Ferkans — Interactive Telecom Tutor

What Can We Achieve Without Telling the Transmitter?

In many wireless systems, the receiver can estimate the channel through pilot symbols, but feeding this information back to the transmitter is costly or infeasible (especially in FDD systems or high-mobility scenarios). This raises a natural question: if only the receiver knows the channel, what is the best rate we can reliably achieve?

The remarkable answer is that for ergodic fading, the capacity with CSIR only is achieved by a simple strategy: transmit at constant power regardless of the channel state and let the decoder, which knows $H$ , handle the varying SNR. No transmitter adaptation is needed for the ergodic rate.

Theorem: Ergodic Capacity with CSIR Only

Consider the scalar fading channel $Y = HX + Z$ with $Z \sim \mathcal{N}(0, N)$ , average power constraint $\mathbb{E}[|X|^2] \leq P$ , and ergodic fading $H$ with $\mathbb{E}[|H|^2] = 1$ . If the receiver knows $H$ (CSIR) but the transmitter does not, the capacity is

$C_{\text{erg}} = \mathbb{E}_H\!\left[\frac{1}{2}\log\!\left(1 + |H|^2 \,\text{SNR}\right)\right],$

where $\text{SNR} = P/N$ . The capacity is achieved by i.i.d. Gaussian inputs $X_i \sim \mathcal{N}(0, P)$ (constant power, independent of $H$ ).

The decoder knows $H$ , so for each fading realization $H = h$ , the channel looks like an AWGN channel with SNR $|h|^2 \cdot \text{SNR}$ . By coding across many fading states, the effective rate averages to $\mathbb{E}[\frac{1}{2}\log(1 + |H|^2 \text{SNR})]$ . The transmitter does not need to know $H$ because it uses constant power — the law of large numbers does the averaging.

Show Hint

Think of the fading channel as a collection of parallel AWGN sub-channels, one for each fading state.

For a given $H = h$ , the mutual information is $\frac{1}{2}\log(1 + |h|^2 P/N)$ . Average over $H$ .

Show that constant-power Gaussian input maximizes $I(X; Y | H)$ for each state, so no power adaptation helps.

Proof

Conditional mutual information

For a given fading realization $H = h$ , the channel is $Y = hX + Z$ with $Z \sim \mathcal{N}(0, N)$ . This is an AWGN channel with gain $h$ . From Chapter 10, the mutual information for input distribution $p_X$ with $\mathbb{E}[|X|^2] \leq P$ is maximized by $X \sim \mathcal{N}(0, P)$ , yielding

$I(X; Y | H = h) = \frac{1}{2}\log\left(1 + \frac{|h|^2 P}{N}\right).$

Notice that the maximizing input distribution $\mathcal{N}(0, P)$ does not depend on $h$ . This is the key observation.

Averaging over fading states

Since the fading is ergodic, the coding theorem for compound channels tells us that the capacity is

$C_{\text{erg}} = \max_{p_X:\, \mathbb{E}[|X|^2] \leq P} \mathbb{E}_H\!\left[I(X; Y | H)\right].$

Since $X \sim \mathcal{N}(0, P)$ maximizes $I(X; Y | H = h)$ for every $h$ simultaneously, it also maximizes the expectation. Therefore

$C_{\text{erg}} = \mathbb{E}_H\!\left[\frac{1}{2}\log\!\left(1 + |H|^2 \,\text{SNR}\right)\right].$

Achievability

Achievability follows from random coding over a block of length $n$ that spans many independent fading realizations. Generate a codebook of $2^{nR}$ codewords, each i.i.d. $\mathcal{N}(0, P)$ . The decoder, knowing $(Y^n, H^n)$ , uses joint typicality decoding.

For each fading state $H_i = h_i$ , the channel behaves as $Y_i = h_i X_i + Z_i$ , and the decoder knows $h_i$ . By the ergodic theorem, $\frac{1}{n}\sum_{i=1}^n \frac{1}{2}\log(1 + |H_i|^2 \text{SNR}) \to \mathbb{E}[\frac{1}{2}\log(1 + |H|^2 \text{SNR})]$ almost surely.

Standard random coding analysis shows that any rate below this limit is achievable with vanishing error probability. $\blacksquare$

,

Theorem: Fading Reduces Ergodic Capacity (Jensen's Inequality)

For the ergodic fading channel with CSIR only,

$C_{\text{erg}} = \mathbb{E}\!\left[\frac{1}{2}\log\!\left(1 + |H|^2 \,\text{SNR}\right)\right] \leq \frac{1}{2}\log\!\left(1 + \mathbb{E}[|H|^2] \cdot \text{SNR}\right) = C_{\text{AWGN}}.$

Equality holds if and only if $|H|^2$ is deterministic (no fading). The gap $C_{\text{AWGN}} - C_{\text{erg}}$ measures the capacity penalty due to fading.

