There is no single, unique way of generating and transmitting OFDM over fiber. Luckily so, since otherwise reading these heaps on OFDM papers would have been really boring. To give an (incomplete) overview of some of the methods, Fig. 1 shows a possible hierarchy that would have made old Aristotle proud$^1$.

At the root “node” OFDM can be separated into optically and electrically multiplexed OFDM. In the electrical version (let’s call that E-OFDM) the multiplexing is usually done by using an inverse FFT in a digital signal processor (DSP). That the inverse FFT can be used for that purpose was shown in this post. In the optical version (O-OFDM) the subchannels are multiplexed optically, generally by combining spectrally overlapping signals in an optical coupler. There are various methods to achieve this, although some only generate “OFDM-like” signals whose subchannels are not all modulated independently.

Electrically multiplexed OFDM (E-OFDM)

The output of an inverse FFT is generally complex-valued even if the inputs were real-valued. This is a consequence of the asymmetry of the signal spectrum which is defined by the inputs to the FFT$^2$. This is shown in Fig. 2. Since there is no such thing as a complex electrical current, the DSP will output two real-valued signals, one for the real part and one for the imaginary part of the OFDM signal. These two output signals can be used to directly drive an optical IQ-modulator which can imprint the complex OFDM signal directly onto a continuous light source. The disadvantage of this approach is that the optical components needed are still quite costly, and people started looking for ways to work with only a single electrical drive signal.

real-valued baseband signals

The most simple way is to adjust the inputs into the inverse FFT in such a way that corresponding positive and negative frequency components are conjugate complex. The resulting output of the inverse FFT must then be real-valued, as illustrated in Fig. 3 and as can be derived from the definition of the real part of a complex number,

$$c_n(t) \exp\bigl(i \omega_n t\bigr) + c_n^*(t) \exp\bigl(-i \omega_n t\bigr) = 2 \Re \bigl\lbrace c_n(t) \exp\bigl(i \omega_n t\bigr) \bigr\rbrace\tag{1}$$

Hence, when $c_n = c_{-n}^* = c_{N-n}^*$ (the latter equality results from the cyclic nature of the N-point DFT, see also footnote 2), the resulting signal will be real-valued. While this is one of the easiest ways to generate real-valued OFDM signals and needs only a single digital-to-analog converter (DAC), it wastes half of the DSP bandwidth. Wasting half the DSP capacity is usually not a good way to set capacity records.

radio-frequency IQ-modulation

A more elegant, though slightly more expensive, way to turn complex-valued signals into something real (I stole that from some pop song), is to use an electrical IQ-modulator which “mixes” the OFDM signal up to an intermediate radio frequency. This is basically the same as the optical IQ-modulator, but wholly in the electrical domain. But we didn’t talk about the optical IQ-modulator yet, either, so here’s the principle of IQ-modulation:

To modulate two electrical drive signals, $b^{re}(t)$ and $b^{im}(t)$ onto an intermediate frequency, we use the setup of Fig. 4. Mathematically,

$$I(t) = b^{re}(t) \cos\bigl(\omega t\bigr) - b^{im}(t) \sin\bigl(\omega t\bigr)\tag{2}$$

Now if you have paid attention so far and always read the post footnotes, this will look familiar. Looking at footnote 2 of the first post, we can write the signal $I(t)$ as

$$I(t) = \Re\Bigl\lbrace \Bigl[b^{re}(t) + i\, b^{im}(t)\Bigr] \exp\bigl( i \omega t \bigr) \Bigr\rbrace$$

which should explain the naming of the variables. Obviously, the signal $I(t)$ must be real-valued and can be used to drive an intensity modulator, such as an electroabsorption modulator or field modulator such as a Mach-Zehnder modulator. However, you could also use it to drive a phase modulator or frequency modulator, and this has all been done, I think. The spectrum of $I(t)$ is shown in Fig. 5. While we have used the full capacity of the DSP, we need two DACs (one for $b^{re}$ and one for $b^{im}$) and a large modulator bandwidth due to the intermediate frequency. And of course the IQ-mixer. The modulator bandwidth is used more efficiently by doing the IQ-upmixing directly onto the optical carrier frequency, without an intermediate frequency, but this requires an optical IQ-modulator.

optical IQ-modulation

The optical IQ-modulator, shown in Fig. 6, does basically the same as the electrical IQ-modulator and as was described with (2), except that in this case the carrier frequency $\omega$ is not intermediate but the optical frequency of the laser used to transmit the OFDM signal. This is usually around 190 THz or so, and the bandwidth of the OFDM signal is just a tiny fraction of the full spectrum. This makes IQ-modulation a bit easier from the technical viewpoint. Here’s why:

To upmix a complex input signal (comprising two real-valued signals) onto a frequency $\omega$, we need a cosine and a sine function of $\omega$, as given in (2). Usually, one would use the same source for both, which outputs, say, the cosine oscillation at $\omega$ (the light amplitude, being a real-world electric field, must always be real-valued - working with the complex exponential function makes one forget this sometimes). To get the sine oscillation, we simply need to shift the phase of the cosine signal by $\pi/2$ (see Fig. 6). The optical path length needed to achieve such a phase shift is

$$L_{\pi/2}(\omega) = \frac{\lambda(\omega)}{4 n} \approx \frac{0.4\,\mu\text{m}}{n}$$

where $\lambda(\omega) = 2\pi \, c / \omega$ is the free-space wavelength and $n$ is the refractive index of the waveguide used. Such small optical path lengths are usually implemented by just heating a waveguide or using the electro-optical effect to change its refractive index $n$ a bit until it fits. Anyway, the point here is that $\lambda_0$ will be very nearly the same for all frequency components of the tiny OFDM band around the central $\omega$ up to a few decimals, so we need not worry about the frequency dependence of $L_{\pi/2}$. With electrical upmixing, on the other hand, the OFDM band may span an octave or more (a band from some $\omega_\text{min}$ to an $\omega_\text{max} \ge 2 \omega_\text{min}$), which means that $L_{\pi/2}$ should also span an octave or more. Even though my RF knowledge is somewhat basic, I think this might be a challenge. But that’s not part of this overview.

Anyway, optical IQ-modulators have been available for a few years now and are, at least in the labs, common equipment. Since there is no longer any redundancy in the (one-sided) optical spectrum – we only consider the right side of the spectra in Fig. 5 when the carrier frequency is optical, as sort of explained around eq. (6) of this post – the spectral efficiency will be higher with the optical IQ-modulator by a factor of 2 over the above approaches, but they are much more expensive than “regular” Mach-Zehnder or electroabsorption modulators. Like almost everywhere, one has to make a trade-off between performance and cost.

