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[SOLVED] Root Mean Square

dwsmith

Well-known member
Feb 1, 2012
1,673
I am reading about the root mean square and Parseval's Theorem but I don't understand how we find $A_0$.

So it says the average $\langle x\rangle$ is zero and the $x_{\text{RMS}} = \sqrt{\langle x^2\rangle}$ where
$$
\langle x^2\rangle = \frac{1}{\tau}\int_{-\tau/2}^{\tau/2}x^2dt
$$
The Fourier expansion of $x(t)$ is
$$
x(t) = \sum_{n = 0}^{\infty}A_n\cos(n\omega t - \delta_n).
$$
Then we obtain an integral of a double sum which simplifies down to
$$
\langle x^2\rangle = A_0^2 + \frac{1}{2}\sum_{n = 1}^{\infty}A_n^2.
$$
Then there is a note that $A_0 = \langle x\rangle$ which is supposed to be zero. How do I find the nonzero $A_0$ value?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I am reading about the root mean square and Parseval's Theorem but I don't understand how we find $A_0$.

So it says the average $\langle x\rangle$ is zero and the $x_{\text{RMS}} = \sqrt{\langle x^2\rangle}$ where
$$
\langle x^2\rangle = \frac{1}{\tau}\int_{-\tau/2}^{\tau/2}x^2dt
$$
The Fourier expansion of $x(t)$ is
$$
x(t) = \sum_{n = 0}^{\infty}A_n\cos(n\omega t - \delta_n).
$$
Then we obtain an integral of a double sum which simplifies down to
$$
\langle x^2\rangle = A_0^2 + \frac{1}{2}\sum_{n = 1}^{\infty}A_n^2.
$$
Then there is a note that $A_0 = \langle x\rangle$ which is supposed to be zero. How do I find the nonzero $A_0$ value?
$A_0$ is the first coefficient of the fourier series transform, which is in your case:

$$A_0 = \frac 1 \tau \int_{-\frac \tau 2}^{\frac \tau 2} (x) dx = \langle x \rangle$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
If $A_n$ and $f_n$ are defined as
$$
A_n = \frac{f_n}{\sqrt{\omega_0^2 -n^2\omega^2)^2+4\beta^2n^2\omega^2}}
$$
and
$$
f_n = \frac{2}{\pi n}\sin\left(\frac{\pi n \Delta\tau}{\tau}\right),
$$
how do I find $A_0$ here since $f_n$ would have a division by 0.
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
If $A_n$ and $f_n$ are defined as
$$
A_n = \frac{f_n}{\sqrt{\omega_0^2 -n^2\omega^2)^2+4\beta^2n^2\omega^2}}
$$
and
$$
f_n = \frac{2}{\pi n}\sin\left(\frac{\pi n \Delta\tau}{\tau}\right),
$$
how do I find $A_0$ here since $f_n$ would have a division by 0.
Yep.
That would mean that $A_0$ is not defined.
So we'd have to assign it a special value.

As a possibility you could choose the limit when n approaches zero.

$$f_0 \overset{def}{=} \lim_{n \to 0} f_n = \lim_{n \to 0} \frac{2}{\pi n}\sin\left(\frac{\pi n \Delta\tau}{\tau}\right) = \frac {2 \Delta\tau} {\tau}$$
$$A_0 \overset{def}{=} \frac{\frac {2 \Delta\tau} {\tau}}{\sqrt{\omega_0^2 -0^2 \cdot \omega^2)^2+4\beta^2 \cdot 0^2 \cdot \omega^2}}$$
I'm not sure how the parenthesis should be balanced, so I'll leave it like this.