# What happens to the standard deviation of a white Gaussian noise after low pass filter?

If I apply an ideal box-car low pass filter on a random signal with white Gaussian noise, what happens to the Gaussian distribution's $$\sigma$$? I know that the auto-correlation function becomes a sinc(), but not sure where to go from here... Thanks!

I assume you are working with discrete-time, since continuous-time white noise has infinite power ($$\sigma^2$$).

First, remember that the power of a stationary process is always equal to the autocorrelation at 0 ($$P_x = R_x[0]$$); and the variance is the autocovariance at 0 ($$\sigma^2_x = \rm{Cov}_x[0]$$). These 2 expressions are equal for processes with 0 mean, which is the case for white noise. So, if you already calculated the autocorrelation, you have the power.

But you don't need to calculate the autocorrelation of the filtered signal. You can use Parseval's property and calculate the variance in frequency domain:

$$\sigma_y^2 = \frac{1}{2\pi}\int_{2\pi} |H(e^{j\theta})|^2 G_x(e^{j\theta}) d\theta = \frac{1}{2\pi}\int_{-\theta_0}^{+\theta_0} 1 \times \sigma^2 d\theta$$ since the ideal lowpass filter is 1 between $$\pm\theta_0$$, and the spectral density of the input signal is $$\sigma^2$$ (white noise).

The result is very easy to calculate, and gives $$\sigma_y^2 = (\theta_0/\pi) \sigma^2$$.

• Thank you! Just what I'm looking for. Mar 16 '20 at 2:26
• Oh BTW, is it necessary that the output signal also follows Gaussian distribution? Mar 16 '20 at 2:29
• Gaussian noise passed through a linear, time-invariant filter stays Gaussian. Mar 16 '20 at 3:35
• @Juancho: Your answer makes perfect because it's a fraction of $\pi$ since white noise has constant variance across frequencies. But I'm very bad in frequency domain so could you define $H(e^{j\theta})$ and $G_{x}(e^{j\theta})$. Thanks. Mar 16 '20 at 4:53
• I am not Juancho, but it seems that H is the filter frequency response, and G is the system frequency response. Mar 18 '20 at 10:11