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How to find the RMS value of a signal comprising of sum of sinusoidals?

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vickyuet

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How to find the rms value of a signal comprising of sum of sinusoidals?
for example: RMS value of signal
i.e
f(t)=10sin(5t)+10cos(20t)+5sin(t)
 

Re: circuit analysis:

By definition the rms value of a periodic signal x(t) with period T is the following:

\[\sqrt{\frac{1}{T}\int_{t_o}^{t_o+T} x^2(t) dt}\]

In your case, T=1, and you can simply allow \[t_o=0\] and integrate the square of your expression from 0 to 1.
 

Re: circuit analysis:

jayc said:
By definition the rms value of a periodic signal x(t) with period T is the following:

\[\sqrt{\frac{1}{T}\int_{t_o}^{t_o+T} x^2(t) dt}\]

In your case, T=1, and you can simply allow \[t_o=0\] and integrate the square of your expression from 0 to 1.

The above formula is totally correct. However in a simple case as urs U can calculate the RMS value by multiplying the sine and cosine amplitudes by √2/2 and suming over the terms. Namely:

(√2/2)×Σ(sine or cosine amplitude)

Regards
 

Re: circuit analysis:

The Joy of Electronics said:
jayc said:
The above formula is totally correct. However in a simple case as urs U can calculate the RMS value by multiplying the sine and cosine amplitudes by √2/2 and suming over the terms. Namely:

(√2/2)×Σ(sine or cosine amplitude)

Regards

I don't think you can just sum the terms because you have to take entire signal
and square it and this produces cross terms. In other words, the squaring operations does not just produce the square of the individual terms. Another note is that the period is not 1, it is actually 2*pi seconds.

I am attaching the waveforms for each component of the waveform and the bottom one is the sum. I get an RMS value of 10.554V. Note that if you just did sqrt(2)/2 times the amplitudes you would get 17.675V which is close but not correct.

Best regards,
v_c
 

Re: circuit analysis:

OR SIMPLY

√∑(RMS of each term)^2


28_1170718674.GIF
 

Re: circuit analysis:

you can use the following formula:
rms=√Σ(each rms value)Λ2
 

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