I tried the following order of operations:

Let's consider the spectral density of the square of white noise with with a limited frequency band. If $S_{x} \! \left( f \right)$ is the spectral density of white noise then $S_{y} \! \left( f \right)$ is the spectral density of the square of this noise. For band-limited white noise, the spectral density of the power $S_{x} \! \left( f \right)$ is constant between $-B$ and $B$ and is equal to $N_{0}$ (spectral power per unit frequency). Then the power spectral density of a square of white noise with a limited frequency band is written as follows:

$$ S_{y} \! \left( f \right) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} S_{x} \! \left( \omega \right) S_{x} \! \left( f - \omega \right) d \omega \tag{1} $$

Substituting the value for $S_{x} \! \left( \omega \right) = N_{0}$ in the range $-B < \omega < B$ (and $0$ otherwise), we obtain:

$$ S_{y} \! \left( f \right) = \frac{1}{2 \pi} = \int_{-B}^{B} N_{0}^2 \, d \omega \tag{2} $$

Simplifying the integral we obtain:

$$ S_{y} \! \left( f \right) = \frac{1}{2 \pi} N_{0}^{2} 2 B = \frac{N_{0}^{2} B}{\pi} \tag{3} $$

But that answer is wrong.

  • 3
    $\begingroup$ Hi, welcome to DSP.SE! Please typeset your equations using $\LaTeX$ rather than adding pictures of your Maths. As for your question the integral you are trying to solve is a convolution integral. What you should actually solve for is: $$S_y(f) = \frac{N_0^2}{2\pi} \int_{-B}^{B} \mathbb{1}_{\{ |f-\omega| \leq B \}} \, d\omega $$ where $\mathbb{1}$ is the indicator function, which is 1 if $|f-\omega| \leq B$ and 0 otherwise. This integral effectively counts the overlap between two bandwidths centered at $f $ and $\omega$, respectively, both with a bandwidth of $2B$. $\endgroup$ Commented Jan 7 at 11:07
  • $\begingroup$ You know that the answer you came up with is wrong. Do you have the correct solution? If so, please add it to your question. $\endgroup$
    – Matt L.
    Commented Jan 7 at 13:51
  • $\begingroup$ Can you tell us where you got your $(1)$ from? It is wrong, and it is not too surprising that further analysis will give the wrong results. Note that the square of any (nontrivial) signal or noise will have positive mean, resulting in an impulse at $\omega=0$ in the PSD. $\endgroup$ Commented Jun 8 at 14:48

2 Answers 2


Let $\{X(t): -\infty < t < \infty\}$ denote a low-pass band-limited WSS white noise process of bandwidth $B$ Hz and total noise power $N_0B$. Thus, the $X(t)$'s are a collection of zero-mean random variables, and the power spectral density of the process is $$S_X(f) = \dfrac{N_0}{2}\operatorname{rect}\left(\dfrac{f}{2B}\right)\tag{1}$$ while its autocorrelation function is $$R_X(\tau) = E[X(t)X(t+\tau)] = N_0B \operatorname{sinc}(2Bt).\tag{2}$$ Note that all the $X(t)$'s are zero-mean random variables with variance $N_0B$ and that $X(t)$ and $X(t+\tau)$ are uncorrelated if and only if $\tau$ is a nonzero integer multiple of $\frac{1}{2B}$. For other values of $\tau$, $X(t)$ and $X(t+\tau)$ are correlated random variables.

Now suppose that $\{Y(t): -\infty < t < \infty\}$ is a related random process such that $Y(t) = X^2(t)$ for all $t, -\infty < t < \infty.$ The OP asks: what is $S_Y(f)$,the PSD of $\{Y(t)\}$? Well, let's determine $R_Y(\tau)$ first. We have that $$R_Y(\tau) = E[Y(t)Y(t+\tau)] = E[X^2(t)X^2(t+\tau)]$$ but generally cannot proceed further because we don't know the joint distribution of $X(t)$ and $X(t+\tau)$. However, when $\{X(t)\}$ is assumed to be a Gaussian white noise process, we have that the random variables $X(t)$ and $X(t+\tau)$ enjoy a jointly Gaussian distribution and it is known (see the section Properties: Higher moments in the Wikipedia link provided) that \begin{align}E[X^2(t)X^2(t+\tau)] &= \operatorname{var}\big(X(t)\big)\operatorname{var}\big(X(t+\tau)\big)+2\operatorname{cov}\big(X(t),X(t+\tau)\big)\\ &= (N_0B)^2 + 2N_0B \operatorname{sinc}(2Bt).\end{align} Applying $(1)$ and $(2)$, we see that $$S_Y(f) = (N_0B)^2\delta(f) + N_0\operatorname{rect}\left(\dfrac{f}{2B}\right)\tag{3}$$ with an impulse at $f=0$ as mentioned in my comment on the main question. Note that the total power has increased from $N_0B$ to $(N_0B)^2 + 2N_0B$ but the PSD still has support $[-B,B]$, that is, squaring has not increased the bandwidth. This is in sharp contrast to what happens with (finite energy) deterministic signals for which squaring generally increases the bandwidth, and so take all the above with a large grain of salt!


We should take this step by step. Let's say we have a noise signal $n(t)$ and it's Fourier Transform $N(\omega)$

Squaring in the time domain is multiplication with itself.

$$q(t) = n^2(t) = n(t) \cdot n(t) \tag{1}$$

Multiplication in time is equivalent to convolution in the frequency domain

$$Q(\omega) = N(\omega) * N(\omega) \tag{2}$$

where $*$ is the convolution operator.

The power spectral density is the magnitude squared of the spectrum

$$S_q(\omega) = |Q(\omega)|^2 = |N(\omega) * N(\omega)|^2 \tag{3}$$

So it's NOT the convolution of the PSD of itself, but you have to convolve the spectra and then square that result to get the final PSD.

  • $\begingroup$ I disagree with this answer. The square of any nontrivial signal, whether deterministic or stochastic, must have a positive mean, and so its PSD must include an impulse at the origin. The convolution (2) will not result in an impulse and so neither will (3) have an impulse. $\endgroup$ Commented Jun 8 at 14:39

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