Optically multiplexed OFDM (O-OFDM)

The optical multiplexing of OFDM signals is actually fairly simple. The demultiplex is somewhat more involved, but we will discuss this another time. The OFDM symbol $C(t)$ was defined in eq. (10) of this post as

$$C(t) = \sum_{k=0}^{N-1} C_k(t) = \sum_{k=0}^{N-1} c_k \cdot \exp\bigl(i\omega_k t\bigr)$$

Hence, if we have a bunch of optical oscillators operating at the various required subcarrier frequencies $\omega_k$, we must “simply” impart the signals $c_k$ onto each of them, synchronously. This last word here is really important, since we need to have all subchannel synced at the receiver to perform the DFT over the OFDM symbol.$^3$ What makes this a little bit difficult is the need for all subcarrier sources to be locked in frequency for a duration of at least a symbol, but likely much longer, and also constant in phase over an OFDM symbol length.

There are several methods to generate the required optical subcarriers. The easiest, “brute-force” method is to use $N$ tunable laser diodes and to lock them to the subcarrier frequencies. However, with each laser having finite phase noise and frequency drift, it is very difficult to keep all phases constant for anything but the shortest OFDM symbols – and short symbols usually means few subchannels, which makes OFDM rather useless. Another approach is to use a single source and expand its output into a frequency comb, separate the comb into its spectral lines, modulate these lines separately, and combine all these subchannel signals to form the OFDM signal. This is shown in Fig. 7. There are again several methods to generate the comb, like mode-locked lasers possibly succeeded by a spectral expansion using e.g. four-wave mixing, or the very nifty recirculating frequency shifter [1], which we shall not discuss at the moment.

To modulate all subchannels independently, an optical transmitter is needed for each subchannel, including the necessary processing electronics, digital-to-analog converters, and modulators. The exact requirements on these parts depend on the modulation format and symbol rate used. For the quadrature amplitude modulation formats, which offer very high spectral efficiency and are easily generated in the DSP in the E-OFDM methods detailed above, the output signals are complex-valued, both in each subchannel and in the combined OFDM channel. To use these in O-OFDM, we need two DACs and an optical IQ-modulator. Not the cheapest solution. But since we need to spend a lot of money on this equipment, we at least get to modulate the subcarriers at much higher symbol rates. We are no longer limited by electronics to Mbaud, but easily can do Gbaud. These high symbol rates and the associated large subchannel spacing, on the other hand, make the OFDM signal very sensitive to group-velocity dispersion (GVD).

In the lab, we usually modulate the same signals onto all even-numbered and all odd-numbered subchannels so that for the experiments one only needs a total of two transmitters, but this does not transfer into “real” communication links.

Another drawback of this approach is the power loss in the $N:1$ subchannel combiner. Such a combiner inherently must have very high losses (reciprocity is the magic word) and only $1/N$ of the power in each subchannel will make it into the OFDM signal. Again in the demo lab, when the number of transmitters is 2 no matter how large $N$, these losses will be around 3 dB, which is acceptable. In a 32-subchannel system, this would be 15 dB at least. Sure, one can always amplify the signals again, but does so with degrading signal-to-noise ratio.

However, O-OFDM is relatively new and the requirements shown here may not be the final word.

1 Please note that this is only one way of structuring the OFDM methods that appeared very sensible to me. Everyone seems to have their own way of doing this and naming things. Fig. 1 is just a reference for the remainder of these posts.

2 At this point, looking at the figures, one may wonder where these negative frequency components come from that make the spectrum either symmetric or not. By definition of the OFDM symbol,

$$C(t) = \sum_{k=0}^{N-1} c_k \cdot \exp\bigl(i\omega_k t\bigr)$$

there are only positive frequency components in the signal. The reason lies in the sampling. Since the $N$-point DFT has an output consisting of $N$ time samples, the highest contained frequency component is, according to Nyquist,

$$f_\text{Nyquist} = \frac{1}{2\,\Delta t} = \frac{N}{2} \Delta f = \frac{\omega_{N/2-1}}{2\pi}$$

where $1/\Delta t$ and $\Delta f$ are the sample rate and subchannel spacing, respectively. It is quite easily imaginable by looking at the counter-clockwise rotating phasor in the complex plane. For all frequencies $\gt \omega_{N/2-1}$, the phasor rotates more than a half circle from one sample to the next and it rather looks like it (slowly) rotating clockwise instead, see Fig. 8. At $\omega_{N}$ it would rotate a full circle between samples and this could not be distinguished from it not rotating at all. This is the underlying reason for the periodicity of the spectrum of a sampled signal and the whole aliasing problem. Anyway, it makes more sense to show anything over $\omega_{N/2-1}$ to have negative frequencies, since it makes the whole symmetry thing and any estimations of required bandwidth easier.

3 One advantage of O-OFDM is that (ideally) we can fix the timing for each subchannel separately (since there is a separate modulator for each subchannel). Hence, in the presence of GVD we can avoid the need for an extra cyclic prefix by delaying the faster subchannels right at the transmitter so that they will arrive simultaneously at the receiver where we can then perform the DFT without the need for GVD compensation or cyclic extensions. This does require, however, that the subchannel symbols are not significantly distorted by GVD which then limits the effectiveness of this approach.

[1] S. Chandrasekhar, X. Liu, B. Zhu, and D. W. Peckham, “Transmission of a 1.2-Tb/s 24-carrier no-guard-interval coherent OFDM superchannel over 7200-km of ultra-large-area fiber,” in European Conference on Optical Communication (ECOC), September 2009, paper PD2.6.

Arc de Triomphe (panorama)

looking down Avenue des Champs-Élysées in the direction of Palais du Louvre (center)...

Arc de Triomphe (panorama) v2

This is another version of the Arc de Triomphe panorama, using the same set of photos, in which the individual shots were stitched together differently to give a more realistic perspective.
I think I like this one better, but with this method the panoramas cannot become too wide since the distortions increase significantly the further one goes to the left and right. That is why the Tour Eiffel is missing in this one...

La Défense (panorama)

This is another one taken from the Arc de Triomphe, this time looking in the opposite direction, down the Avenue de la Grand Armée towards La Défense.

La Tour Eiffel

inside La Tour Eiffel

This was taken going up the stairs inside the southern “leg” of the tower, somewhere below the lowest platform. At night, the whole tower is bathed in yellowish sodium light, inside and out. The round things in the center of the picture are the lamps, illuminating upwards away from me.

