Proof that module of FT of 2 independent signals is sum of modules

Question

I found on these posts (PSD subtraction and PSD of a sum of two stationary real signals) what I expected : that, just like the variance of the sum of 2 independent signals is the sum of the variances of these signals, the psd of 2 independent signals is the sum of the psd's of these signals. I understand the explanation using the fact that the psd is the FT of the correlation function, but I am wondering if this can be prooved only using FT-space calculation. In other words, how to show that, for $x$ and $y$ being 2 independents discrete real signals, then: $$ |FT(x+y)|^2 = XX^* + YY^* + XY^* + X^*Y = |X|^2 + |Y|^2 + XY^* + X^*Y = |X|^2 + |Y|^2 $$ where $X = FT(x)$, $Y=FT(y)$, i.e how to show that : $$XY^* + X^*Y = 0$$ (using DFT notation for eg, x and y being N-lenght signals).

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After a few try, I end up here, for spectrum variable k, DFT on N points : $$(XY^* + X^*Y)[k] = \sum_{i,j}x[i]y[j]2\cos(2\pi k/N(i-j))$$ which really is 2 times the real part of the CSD. But then I don't see how to that this equals 0 using the fact that $x$ and $y$ are independents (which I didn't used so far in my calculation).

Cheers

Answer 1

Assume $x[n]$ and $y[n]$ represent two jointly WSS (wide-sense stationary) discrete-time random processes. Further assume that they have zero means, without losing the generality.

If $x[n]$ and $y[n]$ are independent, then they are also uncorrelated, and since they have zero means, they are also orthogonal; i.e.

$$ E\{ x[n]y^*[n-k]\} = 0 \tag{1}$$

Let $$z[n] = x[n] + y[n] \tag{2} $$ be the sum process, then PSD (power spectral density) of $z[n]$ is

$$ S_z(\omega) = FT\{ r_z[m] \} \tag{3} $$ where $r_z[m]$ is the Auto-Correlation sequence of $z[n]$:

$$ r_z[m] = E\{ z[n]z^*[n-m] \} \tag{4} $$

Based on Eq.1 and Eq.2, expand Eq.4 as:

$$ \begin{align} r_z[m] &= E\{ z[n]z^*[n-m] \} \\ & = E\{ (x[n]+y[n])(x^*[n-m]+y^*[n-m]) \} \\ & = E\{ x[n]x^*[n-m]\} + E\{ y[n]y^*[n-m]\} + \\ &+ E\{ x[n]y^*[n-m]\} + E\{ y^*[n]x[n-m]\} \\ & = r_x[m] + r_y[m] + 0 + 0 \tag{5} \end{align} $$

The zeros on the last line follow because of Eq.1, ($x[n]$ and $y[n]$ are othogonal).

Hence what follows is that the PSD of $z[n]$ is the sum of PSDs of $x[n]$ and $y[n]$ :

$$ S_z(\omega) = S_x(\omega) + S_y(\omega) \tag{6}$$

Everything up to this point is theory based on mathematical (probabilty) model of random processes. What you are after, however, is about the practical computation of the PSD estimation and how the independence of $x[n]$ and $y[n]$ is reflected into tho Fourier calculations.

Assuming we use the periodogram estimate for PSD : $$ \hat{S}_x(\omega) = \frac{1}{N} |X(\omega)|^2 \tag{7}$$

then what we want to show is, as you have also worked out,

$$ \hat{S}_z(\omega) = \hat{S}_x(\omega) + \hat{S}_y(\omega) \tag{8}$$

which implies that

$$ |FT(x+y)|^2 = (X + Y)(X^*+Y^*) = |X|^2 + |Y|^2 + XY^* + X^*Y = |X|^2 + |Y|^2 \tag{9} $$

and Eq.9 implies that:

$$ XY^* + X^*Y = 0 \tag{9}$$

Eventually it's convincing to show that $XY^* = 0$.

To show this is you can use the ergodicty assumption. If $x[n]$ and $y[n]$ are ergodic in their cross-correlation, then the theoretical expectation can be replaced by the practical sums as:

$$E\{x[n]y^*[n-m] \} = \lim_{N \to \infty} \frac{1}{N} \sum_{n=0}^{N-1} x[n]^*y[n-m] \tag{10} $$

then it can be shown that the estimate is equivalent to convolution of the sequences :

$$ \frac{1}{N} \sum_{n=0}^{N-1} x[n]y^*[n-m] = x[n] \star y^*[-n]\tag{11} $$ But from Eq.1 we know that the $x[n]$ and $y[n]$ are orthogonal implying that

$$ x[n] \star y^*[-n] = 0 \tag{12}$$

finally, if this sum is zero then it's Fourier transform also is zero, completing the proof.

$$ x[n] \star y^*[n] = 0 \implies X[k]Y^*[k] = 0 $$

Note that, as MArcus has already commented, the equality holds only for the ideal case of infinite length process realization and for any practical finite length sample sequence, the Eq.s (6,7,8,9,10,12) won't hold true;i.e., they will be approximate).