Ferkans — Interactive Telecom Tutor

Where to Place the Relay?

The performance of a relay-assisted link depends critically on the relay's position. Place it too close to the source and the $R \to D$ link is weak; place it too close to the destination and the $S \to R$ link becomes the bottleneck. For DF relaying, the optimal position balances the source-relay and relay-destination link qualities. For AF, the relay should be closer to the destination to minimise noise amplification. This optimisation has direct practical relevance: operators deploying relay nodes or IAB (integrated access and backhaul) sites must choose locations that maximise the rate gain over direct transmission.

Definition:
Relay Position Parameterisation

Consider a source at position $0$ and destination at position $d_{SD}$ on a line. The relay is placed at position $f \cdot d_{SD}$ , where $f \in (0, 1)$ is the relay position fraction:

$d_{SR} = f \cdot d_{SD}, \qquad d_{RD} = (1-f) \cdot d_{SD}$

The received SNRs (with path-loss exponent $\alpha$ ) are:

$\text{SNR}_{SR} = \text{SNR}_{0} \cdot (f \cdot d_{SD})^{-\alpha} = \text{SNR}_{0} \cdot d_{SD}^{-\alpha} \cdot f^{-\alpha}$ $\text{SNR}_{RD} = \text{SNR}_{0} \cdot d_{SD}^{-\alpha} \cdot (1-f)^{-\alpha}$ $\text{SNR}_{SD} = \text{SNR}_{0} \cdot d_{SD}^{-\alpha}$

The relay position optimisation seeks:

$f^{\star} = \arg\max_{f \in (0,1)} C_{\text{relay}}(f)$

where $C_{\text{relay}}$ is the achievable rate of the chosen relay protocol.

Definition:
Relay Gain over Direct Transmission

The relay gain is the ratio of the relay-assisted rate to the direct transmission rate:

$G_{\text{relay}}(f) = \frac{C_{\text{relay}}(f)}{C_{\text{direct}}}$

where $C_{\text{direct}} = \log_2(1 + \text{SNR}_{SD})$ .

Relaying is beneficial when $G_{\text{relay}} > 1$ , i.e., the relay-assisted rate exceeds the direct rate despite the half-duplex loss. The relay gain depends on:

The path-loss exponent $\alpha$ (higher $\alpha$ favours relaying),
The relay position $f$ ,
The transmit SNR,
The relay protocol (DF, AF, CF).

Theorem: Optimal Relay Position for DF

For half-duplex DF relaying with equal time allocation ( $t = 1/2$ ) and path-loss exponent $\alpha$ , the optimal relay position that maximises the achievable rate satisfies:

$C_{\text{DF}}(f) = \frac{1}{2}\min\!\left\{ \log_2(1 + \gamma f^{-\alpha}),\; \log_2(1 + \gamma + \gamma(1-f)^{-\alpha})\right\}$

where $\gamma = \text{SNR}_{0} \cdot d_{SD}^{-\alpha}$ .

The optimal position $f^{\star}$ equates the two arguments of the min:

$\gamma f^{-\alpha} = \gamma + \gamma(1-f)^{-\alpha}$

which simplifies to:

$f^{-\alpha} = 1 + (1-f)^{-\alpha}$

For $\alpha = 2$ : $f^{\star} \approx 0.382$ (closer to source). For $\alpha = 4$ : $f^{\star} \approx 0.435$ (nearly midpoint). As $\alpha \to \infty$ : $f^{\star} \to 1/2$ (midpoint).

The DF rate is bottlenecked by the weaker of the two links. The optimal position balances them. Because the destination also has the direct link ( $\text{SNR}_{SD}$ ) which strengthens the second min-argument, the relay should be shifted toward the source to compensate. As $\alpha$ increases, the relay links dominate and the optimal position converges to the midpoint $f = 1/2$ .

Proof

Rate expression

The DF achievable rate with relay at fraction $f$ is:

$C_{\text{DF}}(f) = \frac{1}{2}\min\{C_{SR}(f), C_{SD+RD}(f)\}$

where $C_{SR}(f) = \log_2(1 + \gamma f^{-\alpha})$ and $C_{SD+RD}(f) = \log_2(1 + \gamma + \gamma(1-f)^{-\alpha})$ .

Balancing condition

The max-min is achieved when the two arguments are equal:

$1 + \gamma f^{-\alpha} = 1 + \gamma + \gamma(1-f)^{-\alpha}$ $f^{-\alpha} = 1 + (1-f)^{-\alpha}$

This is a transcendental equation in $f$ that depends on $\alpha$ but not on $\gamma$ (the optimal position is independent of the absolute SNR).

Numerical solutions

Solving numerically for several $\alpha$ :

$\alpha$	$f^{\star}$	Interpretation
2	0.382	38% from source
3	0.414	41% from source
4	0.435	44% from source
6	0.455	46% from source

The relay is always closer to the source than the midpoint, converging to $f = 0.5$ as $\alpha \to \infty$ . $\blacksquare$

,

Optimal Relay Placement: DF vs AF Rate Sweep

Watch the relay move from source to destination while DF and AF rates update in real-time, revealing the optimal position.

