Exercises

$\mathbf{C}\mathbf{C}^H = \begin{bmatrix} x_1 & -x_2^* \\ x_2 & x_1^* \end{bmatrix} \begin{bmatrix} x_1^* & x_2^* \\ -x_2 & x_1 \end{bmatrix}$$$[\mathbf{C}\mathbf{C}^H]{11} = |x_1|^2 + |x_2|^2$$[\mathbf{C}\mathbf{C}^H]{12} = x_1 x_2^* - x_2^* x_1 = 0$$[\mathbf{C}\mathbf{C}^H]{21} = x_2 x_1^* - x_1^* x_2 = 0$$[\mathbf{C}\mathbf{C}^H]{22} = |x_2|^2 + |x_1|^2$Therefore$\mathbf{C}\mathbf{C}^H = (|x_1|^2 + |x_2|^2)\mathbf{I}_2$, confirming the orthogonality property.$\blacksquare$

$\Delta\mathbf{C}\,\Delta\mathbf{C}^H = \begin{bmatrix} 2 & 0 & 1 \\ 0 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}$$

The characteristic polynomial is $\det(\lambda\mathbf{I} - \mathbf{A}) = (\lambda - 2)[(\lambda - 2)^2 - 2] + \text{correction}$.

Computing directly: eigenvalues are $\lambda_1 \approx 3.414$, $\lambda_2 = 2$, $\lambda_3 \approx 0.586$. All three are positive, so $r = 3$ (full rank).

(b) Diversity order $= r \times N_r = 3 \times 2 = 6 = N_t N_r$. This codeword pair achieves full diversity.

(c) $\det(\mathbf{A}) = 3.414 \times 2 \times 0.586 = 4$. Coding gain $= \det(\mathbf{A})^{1/r} = 4^{1/3} \approx 1.587$. $\blacksquare$

A rate-1 complex OSTBC for $N_t = 4$ encodes $K = T$ complex symbols over $T$ time slots. Writing each complex symbol as $x_k = a_k + jb_k$, we need $2K = 2T$ real variables. The orthogonality condition $\mathbf{C}\mathbf{C}^H = (\sum |x_k|^2)\mathbf{I}_4$ requires that the real and imaginary component matrices satisfy the Hurwitz-Radon amicability conditions.

For $N_t = 4$, the Hurwitz-Radon number is $\rho(4) = 4$. A complex rate-1 design needs at least $2N_t = 8$ real amicable matrices (one for the real and one for the imaginary part of each of 4 complex symbols).

Since $\rho(4) = 4 < 8$, it is impossible to find 8 mutually amicable $4 \times 4$ real matrices. Therefore, no rate-1 complex OSTBC exists for $N_t = 4$. $\blacksquare$

The maximum rate for $N_t = 4$ is $R = 3/4$, achieved by the rate-3/4 OSTBC that encodes 3 complex symbols in 4 time slots (requiring 6 real amicable matrices, and $\rho(4) = 4$ is sufficient when the design is relaxed to allow non-square code matrices). $\blacksquare$

(a) Each of $N_t = 4$ antennas transmits $\log_2(16) = 4$ bits/symbol. Spectral efficiency $= N_t \times \log_2(M) = 4 \times 4 = 16$ bits/s/Hz.

(b) $M^{N_t} = 16^4 = 65{,}536$ candidate vectors.

(c) MMSE-OSIC requires $N_t$ iterations, each involving an $O(N_t^2 N_r)$ matrix computation (after deflation). Total: $O(N_t^3 N_r) = O(4^3 \times 4) = O(256)$. $\blacksquare$

$\hat{\mathbf{x}}_{\text{ZF}} = (\mathbf{H}^{H}\mathbf{H})^{-1}\mathbf{H}^{H}\mathbf{y} = \mathbf{x} + \underbrace{(\mathbf{H}^{H}\mathbf{H})^{-1}\mathbf{H}^{H}\mathbf{n}}_{\tilde{\mathbf{n}}}$$

$\text{Cov}(\tilde{\mathbf{n}}) = (\mathbf{H}^{H}\mathbf{H})^{-1}\mathbf{H}^{H} \cdot \sigma^2\mathbf{I} \cdot \mathbf{H}(\mathbf{H}^{H}\mathbf{H})^{-1} = \sigma^2 (\mathbf{H}^{H}\mathbf{H})^{-1}$$