JSOVT(MB): La Tour Eiffel

We didn’t have a JSOVT(MB) for quite a while I guess, even though Julia’s been sitting everywhere. Heh. Well, here’s one that fits nicely into the Eiffel Tower series that I got going here. It was actually Julia’s idea to do this, it being her series and all.

Okay, so this was taken on the first platform of the Eiffel tower. We took the stairs up to the second platform, from where you have to take the elevator to the very top. However, after waiting in line for this elevator for about half an hour we were told that our tickets were only good for the stairs and thus the second platform, and we would have to buy “extension tickets” to go all the way up. This would have meant getting in that line again after buying the tickets and after a whole day’s walking through Paris we were too tired at 11pm to do that. So I guess we have to come back to Paris one day to finish the Eiffel Tower...

Arc de Triomphe du Carrousel (panorama)

There is nothing very amazing about this panorama and the only reason that I post it is the amount of work that went into it. It’s the little brother of the Arc de Triomphe that lives between the Louvre and the Tuileries. When we were at the place, there were at least three times as many people there as you can see in this picture. I shot the panorama with a lot of overlap between pictures and was able to remove a lot of the folks in post, as they were moving around. Lots of cutting, though, which took its time.

We didn’t visit the Louvre on this trip, since had we done so we wouldn’t have had time for much else, but it’s gonna happen one day. It’s only a 3-hour train ride...

I’m hoping this might be useful for other folks out there as well (especially the MATLAB wizards), so I’m making it available for download. If you think there’s something wrong or missing, let me know in the comments and I’ll try to fix it.

Download the DFT Sheat Sheet.

If I find the time and motivation, I’m planning on doing something similar for the regular Fourier transform (probably without the images) and for some of the more important trigonometric identities, since I find myself looking this stuff up in Wikipedia all. the. time.

Die Schrift heißt Windsor Light Condensed, und wird auch gerne von Woody Allen benutzt, wie man hört.

Dispersion

We have seen in this post that the DFT can be used at the receiver to demultiplex the OFDM signal. For an OFDM channel with $N$ subchannels, we simply take $N$ samples spread evenly over the symbol length $T$ and perform the DFT on these samples. The $N$ outputs of the DFT then correspond to the signals in each subchannel. This works perfectly in a back-to-back configuration (where we pop the receiver input right onto the transmitter output). Here, the symbols in all subchannels are aligned in time and to the DFT window, just as they are output by the inverse DFT at the transmitter. This is shown in Fig. 1a. Now let’s assume that we send the signal over a dispersive channel – i.e. a channel that introduces different group delays in some form or another. In wireless applications, there is multi-path dispersion, where fractions of the original signal take different paths through the air to get to the receiver, by e.g. being reflected or diffracted. In multi-mode fibers there is mode dispersion, where light in each mode takes a different amount of time to get to the receiver. In single-mode optical fibers, there is (almost) no mode dispersion, but chromatic dispersion causes each frequency component (and thus each subchannel) to have a different propagation velocity. In all cases, parts of the original OFDM symbol appear at the receiver outside the DFT window of length $T$ – either delayed copies or the fast/slow subchannels. This also means that parts of the neighboring symbols appear inside the DFT window. This is shown in Fig. 1b and c.

Fig. 1: When a cyclic prefix becomes necessary; a) illustrates the ideal alignment of symbols in each subchannel at the transmitter output (disregarding spectral overlap for now) b) symbol alignment with two-path propagation, c) symbol alignment with accumulated chromatic dispersion. The DFT window is also shown. Parts of neighboring symbols (hatched) will interfere with calculation of the DFT whenever there is dispersion.

If we let the window $T$ start with the arrival of the fastest subchannel at $t=0$, the slower subchannels would at this moment still carry the remainders of the previous symbol, since they have been delayed. Thus, these subchannels would have a symbol transition within the DFT window, and parts of both symbols would be mixed in the DFT outputs corresponding to the respective subchannels. On the subchannel level, this leads to inter-symbol interference (ISI). Its effect on the constellation diagram of a 16QAM-modulated subchannel is shown in Fig. 2 for various amounts of relative time shift between neighboring subchannels.

Fig. 2: Constellation diagrams of a single OFDM subchannel with various amounts of time shift between neighboring subchannels (cf. Fig. 1c); a) all subchannels aligned as in Fig. 1a, b) 0.01 T delay between neighboring subchannels, c) 0.02 T delay between subchannels. Modulation format within each subchannel was 16QAM, and there were 7 subchannels total in the OFDM channel.

Another way to describe the interference that occurs due to dispersion in OFDM channels is by looking at the DFT spectrum of a single symbol, as we did in this post, in Figs. 3 and 4. If a symbol transition occurs within the DFT window for some subchannel, the DFT spectrum of that subchannel will no longer be a single Dirac delta peak, but will have components that are spread out over many neighboring subchannel “slots.” This is shown in Fig. 3 for different amounts of misalignment between DFT window and subchannel symbols. Due to the nature of this interference it may be called inter-channel interference (ICI).

Fig. 3: Single-symbol OFDM spectra of a single subchannel; a) ideal case with no symbol transition within the DFT window, b) symbol transition at T/4 due to e.g. group velocity dispersion, c) symbol transition at T/2.

How the Cyclic Prefix Works

A way to avoid this is to extend the duration of each symbol in Fig. 1 without increasing the length $T$ of the DFT window at the receiver. This is shown in Fig. 4.

Fig. 4: Illustration of cyclic prefix principle. By making the OFDM symbol longer than the DFT window, the window can be shifted so that there is again no interference from neighboring symbols (hatched) during calculation of the DFT.

Such an extension is easily done since the contribution to the OFDM symbol of a single subchannel $k$ is just a continuous wave with complex amplitude $c_{k}$ at a carrier frequency $\omega_{k}$ – see also equation (10) in the basics post. We simply need to hold amplitude $c_k$ of the subcarrier oscillation constant a bit longer, say $T’$ with $T’ > T$. This reduces the symbol rate from the minimum possible $R = \Delta f = 1/T$ to the somewhat lower $R’ = 1/T’$. Now some people will argue that since

$$\Delta f \ne \frac{1}{T’}$$

we no longer have an OFDM signal and we’re just labeling it OFDM to get funded by using popular buzzwords in proposals. Not so. The subchannels $k$ are still orthogonal over the integration interval $T$, i.e. orthogonality condition (1) from this post still holds. The subchannel signals within that window have not changed (except maybe for a phase shift due to the dispersion – I’ll talk in another post how that can be fixed) and we still need the DFT of length $T$ at the receiver for demultiplexing. If we changed the DFT window length to $T’$, the orthogonality of the subchannels would be lost since (1) would be no longer fulfilled. To paraphrase that, since we are still using the DFT to demultiplex the signal, and the DFT essentially computes the orthogonality integral for each subchannel simultaneously, we have an OFDM signal. The ability to trade effective data rate for dispersion robustness is actually one of the strong points of OFDM and has been made use of for a long time.

