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As shown in the figure below, the Time Delay Integration (TDI, the right side of the figure) is aimed at accumulating multiple exposures of the same (moving) object, effectively increasing the integration time available to collect incident light.

The question is, can we simply multiply a positive value to each pixel within a low-contrast image in order to enhance contrast and to discard TDI technique? (e.g. example)

Someone may argue that pixels may have a risk at saturation after multiplying a contrast factor. Yes, but we can initially scale each pixel value down and multiply a contrast factor later to avoid the saturation issue. I am sure that I have missed something. I will appreciate it if you provide an answer, thanks!

enter image description here

Why is this question relevant to astronomy? Strictly speaking, the question is relevant to Astrophotography. Please refer to What was the first use of time-delay integration in Astronomy? Are there instances before GAIA?

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    $\begingroup$ Multiplying by a constant value (as opposed to using a nonlinear function) won't even "enhance contrast", let alone increase signal-to-noise, which is what longer integrations (whether staring or TDI) do. $\endgroup$ Commented Dec 9, 2022 at 12:28
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    $\begingroup$ relevant astronomy.stackexchange.com/questions/44086 $\endgroup$ Commented Dec 9, 2022 at 21:34
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    $\begingroup$ @uhoh asked you to make the relevance to astronomy, for example describing how the technique is used in astrophotography. That is a post about the use of TDI in astrophotography. Hence it is relevant to how you could improve this question. $\endgroup$ Commented Dec 10, 2022 at 3:29
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    $\begingroup$ @uhoh, thanks for your reminder! I have put a short description into the post! $\endgroup$ Commented Dec 11, 2022 at 9:21
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    $\begingroup$ Shall we at least temporarily ban chatbot-generated posting until it becomes clearer what it means for Stack Exchange? $\endgroup$ Commented Dec 12, 2022 at 21:24

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TDI is aimed at enhancing signal-to-noise ratio (SNR) rather than contrast. The enhanced SNR is proportional to $N^{0.5}$, where $N$ is the number of measurements on the same object.

Proof. Assuming noises are uncorrelated between two observations, the variance of integrated signal-to-noise ratio (SNR) can be simplified as follows:

$\displaystyle\sigma^2_{X+Y}=\sigma^2_{X}+\sigma^2_{Y}+2\text{Cov}(X,Y)=\sigma^2_{X}+\sigma^2_{Y}\sim2\sigma^2_{X}$.

The integrated SNR turns out to be:

$\displaystyle\frac{S_{X+Y}}{\sigma_{X+Y}}\sim\frac{S_{X}+S_{Y}}{2^{0.5}\sigma_{X}}=\left[\frac{(S_{X}+S_{Y})0.5}{\sigma_{X}}\right]2^{0.5}\sim\left(\frac{S_{X}}{\sigma_{X}}\right)2^{0.5}$,

, which is higher than individual SNR by $2^{0.5}$.

Similar conclusion can be drawn when measuring the same object $N$ times, the integrated SNR can be enhanced by a factor of $\sqrt{N}$.

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    $\begingroup$ @uhoh Thanks for your great efforts on maintaining Stack Exchange! I have converted my answer to text/LaTex! $\endgroup$ Commented Dec 11, 2022 at 9:03

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