2

I try to write down the equations that show how constellation symbols (PSK, QAM etc.) are modulated into a time-continuous signal. I read the articles referenced at the end. However, I am still stuck at how the inverse discrete Fourier transform (IDFT) is supposed to produce the subcarriers at frequencies $f_\mathrm C + n f_\mathrm \Delta$, where $f_\mathrm C$ is the carrier frequency and $f_\mathrm \Delta$ is the frequency distance between two adjacent subcarriers.

The IDFT of the constellation symbols $s_n$ of a single OFDM symbol is: $$ x_k = \sum_{n=0}^{N_\mathrm C - 1} s_n \exp\left( \jmath 2 \pi k n / N_\mathrm C \right) $$ $N_\mathrm C$ is the number of subcarriers. I ignored the cyclic prefix.

To my understanding, the analog signal is produced by using a rectangular pulse shape (of length $T$) before upconversion. \begin{align} x(t) &= \sum_{k=0}^{N_\mathrm C - 1} x_k \mathrm{rect}\left( \frac{t - nT}{T} \right) \\\\ &= \sum_{k=0}^{N_\mathrm C - 1} \sum_{n=0}^{N_\mathrm C - 1} s_n \cdot \exp\left( \jmath 2 \pi k n / N_\mathrm C \right) \cdot \mathrm{rect}\left( \frac{t - nT}{T} \right) \end{align}

Converting this to the frequency domain: \begin{align} X(f) &= \sum_{k=0}^{N_\mathrm C - 1} \sum_{n=0}^{N_\mathrm C - 1} s_n \cdot \exp\left( \jmath 2 \pi k n / N_\mathrm C \right) \cdot \mathrm{sinc}(T f) \cdot \mathrm{exp}\left( -\jmath 2 \pi f n T \right) \cdot T \\\\ &= \sum_{k=0}^{N_\mathrm C - 1} \sum_{n=0}^{N_\mathrm C - 1} s_n \cdot \exp\left( \jmath 2 \pi n T k f_\mathrm \Delta \right) \cdot \mathrm{sinc}(T f) \cdot \mathrm{exp}\left( -\jmath 2 \pi f n T \right) \cdot T & \text{since: } f_\mathrm \Delta = 1 / (N_\mathrm C T) \end{align} Here, $\mathrm{sinc}(q) = \sin(\pi q) / (\pi q)$ is the normalized sinc function.

I cannot see how the IDFT exponentials $\exp\left( \jmath 2 \pi n T k f_\mathrm \Delta \right)$ actually change anything besides the phase since they include $n T$ instead of the continuous time $t$. I see that is similar to the Fourier series of some signal $y(t)$ $$ y(t) = \sum_{n=0}^N c_n \exp\left( \jmath 2 \pi n t / T \right) $$ as indicated in this explanation. But, again, I do not see where the switch to continuous time happens for the IDFT exponentials.

What am I missing here?

Related articles:

