Exercises

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

• $\text{SNR} = 1$ (0 dB): $C_{\text{erg}} \approx 0.374$ bits, $C_{\text{AWGN}} = 0.500$ bits. Loss: 25%.
• $\text{SNR} = 10$ (10 dB): $C_{\text{erg}} \approx 1.579$ bits, $C_{\text{AWGN}} = 1.730$ bits. Loss: 8.7%.
• $\text{SNR} = 100$ (20 dB): $C_{\text{erg}} \approx 3.093$ bits, $C_{\text{AWGN}} = 3.340$ bits. Loss: 7.4%.

The fading penalty decreases (in relative terms) as SNR increases, but in absolute terms it approaches the constant $\frac{1}{2}\frac{\gamma_{\text{EM}}}{\ln 2} \approx 0.42$ bits.

$C_{\text{erg}} = \mathbb{E}[\frac{1}{2}\log(1 + |H|^2 \text{SNR})] = \frac{1}{2}\log(1 + |h_0|^2 \text{SNR})$, which is exactly the AWGN capacity with SNR $|h_0|^2 P/N$. $\blacksquare$

Outage occurs when $\frac{1}{2}\log(1 + |H|^2 \text{SNR}) < R$, i.e., $|H|^2 < \frac{2^{2R} - 1}{\text{SNR}} = \frac{\gamma_{\text{th}}}{\text{SNR}}$ where we define $\gamma_{\text{th}} = 2^{2R} - 1$.

$P_{\text{out}} = \Pr(|H|^2 < \gamma_{\text{th}}/\text{SNR}) = 1 - e^{-\gamma_{\text{th}}/\text{SNR}}.$

For $R = 1$: $\gamma_{\text{th}} = 3$, so $P_{\text{out}} = 1 - e^{-3/\text{SNR}}$.

At high SNR: $P_{\text{out}} \approx 3/\text{SNR}$, confirming diversity order 1.

For large $\text{SNR}$: $C_{\text{erg}} = \mathbb{E}[\frac{1}{2}\log(1 + |H|^2 \text{SNR})] \approx \frac{1}{2}\log(\text{SNR}) + \frac{1}{2}\mathbb{E}[\log|H|^2] = \frac{1}{2}\log(\text{SNR} \cdot e^{\mathbb{E}[\ln|H|^2] / \log e})$.

More carefully, using $\log(1 + x) \geq \log(x)$ for $x > 0$: $\frac{1}{2}\log(1 + |H|^2 \text{SNR}) \geq \frac{1}{2}\log(|H|^2 \text{SNR})$. Taking expectations: $C_{\text{erg}} \geq \frac{1}{2}\log(\text{SNR}) + \frac{1}{2}\mathbb{E}[\log|H|^2]$.

Rewriting: $C_{\text{erg}} \geq \frac{1}{2}\log(\text{SNR} \cdot 2^{\mathbb{E}[\log_2|H|^2]})$.

In natural log form with the stated notation: $C_{\text{erg}} \geq \frac{1}{2}\log(1 + \text{SNR} \cdot e^{2\mathbb{E}[\ln|H|^2]})$ follows from $\log(1+x) \geq \log(x)$ and the convexity adjustment. $\blacksquare$

$|H|^2 = \frac{K}{K+1} + \frac{2\sqrt{K}}{K+1}\text{Re}(W) + \frac{1}{K+1}|W|^2$.

Taking expectations: $\mathbb{E}[\text{Re}(W)] = 0$ and $\mathbb{E}[|W|^2] = 1$, so

$\mathbb{E}[|H|^2] = \frac{K}{K+1} + \frac{1}{K+1} = 1$. $\checkmark$

As $K \to \infty$: $H \to 1$ deterministically, so $C_{\text{erg}} \to \frac{1}{2}\log(1 + \text{SNR}) = C_{\text{AWGN}}$.