As we will see, the additional signal processing at the transmitter required for dispersion robustness is very little compared to the processing required for (adaptive) dispersion compensation in a coherent transmitter. However, when including the (inverse) DFT operations in the Tx and Rx in our complexity calculation, the signal processing expenditure is almost equal – this was presented very nicely by Spinnler at ECOC a while back [1].

Generation of the Cyclic Prefix

In most implementations, the inverse DFT at the transmitter is computed using digital signal processors. Since the receiver DFT window length is fixed at $T$ (since this determines the subchannel spacing $\Delta f$), we cannot just make the subchannel symbols a bit longer. In each clock cycle, the inverse DFT is computed over the $N$ subchannel inputs to generate an OFDM symbol, and there is no way to compute additional fractional symbol slots this way. However, since on the subchannel level an OFDM symbol consists of a full number $k$ of (sampled) complex oscillations (see this post), the samples necessary to extend the subchannel symbol in time will actually look like the first few samples of the subchannel symbol due to the periodicity of the sine, cosine, and complex exponential. We can then extend all subchannels simultaneously by copying the first few samples of the whole inverse DFT-computed OFDM symbol to its end, or prepending the final few samples to its beginning, as shown in Fig. 5. This cyclic continuation, which gave the cyclic prefix its name. Since it’s called prefix, I assume that data usually gets prepended but I have no idea why it couldn’t be a cyclic postfix either.

Fig. 5: Illustration of the cyclic postfix. The symbols of Fig. 1b in the basics post are extended cyclically, since each subchannel symbol consists of a full number of sine or cosine oscillations.

With a cyclic prefix/postfix of a length corresponding at least to the offset between the slowest and fastest subchannel, the received subchannel will again be unaffected by any dispersion, as shown in Fig. 6. Due to the different start time of the DFT window (in this case shifted by $0.07 T$) there a phase shift appears in the constellation diagram. This phase shift is different for each subchannel – due to the dispersion – but can be easily corrected.

Fig. 6: Constellation diagram of the received signal with the same dispersion as Fig. 2b, but with sufficient cyclic prefix. Transmission data rate can be traded for received signal quality.

Spectrum

The power spectral density (PSD) distribution also changes. This occurs since the symbol rate within the subchannels is reduced with a cyclic prefix while keeping the subchannel frequency separation $\Delta f$ constant. Namely, the spectrum is no longer square as shown in this post, but acquires dips between the subchannels, as these now overlap less. This is shown in Fig. 7 for cyclic prefixes of 25 and 50 percent of the original symbol length $T$. For large cyclic prefixes, the spectra start looking similar to regular WDM spectra. However, the subchannels still overlap significantly, as shown by the dashed line showing the PSD of a single subchannel. Therefore, the DFT must still be used for demultiplexing instead of simple filtering, as explained before.

Fig. 7: Spectra of OFDM channels with and without cyclic prefixes; a) OFDM spectrum without CP, b) spectrum with CP of T/4, c) spectrum with CP of T/2. Even with large CP, the subchannel spectra still overlap significantly and can only be ideally recovered using the DFT instead of filtering. This post describes how the spectra were calculated.

[1] B. Spinnler, “Complexity of algorithms for digital coherent receivers,” in 35th European Conference of Optical Communication (ECOC), September 2009, paper 7.3.6.

We want to separate the Kerr nonlinearity evolution equation$^1$
$\newcommand{\vect}[1]{\mathbf{#1}}$ $\newcommand{\bra}[1]{\langle #1|}$ $\newcommand{\ket}[1]{|#1\rangle}$ $\newcommand{\braket}[2]{\langle #1|#2\rangle}$
$$\partial_z \ket{u} - i \bar \gamma \braket{u}{u}\ket{u} = 0\tag{A1}$$

with a DWDM signal

$$\ket{u} = \sum_{\nu=1}^N \ket{u_\nu} \exp \bigl[i \Delta \omega_\nu t\bigr]\tag{11}$$

into terms for SPM, XPM, and XPolM (ignoring FWM) that appear within the probe channel $\rho$ at $\Delta\omega_\rho$. The terms remaining from the triple sum which results when inserting (11) into (A1) and setting $\Delta\omega_\rho = 0$ are (cf. (12) and (13) in [1])

$$\partial_z \ket{u_\rho} - i \bar\gamma \Bigl[ \braket{u_\rho}{u_\rho}\ket{u_\rho} + \braket{u_\nu}{u_\nu}\ket{u_\rho} + \braket{u_\nu}{u_\rho}\ket{u_\nu} \Bigr] = 0\tag{A2}$$

These terms were summarized in [1] as

$$ \partial_\zeta\ket{u_\rho} - i \bar\gamma \Bigl[ \underbrace{\braket{u_\rho}{u_\rho}\vphantom{\Bigg|}}_\mathrm{SPM} + \sum_{\nu \ne \rho} \Bigl( \underbrace{\frac{3}{2}\, \braket{u_\nu}{u_\nu}\vphantom{\Bigg|}}_\mathrm{XPM} + \underbrace{\frac{1}{2}\, \vect U_\nu\cdot \vec\sigma\vphantom{\Bigg|}}_\mathrm{XPolM} \Bigr)\Bigr] \ket{u_\rho} = 0\tag{14}$$

However, it is important to notice that the term labeled ‘XPM’ is only the average XPM over all relative polarization states of probe $\rho$ and interferer $\nu$ and the term labeled ‘XPolM’ also contains the portion of XPM which depends on the relative SOPs of $\rho$ and $\nu$. Physically, XPM from co-polarized interferers must be larger than from orthogonal interferers because of the coherent mixing term that appears as a result of the inner product in (A1) or the third term in the square brackets of (A2). To examine this mathematically, we write

$$\ket{u_\nu} = u_\nu \ket{e_1} \quad \text{and} \quad \ket{u_\rho} = u_\rho \braket{e_1}{e_\rho} \ket{e_1} + u_\rho \braket{e_2}{e_\rho} \ket{e_2}\tag{A3}$$

