OFDM loss of E_{b}/N_{0} resulting from cyclic prefix and pilot subcarriers
OFDM systems show a loss of E_{b}/N_{0} due to the cyclic prefix and pilot carriers.
Effect of cyclic prefix
The last {N}_{G} symbols of the OFDM symbol of length N are copied and added as cyclic prefix. As the receiver removes the cyclic prefix its energy is lost. This results in a loss of E_{b}/N_{0}:
{\xdf}_{1}=\frac{N}{N+{N}_{G}}
Where
N Number of subcarriers
{N}_{G} Discrete length of the guard interval
Effect of pilot subcarriers
Pilot carriers form an overhead and do not contribute to the user bit rate.
This results in a loss of E_{b}/N_{0}:
{\xdf}_{2}=\frac{{N}_{SD}}{{N}_{SD}+{N}_{SP}}
Where
{N}_{SD} Number of data subcarriers
{N}_{SP} Number of pilot subcarriers
Overall loss
The overall loss of both cyclic prefix and pilot subcarriers is given by
\xdf=\frac{N}{\left(N+{N}_{G}\right)}\frac{{N}_{SD}}{\left({N}_{SD}+{N}_{SP}\right)}
Where
\xdf=1 for single carrier systems
\xdf\le 1 for OFDM systems
This applies to any modulation scheme. For BPSK and QPSK the BER for AWGN channel is given by:
{p}_{b}=0.5erfc\left(\sqrt{\xdf\frac{{E}_{b}}{{N}_{0}}}\right)
OFDM systems show a loss of E_{b}/N_{0} due to cyclic prefix and pilot carriers, see tutorial.
$$\xdf=\frac{N}{\left(N+{N}_{G}\right)}\frac{{N}_{SD}}{\left({N}_{SD}+{N}_{SP}\right)}$$ |
Example
In this experiment we analyze a specific IEEE 802.11ac OFDM setup. The theoretical BER for AWGN channel is calculated and compared with simulation results.
The following setting of IEEE 802.11ac is used:
VHT- MCS Index | Modulation | $$R$$ | $${N}_{BPSC}$$ | $${N}_{SD}$$ | $${N}_{SP}$$ | Data rate (Mb/s) | |
---|---|---|---|---|---|---|---|
800 ns GI | 400 ns GI | ||||||
1 | QPSK | 1/2 | 2 | 234 | 8 | 58.5 | 65.0 |
For the 400 ns GI setup the discrete length of the guard interval ${N}_{G}$ is 32. The number of subcarriers $N$ is 256. The loss of E_{b}/N_{0} yields to
$$\xdf=\frac{N}{\left(N+{N}_{G}\right)}\frac{{N}_{SD}}{\left({N}_{SD}+{N}_{SP}\right)}=\frac{256}{\left(256+32\right)}\frac{234}{\left(234+8\right)}=\mathrm{0,8595}=-\mathrm{0,658}dB.$$ |
The effective energy per bit to noise power-spectral-density ratio is $\xdf{E}_{b}/{N}_{0}$. This loss results in a BER degeneration as shown in the following table.
$${E}_{b}/{N}_{0}$$ in dB | SC BER | $$\xdf{E}_{b}/{N}_{0}$$ in dB | OFDM BER |
---|---|---|---|
-20 | 4,438E-01 | -20,658 | 4,478E-01 |
-15 | 4,007E-01 | -15,658 | 4,078E-01 |
-10 | 3,274E-01 | -10,658 | 3,392E-01 |
-5 | 2,132E-01 | -5,658 | 2,305E-01 |
0 | 7,865E-02 | -0,658 | 9,491E-02 |
5 | 5,954E-03 | 4,342 | 9,863E-03 |
Start
Let's first check the E_{b}/N_{0} setting.
- The uncoded bitrate is ${R}_{B}=130Mb/s$. That's the simulation's bitrate as it doesn't consider coding.
- The transmission power after the symbol mapper is $1{V}^{2}$. Unused subcarriers reduce the transmission power: ${S}_{S}=\frac{{N}_{SD}+{N}_{SP}}{N}1{V}^{2}=\frac{234+8}{256}=\mathrm{0,9453}{V}^{2}$
$\to $ Measure the transmission power in the simulation!
- The energy per bit yields to ${E}_{b}=\frac{{S}_{S}}{{R}_{B}}=\frac{\mathrm{0,9453}{V}^{2}}{130Mb/s}=\mathrm{7,2716}n{V}^{2}s$
- Noise power spectral density must be set to ${N}_{0}=\mathrm{7,2716}n{V}^{2}/Hz$ for ${E}_{b}/{N}_{0}=1$.
$\to $ Check the value of Power-spectral-density N0 of the block Noise in the simulation (click on Noise).
Measure the bit error rate in the simulation!
Adjust E_{b}/N_{0} (press F11) and measure the corresponding bit error rate.
Next steps
Now select a different OFDM setup: Simulation - Setup (F12). When the guard interval (GI) is increased to 800 ns the loss of E_{b}/N_{0} and BER will increase. Calculate and measure the BER analogous to the table above.
This simulation app implements an OFDM transmission using AWGN channel. The bit error rate is measured for variable E_{b}/N_{0}.
Key | Action |
---|---|
Simulation - Settings (F11) |
Adjust E_{b}/N_{0} and measure the corresponding bit error rate. |
Simulation - Setup (F12) |
Modify the OFDM config and press OK to restart the simulation. |