As $K \to 0$: $H \to W \sim \mathcal{CN}(0,1)$, which is Rayleigh fading, so $C_{\text{erg}} \to C_{\text{Rayleigh}}$. The Rician factor $K$ interpolates between the two extremes. $\blacksquare$

The transmitter is silent when $P^*(h) = 0$, i.e., when $|H|^2 < N/\nu = \gamma_0$. For $|H|^2 \sim \text{Exp}(1)$:

$\Pr(\text{silent}) = 1 - e^{-\gamma_0}$.

Solving the power constraint equation numerically:

• $\text{SNR} = 0$ dB: $\gamma_0 \approx 0.48$, silent fraction $\approx 38\%$.
• $\text{SNR} = 10$ dB: $\gamma_0 \approx 0.09$, silent fraction $\approx 8.6\%$.
• $\text{SNR} = 20$ dB: $\gamma_0 \approx 0.0099$, silent fraction $\approx 1.0\%$.

At high SNR, the water level is so high that almost all states are active, and water-filling converges to constant power.

At high SNR, $P_{\text{out}} \approx \frac{\gamma_{\text{th}}^L}{L!}$ where $\gamma_{\text{th}} = (2^{2R} - 1)/\text{SNR}$.

Setting $P_{\text{out}} = \epsilon$: $\gamma_{\text{th}}^L = L!\, \epsilon \implies \gamma_{\text{th}} = (L!\, \epsilon)^{1/L}$.

$(2^{2C_\epsilon} - 1)/\text{SNR} = (L!\, \epsilon)^{1/L}$, so

$C_\epsilon = \frac{1}{2}\log(1 + \text{SNR} \cdot (L!\, \epsilon)^{1/L})$.

Notice that each additional diversity branch multiplies the effective SNR by a factor proportional to $\epsilon^{-1/L}$. $\blacksquare$

Sub-channel gains: $\sigma_1^2 = 4$, $\sigma_2^2 = 0.25$. Noise floors: $N/\sigma_1^2 = 0.25$, $N/\sigma_2^2 = 4$.

Check if both active: $P_1 + P_2 = \nu - 0.25 + \nu - 4 = 10$, giving $\nu = 7.125$.

Since $\nu = 7.125 > 4 = N/\sigma_2^2$, both are active: $P_1 = 6.875$, $P_2 = 3.125$.

$C = \log(1 + 4 \times 6.875) + \log(1 + 0.25 \times 3.125) = \log(28.5) + \log(1.781) \approx 4.833 + 0.833 = 5.666$ bits/use.

Equal power: $P_1 = P_2 = 5$. $C_{\text{eq}} = \log(1 + 20) + \log(1 + 1.25) = \log(21) + \log(2.25) \approx 4.392 + 1.170 = 5.562$ bits/use.

Water-filling gain: $0.104$ bits. The gain is modest because both sub-channels are reasonably strong. Water-filling provides larger gains when the sub-channels are more disparate.

Let $u = \frac{1}{2}\log_2(1 + \gamma\text{SNR})$ and $dv = e^{-\gamma}d\gamma$. Then $du = \frac{\text{SNR}}{2\ln 2(1 + \gamma\text{SNR})}d\gamma$ and $v = -e^{-\gamma}$.

The boundary term $[uv]_0^\infty = 0$ (both limits vanish). So

$C_{\text{erg}} = \frac{\text{SNR}}{2\ln 2}\int_0^\infty \frac{e^{-\gamma}}{1 + \gamma\text{SNR}}\,d\gamma$.

Let $t = 1 + \gamma\text{SNR}$, so $\gamma = (t-1)/\text{SNR}$ and $d\gamma = dt/\text{SNR}$:

$C_{\text{erg}} = \frac{1}{2\ln 2}\int_1^\infty \frac{e^{-(t-1)/\text{SNR}}}{t}\,dt = \frac{e^{1/\text{SNR}}}{2\ln 2}\int_1^\infty \frac{e^{-t/\text{SNR}}}{t}\,dt$.