The noise variance for stream $k$ is $\sigma^2_k = \sigma^2[(\mathbf{H}^{H}\mathbf{H})^{-1}]_{kk}$. Assuming unit-power symbols ($\mathbb{E}[|x_k|^2] = 1$):

$$\text{SNR}_{k} = \frac{1}{\sigma^2[(\mathbf{H}^{H}\mathbf{H})^{-1}]_{kk}}$$

$\blacksquare$

$\mathbf{H}^{H}\mathbf{H} = \begin{bmatrix} 1.09 & 0.86 \\ 0.86 & 1.69 \end{bmatrix}KATEXPLACEHOLDER0END\mathbf{H}^{H}\mathbf{H} + 0.1\mathbf{I} = \begin{bmatrix} 1.19 & 0.86 \\ 0.86 & 1.79 \end{bmatrix}$$

$\det = 1.19 \times 1.79 - 0.86^2 = 2.1301 - 0.7396 = 1.3905$

$$(\mathbf{H}^{H}\mathbf{H} + 0.1\mathbf{I})^{-1} = \frac{1}{1.3905} \begin{bmatrix} 1.79 & -0.86 \\ -0.86 & 1.19 \end{bmatrix} = \begin{bmatrix} 1.287 & -0.619 \\ -0.619 & 0.856 \end{bmatrix}$$

$$\text{SINR}_1 = \frac{1}{1.287} - 1 = -0.223 \Rightarrow \text{SINR}_1 \approx -0.223 + 1 = 0.777 \Rightarrow \frac{1}{1.287} - 1 \approx -0.223$$

Correction: $\text{SINR}_1 = 1/[(\mathbf{H}^{H}\mathbf{H}+\sigma^2\mathbf{I})^{-1}]_{11} - 1 = 1/1.287 - 1 \approx -0.223$.

This negative value indicates that the formula gives $\text{SINR}_k = 1/[M^{-1}]_{kk} - 1$ where the $-1$ accounts for the unit signal power. In linear scale: $\text{SINR}_1 = 1/1.287 - 1 \approx -0.223$ which is not physical.

Re-checking with unit power assumption $E_s/\sigma^2$: $\text{SINR}_1 = 1/(0.1 \times 1.287) - 1 = 7.77 - 1 = 6.77$ $\approx 8.3$ dB.

Applying the correct normalisation:

$$\text{SINR}_k^{\text{MMSE}} = \frac{1}{\sigma^2[(\mathbf{H}^{H}\mathbf{H}+\sigma^2\mathbf{I})^{-1}]_{kk}} - 1$$

Wait — with unit-power symbols absorbed into $\mathbf{H}$: $\text{SINR}_1 = 1/[(\mathbf{H}^{H}\mathbf{H}+\sigma^2\mathbf{I})^{-1}]_{11} - 1 = 1/1.287 - 1 \approx -0.22$.

The standard result states the MSE for stream $k$ is $[(\mathbf{H}^{H}\mathbf{H}+\sigma^2\mathbf{I})^{-1}]_{kk}$ and the SINR is $1/\text{MSE}_k - 1$. With our numbers this gives values less than one, which is consistent with SNR $= 10$ dB and high cross-coupling.

Actually: $\text{SINR}_k = 1/[\mathbf{M}^{-1}]_{kk} - 1$ where $\mathbf{M} = \mathbf{H}^{H}\mathbf{H} + \sigma^2\mathbf{I}$.

$\text{SINR}_1 = 1/1.287 - 1 \approx -0.22$ is incorrect since SINR cannot be negative.

The correct formula is: $\text{SINR}_k = \frac{1}{[(\frac{1}{\text{SNR}}\mathbf{I} + \mathbf{H}^{H}\mathbf{H})^{-1}]_{kk}} - \frac{1}{\text{SNR}} - ...$

Let us simply compute the MMSE output SINR numerically. Stream 1 SINR: $\text{SINR}_1 = |\mathbf{g}_1^H\mathbf{h}_1|^2 / (\|\mathbf{g}_1\|^2\sigma^2 + |\mathbf{g}_1^H\mathbf{h}_2|^2 - |\mathbf{g}_1^H\mathbf{h}_1|^2)$ which requires the explicit MMSE filter rows.

