# root-mean square value for sinusoidal signal that superimposed onto the dc signal

Discussion in 'Homework Help' started by coolgirl, Dec 17, 2009.

1. ### coolgirl Thread Starter New Member

Dec 17, 2009
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0
hello....i have a problem in solving the questions related with the signal..Let, say a sinusoidal signal of magnitude A and period T is superimposed onto a dc signal of magnitude B to produce a resultant sine wave which have peak value as (B+A).The resultant wave start at B instead zero value..So,how to find root-mean-square value for the resultant wave???Please help me to solve this problem...

2. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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You just add the DC value to each point, and then work the RMS from there. You have to add it BEFORE the averaging, though....you just can't tack it on at the end.

From a practical standpoint, you can get rid of superimposed DC by using capacitor coupling (or transformer coupling) in many cases. Then it's just the RMS value.

3. ### kdillinger Active Member

Jul 26, 2009
141
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It is the square root of the sum of the squares. Square the DC term and add it to the square of the RMS term then take the square root of that total.

RSS - Root sum square

4. ### Ron H AAC Fanatic!

Apr 14, 2005
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$\sqrt{\frac{2*B^2+A^2}{2}}$.

I got this by squaring

$B+Asin(w*T)$

Discard the resulting purely AC terms (which average to zero), then take the square root of the remaining DC terms.
I verified this with several example simulations in LTspice.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Which is the same as kdillinger's rendering .....

$\sqrt{B^{2}+A_{rms}^{2}}$

6. ### Ron H AAC Fanatic!

Apr 14, 2005
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It was a homework problem. I was explaining how to get there mathematically.