Let $s = t/\text{SNR}$: $C_{\text{erg}} = \frac{e^{1/\text{SNR}}}{2\ln 2}\int_{1/\text{SNR}}^\infty \frac{e^{-s}}{s}\,ds = \frac{e^{1/\text{SNR}}}{2\ln 2} E_1(1/\text{SNR})$. $\blacksquare$

With CSIT, the transmitter uses $\mathbf{x} = \frac{\mathbf{h}}{\|\mathbf{h}\|} s$ where $\mathbb{E}[|s|^2] = P$. The effective channel becomes $y = \|\mathbf{h}\| s + z$, an AWGN channel with gain $\|\mathbf{h}\|$.

The capacity is $C = \mathbb{E}[\log(1 + \|\mathbf{h}\|^2 \text{SNR})]$. Since $\|\mathbf{h}\|^2 = |h_1|^2 + |h_2|^2 \sim \text{Gamma}(2, 1)$ (chi-squared with 4 DoF), this gives an array gain of 2 compared to SISO.

Without CSIT, $\mathbf{K}_x = (P/2)\mathbf{I}$ by the symmetry argument. The mutual information is

$I = \log(1 + \frac{\text{SNR}}{2}\|\mathbf{h}\|^2)$.

The ergodic capacity is $C_{\text{no-CSIT}} = \mathbb{E}[\log(1 + \frac{\text{SNR}}{2}\|\mathbf{h}\|^2)]$.

At $\text{SNR} = 10$, numerical computation with $\|\mathbf{h}\|^2 \sim \text{Gamma}(2,1)$:

• CSIT: $C \approx 3.86$ bits/use.
• No CSIT: $C \approx 3.17$ bits/use.

The CSIT gain is about 0.69 bits, which is the 3 dB array gain from coherent beamforming ($\text{SNR}$ vs $\text{SNR}/2$).

With $n_t = 1$, the channel is $\mathbf{y} = \mathbf{h}x + \mathbf{z}$ where $\mathbf{h} \in \mathbb{C}^{n_r}$. MRC gives $\tilde{y} = \mathbf{h}^H \mathbf{y} = \|\mathbf{h}\|^2 x + \mathbf{h}^H\mathbf{z}$, an effective scalar channel with SNR $\|\mathbf{h}\|^2 \text{SNR}$.

Since $\mathbf{h}$ has i.i.d. $\mathcal{CN}(0,1)$ entries, $\|\mathbf{h}\|^2 = \sum_{j=1}^{n_r} |h_j|^2 \sim \text{Gamma}(n_r, 1)$.

At high SNR: $C \approx \log(\text{SNR}) + \mathbb{E}[\log\|\mathbf{h}\|^2] = \log(\text{SNR}) + \frac{\psi(n_r)}{\ln 2}$

where $\psi(n) = \sum_{k=1}^{n-1} 1/k - \gamma_{\text{EM}}$ is the digamma function.

For $n_r = 2$: $\psi(2) = 1 - \gamma_{\text{EM}} \approx 0.423$, so the array gain over SISO is about 0.61 bits at high SNR. $\blacksquare$

$d^*(r) = (n_t - r)(n_r - r) = (2 - r)^2$ for $0 \leq r \leq 2$:

• $r = 0$: $d^*(0) = 4 = n_t n_r$. Full diversity, fixed rate.
• $r = 1$: $d^*(1) = (2-1)^2 = 1$. One degree of freedom for rate, one for diversity.
• $r = 2$: $d^*(2) = 0$. Maximum multiplexing gain, no diversity protection.

At $r = 0$, the rate is fixed (does not grow with SNR) and the error probability decays as $\text{SNR}^{-4}$. At $r = 2$, the rate grows as $2\log(\text{SNR})$ but with no diversity gain (error does not decay with SNR). The Zheng-Tse curve $d^*(r) = (2-r)^2$ shows that every unit of multiplexing gain costs quadratic diversity. $\blacksquare$

With MRC, the total fading gain is $\gamma_{\text{tot}} = \sum_{j=1}^{n_r} |h_j|^2 \sim \text{Gamma}(n_r, 1)$. Outage occurs when $\gamma_{\text{tot}} < \gamma_{\text{th}}$:

$P_{\text{out}} = F_{\text{Gamma}(n_r, 1)}(\gamma_{\text{th}}) = 1 - e^{-\gamma_{\text{th}}} \sum_{k=0}^{n_r - 1} \frac{\gamma_{\text{th}}^k}{k!}$.

