What is Carrier Frequency Offset (CFO) and How It Distorts the Rx Symbols

Three different cases for carrier frequency offset

In Physics, frequency in units of Hz is defined as the number of cycles per unit time. Angular frequency is the rate of change of phase of a sinusoidal waveform with units of radians/second.

    \begin{equation*}         2\pi f = \frac{\Delta \theta}{\Delta t}     \end{equation*}

where \Delta\theta and \Delta t are the changes in phase and time, respectively. A Carrier Frequency Offset (CFO) usually arises due to two reasons:

  1. [Frequency mismatch between the Tx and Rx oscillators] No two devices are the same and there is always some difference between the manufacturer’s nominal specification and the real frequency of that device. Moreover, this actual frequency keeps changing (slightly) with temperature, pressure, age, and some other factors.
  2. [Doppler effect] A moving Tx, Rx or any kind of movement around the channel creates a Doppler shift that creates a carrier frequency offset as well. We will learn more about it during the discussion of a wireless channel in another post.

To see the effect of the carrier frequency offset F_\Delta, consider again a received passband signal consisting of two PAM waveforms in I and Q arms.

(1)   \begin{align*}         r(t) &= v_I(t) \sqrt{2} \cos \Big[2\pi (F_C+F_\Delta) t + \theta_\Delta\Big] - v_Q(t) \sqrt{2}\sin \Big[ 2\pi (F_C+F_\Delta) t + \theta_\Delta \Big]\nonumber\\         &= v_I(t) \sqrt{2} \cos \Big[2\pi F_Ct + 2\pi F_\Delta t + \theta_\Delta\Big] - \nonumber \\ &\hspace{2in}v_Q(t) \sqrt{2}\sin \Big[ 2\pi F_C t+ 2\pi F_\Delta t + \theta_\Delta\Big]     \end{align*}

Here, the carrier offset can be seen as 2\pi F_\Delta t which is changing with time. Let us look into how to find F_{\Delta,\textmd{max}}, the maximum value CFO can take.

The accuracy of local oscillators in communication receivers is defined in terms of ppm (parts per million). 1 ppm is just what it says: 1 out of 10^6 parts. To get a feel of how big or small this number is, 10^6 seconds translate into

    \begin{equation*}             \frac{10^6~ \textmd{seconds}}{24~ \textmd{hours/day} \times 3600 ~\textmd{seconds/hour}} \approx 11.5~ \textmd{days}         \end{equation*}

Hence, 1 ppm is equivalent to a deviation of 1 second every 11.5 days. This might sound entirely harmless but for the purpose of typical wireless communication systems operating at several GHz of carrier frequency F_C and several MHz of symbol rates R_M, it is one of the major sources of signal distortion.

The ppm rating at the oscillator crystal indicates how much its frequency may deviate from the nominal value. Consider an example of a wireless system operating at 2.4 GHz and with \pm 20 ppm crystals, which is a more or less standard rating. The maximum deviation of the carrier frequency at the Tx or Rx can be

    \begin{equation*}             \pm \frac{20}{10^6} \times 2.4 \times 10^9 = \pm 48 \textmd{kHz}         \end{equation*}

However, in the worst case scenario, the Tx can be 20 ppm above (or below) the nominal frequency, while the Rx can be 20 ppm below (or above) the nominal frequency, resulting in the overall difference of 40 ppm between the two. So the worst case CFO due to local oscillator mismatch in this example can be

    \begin{equation*}             \pm 2 \times 48 = \pm 96 \textmd{kHz}         \end{equation*}

Keep in mind that this calculation is for basic precision and the actual frequency may vary depending on environmental factors, mainly the temperature and aging. Finally, a movement anywhere in the channel (whether by Tx, Rx or some other object within that environment) adds Doppler shift which can be up to several hundreds of Hz. Although this Doppler shift is much less than the oscillator generated mismatch, it severely distorts the Rx signal due to the changes it causes in the channel.

Now we turn our attention towards Figure below to differentiate between three possibilities in which F_\Delta can dictate the receiver design.

Three different cases for carrier frequency offset

Case 1: CFO > F_{\Delta,\textmd{max}}

It was illustrated in another article that an input signal at the Rx is filtered by an analog prefilter to remove the out of band noise. Ideally, the frequency response of this prefilter G(F) should be flat within the frequency range

    \begin{equation*}         |F| \le B + F_{\Delta,\textmd{max}}     \end{equation*}

so that the incoming signal can pass through in an undistorted manner (we are assuming a flat wireless channel here as well. Actual wireless channels will be discussed later). The passband width of this prefilter is designed according to the maximum CFO F_{\Delta,\textmd{max}} expected at the Rx.