The function $f(x) = \frac{1}{2}\log(1 + x)$ is strictly concave. By Jensen's inequality, the average of a concave function of a random variable is less than or equal to the concave function of the average. Intuitively, the capacity loss from deep fades is not compensated by the gain from strong fades — the $\log$ saturates, so strong states help less than weak states hurt.

Proof

Apply Jensen's inequality

The function $g(t) = \frac{1}{2}\log(1 + t)$ is strictly concave in $t \geq 0$ (since $g''(t) = -\frac{1}{2\ln 2} \cdot \frac{1}{(1+t)^2} < 0$ ). Applying Jensen's inequality to $t = |H|^2 \cdot \text{SNR}$ :

$\mathbb{E}[g(|H|^2 \text{SNR})] \leq g(\mathbb{E}[|H|^2 \text{SNR}]) = g(\text{SNR}),$

where the last equality uses $\mathbb{E}[|H|^2] = 1$ .

Equality condition

Equality in Jensen's inequality holds if and only if $|H|^2 \text{SNR}$ is a constant a.s., which requires $|H|^2 = 1$ a.s. (no fading). For any non-degenerate fading distribution, the inequality is strict. $\blacksquare$

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The Fading Penalty Is Often Modest

While fading always hurts compared to AWGN (at the same average SNR), the loss is often surprisingly small for ergodic fading:

For Rayleigh fading at $\text{SNR} = 20$ dB, the ergodic capacity is about 90% of the AWGN capacity.
At low SNR, the penalty is negligible because $\log(1 + x) \approx x$ is nearly linear, so Jensen's inequality is nearly tight.
At high SNR, the penalty approaches a constant gap (in bits) rather than growing with $\text{SNR}$ .

The real impact of fading manifests in the outage regime (Section 13.4), where the occasional deep fade causes catastrophic failure for a single codeword.

Example: Ergodic Capacity of Rayleigh Fading Channel

Compute the ergodic capacity of a Rayleigh fading channel ( $H \sim \mathcal{CN}(0,1)$ ) with CSIR only at $\text{SNR} = 10$ dB. Compare with the AWGN capacity.

Solution

Set up the integral

For Rayleigh fading, $|H|^2 \sim \text{Exp}(1)$ with PDF $f(t) = e^{-t}$ for $t \geq 0$ . The ergodic capacity is

$C_{\text{erg}} = \int_0^\infty \frac{1}{2}\log(1 + t \cdot \text{SNR}) \, e^{-t} \, dt.$

Evaluate numerically

At $\text{SNR} = 10$ (i.e., 10 dB), numerical integration gives

$C_{\text{erg}} \approx 1.579 \text{ bits/channel use}.$

The AWGN capacity at the same SNR is

$C_{\text{AWGN}} = \frac{1}{2}\log(1 + 10) \approx 1.730 \text{ bits/channel use}.$

Compare

The fading penalty is $1.730 - 1.579 = 0.151$ bits, or about 8.7% of the AWGN capacity. This confirms that the ergodic capacity loss from Rayleigh fading is modest at moderate SNR.

In closed form, the Rayleigh ergodic capacity can be written using the exponential integral: $C_{\text{erg}} = \frac{1}{2\ln 2} \cdot e^{1/\text{SNR}} \cdot E_1(1/\text{SNR})$ , where $E_1(x) = \int_x^\infty \frac{e^{-t}}{t}\,dt$ .

Ergodic Capacity: Fading vs. AWGN

Compare the ergodic capacity of a Rayleigh fading channel (CSIR only) with the AWGN channel capacity as a function of $\text{SNR}$ . Observe that fading always reduces capacity (Jensen's inequality) but the gap remains modest, especially at low and moderate SNR.

Parameters

Max SNR (dB)30

Upper limit of the SNR range in decibels

Fading distribution

Distribution of the fading gain $|H|^2$

Low-SNR and High-SNR Behavior

The ergodic capacity has clean asymptotic forms:

Low SNR ( $\text{SNR} \to 0$ ): Since $\log(1 + x) \approx x/\ln 2$ for small $x$ ,

$C_{\text{erg}} \approx \frac{\mathbb{E}[|H|^2] \cdot \text{SNR}}{2 \ln 2} = \frac{\text{SNR}}{2 \ln 2} = C_{\text{AWGN}}.$

Fading causes no capacity loss at low SNR. Intuitively, the log is nearly linear, so Jensen's inequality is nearly tight.

