Ferkans — Interactive Telecom Tutor

The PAPR Challenge

The OFDM time-domain signal is the superposition of $N$ independent subcarrier waveforms. When these waveforms add constructively, the instantaneous power can be much larger than the average power. This high peak-to-average power ratio (PAPR) forces the power amplifier (PA) to operate with large back-off, reducing efficiency and increasing cost — a fundamental challenge for OFDM transmitters.

Definition:
Peak-to-Average Power Ratio (PAPR)

For a continuous-time OFDM signal $x(t) = \sum_{k=0}^{N-1} X[k] e^{j2\pi k\Delta f\, t}$ , the PAPR is defined as:

$\text{PAPR} = \frac{\max_{0 \leq t < T_{\text{sym}}} |x(t)|^2}{\mathbb{E}[|x(t)|^2]} = \frac{\|x\|_\infty^2}{\|x\|_2^2 / T_{\text{sym}}}$

For the discrete-time version with $N$ samples:

$\text{PAPR} = \frac{\max_{0 \leq n \leq N-1} |x[n]|^2}{\frac{1}{N}\sum_{n=0}^{N-1} |x[n]|^2}$

PAPR is often expressed in dB: $\text{PAPR}_{\text{dB}} = 10 \log_{10}(\text{PAPR})$ .

The maximum possible PAPR is $N$ (or $10\log_{10} N$ dB), occurring when all $N$ subcarriers add in phase.

Definition:
CCDF of PAPR

The complementary cumulative distribution function (CCDF) of PAPR is:

$\text{CCDF}(\gamma) = \Pr(\text{PAPR} > \gamma)$

The CCDF is the standard metric for comparing PAPR reduction techniques. A common reference point is the PAPR value at $\text{CCDF} = 10^{-3}$ (i.e., the PAPR exceeded with probability $0.1\%$ ).

Theorem: PAPR Distribution for Large N

For an OFDM signal with $N$ subcarriers carrying i.i.d. data symbols, when $N$ is large, the time-domain samples $x[n]$ are approximately i.i.d. complex Gaussian by the central limit theorem. The CCDF of the discrete-time PAPR is approximately:

$\Pr(\text{PAPR} > \gamma) \approx 1 - (1 - e^{-\gamma})^N$

For the continuous-time signal (with oversampling factor $J$ ), the approximation becomes:

$\Pr(\text{PAPR} > \gamma) \approx 1 - (1 - e^{-\gamma})^{\alpha N}$

where $\alpha \approx 2.8$ is an empirical correction factor for $J = 4$ oversampling.

Each of the $N$ time-domain samples has a Rayleigh-distributed envelope. The probability that the maximum of $N$ independent Rayleigh samples exceeds a threshold $\gamma$ follows from order statistics: the CDF of the maximum is the product of individual CDFs.

Proof

Gaussian approximation

By the CLT, $x[n] = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} X[k] e^{j2\pi kn/N}$ is approximately $\mathcal{CN}(0, \sigma_x^2)$ for large $N$ , where $\sigma_x^2 = \mathbb{E}[|X[k]|^2]$ .

Per-sample peak probability

Since $|x[n]|^2 / \sigma_x^2$ is exponentially distributed with unit mean:

$\Pr\!\left(\frac{|x[n]|^2}{\sigma_x^2} > \gamma\right) = e^{-\gamma}$

$\Pr\!\left(\frac{|x[n]|^2}{\sigma_x^2} \leq \gamma\right) = 1 - e^{-\gamma}$

Maximum over N samples

Assuming approximate independence of the $N$ samples:

$\Pr(\text{PAPR} \leq \gamma) = \Pr\!\left(\max_n \frac{|x[n]|^2}{\sigma_x^2} \leq \gamma\right) \approx (1 - e^{-\gamma})^N$

Therefore $\Pr(\text{PAPR} > \gamma) \approx 1 - (1 - e^{-\gamma})^N$ . $\blacksquare$

,

Definition:
Clipping

Clipping is the simplest PAPR reduction technique. The signal amplitude is hard-limited to a maximum value $A_{\max}$ :

$\tilde{x}[n] = \begin{cases} x[n] & |x[n]| \leq A_{\max} \\ A_{\max} \cdot e^{j\angle x[n]} & |x[n]| > A_{\max} \end{cases}$

The clipping ratio is defined as $\text{CR} = A_{\max}^2 / P_{\text{avg}}$ , often in dB. Clipping is a nonlinear operation that introduces in-band distortion (increased BER) and out-of-band radiation (spectral regrowth), typically followed by filtering.

Definition:
Selected Mapping (SLM)

Selected mapping (SLM) generates $U$ candidate OFDM symbols by multiplying the data vector $\mathbf{X}$ element-wise with $U$ different phase sequences $\mathbf{P}^{(u)} = [e^{j\theta_0^{(u)}}, \ldots, e^{j\theta_{N-1}^{(u)}}]^T$ :

$\mathbf{X}^{(u)} = \mathbf{P}^{(u)} \odot \mathbf{X}, \qquad u = 1, \ldots, U$

The candidate with the lowest PAPR is transmitted. The index $u^*$ of the selected sequence must be communicated to the receiver as side information ( $\lceil \log_2 U \rceil$ bits).

With $U$ candidates, the CCDF improves to:

$\Pr(\text{PAPR}_{\text{SLM}} > \gamma) \approx \left[1 - (1 - e^{-\gamma})^N\right]^U$

PAPR Build-Up from Subcarrier Superposition

Visualise the instantaneous power

|x(t)|^2

of an OFDM signal. When the

N

subcarrier phasors align constructively, the power spikes far above the average — illustrating why OFDM has high PAPR and why power amplifier back-off is needed.