with

$$\braket{e_n}{e_m} = \delta_{nm} \quad n,m \in \lbrace 1,2\rbrace$$

so that $\ket{e_1}$ and $\ket{e_2}$ form an orthonormal basis in Jones space which is determined by the SOP of $\ket{u_\nu}$. What we have done in (A3) is to separate the probe field into a part that is co-polarized with $\ket{u_\nu}$ and a part that is orthogonal to it by using the projection operators $\ket{e_1}\bra{e_1}$ and $\ket{e_2}\bra{e_2}$ on $\ket{u_\rho}$. We now expand the terms of (A2) that contain the cross-channel nonlinearities by using (A3) and have

$$\begin{aligned}
&\braket{u_\nu}{u_\nu}\ket{u_\rho} + \braket{u_\nu}{u_\rho}\ket{u_\nu}\vphantom{\frac{3}{2}}\\
&\quad = u_\nu^* \, u_\nu \, u_\rho \braket{e_1}{e_\rho} \braket{e_1}{e_1}\ket{e_1} + u_\nu^* \, u_\nu \, u_\rho \braket{e_2}{e_\rho} \braket{e_1}{e_1}\ket{e_2}\\
&\quad\quad +\, u_\nu^* \, u_\nu \, u_\rho \braket{e_1}{e_\rho} \braket{e_1}{e_1}\ket{e_1} + u_\nu^* \, u_\nu \, u_\rho \braket{e_2}{e_\rho} \braket{e_1}{e_2}\ket{e_1}\vphantom{\frac{3}{2}}\\
&\quad = 2 u_\nu^* \, u_\nu \, u_\rho \braket{e_1}{e_\rho} \ket{e_1} + u_\nu^* \, u_\nu \, u_\rho \braket{e_2}{e_\rho} \ket{e_2}\\
&\quad = \frac{3}{2} u_\nu^* \, u_\nu \, \ket{u_\rho} + \frac{1}{2} u_\nu^* \, u_\nu \, u_\rho \Bigl[ \braket{e_1}{e_\rho} \ket{e_1} - \braket{e_2}{e_\rho }\ket{e_2} \Bigr]\tag{A4}
\end{aligned}$$

Hence, the co-polarized part of the probe channel experiences twice as much phase shift as the orthogonal part (cf. the second equality above). In the last line of (A4) we have separated the phase shifts into a mean and a differential part – we could also have obtained that relation by doing the $\sigma$-expansion of (14) and using (A3). The differential phase shift on one hand gives rise to XPolM (and can be regarded as a nonlinear birefringence) and on the other hand causes the XPM to become dependent on the relative SOP between probe and interferer. To clearly separate both effects, we expand the differential using the product rule and obtain

$$\partial_z \ket{u_\rho} = \partial_z u_\rho \cdot \ket{e_\rho} + u_\rho \cdot \partial_z \ket{e_\rho}\tag{A5}$$

where $\ket{e_\rho}$ denotes the SOP of the probe. Thus by separating (A4) into a part aligned with $\ket{e_\rho}$ (which leads to no SOP change and is thus pure XPM) and a part that is orthogonal to it (leading to a change of the SOP only) we can achieve this separation. For the former we use the projection operator for the SOP of the probe, $\ket{e_\rho}\bra{e_\rho}$, and for the latter its orthogonal equivalent $\ket{e’_\rho}\bra{e’_\rho}$, with $\ket{e_\rho}$ and $\ket{e’_\rho}$ forming an orthonormal basis. We have

$$\begin{aligned}
\partial_z u_\rho \cdot \ket{e_\rho} &= i \bar \gamma u_\rho^* u_\rho \ket{u_\rho} + i \bar \gamma \, \frac{3}{2} u_\nu^* \, u_\nu \, \ket{u_\rho}\\
&\quad +\, i \bar \gamma \, \frac{1}{2} u_\nu^* \, u_\nu \Bigl[ \braket{e_1}{e_\rho} \braket{e_\rho}{e_1} - \braket{e_2}{e_\rho } \braket{e_\rho}{e_2} \Bigr] \ket{u_\rho}
\end{aligned}\tag{A6}$$

which is the same as

$$\begin{aligned}
\partial_z u_\rho &= i \bar \gamma u_\rho^* u_\rho \, u_\rho + i \bar \gamma \, \frac{3}{2} u_\nu^* \, u_\nu \, u_\rho\\
&\quad +\, i \bar \gamma \, \frac{1}{2} u_\nu^* \, u_\nu \Bigl[ \bigl|\braket{e_1}{e_\rho}\bigr|^2 - \bigl|\braket{e_2}{e_\rho }\bigr|^2 \Bigr] u_\rho
\end{aligned}\tag{A7}$$

which Karlsson and Sunnerud in [2, eq. (10)] have written as

$$\partial_z u_\rho = i \bar \gamma u_\rho^* u_\rho \, u_\rho + i \bar \gamma \, \frac{3 u_\nu^* \, u_\nu + \hat{\vect{S}}_\rho \cdot \vect{S}_\nu}{2} u_\rho$$

which has been adapted to current notation and where $\vect{S}$ is a Stokes vector and $\hat{\vect{S}}$ is an SOP (Stokes vector normalized to unit length). One can show if one so desires that these two are equivalent. Karlsson and Sunnerud’s expression requires the knowledge of the associated Stokes vectors whereas (A7) is given completely in Jones coordinates. The corresponding expression for the SOP change is

$$\partial_z \ket{e_\rho} = i \bar \gamma \, \frac{1}{2} u_\nu^* \, u_\nu \Bigl[ \braket{e_1}{e_\rho} \braket{e’_\rho}{e_1} - \braket{e_2}{e_\rho} \braket{e’_\rho}{e_2} \Bigr] \ket{e’_\rho}\tag{A8}$$

This is zero whenever $\ket{e_\rho} = \ket{e_1}$ or $\ket{e_\rho} = \ket{e_2}$, in agreement with the Stokes space description (15) in the original paper [1].

Again, I’d like to thank Chongjin, Curtis and Alexei for pointing this out to me.

1 I will omit the attenuation term that appears in the paper consequently throughout to make the equations shorter. Also, any equations whose number does not begin with A correspond to the equation in [1] with the same number.

[1] M. Winter, C.-A. Bunge, D. Setti, K. Petermann, “A statistical treatment of cross-polarization modulation in DWDM systems,” Journal of Lightwave Technology, vol. 27, no. 17, pp. 3739–3751, Sep 2009.
[2] M. Karlsson and H. Sunnerud, “Effects of nonlinearities on PMD-induced system impairments,” Journal of Lightwave Technology, vol. 24, no. 11, pp. 4127–4137, Nov 2006.

