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In Orthogonal frequency division multiplexing (OFDM), data symbols are modulated using an N pt-IFFT operation to produce ONE output OFDM

Assume we have total of $N$ QAM symbols which are input to the IDFT operation.

1) Is the output time domain samples $X_k$ also of length $N$ in length?

2) If ONE data QAM symbol has symbol duration $T$, is ONE OFDM symbol of symbol duration $T$ or is one SAMPLE of the OFDM symbol of duration $T$?

Thanks looking forward for your answers

Update

One example of OFDM modulator is the one used in LTE, in particular the subcarrier bandwidth is 15 KHz, and the OFDM symbol (and not one sample of the OFDM symbol) is 1/15k = 66.67 $\mu $ sec which is equal to the transmission time of one symbol (for example 1 QAM) over one subarrier. If my understanding is correct than the answer provided below is wrong. Can somoeone please double check?

Thanks

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3 Answers 3

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  1. The number of ouput samples $M$ of the IDFT is solely determined by the IDFT order $N$, where $M=N$. Whether some of the IDFT inputs (= subcarriers) remain unmodulated (i.e. are set to zero) is irrelevant for $M$.
  2. In communications, the symbol duration is defined as the time interval, in which the modulation of the information carrier is constant. This concept is applied to subcarriers in OFDM. Let $T_\mathrm s=1/f_\mathrm s$ be the sampling period, then the OFDM symbol duration without guard interval is $T_1=NT_\mathrm s$. If a cyclic prefix of $G$ samples is inserted, the OFDM symbol duration including the guard interval is $T_2=(N+G)T_\mathrm s$. Some literature defines the OFDM symbol duration as $T_1$, some as $T_2$. The subcarrier bandwidth in OFDM is $f_\mathrm{sc}=f_\mathrm s / N$. With the definition of $T_1$ it follows that $1/f_\mathrm{sc}=N/f_\mathrm s=NT_\mathrm s=T_1$. Before cyclic prefix insertion, a subcarrier has a constant QAM modulation during $T_1$. After cyclic prefix insertion this duration is increased to $T_2$.
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  • $\begingroup$ Thanks! Regarding your definition of what a symbol duration is, you said its the interval over which information carrier has a constant value, I am just wondering usually the information carrier is $+cos(\omega t)$ for example if BPSK modulated for $t\leq T_s$ and cosine funciton is not constant over Ts. How does this definition work then $\endgroup$
    – Tyrone
    Commented May 26, 2015 at 18:02
  • $\begingroup$ Good point. I should say: "is modulated with a constant value". Will edit... $\endgroup$
    – Deve
    Commented May 27, 2015 at 8:28
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  1. The output time domain samples are also of same length $N$. Your figure shows a part of a OFDM receiver. At the OFDM transmitter, IFFT is used.

  2. One OFDM symbol duration is $NT$ and obviously one sample of the OFDM symbol has a duration $T$.

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  • $\begingroup$ i am not sure this is correct. i believe one ofdm symbol is of duration $T$ $\endgroup$
    – Tyrone
    Commented May 26, 2015 at 1:20
  • $\begingroup$ @Tyrone, $\frac{1}{NT}$ is the sub-carrier bandwidth and hence $NT$ is the OFDM symbol duration (without cyclic prefix included). $\endgroup$
    – Oliver
    Commented May 26, 2015 at 1:32
  • $\begingroup$ in LTE the subcarrier bandwidth is 15 kHz. The OFDM symbol time (and not sample) is 1/15kHZ = 66.67 microsec can you please double check. its major confusion for me $\endgroup$
    – Tyrone
    Commented May 26, 2015 at 1:35
  • $\begingroup$ @Tyrone Please see the article risorse.dei.polimi.it/dsp/tlc/PDF-files/pilotbased.pdf $\endgroup$
    – Oliver
    Commented May 26, 2015 at 8:34
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Assume we have total of N QAM symbols which are input to the IDFT operation.

1) Is the output time domain samples Xk also of length N in length?

No. Remember IFFT is a radix-2 operation. The output of the IFFT is next nearest power of two to the number of QAM symbols. For example for 6 Resource block configuration (1.4 MHz) there are 72 subcarriers. The number of IFFT bins and hence the no. of IFFT output samples for this configuration is 128 with 128-72 is zero padded samples. 66.6 μ sec is useful symbol duration corresponding to the IFFT output. We also have to add the cyclic prefix to this useful symbol to produce the final OFDM symbol as shown in the following figure [1]

enter image description here

2) If ONE data QAM symbol has symbol duration T, is ONE OFDM symbol of symbol duration T or is one SAMPLE of the OFDM symbol of duration T?

One QAM symbol duration is 1/BW where BW is the transmission bandwidth.

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  • $\begingroup$ thank you very much. regarding question 2 you answered regarding the QAM symbol duration what about the OFDM symbol? thanks again. $\endgroup$
    – Tyrone
    Commented May 26, 2015 at 5:16
  • $\begingroup$ OFDM useful symbol duration is 1/15KHz as shown in the answer for question 1. $\endgroup$
    – Naveen
    Commented May 26, 2015 at 5:20
  • $\begingroup$ ok great and thanks. Just wondering why is the QAM symbol duration 1/tranmissionbandwidth. Although it is transmitted over one subchannel of much smaller bandwidth. i.e i would have thought that the duration of QAM is also 1/15k @Naveen $\endgroup$
    – Tyrone
    Commented May 26, 2015 at 15:58
  • $\begingroup$ Good question. Actually, I'm not certain about this. My assumption was that the switch rate for the serial-parallel converted should be same as sampling rate(Ts=32.7 ns) and thus the QAM symbol rate proportional to the Bandwidth. 1/15KHz looks too slow and would cause a considerable overhead even before IFFT, CP processing and OFMD symbol transmission stages. @Deve Please correct me if I'm wrong. $\endgroup$
    – Naveen
    Commented May 26, 2015 at 17:12
  • $\begingroup$ @Naveen The "QAM symbol rate" is not a common concept in OFDM and we should thus define what we mean by that. Fact is, that the modulation (by e.g. a QAM symbol) of every subcarrier is constant during the OFDM symbol duration. After that duration, its modulation is switched to the next QAM symbol. So saying that the QAM symbol duration is equal to the OFDM symbol duration makes sense. In my answer I have assumed this definition which leads to a QAM symbol duration of N/BW. $\endgroup$
    – Deve
    Commented May 27, 2015 at 8:45

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