  1. How are subcarrier symbols modulated into an analog OFDM signal? (asked by me)
  2. How to convert discrete-time QAM pulses to a continuous-time signal using the IFFT (for OFDM)?
  3. IFFT and DAC relationship in OFDM Transmitter
Cornelius
  • 23
  • 4
  • You still have the misconception that you do any specific pulse shaping (other than the inherent limited sum length of the IFFT), and that's probably what's confusing you here. There's no "rectangular pulse shaping before upconversion" done here; that would be very bad. I tried to clean up with such misconceptions in my answer to your other question. – Marcus Müller Feb 08 '24 at 15:23
  • Yes, there must be something wrong in how I think about these things: The output of the IDFT is just an array of complex numbers. I thought these numbers would be, for some time interval (--> therefore the rect function), multiplied by the carrier - one after the other. Is this wrong? – Cornelius Feb 08 '24 at 15:39
  • 1
    The output of the IDFT is just an array of complex numbers. right. multiplied by the carrier - one after the other. Is this wrong? Yes, that's wrong. They first go through a reconstruction filter as part of the digital-to-analog conversion process. However, that one analog carrier is totally irrelevant to the existence of the spectrum; it just shifts the whole baseband spectrum up. – Marcus Müller Feb 08 '24 at 15:57
  • @Cornelius Think about OFDM with one subcarrier and no upconversion. You start with a vector of $N$ zeros, except for one element $X_n$, at position $n$, which is complex. Perform the IDFT of the vector and give the result to a DAC. Can you calculate the result? It will be a sine wave whose amplitude and phase are determined by $X_n$ and frequency determined by $n$. Now add another element $X_m$ and repeat. I think this should clarify things for you. – MBaz Feb 08 '24 at 16:37
  • Does it help at all to observe how a cosine is composed of two exponentials as given by Euler’s formula: $2\cos(\omega t) = e^{j\omega t} + e^{-j\omega t} $, such that each exponential is simply a spinning phasor in time on the complex plane, and a single tone in the DFT. – Dan Boschen Feb 08 '24 at 16:51
  • @MBaz I understand that the output of the IDFT can be seen as a sampled mixture of (complex) sinusoids. You wrote: "give the result to a DAC", which - to my understanding - means using reconstruction filter or a pulse shape (equivalent). In my formulae I used a rectangular pulse shape. But according to Marcus Müller this is not correct. - My goal: I would like to construct the analog signal (baseband) and then to see how it looks like. – Cornelius Feb 08 '24 at 17:40
  • @Cornelius It's not actually a pulse shape in the traditional meaning. It's just that the DAC output is time limited (you feed it only $N$ samples). I guess you can think of it as the shape of the entire OFDM symbol of $N$ samples -- just don't try to apply it to each sample! – MBaz Feb 08 '24 at 18:03
  • @MarcusMüller What type of reconstruction filter does one use here? – Cornelius Feb 09 '24 at 09:31
  • @Cornelius no, a pulse shape and a reconstruction filter are not equivalent! – Marcus Müller Feb 09 '24 at 11:58
  • 1
    @Cornelius this is all very basic theory of continuous-time/discrete-time signal theory, i.e., of sampling, so I'm almost certain you have had a textbook that explains that in more depth than we could do here in the comments. For me, this was in the 3rd semester course "signals and systems", if you had a similar one. However, quickly: The reconstruction filter is everything that happens between a DAC emitting pulses with the desired amplitude and the perfectly bandlimited to $f_s/2$ time-continuous waveform. The "perfectly bandlimited" implies that if you multiply that filter's frequency … – Marcus Müller Feb 09 '24 at 12:02
  • …response with a rectangle from $-f_s/2$ to $+f_s/2$, it mustn't change. And the "reconstruction" aspect means that in time-domain, it needs to have $h(t=0)\equiv 1$, and $h(t=nT)=0,\quad T=1/f_s,, n=\mathbb Z \setminus{0}$, otherwise other sample instants would have an influence on the value at any given sample instant. The obvious choice here becomes the $h(t)=\operatorname*{sinc}(\pi f_s t)=\begin{cases} \sin(\pi tf_s)/(\pi f_s t ) & t\ne 0\ 1 & t = 0.\end{cases}$ – Marcus Müller Feb 09 '24 at 12:07
  • Idyllically, the Fourier transform of that function already is that rectangular function that we demanded multiplying with changes nothing about itself, and since both for the regions where it's 0, 0·0 = 0, and for the regions where it's 1, 1·1=1, and there's no other regions in a rectangular function, the criteria needed for this to be a sensible reconstruction filter are fulfilled. (We can go on to show that it's the only smooth filter that fits the requirements, but, again, not our job here to do basic DSP derivations that exist in a hundred textbooks) – Marcus Müller Feb 09 '24 at 12:20
  • @MarcusMüller Thank you for your patience. It wasn't my intention to annoy you. I am sorry if I did. :-/ – Cornelius Feb 09 '24 at 12:59
  • 1
    @Cornelius you didn't, don't worry! Thing is just that you're really just in need of refreshing of basics, so you can stop adding "magic" elements to OFDM that aren't there, as I honestly think this might be a bit frustrating to you, seeing this is your second question with this problem :) – Marcus Müller Feb 09 '24 at 13:07

0 Answers0