015
15
Hi,

WOW, a resounding *YES*, very good work there :) for the 4.6 result.
That was the second way to do it.

But why did you divide by 2 in the second part?
Note that you must get the same result (4.6) if you do:
sqrt(DC^2+AC^2)
When i insert the value I calculated in sqrt(DC^2+AC^2), I do get 4.6368 so it seems like I got the right answer. But im still not sure if set to DC, will it show -3 or 3?
 

MrAl

Joined Jun 17, 2014
7,761
When i insert the value I calculated in sqrt(DC^2+AC^2), I do get 4.6368 so it seems like I got the right answer. But im still not sure if set to DC, will it show -3 or 3?

The RMS value of -3v DC is still 3 volts. RMS will be positive because the root of the mean of the square will always be positive.

I see what you did now. You put the '2' outside of the square because sqrt(2)^2=2.
I would have written it like this for clarity:
ACpart=sqrt[(3/sqrt(2))^2+(-4/sqrt(2))^2]

but the way you did it was ok too.
 
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