6.4  DMT/OFDM rate estimation

OFDM can carry information bits in all $N$ tones while DMT can use $K\; =\; N∕2\; +\; 1$ tones (the DC and Nyquist tones must be real PAM values while the others $N∕2\; -\; 1$ can be complex QAM symbols).¹

Assuming the channel for a given tone is not frequency-selective (it is approximately flat within the tone spacing $\mathrm{\Delta }f$) and the transmission on this tone can be modeled as an AWGN channel, the Shannon capacity is given by Eq. (4.53) and Eq. (4.60)

The Shannon capacity indicates an upper bound on bit rate but coding schemes have to control the bit rate and SNR to achieve a given symbol error probability $P_e$. This is a non-linear relation that depends on the adopted code. For uncoded QAM (and few other modulations), one can use the convenient gap approximation, which specifies, for a given ${\mathrm{\text{SNR}}}_k$, how many bits can be loaded at tone $k$ for obtaining a given $P_e$. The goal is to keep $P_e$ constant over all tones. As discussed in Section 5.8, the gap approximation is

where $\Gamma$ denotes the gap or SNR-gap to capacity. The gap is typically specified in $10{log}_{10}(\Gamma )$ dB and converted to linear scale to be used in Eq. (6.3). There are convenient expressions to obtain the gap $\Gamma \; =\; f(P_e)$ as a function of the target $P_e$ for simple schemes such as uncoded QAM. When more efficient coding is used, the gap should be modified. Alternatively, if channel coding is used, the SNR can be multiplied by the net coding gain ${\Gamma }_c$ to obtain

In practice, it may be convenient to increase the robustness of the system by some target margin ${\gamma }_t$, given that the $\mathrm{\text{SNR}}$ can deteriorate and, consequently, the errors surpass the target $P_e$. Hence, instead of using Eq. (6.4), a margin ${\gamma }_t>\; 1$ on tone $k$ can be imposed to find a smaller number of bits according to

where $\zeta \; =\; (\Gamma {\gamma }_t)∕{\Gamma }_c$.

The bitloading is normally executed during the modem initialization stage. Afterwards, the margin can change over time due to noise variations and the number of bits per tone may need to be adjusted. The current margin (or noise margin) is denoted as $\gamma$ and may be different from the target ${\gamma }_t$.

When dealing with Eq. (6.5) it is convenient to deal with the quantities in dB, such that

where

is the amount in dB by which the SNR should be reduced before using Eq. (4.53) to obtain the number of bits per tone when one aims at operating at $P_e$. In summary, after the system designer specified $\Gamma$ and ${\Gamma }_c$, the margin $\gamma$ is the amount by which the SNR on the channel may be lowered before performance degrades to a probability of error greater than the target error probability used when calculating the gap.

From Eq. (6.5) with ${\Gamma }_c=\; 1$ (no coding gain), the margin can be written as

where $\tilde{b}$ represents the number of bits actually used.

Dealing with both margin $\gamma$ and gap $\Gamma$ seems redundant because the margin could be incorporated into the gap by redefining it as $\widehat{\Gamma }=\; \Gamma \gamma$. However, they play different roles in the design and operation of DSL systems. The gap is determined by the required $P_e$ and is a fixed value, while the current margin $\gamma$ varies over time according to Eq. (6.6).

From Eq. (6.3) and Eq. (6.5), the current margin for a given tone $k$ can be written as

where $b_k$ is given by Eq. (6.3) and ${\tilde{b}}_k$ is the number of bits actually used. A DSL system tries to keep all tones with a margin close to the target, i. e., ${\gamma }_k\approx {\gamma }_t$.

To observe an example of using margins, assume a QAM tone (the subscript $k$ will be omitted) with $\mathrm{\text{SNR}}=\; 10^4$ (40 dB) and targets $P_e=\; 10^{-9}$ and ${\gamma }_t=\; 100$ (20 dB). According to Table 5.2, the gap is $\Gamma \; =\; 12.8924$ (11.1 dB) and Eq. (6.3) leads to $b\; =\; 9.6$. This means that, if the system manages to operate with a non-integer $\tilde{b}=\; 9.6$, the margin would be 0 dB. A value $\tilde{b}>\; 9.6$ means the margin is negative and the error is higher than $P_e$. Imposing ${\gamma }_t=\; 100$ leads to $\widehat{b}=\; 3.13$. But if the tone is loaded with $\tilde{b}=\; 5$ bits, Eq. (6.6) or, alternatively, Eq. (6.7), indicates that the margin is $\gamma \; =\; 25.021$ (${\gamma }_{\mathrm{\text{dB}}}\approx \; 14$) instead of the target ${\gamma }_t=\; 100$.

The convenience of working with dB can be seen in the following reasoning. The values ${\gamma }_t=\; 20$ dB and ${\Gamma }_{\mathrm{\text{dB}}}=\; 11.1$ dB dictate that the original ${\mathrm{\text{SNR}}}_{\mathrm{\text{dB}}}=\; 40$ is lowered to $40\; -\; 20\; -\; 11.1\; =\; 8.9$ and only this amount is effectively used to calculate the number of bits. More generally, one has

where

Given that there are $K$ tones to transmit information at, the total number of bits per DMT symbol is

Hereafter, for simplicity, it is assumed that all tones can carry QAM symbols such Eq. (4.53) applies with $D\; =\; 2$.

In multicarrier, the term symbol rate often refers to the DMT symbol rate $R_{\mathrm{\text{dmt}}}=\; 1∕T_{\mathrm{\text{dmt}}}$, where $T_{\mathrm{\text{dmt}}}$ is the time interval to transmit a complete DMT symbol (that includes several QAM symbols). Therefore, the DMT data rate in bits per second (bps) is

which can be expanded using Eq. (6.5) and ${\zeta }_{\mathrm{\text{dB}}}=\; 10{log}_{10}\zeta$ into

Taking in account the CP samples leads to

such that

If there is no CP ($L_{\mathrm{\text{cp}}}=\; 0)$, $T_{\mathrm{\text{dmt}}}=\; N{T}_s$ and ${\mathrm{\Delta }}_f=F_s∕N\; =\; 1∕(N{T}_s)\; =\; 1∕T_{\mathrm{\text{dmt}}}=R_{\mathrm{\text{dmt}}}$. In this specific case, Eq. (6.12) can be written as

But in the general case, dividing both sides of Eq. (6.13) by $N$ and combining with ${\mathrm{\Delta }}_f=F_s∕N$ leads to

Substituting Eq. (6.15) into Eq. (6.12) leads to

which can be written as

Besides the CP, DSL has additional overhead (for example, to organize the information in frames for transmission) and it is convenient to define the framing overhead $\nu \; \triangleq \; 1\; -\; (R∕\widehat{R})$ such that

When the overhead consists solely of the cyclic prefix samples it is given by