The power spectral density (PSD) of the OFDM channel is shown below in logarithmic dB units.$^1$ The spectrum is almost rectangular, and the signal power is spread very evenly over the used bandwidth (similar to white noise), which is one of the advantages of OFDM. The first out-of-band sidelobes are always ~10.5 dB below the peak PSD value$^2$, which are an artifact of the sinc-shaped subchannel spectrum. The width of these sidelobes decreases with the spectral width of each subchannel and thus with increasing subchannel count, as can be seen in the figure.

This nearly rectangular shape also allows multiple WDM OFDM channels to be located spectrally close to each other. However, if not filtered away (and thus affecting a number of the outside channels) the considerable width of the out-of-band OFDM spectrum imposes a minimum spectral distance – unless of course their subchannels are mutually orthogonal (i.e. all subchannels of all OFDM channels fulfill the orthogonality condition). This has been dubbed orthogonal band multiplexed OFDM and was published e.g. in [2]. The advantage of this approach is that each OFDM band can be generated by a transmitter that needs only a fraction of the bandwidth required to generate the full-width OFDM band. At the receiver, optical filters can separate the different bands with each band being received in a different low-bandwidth circuit.

Fig. 3 compares the (unfiltered) spectra of an OFDM channel and a comparable (i.e. same total bit rate) “regularly” modulated channel. Both are assumed to carry the same modulation format and have rectangular pulses. The single channel can be seen as the limit of Fig. 1 for a small number of subchannels. Thus, the sidelobes consume a considerable part of the spectrum. However, these sidelobes can be filtered very generously, whereas such filtering is detrimental for the orthogonality required for the OFDM subchannels (the subchannel symbols must remain constant over the DFT window length $T$).

Finally, here is a picture of an “OFDM spectrum” often seen in the literature (albeit usually more colorful):

This is a graph­i­cal super­po­si­tion of mul­ti­ple sub­chan­nel spec­tra. It is the cor­rect continuous Fourier transform for a single symbol of each sub­chan­nel by itself, but the com­pos­ite spec­trum of the OFDM sig­nal will not just be the sum of all these curves. The Fourier spectrum of more symbols of a single subchannel would not even look like this, but would instead be a single delta peak in the limit of infinite symbols (see footnote 1 for a derivation). When deal­ing with ran­dom sequences you have to work with the PSD instead of the direct spec­trum. Furthermore, the PSD is real-valued and has no neg­a­tive val­ues. The curves in Fig. 4 cor­re­spond also to the trans­fer func­tions of a DFT. With­out going into too much math­e­mat­i­cal detail, the sinc-shaped spec­trum is a result of the rec­tan­gu­lar time win­dow of the DFT in e.g. (12) which is then frequency-shifted by the expo­nen­tial term. Since the DFT trans­fer func­tion is roughly the same as the square root of the PSD of each sub­chan­nel,$^3$ the DFT is the so-called matched fil­ter to ide­ally demul­ti­plex OFDM signals.

Update (August 31, 2010): The derivation in footnote 1 is more or less completely redone, without using convolution. Should be clearer now. Also Fig. 4 is now embedded as a Flash object, which makes the file smaller and is less work, actually.

Update (September 1, 2010): All figures in this post are now Flash objects.

1 Working out the continuous spectrum of a randomly modulated subchannel or channel is actually a bit tricky. Instead of looking at a single OFDM symbol, as in (10) from the basics post, we need to look at the sequence of all symbols in a single subchannel $k$. This can be written as

$$C_{k}(t) = \sum_{m = -\infty}^{\infty} c_{mk} \cdot \Pi_{T}(t-mT) \exp\bigl(i \omega_{k} [t-mT]\bigr)$$

where $c_{mk}$ is a random variable and $\Pi_{T}(t)$ is a rectangular window of width $T$, centered on $t=0$. The exponential function describes the (unmodulated) OFDM symbol shape in subchannel $k$. Because of the linearity of the Fourier transform, we can transform each term of the sum separately and have

$$\tilde C_{k}(\omega) = \sum_m \tilde C_{mk}(\omega)$$

with

$$ \tilde C_{mk}(\omega) = c_{mk} \, \mathrm{sinc}\Bigl(\frac{\omega - \omega_k}{2}\Bigr) \exp\bigl(-i \omega m T\bigr)$$

where we used both the time-shift and frequency-shift properties of the Fourier transform. Now the exponential term will add a rapidly varying phase to the spectrum for all $\omega \ne 2\pi n / T$ with integer $n$, and the sinc-function has zeros at all $\omega = 2\pi n / T$ except for $n=k$. The phases will generally be different for all $m$ so that when superposing infinitely many single-symbol continuous spectra, as required by the sum above, we will have something like a random walk with very many steps in the complex plane. This means that $\tilde C_{mk}(\omega)$ will have a zero average, but $\bigl|\tilde C_{mk}(\omega)\bigr|$ will on average increase with the square root of the number of steps. For each $\omega$, the random walk will be different due to the $\omega$-dependent phase term above. And that’s without assuming anything about the statistics of the $c_{mk}$. We could obtain something meaningful out of this by ensemble-averaging $\bigl|\tilde C_{mk}(\omega)\bigr|$, which would give us something proportional to the sinc-function after sufficiently long averaging (actually, its absolute value), but hardly anything like Fig. 4.

We can make use of the known autocorrelation properties of the random variable $c_{mk}$ to more easily get an equally useful quantity, the power spectral density (PSD), which is defined for a single subchannel as

$$P_k(\omega) = \tilde C_{k}(\omega) \, \tilde C_{k}^*(\omega)$$

The PSD tells us how much power (or energy, depending how you look at it) is contained in an infinitesimal spectral slice $d\omega$ and is thus quite useful. Inserting $\tilde C_{k}(\omega)$ from above, we have

$$P_k(\omega) = \mathrm{sinc}^2\Bigl(\frac{\omega - \omega_k}{2}\Bigr) \sum_m \sum_n c_{mk} c_{nk}^* \exp\bigl(-i \omega \bigl[m - n\bigr] T\bigr)$$

The term $\sum_n c_{mk} c_{nk}^*$ corresponds to the autocorrelation of the time-discrete variable $c_{mk}$ which is zero for all $n \ne m$ for a random variable. Hence we have

$$P_k(\omega) = \mathrm{sinc}^2\Bigl(\frac{\omega - \omega_k}{2}\Bigr) \sum_m c_{mk} c_{mk}^*$$

The PSD of a single subchannel is thus sinc$^2$-shaped and scales with the energy $\sum_m c_{mk} c_{mk}^*$ in the signal, as it should. If we now add more subchannels $l$ with independent random data, the PSD will be simply the sum of the individual subchannel PSDs (the reason being that the cross-correlation between $c_{k}$ and $c_{k}$ is zero for $k\ne l$ and so are the mixing terms that would appear in the product term of the above equation for multiple subchannels). This is what is shown in Fig. 1.