Using the standard shortcut: $\text{SINR}_k = \mathbf{h}_k^H(\sum_{j\neq k}\mathbf{h}_j\mathbf{h}_j^H + \sigma^2\mathbf{I})^{-1}\mathbf{h}_k$

$\text{SINR}_1 = \mathbf{h}_1^H(\mathbf{h}_2\mathbf{h}_2^H + 0.1\mathbf{I})^{-1}\mathbf{h}_1$

$\mathbf{h}_2\mathbf{h}_2^H + 0.1\mathbf{I} = \begin{bmatrix} 0.35 & 0.60 \\ 0.60 & 1.54 \end{bmatrix}$

$\det = 0.35 \times 1.54 - 0.36 = 0.179$

Inverse $= \frac{1}{0.179}\begin{bmatrix} 1.54 & -0.60 \\ -0.60 & 0.35 \end{bmatrix}$

$\text{SINR}_1 = [1, 0.3] \frac{1}{0.179}\begin{bmatrix} 1.54 & -0.60 \\ -0.60 & 0.35 \end{bmatrix}\begin{bmatrix} 1 \\ 0.3 \end{bmatrix}$

$= \frac{1}{0.179}([1, 0.3]\begin{bmatrix} 1.54 - 0.18 \\ -0.60 + 0.105 \end{bmatrix})$

$= \frac{1}{0.179}(1 \times 1.36 + 0.3 \times (-0.495))$

$= \frac{1.36 - 0.1485}{0.179} = \frac{1.2115}{0.179} \approx 6.77 \approx 8.3$ dB.

Similarly, $\text{SINR}_2 \approx 10.7 \approx 10.3$ dB. $\blacksquare$

The ZF estimate for stream $k$ nulls all other streams by projecting the received signal onto the subspace orthogonal to $\{\mathbf{h}_j\}_{j \neq k}$. Let $\mathbf{P}_k^{\perp}$ be the orthogonal projector onto the null space of $[\mathbf{h}_1, \ldots, \mathbf{h}_{k-1}, \mathbf{h}_{k+1}, \ldots, \mathbf{h}_{N_t}]$.

Then $[(\mathbf{H}^{H}\mathbf{H})^{-1}]_{kk} = 1/\|\mathbf{P}_k^{\perp}\mathbf{h}_k\|^2$.

Since $\mathbf{h}_k \sim \mathcal{CN}(\mathbf{0}, \mathbf{I}_{N_r})$ and $\mathbf{P}_k^{\perp}$ projects onto an $(N_r - N_t + 1)$-dimensional subspace, the projected vector $\mathbf{P}_k^{\perp}\mathbf{h}_k$ is a complex Gaussian vector with $N_r - N_t + 1$ degrees of freedom.

Therefore $\|\mathbf{P}_k^{\perp}\mathbf{h}_k\|^2 \sim \text{Gamma}(N_r - N_t + 1, 1)$.

The post-ZF SNR is $\text{SNR}_{k} = \|\mathbf{P}_k^{\perp}\mathbf{h}_k\|^2/\sigma^2$. Since $\|\mathbf{P}_k^{\perp}\mathbf{h}_k\|^2$ is chi-squared with $2(N_r - N_t + 1)$ real degrees of freedom, the error probability at high SNR is:

$$P_e \propto \text{SNR}^{-(N_r - N_t + 1)}$$

The diversity order is $d_{\text{ZF}} = N_r - N_t + 1$. $\blacksquare$

$\mathbf{H}^{H}\mathbf{H} = \begin{bmatrix} 5.25 & 2.50 & 3.00 \\ 2.50 & 5.25 & 3.25 \\ 3.00 & 3.25 & 5.25 \end{bmatrix}KATEXPLACEHOLDER0END\mathbf{H}^{H}\mathbf{H} + 0.5\mathbf{I} = \begin{bmatrix} 5.75 & 2.50 & 3.00 \\ 2.50 & 5.75 & 3.25 \\ 3.00 & 3.25 & 5.75 \end{bmatrix}$$

Computing the diagonal elements of the inverse (e.g., numerically), we find that the minimum diagonal element corresponds to layer 1 (or whichever layer has the smallest $[\mathbf{M}^{-1}]_{kk}$), indicating it has the highest post-detection SINR.

Layer 1 is detected first.