High SNR ( $\text{SNR} \to \infty$ ): We have $\frac{1}{2}\log(1 + |H|^2 \text{SNR}) \approx \frac{1}{2}\log(\text{SNR}) + \frac{1}{2}\log(|H|^2)$ , so

$C_{\text{erg}} \approx \frac{1}{2}\log(\text{SNR}) + \frac{1}{2}\mathbb{E}[\log |H|^2].$

The gap to AWGN capacity is $-\frac{1}{2}\mathbb{E}[\log |H|^2]$ , which is a constant independent of $\text{SNR}$ . For Rayleigh fading, $\mathbb{E}[\log_2 |H|^2] = -\gamma_{\text{EM}}/\ln 2 \approx -0.833$ bits, where $\gamma_{\text{EM}} \approx 0.5772$ is the Euler-Mascheroni constant. The high-SNR gap is thus about 0.42 bits.

Common Mistake: Constant Power Is Optimal Only for Ergodic Capacity

Mistake:

Concluding from the CSIR ergodic capacity result that constant-power transmission is always optimal, regardless of the performance metric.

Correction:

Constant power is optimal for ergodic capacity because the mutual information $I(X; Y | H = h)$ is maximized by $X \sim \mathcal{N}(0, P)$ for every $h$ simultaneously. But for other metrics — such as minimizing outage probability, maximizing delay-limited capacity, or optimizing throughput with hybrid ARQ — power adaptation with CSIT can provide significant gains. The next section shows that water-filling over fading states increases ergodic capacity when CSIT is available.

Quick Check

For the ergodic fading channel with CSIR only, why does constant-power transmission achieve capacity?

Because the transmitter has no information to adapt to

Because $X \sim \mathcal{N}(0, P)$ maximizes $I(X; Y | H = h)$ for every $h$ , so no state-dependent power control can help

Because Jensen's inequality forces equal power allocation

Because the channel is memoryless

Correction:

Because

X \sim \mathcal{N}(0, P)

maximizes

I(X; Y | H = h)

for every

h

, so no state-dependent power control can help

The optimal input for each AWGN sub-channel (with gain $h$ ) is Gaussian with power $P$ , regardless of $h$ . Since the same input distribution is optimal for every fading state, there is nothing to be gained by adapting the power to $H$ .

⚠️Engineering Note

Channel Estimation Overhead Reduces Effective Rate

The ergodic capacity formula assumes perfect CSIR, but in practice the receiver must estimate the channel from pilot symbols. In an OFDM system with $T$ time slots per coherence interval and $F$ frequency bins per coherence bandwidth, a block of $TF$ resource elements is available. Of these, $\tau$ must be used for pilots, leaving $(TF - \tau)$ for data.

The effective throughput is approximately

$R_{\text{eff}} \approx \frac{TF - \tau}{TF} \cdot C_{\text{erg}}(\text{SNR}_{\text{eff}}),$

where $\text{SNR}_{\text{eff}}$ accounts for the estimation error. For rapidly varying channels (small $T$ ), the overhead can consume a significant fraction of the resources.

Practical Constraints

•
Pilot density must satisfy $\tau \geq n_t$ to estimate all transmit dimensions
•
Channel estimation error adds an irreducible noise floor proportional to $1/\text{SNR}_{\text{pilot}}$
•
In 5G NR, DMRS patterns allocate 1-4 OFDM symbols per slot for channel estimation

📋 Ref: 3GPP TS 38.211, Section 7.4.1

Ergodic capacity

The maximum achievable rate for reliable communication over a fading channel when the codeword spans many independent fading realizations. Given by $C_{\text{erg}} = \mathbb{E}_H[\frac{1}{2}\log(1 + |H|^2 \text{SNR})]$ for the scalar fading channel with CSIR.

Related: CSIR, Rayleigh fading

Key Takeaway

The ergodic capacity with CSIR only is $C_{\text{erg}} = \mathbb{E}[\frac{1}{2}\log(1 + |H|^2 \text{SNR})]$ , achieved by constant-power Gaussian transmission. Fading always reduces capacity compared to AWGN (by Jensen's inequality), but the loss is modest for ergodic channels. The transmitter does not need to know the channel to achieve the ergodic rate — CSIR alone suffices.

Ergodic Capacity: AWGN vs Rayleigh Fading

Side-by-side comparison of ergodic capacity curves for the AWGN channel and the Rayleigh fading channel (CSIR only). The animation highlights the fading penalty predicted by Jensen's inequality — fading always hurts, but the loss is modest at practical SNR values.