2 Armstrong in [1] claims that the sidelobes are 13 dB below the peak, which is only true for a single subchannel (a sinc function), but not for the superposition of multiple subchannels, where all sidelobes add.

3 Actually, the DFT transfer function must be periodic as a result of the sampling in the time domain. The sinc-shaped transfer function of the Fourier transform thus overlaps with its images in the neighboring periods, which is the same as “wrapping around” at the period edges. This alters the function shape at the edges of the frequency window a bit. This is shown in Fig. 5 for the center channel, which is largely unaffected, and an edge channel, which differs significantly on the opposite edge. However, for a single OFDM symbol, the transfer function zeros will still be at the right places, and everything is okay. We just need to take care that there is no signal at the next transfer function maximum (shown dashed) because this will interfere with our subchannel of interest. <!-- Therefore, the OFDM signal is usually low-pass filtered before sampling. To allow for less steep filter transfer functions and to distort the subchannels as little as possible by this filtering, the outermost subchannels are often left empty. That the DFT order then is higher than the number of OFDM subchannels is also referred to as oversampling, since we have more sample points (equal to the DFT order) than necessary to distinguish all subchannels; see also this post. -->

[1] J. Armstrong, “OFDM for optical communications,” Journal of Lightwave Technology, vol. 27, no. 3, pp. 189-204, February 2009. http://dx.doi.org/10.1109/JLT.2008.2010061

[2] W. Shieh, Q. Yang, and Y. Ma, “107 Gb/s coherent optical OFDM transmission over 1000-km SSMF fiber using orthogonal band multiplexing,” Optics Express, vol. 16, no. 9, pp. 6378-6386, April 2008. http://dx.doi.org/10.1364/OE.16.006378

Gestern Abend wurden die Ergebnisse einer nicht-repräsentativen (immerhin haben sie das erwähnt – wahrscheinlich Textblöcke bei der Wahlredaktion geklaut) Umfrage präsentiert, die verglich ob Frauen oder Männer mehr Sexualpartner haben. Dabei kam heraus, dass Männer im Schnitt mit 18 verschiedenen Partnern geschlafen haben, aber Frauen nur mit 7. Das sollte einen eigentlich stutzig machen. Ich zumindest hab’ mich gefragt, mit wem die ganzen Männer denn schlafen. Für jeden Mann, der mit einer Frau schläft, muss es doch eine Frau geben, die mit einem Mann schläft. Ich hab’ das mal aufgemalt.

Beziehungen zwischen Männern und Frauen

Es gibt also eine bestimmte Anzahl von Kontakten (hier 10) zwischen verschiedenen Männern und verschiedenen Frauen. Wenn die Gesamtanzahl der Männer und Frauen etwa gleich wäre, müsste die Anzahl von Kontakten über die Gesamtzahl von Männern oder Frauen gemittelt ja gleich sein (hier jeweils 10/7 = 1.43 Kontakte pro Person). Nur wenn es deutlich weniger Männer als Frauen gäbe, würde sich ein Ungleichgewicht einstellen. Ich hab’ das mal aufgemalt.

Beziehungen zwischen Männern und Frauen

Die 8 Kontakte führen zu 8/3 = 2.67 Kontakten pro Mann und zu 8/7 = 1.14 Kontakten pro Frau, ungefähr das Verhältnis aus der Reportage.

Die interessanten Informationen, die ich aus dem Bericht mitnehme, sind also nicht, dass Männer mit mehr Frauen schlafen als umgekehrt, sondern dass

• es ungefähr 2.5-mal so viele Frauen wie Männer gibt (zumindest in der Zielgruppe der Sendung)
• und/oder sehr viel mehr Männer schwul als Frauen lesbisch sind (diese Verbindungen fehlen in meinen Grafiken)
• und/oder dass Männer gerne über die Anzahl ihrer Liebschaften lügen

Leider haben es die “Journalisten” verpasst, diese Schlussfolgerungen in ihrer Sendung zu ziehen. RTL halt.

Es gehörte natürlich zum Pflichtprogramm, mal auf das Hamburger Wahrzeichen zu klettern und in luftiger Höhe ein paar YOTOT(MC)* Bilder zu knipsen. Übrigens, wenn man die 400-irgendwas Stufen des Michel auf sich nimmt, kann man die halbe Stunde Wartezeit am Fahrstuhl überspringen – billiger wird der Spaß dadurch zwar nicht, aber im Vergleich zu manch anderen Kirchen sind die Treppen geradezu kommod. Wenn man sich dann erstmal durch die Horden von Gleichgesinnten zu den Plätzen mit Aussicht durchgekämpft hat, muss man nur noch seine Kamera durch die Gitter stecken und kann ein paar erstklassige Panoramen schießen, mit fantastischem Blick über den Hafen auf der einen Seite und die Innenstadt auf der anderen. Bis man weggedrängelt wird...

Hier nun, ohne noch mehr Gelaber, die Bilder:

YOTOT(MC): Saint Michaelis, south-ish view (panorama)

YOTOT(MC): Saint Michaelis, west-ish view (panorama)

YOTOT(MC): Saint Michaelis, north-ish view (panorama)

YOTOT(MC): Saint Michaelis, east-ish view (panorama)

More tk.

* Yaisog on Top of Things (Mostly Churches)

This post continues the introduction to optical OFDM that I started here.

In this post, we’ll discuss what an OFDM symbol looks like and show, starting from the orthogonality condition, that the discrete Fourier transform can be used to demultiplex an OFDM channel into its subchannels. Using the Fourier transform, we also take a look at the OFDM spectrum.

Basically OFDM is just plain old frequency-division multiplexing (FDM) with the orthogonality condition (7), namely

$$\Delta f = \frac{1}{T}$$

where $\Delta f$ is the subcarrier separation and $T$ is the symbol duration. Since the symbol rate $R = 1/T$, we can also write $\Delta f = R$.

In (O)FDM, multiple signals are transmitted using different carrier frequencies. This is just like piano music, where each tone represents a subchannel and the notes correspond to data modulation – in this case simple on-off keying (either a piano key is pressed or not). Mathematically, we can write this for a piano with $N$ keys (or an OFDM channel with $N$ subchannels) as

$$C(t) = \sum_{k=0}^{N-1} C_k(t) = \sum_{k=0}^{N-1} c_k \cdot \exp\bigl(i\omega_k t\bigr)\tag{10}$$

where the $f_k = \omega_k/2\pi$ fulfill the orthogonality condition above. Again, the $c_k$ are the actual encoded data and the $C_k(t)$ are the subchannel symbols. The $C_k(t)$ can either be numbers in a processor that we use to generate our (O)FDM signal or electrical / optical field quantities which can simply be superposed as in (10) by combining the fields from multiple sources. The figure below shows what the subchannel symbols (actually, their real part) look like for the first few $k$, and also a superposition of some such symbols.

