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This question refers to the following excerpt from Computer Networks by Andrew Tanenbaum:

The amount of information that an electromagnetic wave can carry is related to its bandwidth. With current technology, it is possible to encode a few bits per Hertz at low frequencies, but often as many as 8 at high frequencies, so a coaxial cable with a 750 MHz bandwidth can carry several gigabits/sec. From Fig. 2-11 it should now be obvious why networking people like fiber optics so much.

Fig. 2-11 basically just shows the electromagnetic spectrum with "Fiber optics" spanning from \$10^{14}\$ to \$10^{15}\$ Hz, i.e. at substantially higher frequencies than other media (like twisted pair, coax or FM radio).

I understand that a signal's bandwidth corresponds to the "speed" at which it can be modulated, thereby influencing the amount of information that it can transmit. But why does the absolute value of the carrier frequency matter in this case? Why, for example, is a frequency band of 1 kHz more "valuable" if it is located around 100 MHz rather than around 500 Hz (i.e. the baseband)? Wouldn't this also mean that, for example, using a blue instead of a red laser as optical transmitter results in an increased transmission rate?

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    \$\begingroup\$ The carrier frequency doesn't matter. But (10^15) - (10^14) is the bandwidth. That is a big number which represents massive bandwidth. Data transmission rate is determined by signal to noise ratio and bandwidth. With a very high SNR, you can transmit a lot of data over a narrow bandwidth. Of course there are some practical limitations. Google Shannon Limit. \$\endgroup\$
    – user57037
    Commented May 18, 2016 at 3:09
  • \$\begingroup\$ I guess I misread the question. But when Tanenbaum said that it should be obvious why networking people like fiber-optics so much, he may have meant that it was because of the bandwidth. The available bandwidth between 10^15 and 10^14 is 9^14 Hz. \$\endgroup\$
    – user57037
    Commented May 18, 2016 at 3:15

3 Answers 3

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Tannenbaum is alluding to secondary effects, such as intersymbol interference, that reduce the effective signal-to-noise ratio.

Intersymbol interference is created by nonlinearities in the phase/frequency response of the channel. These irregularities are generally proportional to the carrier frequency — for example, if a channel centered at 1 MHz has a quality factor (Q) of 50, it will have a 3-dB bandwidth of 20 kHz. But a similar channel centered at 100 MHz will have a bandwidth of 2 MHz.

If you need a bandwidth of 20 kHz for your signal, the 100-MHz channel will have a flatter response over any 20 kHz segment than the 1-MHz channel, reducing the phase distortion.

This is just one way of thinking about it. In reality, there are many factors that affect the flatness of any given communication channel — and the analog circuitry used to interface to it. It's just that in general, it's easier to keep distortions of this type low if the bandwidth is a smaller fraction of the carrier frequency.

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    \$\begingroup\$ Thank you! So let me see if I got that correctly: When considering the Shannon-Hartley theorem \$C=B \log_2\left(1+SNR\right)\$, then secondary effects are responsible for a reduced SNR in certain situations, impairing the transmission rate? And this is just an additional benefit of fiber-optic communication (on top of immunity to EM interference, cheaper cables, reduced attenuation, ...)? \$\endgroup\$
    – user76520
    Commented May 18, 2016 at 6:13
  • \$\begingroup\$ Yes, that's the general idea. \$\endgroup\$
    – Dave Tweed
    Commented May 18, 2016 at 11:28
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The term bandwidth originally designated the limits of operation of radio transmission which varied mode of operation. Broadcast FM for example uses 200 khz around the carrier frequency. Within these limits there were three modes used frequency modulation, phase modulation, and wide band double sideband amplitude modulation.

Today use of the term bandwidth refers to the data rate and amount of information encoded on rf or optical signal. With coaxial cable as you increase the frequency of operation you also increase signal loss and with long runs packet loss. With Fiber Optics this is less of an issue. The color of a laser is similar to radio frequencies. Red is the lower frequency and blue is the higher. The higher frequency blue would have more bandwidth capability. As for all the digital data technical details nuances I defer to Mr Tweeds expertise. I am more into the telecommunications aspect of it.

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  • \$\begingroup\$ So when you're saying that "blue would have more bandwidth capability", are you referring to bandwidth as a measure for the data rate and the reason for this are the already mentioned secondary effects? Or do you mean the analog bandwidth, i.e. are we able to modulate blue light with a higher frequency than red light? \$\endgroup\$
    – user76520
    Commented May 18, 2016 at 6:25
  • \$\begingroup\$ I would think the blue light having smaller wavelength would be transmit more packets of info per given size of fiber optic cable . Kinda like the idea with Blu Ray and DVD. Same disc size and because the blue wavelength is shorter there are more tracks on a blu-ray disc. 4.6 gb DVD 25 GB blu-ray. Using a Radio analogy Red is 1MHZ and blue is like 1 GHZ , higher frequency more room to play... \$\endgroup\$
    – Old_Fossil
    Commented May 18, 2016 at 6:43
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    \$\begingroup\$ The difference between the frequency/wavelength for red and blue light is less than 2:1 -- 700 nm vs. 400 nm, respectively. The 10:1 range mentioned by Tannenbaum includes a large amount of IR bandwidth, and perhaps some UV as well -- 3000 nm thru 300 nm. But the bandwidth limit for any optical carrier comes from the electrical interfaces on the transmitting and receiving sides, not from any property of the light path. The key advantage of optical fiber is that you can use many carriers (wavelength-division multiplexing), each having the maximum bandwidth that the electronics can support. \$\endgroup\$
    – Dave Tweed
    Commented May 18, 2016 at 11:34
  • \$\begingroup\$ I wrote this a 4AM ..I may have omitted other things as well. modes of transmission and equipment play a part as well. \$\endgroup\$
    – Old_Fossil
    Commented May 18, 2016 at 14:32
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Actually I think you are missing the point. Tannenbaum, as far as I can see, is not saying that 1kHz of bandwidth is more valuable at higher frequencies. I think he is saying that the available bandwidth in a fiberoptic cable is much larger than the available bandwidth in a copper cable.

The progression is MHz, GHz, terahertz, petahertz. If the range for fiberoptic cable is 100 terahertz to 1 petahertz (as you mention in your question), that means the bandwidth is 1 PHz - 100 THz = 900 THz. That is a lot of bandwidth.

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