However, if the CFO is greater than F_{\Delta,\textmd{max}}, then much of the actual intended signal will be filtered out by the analog prefilter and the Rx sampled signal will not even closely resemble the Tx signal as a linear function of Tx data. No amount of signal processing can then recover the signal. Since it is an outcome of poor system design, the only remedy is to redesign the system (particularly the analog frontend) with more accurate estimates and margins for CFO and other such random distortions.

Case 2: 15\% of R_M < CFO < F_{\Delta,\textmd{max}}

Since CFO < F_{\Delta,\textmd{max}}, the Rx signal is within the passband of the analog prefilter G(F) and suffers no distortion. However, remember from a previous post that to maximize the SNR, the Rx signal must be passed through a matched filter. This is not possible in this case because much of the Rx signal bandwidth does not sufficiently overlap with the matched filter due to the CFO being greater than 15\% of symbol rate R_M. If applied, it would remove a significant portion of the incoming signal energy.

Since Rx signal cannot be matched filtered without distorting the signal, and the signal cannot be downsampled to 1 sample/symbol without matched filtering, it is easy to deduce that more than 1 sample/symbol (say, L) is required to trace the frequency offset.

Another route to understand the rationale behind the factor 15\% of the symbol rate R_M is by evaluating the relation

    \begin{equation*}         F_{\Delta} = 0.15 R_M \quad \Rightarrow \quad T_\Delta \approx 6.67~T_M     \end{equation*}

For CFO > 15\% of R_M, an offset cycle gets completed in less than 7 symbol times, which induces a phase rotation of more than 50^\circ on one symbol. Therefore, it cannot be downsampled to 1 sample/symbol for frequency tracking.

Case 3: CFO < 15\% of R_M

When the signal is rotated by less than 15\% R_M, an offset cycle gets completed in less than 7 symbols, and hence the effect of rotation on one symbol, although still significant, can now be tracked at symbol rate, or 1 sample/symbol. In other words, matched filtering keeps most of the signal intact and as a result, symbol boundaries can be marked first (a problem known as symbol timing synchronization which we discuss in another article) before carrier frequency is estimated at the ISI-free symbol-spaced samples.

In the absence of noise, the sampled version of this mismatch in Eq (1) becomes 2\pi F_\Delta nT_S. Now Eq (1) is very similar to phase rotation equation, and hence we can use directly use that result with proper substitution. The expression for the symbol-spaced samples in the presence of CFO F_{\Delta} can be obtained after replacing sample time index n with the symbol time index m.

    \begin{equation*}       \begin{aligned}         z_I(mT_M) &= a_I[m] \cos 2\pi F_\Delta mT_M - a_Q[m]\sin 2\pi F_\Delta mT_M       \\         z_Q(mT_M) &= a_I[m] \sin 2\pi F_\Delta mT_M + a_Q[m]\cos 2\pi F_\Delta mT_M       \end{aligned}     \end{equation*}

In the above equation,

    \begin{equation*}         2\pi F_\Delta mT_M = 2\pi \frac{F_\Delta}{R_M} m = 2\pi F_0 m     \end{equation*}

where the F_0 is defined as the normalized Carrier Frequency Offset (nCFO): CFO normalized by the symbol rate.

    \begin{equation*}           F_0 = \frac{F_\Delta}{R_M}     \end{equation*}

This normalization is very important as we saw in the last subsection. The resulting expression takes the form

    \begin{equation*}       \begin{aligned}         z_I(mT_M) = a_I[m] \cos 2\pi F_0m - a_Q[m]\sin 2\pi F_0m       \\         z_Q(mT_M) = a_I[m] \sin 2\pi F_0m + a_Q[m]\cos 2\pi F_0m       \end{aligned}     \end{equation*}

In polar form, this expression can be written as

    \begin{equation*}         \begin{aligned}           |z(mT_M)| &= \sqrt{a_I^2[m] + a_Q^2[m]} \\           \measuredangle z(mT_M) &= \measuredangle \Big(a_Q[m],a_I[m]\Big) + 2\pi F_0 m         \end{aligned}     \end{equation*}

Notice from the above equation that a carrier frequency offset of F_\Delta keeps the magnitude unchanged but continually rotates the desired outputs a_I[m] and a_Q[m] on the constellation plane. This is drawn for symbol-spaced samples in the scatter plot of Figures below for a 4-QAM and 16-QAM constellation.

A 4-QAM and a 16-QAM constellation spinning in circles due to a carrier frequency offset (CFO)

Due to this reason, a Carrier Frequency Offset (CFO), or F_{\Delta}, in the received signal spins the constellation in a circle (or multiple circles for higher-order modulations). This is a natural outcome since the angular frequency is defined as the rate of change of phase.

The above results are summarized in Table below.
Carrier frequency offset (CFO) classification

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