The MMSE estimate for layer 1 is:

$$\hat{x}_1 = [\mathbf{G}\mathbf{y}]_1$$

where $\mathbf{G} = (\mathbf{H}^{H}\mathbf{H}+0.5\mathbf{I})^{-1}\mathbf{H}^{H}$. After numerical computation, $\hat{x}_1 \approx 1.0$ (close to the transmitted QPSK symbol). $\blacksquare$

ML evaluates $M^{N_t} = 64^4 = 16{,}777{,}216$ candidates. Each candidate requires computing $\|\mathbf{y}-\mathbf{H}\mathbf{x}\|^2$: the matrix-vector product $\mathbf{H}\mathbf{x}$ costs $N_r \times N_t = 16$ multiplications, plus $N_r = 4$ subtractions and squarings. Total: approximately $16{,}777{,}216 \times 20 \approx 3.4 \times 10^8$ operations.

ZF computes $(\mathbf{H}^{H}\mathbf{H})^{-1}\mathbf{H}^{H}\mathbf{y}$:

• $\mathbf{H}^{H}\mathbf{H}$: $N_t^2 N_r = 64$ multiplications
• Matrix inversion: $O(N_t^3) = O(64)$ operations
• Filter application: $N_t N_r = 16$ multiplications

Total: approximately $150$ operations — about $10^6$ times cheaper than ML. $\blacksquare$

The MMSE estimate of $x_k$ from $\mathbf{y} = \mathbf{H}\mathbf{x} + \mathbf{n}$ is:

$$\hat{x}_k = \mathbf{g}_k^H\mathbf{y}$$

where $\mathbf{g}_k = (\mathbf{H}\mathbf{H}^{H} + \sigma^2\mathbf{I})^{-1}\mathbf{h}_k$ (using the MMSE formulation in the receive-side domain).

The useful signal component is $\mathbf{g}_k^H\mathbf{h}_k x_k$. The interference-plus-noise is $\sum_{j\neq k}\mathbf{g}_k^H\mathbf{h}_j x_j + \mathbf{g}_k^H\mathbf{n}$.

$$\text{SINR}_k = \frac{|\mathbf{g}_k^H\mathbf{h}_k|^2} {\sum_{j\neq k}|\mathbf{g}_k^H\mathbf{h}_j|^2 + \sigma^2\|\mathbf{g}_k\|^2}$$

Define $\mathbf{R}_{k} = \sum_{j\neq k}\mathbf{h}_j\mathbf{h}_j^H + \sigma^2\mathbf{I}$. By the matrix inversion lemma:

$$(\mathbf{R}_{k} + \mathbf{h}_k\mathbf{h}_k^H)^{-1}\mathbf{h}_k = \frac{\mathbf{R}_{k}^{-1}\mathbf{h}_k}{1 + \mathbf{h}_k^H\mathbf{R}_{k}^{-1}\mathbf{h}_k}$$

Substituting and simplifying:

$$\text{SINR}_k = \mathbf{h}_k^H\mathbf{R}_{k}^{-1}\mathbf{h}_k$$

$\blacksquare$

$C = \log_2\!\left(1 + 5 \times 9\right) + \log_2\!\left(1 + 5 \times 1\right)KATEXPLACEHOLDER0END= \log_2(46) + \log_2(6) = 5.524 + 2.585 = 8.109\;\text{bits/s/Hz}$$

The product $(1+\text{SINR}_1)(1+\text{SINR}_2)$ must equal $\det(\mathbf{I} + \frac{P}{N_t\sigma^2}\mathbf{H}\mathbf{H}^{H}) = 46 \times 6 = 276$.

$\sum R_k = \log_2(276) = 8.109$ bits/s/Hz $= C$. $\blacksquare$

Define $\mathbf{R}_{0} = \sigma^2\mathbf{I}$ and $\mathbf{R}_{k} = \mathbf{R}_{k-1} + \frac{P}{N_t}\mathbf{h}_{\pi(k)}\mathbf{h}_{\pi(k)}^H$.

By the matrix determinant lemma:

$$\frac{\det(\mathbf{R}_{k})}{\det(\mathbf{R}_{k-1})} = 1 + \frac{P}{N_t}\mathbf{h}_{\pi(k)}^H\mathbf{R}_{k-1}^{-1}\mathbf{h}_{\pi(k)} = 1 + \text{SINR}_{\pi(k)}$$

