Relationship between Fourier coefficients and power spectral density

[Skaiserollz89]

Homework Statement

Show that $\langle|c_{n,m}|^2\rangle=\Phi(f_{x_n},f_{y_m})\Delta f_{x_n} \Delta f_{y_m}$, where
$\langle|c_{n,m}|^2\rangle$ is the ensemble average of fourier coefficients of a phase screen given by $\phi(x,y)=\sum\sum c_{n,m}e^{i2\pi(f_{x_n}x+f_{y_m})}$, and $\Phi(f_{x_n},f_{y_m})$ is the phase power spectral density is the squared-magnitude of the fourier transform of $\phi(x,y)$.

Relevant Equations

$$\langle|c_{n,m}|^2\rangle=\Phi(f_{x_n},f_{y_m})\Delta f_{x_n} \Delta f_{y_m}$$

$$\phi(x,y)=\sum\sum c_{n,m}e^{i2\pi(f_{x_n}x+f_{y_m}y)} $$

Parseval's Theorem:
$$\int\int_{-\infty}^{+\infty}|\phi(x,y)|^2 \,dx\,dy=\int\int_{-\infty}^{+\infty}\Phi(f_{x_n},f_{y_m}) \,df_x\,df_y.$$

Here, $\Phi(f_{x_n},f_{y_m})=|\mathscr{F(\phi(x,y))}|^2$ is the Power Spectral Density of $\phi(x,y)$ and $\mathscr{F}$ is the Fourier transform operator.

Parseval's Theorem relates the phase $\phi(x,y)$ to the power spectral density $\Phi(f_{x_n},f_{y_m})$ by

$$\int\int_{-\infty}^{+\infty}|\phi(x,y)|^2 \,dx\,dy=\int\int_{-\infty}^{+\infty}\Phi(f_{x_n},f_{y_m}) \,df_x\,df_y$$

Substitution of $\phi(x,y)=\sum\sum c_{n,m}e^{i2\pi(f_{x_n}x+f_{y_m}y)}$ into the left side of Parsevals Theorem yields

$$|c_{n,m}|^2=\int\int_{-\infty}^{+\infty}\Phi(f_{x_n},f_{y_m}) \,df_x\,df_y$$.In $\langle|c_{n,m}|^2\rangle=\Phi(f_{x_n},f_{y_m})\Delta f_{x_n} \Delta f_{y_m}$, $\langle|c_{n,m}|^2\rangle$ represents the ensemble average of $|c_{n,m}|^2$ and $\Delta f_{x_n}$ and $\Delta f_{y_m}$ represent the frequency bin widths, so we need to consider the discrete nature of the Fourier transform I think.

In the continuous case, the expression $|c_{n,m}|^2$ represents the contribution of the continuous frequency range within the integral to the squared magnitude of the Fourier coefficient $c_{n,m}$. However, in the discrete case, we can approximate this integral as a sum over the frequency bins.

Thus, we can rewrite the equation as:

$$ |c_{n,m}|^2 \approx \sum_{f_{x_n}} \sum_{f_{y_m}} \Phi(f_{x_n}, f_{y_m}) \Delta f_x \Delta f_y$$where $\Delta f_{x_n}$ and $\Delta f_{y_m}$ represent the width of the frequency bins.

By approximating $\Phi(f_{x_n}, f_{y_m})$ as constant within each sum, it can be moved out in front of the double sum. However, it's important to note that this is an approximation and assumes that $\Phi(f_{x_n}, f_{y_m})$ doesn't vary significantly within each sum.

$$ |c_{n,m}|^2 \approx \Phi(f_{x_n}, f_{y_m})\sum_{f_{x_n}} \sum_{f_{y_m}}\Delta f_x \Delta f_y$$From here, I am not really sure what additional simplifications I can make to get to the result that

$$ \langle|c_{n,m}|^2\rangle=\Phi(f_{x_n},f_{y_m})\Delta f_{x_n} \Delta f_{y_m} $$

but it seems as though the ensemble average has something to do with collapsing the summation.
Any additional help here, or suggestions would be much appreciated.

 

[member38644]

One way to approach this problem is by considering the concept of ergodicity. In signal processing, ergodicity refers to the idea that the statistical properties of a signal can be determined by a single realization of the signal, as long as the signal is stationary (its statistical properties do not change over time).

In this case, we can consider our signal to be the function $\phi(x,y)$ and its Fourier transform $\mathscr{F}[\phi(x,y)]$. By assuming ergodicity, we can say that the ensemble average of $|c_{n,m}|^2$ is equivalent to the time average of the squared magnitude of the Fourier transform at a specific point in space.

Using this idea, we can rewrite the ensemble average as:

$$\langle|c_{n,m}|^2\rangle = \frac{1}{T} \int_0^T |\mathscr{F}[\phi(x,y)]|^2 dt$$

where T represents the total time duration of the signal.

We can then apply Parseval's Theorem to this time average, which gives us:

$$\frac{1}{T} \int_0^T |\mathscr{F}[\phi(x,y)]|^2 dt = \frac{1}{T} \int_0^T \int\int_{-\infty}^{+\infty} \Phi(f_{x_n},f_{y_m}) df_x df_y dt$$

Since $\Phi(f_{x_n},f_{y_m})$ is constant with respect to time, we can move it out of the time integral, giving us:

$$\frac{1}{T} \int_0^T \int\int_{-\infty}^{+\infty} \Phi(f_{x_n},f_{y_m}) df_x df_y dt = \int\int_{-\infty}^{+\infty} \Phi(f_{x_n},f_{y_m}) df_x df_y$$

Now, we can equate this result with the previous expression for the ensemble average, giving us:

$$\int\int_{-\infty}^{+\infty} \Phi(f_{x_n},f_{y_m}) df_x df_y = \langle|c_{n,m}|^2\rangle$$

Comparing this with the previous equation we derived, we can see