Hi guys, i have to learn from some papers this theory and im not sure i understand what`s going on with these meters and i want if you have some time to tell me if i solved correctly these problems.
1. With a DC voltmeter we want to measure the folowing signal: x(t)=4+3sin(w*t)+2sin(w*t)[V]
A DC voltmeter will see only the continue component( im not sure if it has this name in english, sorry if i confused you) of this signal so it will display the value: 4V.
2.Measure the following signal x(t)=2*sqrt(2)*sin(2wt)[V] with an AC voltmeter.
In my papers it says that the value displayed in this case is the peak voltage*0.707.
So, the value is 2V.
3. Here i have a problem. Again we use an AC voltmeter to measure a signal but in this case it has a continue part.
x(t)=3+2*sqrt(2)*sin(wt).
I dont know here if the peak voltage is (3+2*sqrt(2)) --- ( i think this is the right peak voltage) or 2*sqrt(2)).
All in all, i`d solve this exercise like this: U=(3+2*sqrt(2))*0.707=4.12V.
4.Calculate the mean voltage for the signal s(t)=2*sin^2(wt)
Mean voltage is the continue part ( if the signal doesnt have a continue part the mean voltage is 0) of the signal so sin^2(wt)=1/2(1-cos(wt))
s(t)=1-cos(wt).
The continue part is 1, so the mean voltage is 1V.

 

Instead of "continue component" you might use "DC offset" or something like that.

With respect to part (3) of your question it will depend on whether your AC voltmeter is a "true RMS" responding meter type or one which works on a rectifier + scaled reading principle or another of the various AC meter types in use.

The very simple digital & (analog) moving coil multi-meters on AC voltage ranges respond to the average value of the rectified (half or full-wave) input voltage. The reading is then scaled on the presumption that the AC input is purely sinusoidal with no DC offset.

Do you have any guidance on what the proposed AC volt-meter might actually be?