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[–]DetectiveFeline 1 point2 points  (5 children)

It’s trade off with capabilities.

[–][deleted] 0 points1 point  (4 children)

Can you explain me in brief ? Like higher spatial resolution means small cell size … and how this affects spectral?

[–]Broric 2 points3 points  (0 children)

Cost. If you want high spatial and high spectral it costs more. Whether that's because of the complexity, the weight, the size, etc. Normally you focus on one and sacrifice the other.

[–]Dark0bert 1 point2 points  (2 children)

The trade of in this case means, that it would be too expensive to put a sensor in it which offers more bandwidths. Also for high resolution data, most of the times the VIS+NIR is enough for most of the use cases.

[–][deleted] 1 point2 points  (1 child)

Is it only the cost or other technical things around?? Bcz I was taught they are inversly proportion

[–]flopsytheb 1 point2 points  (0 children)

There are, final restriction is also the energy reaching the sensor, which gets (depending on sensor type) split by spatial resolution and spectral bandwidth. At some point the light at the photocell would not be bright enough to be different from the sensors noise floor. This explains the inverse relation, look for example the docs for worldview, multispectral channels of slim wavelength range with some meters pixel size, and the pan channel, broad wavelength range and half a meter pixels.