Find $A_0$ in Root Mean Square & Parseval's Theorem

In summary, the root mean square is a measure of how often a value is repeated and the Parseval's Theorem states that this measure is equal to the sum of the squares of the individual values.
  • #1
Dustinsfl
2,281
5
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?
 
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  • #2
dwsmith said:
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:

$$A_0 = \frac 1 \tau \int_{-\frac \tau 2}^{\frac \tau 2} (x) dx = \langle x \rangle$$
 
  • #3
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:
  • #4
dwsmith said:
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.
 
  • #5


In order to find the value of $A_0$, we can use Parseval's Theorem, which states that the total power of a signal can be calculated by summing the squares of the Fourier coefficients. In this case, we can use the formula $\langle x^2\rangle = A_0^2 + \frac{1}{2}\sum_{n = 1}^{\infty}A_n^2$ to solve for $A_0$. Since we know that the average $\langle x\rangle$ is zero, this means that $A_0 = \langle x\rangle = 0$. Therefore, the nonzero $A_0$ value is zero, which is consistent with the information given in the problem. In summary, we do not need to find a nonzero $A_0$ value because it is already known to be zero based on the given information.
 

FAQ: Find $A_0$ in Root Mean Square & Parseval's Theorem

What is the significance of finding $A_0$ in Root Mean Square?

Finding $A_0$ in Root Mean Square is important because it allows us to measure the average power of a periodic signal. This value is crucial in understanding the overall behavior and characteristics of a signal.

How is $A_0$ calculated in Root Mean Square?

$A_0$ in Root Mean Square is calculated by finding the average value of the squared signal over one period and then taking the square root of that value. This can be represented by the equation $A_0 = \sqrt{\frac{1}{T}\int_{0}^{T}f^2(t)dt}$, where T is the period of the signal.

What is the relationship between $A_0$ and Parseval's Theorem?

Parseval's Theorem states that the total power of a signal can be calculated by taking the squared sum of all of its Fourier coefficients. Since $A_0$ is the first Fourier coefficient, it is included in this calculation and therefore has a direct relationship with Parseval's Theorem.

Can $A_0$ be negative in Root Mean Square?

No, $A_0$ cannot be negative in Root Mean Square. This is because the squared value of a signal is always positive, and taking the square root of the average of these values will also result in a positive value.

How is $A_0$ used in practical applications?

$A_0$ in Root Mean Square is commonly used in signal processing and analysis, as it provides valuable information about the average power of a signal. It can also be used to calculate other important parameters, such as the root mean square value and the peak-to-peak amplitude of a signal.

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