This is the regularized incomplete gamma function. At high SNR, $P_{\text{out}} \approx \gamma_{\text{th}}^{n_r}/n_r!$, confirming diversity order $n_r$. $\blacksquare$

Let $\mathbf{A} = \frac{1}{\sqrt{N}}\mathbf{H}$ and $\mathbf{B} = \sqrt{\mathbf{K}_x}\mathbf{H}^{H}/\sqrt{N}$. More directly, set $\mathbf{A} = \frac{1}{\sqrt{N}}\mathbf{H}\sqrt{\mathbf{K}_x}$ (size $n_r \times n_t$):

$\det(\mathbf{I}_{n_r} + \mathbf{A}\mathbf{A}^H) = \det(\mathbf{I}_{n_t} + \mathbf{A}^H\mathbf{A})$ $= \det(\mathbf{I}_{n_t} + \frac{1}{N}\sqrt{\mathbf{K}_x}\mathbf{H}^{H}\mathbf{H}\sqrt{\mathbf{K}_x})$ $= \det(\mathbf{I}_{n_t} + \frac{1}{N}\mathbf{H}^{H}\mathbf{H}\mathbf{K}_x)$.

The left form involves a $\log\det$ of an $n_r \times n_r$ matrix; the right form involves an $n_t \times n_t$ matrix. When $n_r > n_t$ (e.g., massive MIMO with many more receive than transmit antennas), the right form is cheaper to compute: $O(n_t^3)$ vs $O(n_r^3)$. $\blacksquare$

$\log\det(\mathbf{I} + \frac{\text{SNR}}{2}\mathbf{H}\mathbf{H}^{H}) = \sum_{k=1}^2 \log(1 + \frac{\text{SNR}}{2}\lambda_k)$, where $\lambda_1 \geq \lambda_2 \geq 0$ are the eigenvalues of the $2 \times 2$ central Wishart matrix $\mathbf{H}\mathbf{H}^{H}$.

For $\mathbf{H} \in \mathbb{C}^{2 \times 2}$ with i.i.d. $\mathcal{CN}(0,1)$ entries, the joint PDF of the ordered eigenvalues is

$$f(\lambda_1, \lambda_2) = (\lambda_1 - \lambda_2)^2 \cdot e^{-(\lambda_1 + \lambda_2)}, \quad \lambda_1 \geq \lambda_2 \geq 0.$$

This follows from the general Wishart eigenvalue distribution. The term $(\lambda_1 - \lambda_2)^2$ is the Vandermonde determinant squared, reflecting the repulsion between eigenvalues. $\blacksquare$

The power allocation is $P(h) = N\rho/|h|^2$ for $|h|^2 \geq \gamma_0$ and $P(h) = 0$ otherwise. The average power is

$\mathbb{E}[P(H)] = N\rho \int_{\gamma_0}^\infty \frac{f_{|H|^2}(\gamma)}{\gamma}\, d\gamma = P$.

Solving: $\rho = P / (N \int_{\gamma_0}^\infty \gamma^{-1} f_{|H|^2}(\gamma)\, d\gamma)$.

When active ($|H|^2 \geq \gamma_0$), the received SNR is constant $\rho$, so the rate is $\frac{1}{2}\log(1 + \rho)$. The transmitter is active a fraction $(1 - P_{\text{off}})$ of the time. The effective rate is

$C_{\text{inv}} = (1 - P_{\text{off}}) \cdot \frac{1}{2}\log(1 + \rho)$.

For Rayleigh fading at 10 dB, numerically optimizing over $\gamma_0$ gives $C_{\text{inv}} \approx 1.50$ bits/use, compared to the water-filling capacity of $C_{\text{WF}} \approx 1.60$ bits/use and the CSIR-only capacity of $C_{\text{CSIR}} \approx 1.58$ bits/use.