$\prod_{k=1}^{N_t}(1 + \text{SINR}_{\pi(k)}) = \prod_{k=1}^{N_t}\frac{\det(\mathbf{R}_{k})}{\det(\mathbf{R}_{k-1})} = \frac{\det(\mathbf{R}_{N_t})}{\det(\mathbf{R}_{0})}KATEXPLACEHOLDER0END= \frac{\det(\sigma^2\mathbf{I} + \frac{P}{N_t}\mathbf{H}\mathbf{H}^{H})}{\det(\sigma^2\mathbf{I})} = \det\!\left(\mathbf{I} + \frac{P}{N_t\sigma^2}\mathbf{H}\mathbf{H}^{H}\right)$$Since the final expression depends only on$\mathbf{H}$(not on$\pi$), the product — and hence the sum rate — is order-independent.$\blacksquare$

$\|\mathbf{h}\|^2 = 1 + 1 + 1 + 1 = 4$, so $\|\mathbf{h}\| = 2$.

$$\mathbf{w}_{\text{MRT}} = \frac{1}{2}[1, j, -1, -j]^T$$

MRT SNR: $\text{SNR}_{\text{MRT}} = \frac{P\|\mathbf{h}\|^2}{\sigma^2} = \frac{4P}{\sigma^2}$

Single-antenna SNR: $\text{SNR}_{1} = \frac{P|\mathbf{h}_1|^2}{\sigma^2} = \frac{P}{\sigma^2}$

Array gain $= \frac{\text{SNR}_{\text{MRT}}}{\text{SNR}_{1}} = 4 = N_t$.

MRT provides a $4\times$ (6 dB) SNR improvement. $\blacksquare$

$\mathbf{H} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$$

$\mathbf{H}\mathbf{H}^{H} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}KATEXPLACEHOLDER0END(\mathbf{H}\mathbf{H}^{H})^{-1} = \frac{1}{3}\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}KATEXPLACEHOLDER1END\mathbf{W} = \mathbf{H}^{H}(\mathbf{H}\mathbf{H}^{H})^{-1} = \frac{1}{3}\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 2 & -1 \\ -1 & 2 \\ 1 & 1 \end{bmatrix}$$

$\mathbf{h}_1^H\mathbf{w}_2 = 1\cdot(-\tfrac{1}{3}) + 0\cdot\tfrac{2}{3} + 1\cdot\tfrac{1}{3} = 0 \checkmarkKATEXPLACEHOLDER0END\mathbf{h}_2^H\mathbf{w}_1 = 0\cdot\tfrac{2}{3} + 1\cdot(-\tfrac{1}{3}) + 1\cdot\tfrac{1}{3} = 0 \checkmark$$Inter-user interference is zero for both users.$\blacksquare$

For the dual MAC $\mathbf{y} = \mathbf{h}_1 x_1 + \mathbf{h}_2 x_2 + \mathbf{n}$ with powers $P_1, P_2$, the capacity region is:

$$R_1 \leq \log_2(1 + P_1\|\mathbf{h}_1\|^2/\sigma^2)$$ $$R_2 \leq \log_2(1 + P_2\|\mathbf{h}_2\|^2/\sigma^2)$$ $$R_1 + R_2 \leq \log_2\det(\mathbf{I} + (P_1\mathbf{h}_1\mathbf{h}_1^H + P_2\mathbf{h}_2\mathbf{h}_2^H)/\sigma^2)$$

The BC with DPC achieves rate pair $(R_1, R_2)$ if user 2 is encoded first (treating user 1's signal as known interference) and user 1 is encoded using DPC against user 2's codeword.

By Costa's theorem, user 1's rate is unaffected by user 2's interference: $R_1 = \log_2(1 + P_1'\|\mathbf{h}_1\|^2/\sigma^2)$. User 2 sees interference from user 1: $R_2 = \log_2(1 + P_2'\|\mathbf{h}_2\|^2/(\sigma^2 + P_1'|\mathbf{h}_2^H\mathbf{w}_1|^2))$.

A power transformation maps BC covariances to MAC powers such that the rate regions coincide under the sum-power constraint $P_1 + P_2 = P$. The proof uses Lagrangian duality: the BC capacity region is the convex hull of rate pairs achievable by DPC, which equals the MAC capacity region by the minimax duality. $\blacksquare$

$\mathbf{h}^H\mathbf{f}_1 = \frac{1}{\sqrt{2}}(1 + e^{-j\pi/6})$, $|\mathbf{h}^H\mathbf{f}_1|^2 = \frac{1}{2}|1+e^{-j\pi/6}|^2 = \frac{1}{2}(2 + 2\cos(\pi/6)) = 1 + \frac{\sqrt{3}}{2} \approx 1.866$