Fig. 1a: (click to enlarge) subchannel symbols A₀ through A₅, each encoding the data c_k = 1, where A = Re{C} (see the previous post); the symbols differ by the number of full oscillations they describe. Also shown is the OFDM symbol for c₁ = c₃ = c₄ = 1 (bottom right). The symmetric shape of each symbol is a result of the c_k being real-valued, a property of the Fourier transform.

Fig. 1b: (click to enlarge) subchannel symbols with encoded data, c₁ = 0.5 (amplitude modulation), c₂ = -1 (inversion), c₃ = i (phase shift), and the OFDM symbol resulting from their superposition. The symbol is no longer symmetric due to the complex-valued spectrum resulting from c₃.

For music, a trained ear will be able to differentiate the tones and tell which notes were played, or “transmitted.” Similarly, a “trained” FDM receiver will be able to demultiplex the (O)FDM signal. How does it do that? It can use the known waveform of the transmitted subchannel symbols (that is his training) and compare the signal $C(t)$ to this waveform using the orthogonality property of the subchannels,

$$b_n =\frac{1}{T} \intop_0^T C(t) \cdot \exp\bigl(-i \omega_n t\bigr) \quad \text{with} \quad b_n = \begin{cases} c_k & n = k \\ 0 & n \ne k\end{cases} \tag{11}$$

which is the orthogonality definition (1) that was given in part 1. The integral will be zero for all subchannels $k$ with $k \ne n$ and will yield the coefficient $c_k$ when $n = k$. This is illustrated below for $C(t)$ of Fig. 1 (bottom right) and $n = \lbrace 1,2 \rbrace$. For $n=1$, the integral (area under the curve) is clearly greater than zero, while for $n=2$ it is (not so clearly) zero.

Fig. 2: (click to enlarge) illustrates the mechanism of extracting the original data from the OFDM symbol C according to (11); the coefficient b_n corresponds to the area under the curve C · C_n, shown gray for n = 1 (top) and n = 2 (bottom).

The integral (11) looks suspiciously similar to the Fourier transform (FT) integral, except for its integral limits, which are $-\infty$ and $\infty$ in the case of the FT. More exactly, this particular integral is called a short-time Fourier transform (STFT, with a rectangular window), which is often used for spectrograms, albeit usually with a different window function. Since we are only interested in the Fourier coefficients at the $N$ discrete frequencies $\omega_n$, we can use the discrete Fourier transform (DFT) of order $N$ to calculate the $b_n$, which is a simple sum instead of an integral:

$$b_n = \frac{1}{N} \sum_{m=0}^{N-1} C(t_m) \exp(-i \omega_n t_m) \tag{12}$$

where the $C(t_m)$ are, per definition of the DFT, samples of the received signal at $N$ equidistant points within the transformation interval $T$, so that $t_m = m \cdot T/N$. Thus, to differentiate $N$ subchannels, we need to sample the incoming OFDM signal with a sample rate that is (at least) $N$ times the symbol rate.

If we write the subchannel carrier frequencies as $\omega_n = 2 \pi n\cdot \Delta f = 2 \pi n / T$, we can rewrite (12) as

$$b_n = \frac{1}{N} \sum_{m=0}^{N-1} C\biggl(m\frac{T}{N}\biggr) \exp\biggl(-i 2 \pi \, \frac{n\cdot m}{N}\biggr)\tag{13}$$

which is simply a multiplication of the signal samples with some phase terms. A quick algorithm that makes use of the various redundancies in calculating (8) for all $n$ simultaneously is called the fast Fourier transform (FFT). It is this algorithm that is used in actual OFDM receivers to demultiplex the OFDM channel into its subchannels, after the incoming signal has been sampled and digitized.

We can see a couple of interesting properties of an OFDM signal by looking at its spectrum. To obtain this, we need only solve the Fourier transform (6) for all $\omega$, not just the $\omega_n$. We’ll do it for a single symbol of a single subchannel first, with $C(t) = c_n \exp (i \omega_n t)$. The resulting, sinc-shaped, spectrum is shown below.$^1$ The sinc function shape is a result of the rectangular envelope of the pulses that are used to modulate the subchannel, which lead to the rectangular window in the STFT (11). Clearly (hopefully), the spectrum has zeros at all $\omega_k$ except for $k = n$. Hence, the DFT spectrum, which only gives the spectral content at the $\omega_k$, will only have a single peak. This value of this peak is (at least proportional to) the desired output $c_n$.

Fig. 3: (click to enlarge) spectra of a single subchannel, n = 2; continuous Fourier transform (top) and discrete Fourier transform (bottom).

Since the Fourier transform is linear, we can determine the spectrum of a single symbol of the whole channel by simply adding the (complex) spectra of all subchannels. This is shown below for PAM-modulated subchannels$^2$ (in which the $c_n$ are real-valued).

Fig. 4: (click to enlarge) waveforms (top) and spectra (bottom) of the OFDM symbol with c₁ = c₄ = 1, c₂ = 0.5, c₃ = 0.75, and c₅ = 0.25; continuous (left) and discrete (right) case. Note that the spectrum of the N-point DFT is periodic with period N, as has been hinted at.

Initially, here was a paragraph on the long-time spectrum of an OFDM channel (encompassing more than a single symbol), but I think I’ll turn this into a post of its own, eventually. I guess this is quite enough for a first entry on OFDM. Hope it wasn’t too confusing. By the way, the graphics look a bit better this time, because I made the actual curves with Mathematica which is really great for such stuff...

Note Equation (10) describing the transmitted symbol $C(t)$ also almost looks like an inverse DFT, except that the functions which are summed over are continuous. Since we only need $N$ samples of the symbol for demultiplexing, we could just generate those $N$ samples using the inverse DFT. This is what is usually done in DSP processors at the transmitter. These samples are then converted to analog waveforms approximating (10) using a digital-to-analog converter (DAC) and sent over the fiber.
1 To obtain the real-valued sinc function in this transformation, we need to change the transformation interval boundaries to $(-T/2, T/2)$ (Hermitian symmetric functions have a real-valued Fourier transform). Otherwise there will be a linear phase superposed on the sinc as a result of the time shift.
2 PAM stands for pulse amplitude modulation