# Why the RMS of a PSD curve is the root of the area below

I will try to explain what is my level of understanding of this problem, please correct me if I'm wrong:

1. RMS is the Root Mean Square, it represent the mean value of the input signal.
2. PSD is the measurement of the responses that shows me at which frequencies most of the energy is concentrated.
3. The area below a curve is the integration of that function.

My situation is that several random vibration tests are performed. These tests are called random tests because of the input signal. In contrast to a sine test where the structure is excited with a sinusoidal input, only one frequency is excited at a time, here 'all' frequencies are excited at the same time.

In this case PSD is measured in ${{g^2}/{Hz}}$ and RMS in ${g_{RMS}}$. Armed with that it easy to see that if you multiply PSD per the frequency range and you take the root of the result you will get something in ${g}$'s, but I don't know how exactly derive the famous relation:

$${g_{RMS}=\sqrt{\int_{f_1}^{f_2}PSD(f)df}}$$

The understanding that I have is very basic and it would be great if someone give me a clear idea of the relations among the RMS, PSD and the real signal. Thank you very much.

In the figure I have plotted a standard random signal.

• Can you point to a paper or book you're referencing? Some of these definitions seem a bit odd. – Phonon Oct 21 '13 at 22:07
• It would be nice to have a paper or some basics of this subject! I know that relation only because of the software that we are using, but I can't understand why is that. That's why I need a bit of help. – Sturm Oct 21 '13 at 23:23
• This should really be a comment, but I don't have enough reputation to add a comment it seems. I too would have referred to Parseval's Theorem, or the unitarity of the Fourier transform. However, there's one confusing aspect about your question which makes me think that something's fishy. The RMS value is not a quadratic measure, it's the root of a quadratic measure. So the left hand side of your equation should either carry a square or the right hand side a square root over the entire integral. So I guess you either mean the mean square measure (no root!) or you have made some other mista – Jazzmaniac Oct 22 '13 at 9:50

I think this is simply an aspect of Parseval's Theorem (e.g. click me)

It simply says: sum of squares in the time time domain equals sum of squares in the frequency domain. Substitute "sum" for "integral" if using the analog domain. In other words: total energy in the time domain equals total energy in the frequency domain.

This can easily reproduce your formula, $g_{RMS}$ represents the time domain energy and the integral on the right represents the energy in the frequency domain. The exact scaling depends on the details like length of the signal, periodicity, sample rate, etc.

• Beat me to the punch! :-) – Peter K. Oct 21 '13 at 23:34
• Would anybody like to comment the issue I brought up in my answer below? – Jazzmaniac Oct 22 '13 at 13:43

I think you mean why is the rms of the voltage signal in the time domain the square root of the area below the PSD. This is because the area below the PSD is the average power and average power is the square of V RMS. The RMS of the PSD curve itself is in W/Hz and is not the square root of the area under it which would be in $$\sqrt{W}$$ i.e. $$V$$

Firstly, $$X(2\pi)$$ is the complex coefficient at the frequency of $$2\pi$$ i.e. the sum of [the area under curve of [all real and imaginary (co)sinusoids at the frequency $$2\pi$$ with their individual phases and amplitudes times a complex sinusoid at frequency $$2\pi$$]] in $$Vs$$

$$\frac{1}{2\pi}\int _{-\infty}^{\infty}|X(\omega)|\,d\omega$$

Is the (continuous) sum of the magnitudes of the complex coefficients of all frequency components of the signal, which results in the average complex magnitude divided by the time interval $$[-\infty, \infty]$$ in $$V$$. This is the average voltage of [all real and imaginary (co)sinusoids in the signal with their individual phases and amplitudes and frequencies times a complex sinusoid at those frequencies]

Squaring this you get a power value but it's a meaningless one.

