Aufgaben:Exercise 4.3: Subcarrier Mapping

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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:

• Discrete Fourier Transform   ⇒   $\rm DFT$,

• Inverse Discrete Fourier Transform   ⇒   $\rm IDFT$.

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:

• The task belongs to the chapter  The Application of OFDMA and SC-FDMA in LTE.

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

Solution

(1)  Proposed solution 2 is correct:

• Both arrangements show Single Carrier Frequency Division Multiple Access (SC–FDMA), recognisable by the DFT and IDFT blocks.
• The advantage over Orthogonal Frequency Division Multiple–Access (OFDMA) is the more favourable Peak–to–Average Power–Ratio (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)  Proposed 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 $K = 12$.