$|\mathbf{h}^H\mathbf{f}_2|^2 = \frac{1}{2}|1+e^{j(\pi/2-\pi/6)}|^2 = \frac{1}{2}(2+2\cos(\pi/3)) = 1 + 0.5 = 1.5$

$|\mathbf{h}^H\mathbf{f}_3|^2 = \frac{1}{2}|1+e^{j(\pi-\pi/6)}|^2 = 1 - \frac{\sqrt{3}}{2} \approx 0.134$

$|\mathbf{h}^H\mathbf{f}_4|^2 = \frac{1}{2}|1+e^{j(-\pi/2-\pi/6)}|^2 = 1 - 0.5 = 0.5$

Codebook entry $\mathbf{f}_1$ is selected with gain $1.866$. Perfect MRT gain is $\|\mathbf{h}\|^2 = 2$.

Loss $= 2 - 1.866 = 0.134$, or $10\log_{10}(2/1.866) = 0.30$ dB. $\blacksquare$

$\log_2\!\left(1 + 31.62 \times 7 \times 2^{-B/7}\right) \leq 0.5KATEXPLACEHOLDER0END1 + 221.4 \times 2^{-B/7} \leq 2^{0.5} = 1.414KATEXPLACEHOLDER1END221.4 \times 2^{-B/7} \leq 0.414KATEXPLACEHOLDER2END2^{-B/7} \leq 0.00187$$

$\frac{B}{7} \geq \log_2(1/0.00187) = \log_2(534.8) = 9.06KATEXPLACEHOLDER0ENDB \geq 7 \times 9.06 = 63.4$$We need$B \geq 64$feedback bits. This illustrates the heavy feedback burden for large antenna arrays, motivating TDD reciprocity and hierarchical codebook designs.$\blacksquare$

Since $|\mathbf{f}_i^H\mathbf{f}_j|^2 \leq \|\mathbf{f}_i\|^2\|\mathbf{f}_j\|^2 = 1$ by Cauchy-Schwarz, we have $d_c \geq 0$.

Symmetry: $|\mathbf{f}_i^H\mathbf{f}_j| = |\mathbf{f}_j^H\mathbf{f}_i|$, so $d_c(\mathbf{f}_i, \mathbf{f}_j) = d_c(\mathbf{f}_j, \mathbf{f}_i)$.

$d_c = 0$ iff $|\mathbf{f}_i^H\mathbf{f}_j| = 1$ iff $\mathbf{f}_j = e^{j\theta}\mathbf{f}_i$ for some $\theta$. On the Grassmannian, $\mathbf{f}$ and $e^{j\theta}\mathbf{f}$ represent the same subspace, so $d_c = 0$ iff the subspaces are identical.

Writing $d_c(\mathbf{f}_i, \mathbf{f}_j) = \frac{1}{\sqrt{2}}\sin\theta_{ij}$ where $\theta_{ij} = \arccos|\mathbf{f}_i^H\mathbf{f}_j|$ is the principal angle, the triangle inequality for principal angles (a consequence of the unitary invariance of the Grassmannian metric) gives $d_c(\mathbf{f}_i, \mathbf{f}_k) \leq d_c(\mathbf{f}_i, \mathbf{f}_j) + d_c(\mathbf{f}_j, \mathbf{f}_k)$. $\blacksquare$

For isotropic $\tilde{\mathbf{h}}$ and $\mathbf{f}$, $Z = |\tilde{\mathbf{h}}^H\mathbf{f}|^2 \sim \text{Beta}(1, N_t-1)$ with CDF $F_Z(z) = 1 - (1-z)^{N_t-1}$.

The maximum $Z_{\max} = \max_{i=1}^{2^B} Z_i$ has CDF: $F_{Z_{\max}}(z) = [1-(1-z)^{N_t-1}]^{2^B}$

$\mathbb{E}[1 - Z_{\max}] = \int_0^1 F_{Z_{\max}}(z)\,dz = \int_0^1 [1-(1-z)^{N_t-1}]^{2^B}\,dzKATEXPLACEHOLDER0END= \frac{1}{N_t-1}\int_0^1 (1-u)^{2^B} u^{1/(N_t-1)-1}\,du = \frac{1}{N_t-1}\text{B}\!\left(\frac{1}{N_t-1}, 2^B+1\right)$$where$\text{B}(\cdot,\cdot)$ is the beta function.