The sum of the average voltage of the product of each real and imaginary (co)sinusoid in the signal with their individual phases and amplitudes at a specific frequency times a complex sinusoid at that frequency is:

$$\lim_{T\to\infty}\frac{1}{2T}|X_T(\omega)|$$

The ASD is this sum of the average voltage at a specific frequency divided by the square root of the bandwidth:

$$A_{xx}(\omega) = \lim_{T\to\infty}\frac{1}{\sqrt{2T}}|X_T(\omega)|$$

Anyway,

$$E = \frac{1}{2\pi}\int _{-\infty}^{\infty}|X(\omega)|^2\,d\omega \equiv \int_{-\infty}^{\infty} |x(t)|^2~dt$$

Is the (continuous sum) of the absolute square of the complex coefficients of the components of the signal. I.e. you get the average of the absolute square of the complex coefficients divided by the time interval $$[-\infty, \infty]$$ in $$Ws^2$$. It is the energy of the signal.

$$P_{xx}(\omega) = \lim_{T\to\infty} \frac{1}{2T}|X_T(\omega)|^2$$

Is the absolute square of the magnitude of the total complex coefficient at that particular frequency divided by the time interval $$[-\infty, \infty]$$. It is the PSD, like $$E$$ it is also in $$Ws$$ or more appropriately $$W/Hz$$. The difference with $$E$$ is that $$E$$ is the average $$Ws^2$$ divided by $$s$$ whereas PSD is the $$Ws^2$$ at a specific frequency divided by $$s$$

$$\lim_{T\to\infty} \frac{1}{4T^2}|X_T(\omega)|^2$$

Is the sum of the square of the average voltage of the product of each real and imaginary (co)sinusoid in the signal with their individual phases and amplitudes at a specific frequency times a complex sinusoid at that frequency.

$$P_{\text{Avg}} = \frac{1}{2\pi}\int _{-\infty}^{\infty} \lim_{T\to\infty} \frac{1}{2T}|X_T(\omega)|^2 \,d\omega \equiv \lim _{T\to \infty }{\frac {1}{2T}}\int _{-T}^{T}|x(t)|^{2}\,dt$$

Is the continuous sum of [the absolute square of the magnitude of the complex coefficient at that particular frequency divided by the time interval $$[-\infty, \infty]$$] i.e. it is the average energy of the complex coefficients of the signal divided by the time interval $$[-\infty, \infty]$$, which is the same as the average power of the signal.

Take $$5+\cos(\omega t)$$. The average power of the signal in the time domain is $$25.5W$$ (the average of $$(5+\cos(\omega t))^2$$). The frequency domain transform is $$π (δ(ω - 1) + 10 δ(ω) + δ(ω + 1))$$. $$|X(2\pi)|^2$$ has the units $$V^2s^2$$. One of the frequency domain impulses has a magnitude of $$V\times 2T$$. When this is squared, it can be represented as an impulse with a magnitude of $$V \times 2T \times V \times 2T$$. An impulse of this magnitude would have an area of $${V}^2\times 2T$$. When this area, which expands with $$T$$ in either direction is divided by $$2T$$, and the limit to infinity of this relationship is taken, you get the $$V^2$$, which is $$\frac{2\pi(0.5^2 + 5^2 + 0.5^2)}{2\pi} = 25.5W$$.

The complex coefficient itself is the average real and the average imaginary voltage of [all imaginary and real (co)sinusoids in the signal at that frequency (which have their own individual phases and amplitudes) times the complex sinusoid at that frequency] times the integration interval.

The continuous-time periodogram is (where $$2T$$ is the length of the symbol):

$$S_{xx}(\omega) = \frac{1}{2T}|X_T(\omega)|^2$$

Which is used for continuous aperiodic signals that have a certain pulse width where the rest of the signal to infinity is 0, like an incoming OFDM symbol while it is in the analogue domain. The interval of integration in this case is finite and the frequency domain presents sincs at those frequencies instead: $$A\tau\frac{\sin\left(f\pi \tau\right)}{f\pi\tau}$$, where $$A\tau$$ on the real sinc is the sum of the real parts of [the areas in $$Vs$$ under all the real and imaginary cosinusoids and sinusoids in the pulse window at that frequency times a complex sinusoid of that frequency] and the $$A\tau$$ on the imaginary sinc is the sum of the imaginary parts.

