# A zero mean random signal is uniformly distributed between limits –a and a and its mean square value is equal to its variance

A zero mean random signal is uniformly distributed between limits –a and a and its mean square value is equal to its variance. What is the r.m.s value of the signal?

• The square root of this "mean square", by definition, since "RMS" is the square root of the mean square. Feb 14, 2013 at 14:33
• how do we solve it. Feb 14, 2013 at 14:35

This looks like homework, so here are some pointers :-

Firstly, you need to find the probability density function P(x) for the signal, given that :-

$\int\limits_{-a}^{+a} P(x)\,\mathrm dx=1$

The variance is given by :-

$\int\limits_{-a}^{+a} x^2P(x)\,\mathrm dx$

since the mean is zero.
The rms is the square-root of this by definition (as Olin says), so you have your answer in terms of a.

Q: A zero mean random signal is uniformly distributed between limits –a and a and its mean square value is equal to its variance. What is the r.m.s value of the signal?

A: $$V_{RMS}=\frac{a}{\sqrt3}$$

Proof: The RMS value of a signal is equal to the square root of the variance, only when the signal mean is zero. In this special case, RMS <=> Standard Deviation. Assuming this signal as a continuous random variable with uniform distribution between -a and a.

$$\sigma^2=\int_{-\infty}^\infty(x-\mu)^2f(x)dx$$

Doing the Mean µ = 0 and Probability Density Function f(x)= 1/2a:

$$\sigma^2=\int_{-a}^ax^2\frac{1}{2a}dx$$

$$\sigma^2=\frac{a^2}{3}$$ $$\sigma= V_{RMS} =\frac{a}{\sqrt3}$$

Remarkably is the same RMS value obtained for the triangular and sawtooth waveforms, both defined between voltages -a and a.

Or in peak-to-peak terms:

$$V_{RMS} = \frac{2a}{\sqrt{12}}$$

This result is useful when deriving the famous expression for the maximum ADC Signal-To-Noise-Ratio (for non Sigma-Delta types):

$$SNR (dB)=6.02n+1.76$$ where 2a = 1 LSB and n is the ADC number of bits.