Exercise 4.3: Subcarrier Mapping

From LNTwww


Two SC–FDMA arrangements

The diagram shows two transmission schemes that play a role in connection with  Long Term Evolution  $\rm (LTE)$.  These block diagrams are referred here neutrally as  "arrangement  $\rm A$"  or  "arrangement  $\rm B$".

  • The light grey blocks represent the transition from the time to the frequency domain.
  • The dark grey blocks represent the transition from the frequency to the time domain.


We refer here to the following links:


For the number of interpolation points of DFT and IDFT, realistic numerical values of  $K = 12$  and  $N = 1024$  are assumed.

  • The value  $K = 12$  results from the fact that the symbols are "mapped" to a certain bandwidth by the  "subcarrier mapping".  The smallest addressable block for LTE is  $180 \ \rm kHz$.  With  the subcarrier spacing of  $15 \ \rm kHz$   the value  $K = 12$  results.
  • With the number  $N$  of interpolation points of the IDFT $($with arrangement $\rm A)$ , up to  $J = N/K$  users can thus be served simultaneously.  For subcarrier mapping, there are three different approaches with DFDMA, IFDMA and LFDMA.
  • The first two users are shown in green and turquoise in the diagram.  In subtask  (5)  you are to decide whether the sketch applies to DFDMA, IFDMA or LFDMA.




Note:


Questions

1

What do the outlined arrangements on the information page apply to?

For the LTE downlink,
for the LTE uplink.

2

Which units are shown on the information page?

Arrangement  $\rm A$  shows the transmitter of the LTE uplink.
Arrangement  $\rm B$  shows the receiver of the LTE uplink..
Both models apply equally to the transmitter and receiver.

3

Which blocks not shown are still required?

Before arrangement  $\rm A$  you need a serial-parallel converter.
After arrangement  $\rm B$  you need a parallel-serial converter.

4

How many users  $(J)$  can be served simultaneously with  $K = 12$  and  $N = 1024$ ?

$J \ = \ $

5

Which mapping is the graphic on the information page based on?

Distributed Mapping  (DFDMA),
Interleaved Mapping  (IFDMA),
Localized Mapping  (LFDMA).

6

Which DFT (IDFT) can be realised as FFT (IFFT)?

The DFT in the left area of arrangement  $\rm A$.
The IDFT in the right-hand area of arrangement  $\rm A$.
The DFT in the left-hand area of arrangement  $\rm B$.
The IDFT in the right-hand area of arrangement  $\rm B$.


Solution

(1)  Proposed solution 2 is correct:

  • Both arrangements show "Single Carrier Frequency Division Multiple Access"  $\text{(SC–FDMA)}$, recognisable by the DFT and IDFT blocks.
  • The advantage over "Orthogonal Frequency Division Multiple–Access"  $\text{(OFDMA)}$  is the more favourable Peak–to–Average Power–Ratio  $\text{(PAPR)}$.
  • A large PAPR means that the amplifiers must be operated below the saturation limit and thus at poorer efficiency in order to prevent excessive signal distortion.
  • A lower PAPR also means longer battery life, an extremely important criterion for smartphones.
  • This is why SC-FDMA is used in the LTE uplink.  For the downlink, the aspect mentioned here is less significant.


(2)  Proposed solutions 1 and 2 are correct:

  • While in OFDMA the data symbols to be transmitted directly generate the various subcarriers, in SC-FDMA a block of data symbols is first transformed into the frequency domain using DFT.
  • To be able to transmit multiple users, $N > K$ must apply.  An input block of a user thus consists of $K$ bits.  It is thus obvious that arrangement $\rm A$  applies to the transmitter.
  • Arrangement  $\rm B$,  on the other hand, describes the receiver of the LTE uplink and not the transmitter.


(3)  Both statements are correct:

  • The measures are necessary to be able to process a continuous bit stream at the transmitter,
  • or to ensure a continuous bit stream at the receiver as well.


(4)  The DFT also generates  $K$  spectral values from  $K$  input values.

  • The subcarrier mapping does not change anything.
  • Further users also occupy  $K$  (bits) of the total of  $N$  (bits).
  • Thus  $J = N/K = 1024/12 = 85.333$   ⇒   $J \ \underline{= 85}$  users can be supplied.


(5)  Proposed solution 3 is correct:

  • The graph conforms to the current 3gpp specification, which provides for "Localized Mapping".
  • Here, the  $K$  modulation symbols are assigned to adjacent subcarriers.


(6)  Solutions 2 and 3 are correct:

  • The realisation of DFT or IDFT as an (inverse) "Fast Fourier Transform" is only possible if the number of interpolation points is a power of two.
  • For example, for  $N = 1024$, but not for  $K = 12$.