As the relay position fraction

f

sweeps from 0.1 to 0.9, the DF rate peaks near

f \approx 0.4

(relay slightly closer to source) due to the direct link assistance at the destination.

Relay Position Optimisation

Explore how the relay position affects the achievable rate for DF, AF, and the cut-set bound. Drag the relay position slider to find the optimal placement. Observe that the optimal DF position is slightly closer to the source, while AF prefers the relay closer to the destination. Higher path-loss exponents increase the relay gain and shift the optimal position toward the midpoint.

Parameters

Transmit SNR (dB)20

d_{SD}

(source-destination distance)2

Relay position fraction

f

0.5

Path-loss exponent

\alpha

3

Example: Optimal Relay Placement

A source and destination are separated by $d_{SD} = 3$ km in an urban environment with $\alpha = 4$ . The transmit SNR at 1 km reference distance is 30 dB.

(a) Compute the direct link SNR and rate. (b) Compute the DF rate at $f = 0.5$ (midpoint). (c) Compute the DF rate at the optimal $f^{\star} \approx 0.435$ . (d) Determine the minimum $\alpha$ for which relaying outperforms direct transmission.

Solution

Direct link

(a) $\gamma = \text{SNR}_{SD} = 1000 \times 3^{-4} = 1000/81 = 12.35$ (10.9 dB).

$C_{\text{direct}} = \log_2(1 + 12.35) = \log_2(13.35) = 3.74$ bits/s/Hz.

DF at midpoint

(b) $\text{SNR}_{SR} = 12.35 \times 0.5^{-4} = 12.35 \times 16 = 197.5$ . $\text{SNR}_{RD} = 197.5$ (by symmetry).

$C_{\text{DF}} = \frac{1}{2}\min\{\log_2(198.5), \log_2(1+12.35+197.5)\}$ $= \frac{1}{2}\min\{7.63, 7.72\} = 3.82$ bits/s/Hz.

DF at optimal position

(c) At $f = 0.435$ : $\text{SNR}_{SR} = 12.35 \times 0.435^{-4} = 12.35 \times 27.98 = 345.5$ . $\text{SNR}_{RD} = 12.35 \times 0.565^{-4} = 12.35 \times 9.82 = 121.3$ .

$C_{\text{DF}} = \frac{1}{2}\min\{\log_2(346.5), \log_2(134.6)\}$ $= \frac{1}{2} \times 7.07 = 3.54$ bits/s/Hz.

The midpoint actually yields a slightly higher rate here; the optimal position matters more at lower SNR.

Minimum $\alpha$ for relay gain

(d) Relaying helps when $C_{\text{DF}} > C_{\text{direct}}$ . The half-duplex factor of $1/2$ requires the relay to at least double the effective rate: roughly $f^{-\alpha} > 2$ .

At $f = 0.5$ : $0.5^{-\alpha} > 2$ requires $\alpha > 1$ . Since $\alpha > 2$ always in practice, relaying at the midpoint is always beneficial in terms of the SNR gain. But the rate gain requires $\alpha > 2$ to overcome the half-duplex penalty. $\blacksquare$

Quick Check

For decode-and-forward relaying, why is the optimal relay position slightly closer to the source rather than at the exact midpoint?

Because the relay needs a strong signal from the source to decode reliably

Because the destination combines the direct and relay signals, so the $R \to D$ link does not need to be as strong as the $S \to R$ link

Because path loss is stronger closer to the destination

Because the source transmits at higher power than the relay

Correction:

Because the destination combines the direct and relay signals, so the

R \to D

link does not need to be as strong as the

S \to R

link

At the destination, MRC combines the direct path $S \to D$ and the relay path $R \to D$ , so the effective rate $\log_2(1 + \text{SNR}_{SD} + \text{SNR}_{RD})$ is boosted by the direct link. This means $\text{SNR}_{RD}$ does not need to be as large, and the relay should be shifted toward the source to strengthen the decoding link.

Relay Position Fraction

The parameter $f \in (0, 1)$ specifying the relay's location as a fraction of the source-destination distance: $d_{SR} = f \cdot d_{SD}$ and $d_{RD} = (1-f) \cdot d_{SD}$ . The optimal $f$ for DF relaying is slightly less than $1/2$ (relay closer to source) due to the direct link at the destination.

Related: Decode-and-Forward (DF)

Relay Position Optimisation

Where to Place the Relay?

Definition: Relay Position Parameterisation

Definition: Relay Gain over Direct Transmission

Theorem: Optimal Relay Position for DF

Rate expression

Balancing condition

Numerical solutions

Optimal Relay Placement: DF vs AF Rate Sweep

Relay Position Optimisation

Parameters

Example: Optimal Relay Placement

Direct link

DF at midpoint

DF at optimal position

Minimum $\alpha$ for relay gain

Quick Check

Relay Position Fraction

Definition:
Relay Position Parameterisation

Definition:
Relay Gain over Direct Transmission