- #1

Skaiserollz89

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- 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.

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.