Time-domain power of a 16-subcarrier OFDM symbol showing peak power far exceeding the average (green dashed line).

PAPR CCDF Comparison

Compare the CCDF of PAPR for different numbers of subcarriers and PAPR reduction techniques. Observe how PAPR grows logarithmically with $N$ and how SLM with multiple candidates shifts the CCDF to the left.

Parameters

Number of subcarriers

N

64

SLM candidates

U

1

Show clipping effect

Clipping ratio (dB)6

Example: PAPR of a 4-Subcarrier OFDM Signal

Consider an OFDM signal with $N = 4$ subcarriers and data symbols $X[k] = 1$ for all $k$ (all subcarriers carry the same unit-energy symbol).

(a) Compute the time-domain samples $x[n]$ .

(b) Calculate the PAPR.

(c) What is the maximum possible PAPR for $N = 4$ ?

Solution

Time-domain samples via IDFT

$x[n] = \frac{1}{\sqrt{4}} \sum_{k=0}^{3} 1 \cdot e^{j2\pi kn/4} = \frac{1}{2} \sum_{k=0}^{3} e^{j\pi kn/2}$ $For$ n = 0 $:$ x[0] = \frac{1}{2}(1 + 1 + 1 + 1) = 2 $For$ n = 1 $:$ x[1] = \frac{1}{2}(1 + j - 1 - j) = 0 $For$ n = 2 $:$ x[2] = \frac{1}{2}(1 - 1 + 1 - 1) = 0 $For$ n = 3 $:$ x[3] = \frac{1}{2}(1 - j - 1 + j) = 0$

PAPR calculation

Peak power: $|x[0]|^2 = 4$

Average power: $\frac{1}{4}(4 + 0 + 0 + 0) = 1$

$\text{PAPR} = \frac{4}{1} = 4 = 6.02\;\text{dB}$

Maximum possible PAPR

The maximum PAPR for $N = 4$ is $N = 4$ ( $6.02$ dB), which occurs when all subcarrier signals add in phase at some time instant. This is exactly the case here: $X[k] = 1$ for all $k$ causes all subcarriers to align at $n = 0$ .

In general, $\text{PAPR}_{\max} = N = 10\log_{10}(N)$ dB. $\blacksquare$

Common Mistake: Clipping Without Filtering

Mistake:

Applying hard clipping to reduce PAPR without subsequent filtering, assuming the clipped signal still satisfies spectral mask requirements.

Correction:

Clipping is a nonlinear operation that spreads the signal spectrum, causing out-of-band emissions (spectral regrowth). The clipped signal will violate the transmit spectral mask. Always follow clipping with a band-limiting filter to suppress out-of-band radiation. Note that filtering may cause some peak regrowth, so iterative clipping-and-filtering is often employed.

Quick Check

For an OFDM system with $N = 256$ subcarriers, what is the maximum possible PAPR in dB?

$10\log_{10}(256) = 24.1$ dB

$10\log_{10}(128) = 21.1$ dB

$20\log_{10}(256) = 48.2$ dB

$3$ dB regardless of $N$

Correction:

10\log_{10}(256) = 24.1

dB

The maximum PAPR is $N$ (linear), which occurs when all subcarriers add coherently. In dB: $10\log_{10}(256) = 24.1$ dB.

🚨Critical Engineering Note

Power Amplifier Back-Off and Efficiency

A power amplifier (PA) is most efficient when operating near its saturation point (maximum output power). However, OFDM's high PAPR forces the PA to operate with significant input back-off (IBO) to avoid clipping the signal peaks:

$\text{IBO} \geq \text{PAPR}_{\text{target}} \quad (\text{in dB})$

For OFDM with $N = 1024$ subcarriers, PAPR at CCDF $= 10^{-3}$ is approximately 10.5 dB, requiring $\geq 10.5$ dB IBO. This reduces the PA's power efficiency from a theoretical maximum of 60--70% (class B) to only 5--10%.

The cost implications are severe: in a macro base station with $P_{\text{out}} = 40$ W per antenna, the PA consumes 400--800 W of DC power. For 64-antenna massive MIMO, this totals 25--50 kW per sector — a major component of the total site power budget and operational expenditure.

This is why PAPR reduction techniques (SLM, clipping, DFT precoding) and PA linearisation (digital pre-distortion, DPD) are active research areas with direct commercial impact.

Practical Constraints

•
Typical PA efficiency with 10 dB back-off: 5-10% (class A/B), 15-20% (Doherty)
•
DFT-s-OFDM (SC-FDMA) reduces required back-off by 2-4 dB
•
Digital pre-distortion (DPD) can reduce required back-off by 2-3 dB

Key Takeaway

PAPR is the price of parallel subcarriers. The superposition of $N$ independent subcarrier waveforms can produce peaks up to $N$ times the average power. At CCDF $= 10^{-3}$ , typical PAPR is 10--12 dB for $N = 1024$ , requiring massive PA back-off. This is the principal reason LTE uplink uses SC-FDMA instead of OFDM — the mobile handset cannot afford the power inefficiency.

PAPR

Peak-to-Average Power Ratio — the ratio of the peak instantaneous power to the average power of the OFDM signal. High PAPR requires large power amplifier back-off, reducing efficiency.

SLM

Selected Mapping — a PAPR reduction technique that generates multiple candidate OFDM symbols using different phase rotations and transmits the one with the lowest PAPR.

Tone Reservation

A PAPR reduction technique that reserves a set of subcarriers (not used for data) and optimises their values to minimise the peak power of the time-domain signal.

Peak-to-Average Power Ratio (PAPR)