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I am quite new to the field of optical communication, so perhaps this question is so easy.

In reading some papers I have faced to the concept of optical bandwidth. For example, they say the symbol rate is \$25~G~\frac{\mbox{sym}}{\mbox{sec}}\$ and as a result, the optical bandwidth is \$13.33~GHz\$. Note that that have not spoken anything about the waveform being used.

In electrical domain, the bandwidth is related to the waveform being used. The symbol rate determines the duration of each waveform which affects the electrical bandwidth directly. Can anyone explain how it is related to the optical bandwidth?

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  • \$\begingroup\$ Can you quote the relevant sections of the paper where you read this? \$\endgroup\$ – The Photon Jun 8 '17 at 20:22
  • \$\begingroup\$ @ThePhoton The AWG operates at a sampling rate of 25 GSa/s, leading to the optical bandwidth of 13.33 GHz \$\endgroup\$ – CLAUDE Jun 8 '17 at 20:29
  • \$\begingroup\$ Can you link to the actual paper, or give a citation so people can look it up and see the context? I'm not aware of any rule that bandwidth is 53.32% of baud rate. I typically use a rough rule of thumb that it's about 50% of baud rate. Probably there is some context in the paper that explains why they use such a specific alternate rule. \$\endgroup\$ – The Photon Jun 8 '17 at 21:51
  • \$\begingroup\$ @ThePhoton osapublishing.org/ol/abstract.cfm?uri=ol-39-11-3110 \$\endgroup\$ – CLAUDE Jun 8 '17 at 22:29
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In the paper you have supplied, the optical carrier is modulated with an electrical signal generated by an AWG. The bandwidth of the resulting optical signal will be identical to that of the electrical signal of the AWG and do indeed depend on the waveform.

The paper further states that the AWG employs a sample rate (not symbol rate) of 25 GSa/s. The generated electrical signal is an OFDM signal, which in this case uses an FFT length of 4096, yielding as many subcarriers. However, only 2184 subcarriers are used.

The spacing \$\Delta f\$ between two adjacent subcarriers is $$ \Delta f = \frac{f_s}{N} $$ with \$f_s\$ the sample rate and \$N\$ the FFT length used. (see this dsp.stackexchange answer)

In this case, the subcarrier spacing is thus approximately 6.1 MHz. When 2184 adjacent subcarriers are used, the spacing between the first and last subcarrier is \$2183 \cdot 6.1\text{ MHz} \approx 13.3\text{ GHz}\$. With the bandwidth of a subcarrier negligible compared to this, the bandwidth of the resulting optical signal is also approximately 13.3 GHz.

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