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why optical data is of high bandwidth that that of electrical copper data bandwidth when we use electrical signal to achieve optical data modulation ?

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  • \$\begingroup\$ you can transmit multiple colors and use optical filters at the receiver to separate the signals .... each of the signals could be running at maximum copper data bandwidth \$\endgroup\$ – jsotola Jan 6 at 21:39
  • \$\begingroup\$ yes but copper too can transmit data at different frequencies like optical colors (different frequencies ) and separate different frequencies at the receiver. so, that highlight the question ( for a small length say 1 meter between say two computer nodes which type of communication regarding the high bandwidth and data speed and density excluding cost and optical fibers length advantages. \$\endgroup\$ – dell pc Jan 6 at 22:05
  • \$\begingroup\$ you're ignoring the fact that the different frequencies that e.g. a coax cable can carry are limited, e.g. your average consumer-grade transmission line will cut off at some GHz. So, let's say your electrically usable bandwidth is 1 GHz. In optics, you're doing the same - conducting an electromagnetic wave in a waveguide – but you're doing it potentially with 10s of GHzes of bandwidth. \$\endgroup\$ – Marcus Müller Jan 6 at 22:55
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The problem here is not the feasibility of the modulation, but the ability for cables to carry high speed signals.

Especially with long length, it is much easier (and economical) to carry a high speed signal on en optical fiber than on an electric cable.

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  • \$\begingroup\$ yes, but exclude the length ( suppose the fiber-optic and copper cable are connecting two nodes of one meter apart ) which will provide the highest-bandwidth of data transmission and why? \$\endgroup\$ – dell pc Jan 6 at 21:56
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    \$\begingroup\$ attenuation isn't even the main problem. The main problem is that if you send a signal that covers e.g. 1 to 1000 MHz, then that signal will have about twice the bandwidth of its center frequency, and thus, the cable will have very different delays for the low and the high frequencies within that band. That means that you try to send 999 Mbaud, but since every symbol you send contains all the frequencies, the low frequencies of the next symbol "overtake" the high frequencies of the previous one – you get inter-symbol interference, and a mishmash on the output that you can't pick apart. \$\endgroup\$ – Marcus Müller Jan 6 at 23:13
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    \$\begingroup\$ now, if you now send the same 999 MHz bandwidth on an optical carrier frequency (some hundred Terahertz), the bandwidth is a tiny fraction of the center frequency – and the frequency-selective effects on the medium don't really matter. Hence, typically, high bandwidth gets easier the higher your carrier frequency becomes. \$\endgroup\$ – Marcus Müller Jan 6 at 23:14
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    \$\begingroup\$ @MarcusMüller, and yet in a fiber optic transceiver, for example, you still have to deliver, for example, 100 Gbps from host connector to the laser, over a distance of a few cm. And such things can be done. \$\endgroup\$ – The Photon Jan 7 at 0:34
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    \$\begingroup\$ @ThePhoton sure! But: a) if "short" means "shorter than a wavelength", things are relatively benign as long as you can match lengths and impedances, b) well, even local buses that are serial and high speed need some form of link training/equalization, which is a lot of hassle. \$\endgroup\$ – Marcus Müller Jan 7 at 8:30

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