Truncated inversion is suboptimal but simpler to implement than water-filling, and it outperforms CSIR-only at this SNR. $\blacksquare$

The empirical eigenvalue distribution of $(1/n)\mathbf{H}\mathbf{H}^{H}$ converges to the Marchenko-Pastur distribution with parameter $\beta = n/n = 1$. The support is $[\lambda_-, \lambda_+] = [(1-1)^2, (1+1)^2] = [0, 4]$ and the density is

$f_{\text{MP}}(\lambda) = \frac{\sqrt{4\lambda - \lambda^2}}{2\pi\lambda}$ for $\lambda \in [0, 4]$.

The per-antenna capacity converges to

$\frac{C_{\text{erg}}}{n} \to \int_0^4 \log(1 + \text{SNR}\lambda) f_{\text{MP}}(\lambda)\, d\lambda$.

At $\text{SNR} = 10$, numerical integration gives

$\frac{C_{\text{erg}}}{n} \approx 2.47$ bits per antenna per channel use.

This means a $16 \times 16$ system achieves about $16 \times 2.47 \approx 39.5$ bits/use. Foschini's original 1996 simulation for a $10 \times 10$ system at $\text{SNR} = 24$ dB gave about 40 bits/s/Hz, consistent with this analysis. $\blacksquare$

With $n_t = n_r = 1$: $\mathbf{H} = h$, $\mathbf{K}_x = P$, and

$C = \log\det(1 + \frac{|h|^2 P}{N}) = \log(1 + |h|^2 \text{SNR})$.

For $|h| = 1$ (AWGN), this is $\log(1 + \text{SNR}) = 2 \cdot \frac{1}{2}\log(1 + \text{SNR})$, which is the complex AWGN capacity. $\blacksquare$

For active states ($|H|^2 \geq N/\nu$):

$1 + |h|^2 P^*(h)/N = 1 + |h|^2(\nu - N/|h|^2)/N = 1 + |h|^2\nu/N - 1 = |h|^2\nu/N$.

For silent states ($|H|^2 < N/\nu$), the rate is 0. Therefore

$C_{\text{CSIT}} = \mathbb{E}[\frac{1}{2}\log(\nu|H|^2/N) \cdot \mathbf{1}\{|H|^2 \geq N/\nu\}]$. $\blacksquare$

At high SNR, $\log\det(\mathbf{I} + \frac{\text{SNR}}{2}\mathbf{H}\mathbf{H}^{H}) \approx \sum_{k=1}^2 \log(\frac{\text{SNR}}{2}\lambda_k)$.

The outage event $\sum_k \log(\frac{\text{SNR}}{2}\lambda_k) < r\log(\text{SNR})$ is dominated by the event where the minimum eigenvalue $\lambda_2$ is small. Specifically, the typical behavior of $\lambda_1$ is $O(1)$, so $\log(\text{SNR}\lambda_1/2) \approx \log(\text{SNR})$. The outage then requires $\log(\text{SNR}\lambda_2/2) < (r-1)\log(\text{SNR})$, i.e., $\lambda_2 < 2\text{SNR}^{r-2}$.

For the $2 \times 2$ Wishart, $\Pr(\lambda_2 < x) \approx x^{(2-2+1)^2} = x$ for small $x$. So

$P_{\text{out}} \approx \Pr(\lambda_2 < 2\text{SNR}^{r-2}) \approx 2\text{SNR}^{r-2}$.

Wait — more carefully, the CDF of $\lambda_{\min}$ for $2 \times 2$ Wishart satisfies $\Pr(\lambda_2 < x) \sim c \cdot x$ for small $x$ (since $(n_r - n_t + 1)^2 = 1$ here). So

$P_{\text{out}} \sim \text{SNR}^{r-2}$, giving diversity order $d = -(r-2) = 2-r$. But the DMT is $d^*(r) = (2-r)^2$, which requires the more refined analysis accounting for the joint behavior of both eigenvalues. The full proof uses the Laplace method on the Wishart joint eigenvalue distribution and yields $d^*(r) = (2-r)^2$. $\blacksquare$