The discrete-time finite sample periodogram is typically:

$$S_{xx}(\omega) = \frac{1}{N}\left|\sum_{n=0}^{N-1} x_n e^{-i\omega n \Delta t}\right|^2$$

This can be reformulated as:

$$S_{xx}(\omega) = \frac{1}{T}\left|\sum_{n=0}^{N-1} x_n e^{-i\omega n \Delta t} \Delta t\right|^2$$

Which is a sum of the real parts (and a separate sum of the imaginary parts) of [a riemann sum approximation of the area under the envelope created by [the samples of each real and imaginary (co)sinusoid in the signal at the frequency inserted into the function (with their own amplitudes and phases) times the complex sinusoid at that frequency]], and the output is a complex number with a real part of the result of the real sum and an imaginary part of the result of the imaginary sum, whose complex modulus is then squared and divided by the number of samples.

The power of a finite sample discrete-time signal is

$$P_{\text{Avg}}= \frac{1}{N}\sum_{m=0}^{N-1} \hat{S}_{xx}(m\Delta \omega) =\frac{1}{N}\sum_{m=0}^{N-1} \frac{1}{N}\left|\sum_{n=0}^{N-1} x_n e^{-im\Delta\omega n \Delta t}\right|^2$$

$$\equiv \frac{1}{N^2}\sum_{m=0}^{N-1} \left|X[m]\right|^2 \equiv \frac{1}{N}\sum_{n=0}^{N-1} \left|x[n]\right|^2$$

For a wide-sense stationary complex random process:

$$P_{xx}(\omega) = \lim_{T\to\infty} \frac{1}{2T}\mathbf{E}\left[|X_T(\omega)|^2\right] = \lim_{T\to\infty} \frac{1}{2T} \mathbf{E} \left[\int_{-T}^{T} x^*(t) e^{i\omega t}\, dt \int_{-T}^{T} x(t') e^{-i\omega t'}\, dt' \right] = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \int_{-T}^{T} \mathbf{E}\left[x^*(t) x(t')\right] e^{i\omega (t-t')}\, dt\, dt'.$$

So we can transform this into:

$$\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \int_{-T}^{T} \mathbf{E}\left[x^*(p) x(p+\tau)\right] e^{-i\omega \tau}\, dp\, d\tau.$$

i.e.

$$\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \int_{-T}^{T} R_{xx}(\tau) e^{-i\omega \tau}\, dp\, d\tau.$$

By substituting $$\tau = t'-t$$ and $$p = t$$ and the determinant of the jacobian matrix that transforms $$dt~dt'$$ to $$dp~d\tau$$ is $$1$$

The expected value removes the dependency on $$p$$ because it's a constant:

$$\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \int_{-T}^{T} R_{xx}(\tau) e^{-i\omega \tau}\, dp\, d\tau.$$

$$=\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} 2TR_{xx}(\tau) e^{-i\omega \tau}\, d\tau.$$ $$=\lim_{T\to\infty} \int_{-T}^{T} R_{xx}(\tau) e^{-i\omega \tau}\, d\tau.$$ $$=\int_{-\infty}^\infty R_{xx}(\tau) e^{-i\omega\tau}\,d \tau=R_{xx}(\omega)$$

The continuous time periodogram is (where $$2T$$ is the length of the symbol):

$$S_{xx}(\omega)=\mathbf{E}\left[\frac{|X_T(\omega)|^2}{2T}\right] = \int_{-T}^{T} R_{xx}(\tau) e^{-i\omega \tau}\, d\tau.$$

Where we need the PDF to compute the ensemble average $$\mathbf{E}$$

The discrete time periodogram is:

$$S_{xx}(\omega) = \mathbf{E}\left[\frac{1}{N}\left|\sum_{n=0}^{N-1} x_n e^{-i\omega n \Delta t}\right|^2\right] = \sum_{\tau=0}^{N-1} R_{xx}(\tau) e^{-i\omega \tau}$$

Where we need the PDF to compute the ensemble average $$\mathbf{E}$$ but if $$N$$ is ergodic then a time average can be used. You would get:

$$\lim_{N\to\infty} \frac{1}{N-\tau}\sum_{\tau=0}^{N-1}\sum_{n=0}^{N-1-\tau} x[n]x^*[n+\tau] e^{-i\omega \tau}$$

The estimator of the periodogram of a WSS uses one record:

$$\hat{S}_{xx}(\omega) = \frac{1}{N}\left|\sum_{n=0}^{N-1} x_n e^{-i\omega n \Delta t}\right|^2$$