Exercises

$d^*(k) = (n_t - k)(n_r - k)$ for integer $k \in \{0, 1, \ldots, \min(n_t, n_r)\}$; linearly interpolated between.

$d^*(0) = 9$, $d^*(1) = 4$, $d^*(2) = 1$, $d^*(3) = 0$.

$d_{\max} = 9 = n_t n_r$; $r_{\max} = 3 = \min(n_t, n_r)$.

$\log(5 \text{SNR}^{3}) / \log \text{SNR} = \log 5 / \log \text{SNR} + 3 \to 3$ as $\text{SNR} \to \infty$. So $5 \text{SNR}^{3} \doteq \text{SNR}^{3}$. Similarly $\log(\text{SNR}^{2} (\log \text{SNR})^{10}) / \log \text{SNR} = 2 + 10 \log\log\text{SNR}/\log\text{SNR} \to 2$.

$\text{SNR}^{2} + \text{SNR}^{1.5} = \text{SNR}^{2} (1 + \text{SNR}^{-0.5}) \doteq \text{SNR}^{2}$.

$\log e^{-\text{SNR}} / \log \text{SNR} = -\text{SNR} / \log \text{SNR} \to -\infty$, so the exponent is $-\infty$, i.e., faster than any polynomial decay.

$(n_t - k)(n_r - k) = (n_r - k)(n_t - k)$ by commutativity of multiplication. Also $\min(n_t, n_r) = \min(n_r, n_t)$, so the corner-point set $\{k : 0 \le k \le \min(n_t, n_r)\}$ is identical for $(n_t, n_r)$ and $(n_r, n_t)$.

A piecewise-linear function is uniquely determined by its corner values; since both configurations have the same corner values at the same $k$, the interpolated curves coincide. $\blacksquare$

$d^*(0) = 4$, $d^*(1) = 1$, $d^*(2) = 0$.

$r = 0.5$ lies in the segment $[0, 1]$. Interpolate: $d^*(0.5) = 4 + (1 - 4) \cdot 0.5 = 4 - 1.5 = 2.5$.

Any code at $r = 0.5$ has $d^* \le 2.5$. A DMT-optimal code (e.g., time-sharing Alamouti and rank-1 V-BLAST, or the Golden code) would achieve this bound. A sub-optimal code (e.g., Alamouti with QAM growth) gives $d = 2$ ($= 4 \cdot 0.5$) — 0.5 units below the DMT.

$d_{\rm ML}(1.5) = 3 (1 - 1.5/3) = 3 \cdot 0.5 = 1.5$.

$r = 1.5$ is in segment $[1, 2]$. Interpolate between $(1, 4)$ and $(2, 1)$: $d^*(1.5) = 4 + (1 - 4) \cdot 0.5 = 2.5$.

V-BLAST-ML achieves $d_{\rm ML} = 1.5$; Zheng-Tse bound is $d^*(1.5) = 2.5$. Gap = $1$ unit of diversity exponent — which at high SNR is $\sim 3$ dB per decade of BER improvement. A DMT-optimal code (CDA / Golden-like on $3 \times 3$) closes this gap.

$R(\text{SNR}) = \log_2 M$ constant, so $r = 0$. Alamouti achieves full diversity $d = 2 n_r$ (Chapter 11).

Alamouti reduces to a single scalar channel with effective SNR $X \cdot \text{SNR} / 2$ where $X = \sum_{i,j} |h_{ij}|^2 \sim \chi^2_{2 \cdot 2 n_r} / 2$ (chi-square with $4 n_r$ real dof, i.e. $2 n_r$ complex dof). Outage event: $\log_2(1 + X \text{SNR} / 2) < r \log_2 \text{SNR}$, i.e. $X < \text{SNR}^{r - 1}$.

$P(X < \text{SNR}^{r-1}) \doteq \text{SNR}^{(r-1) \cdot 2 n_r}$ (chi-square CDF exponent at origin). So $d_{\rm Alamouti}(r) = 2 n_r (1 - r)^+$ for $r \in [0, 1]$, zero beyond.

$d_{\rm Alamouti}(r) = 2 n_r (1 - r)^+$ is a straight line (chord) from $(0, 2 n_r)$ to $(1, 0)$ — strictly below the DMT curve for $r \in (0, 1)$. $\blacksquare$

After ZF on $\mathbf{H}$, the $k$-th stream sees SNR $\text{SNR}_{k} = \text{SNR} / [(\mathbf{H}^{H}\mathbf{H})^{-1}]_{kk}$. The diagonal entry $[(\mathbf{H}^{H}\mathbf{H})^{-1}]_{kk}$ is inversely chi- square distributed with $2(n_r - n_t + 1)$ dof — equivalently, $1/[\cdot]_{kk}$ is chi-square with $2(n_r - n_t + 1)$ dof.

With effective diversity order $n_r - n_t + 1$ per stream and per-stream rate $(r/n_t) \log_2 \text{SNR}$, the per-stream outage exponent is (by Example $n_t = n_r = 1$)" data-ref-type="example">EDMT of the Scalar Rayleigh Channel ($n_t = n_r = 1$)) $(n_r - n_t + 1)(1 - r/n_t)^+$.

The post-ZF effective channels on different streams are statistically correlated but share the same outage exponent. The block error event is the union $\bigcup_{k=1}^{n_t} \{\text{stream } k \text{ in outage}\}$; by union bound, $$P_e \le n_t \cdot P(\text{stream 1 in outage}) \doteq \text{SNR}^{-(n_r - n_t + 1)(1 - r/n_t)^+}$$ (the factor $n_t$ is invisible to $\doteq$). So $d_{\rm ZF}(r) = (n_r - n_t + 1)(1 - r/n_t)^+$. $\blacksquare$

Both: $\min = 2$. Corners $(0, 6), (1, 2), (2, 0)$. $r = 1.5$ in segment $[1, 2]$: $d^*(1.5) = 2 + (0 - 2) \cdot 0.5 = 1$.

$\min = 3$. Corners $(0, 12), (1, 6), (2, 2), (3, 0)$. $r = 1.5$ in segment $[1, 2]$: $d^*(1.5) = 6 + (2 - 6) \cdot 0.5 = 4$.

More antennas ($3 \times 4$ vs $2 \times 3$) increase every corner value multiplicatively in the product $n_t n_r$. At $r = 1.5$, going from $(2, 3)$ to $(3, 4)$ raises $d^*$ from $1$ to $4$ — four times the reliability exponent at the same multiplexing level. Each additional antenna pair contributes significant diversity buffer.

$L = 2$, $n_t + n_r - 1 = 5 > L$, so DMT is truncated at $r = L = 2$. $r_{\max}^{\rm eff} = 2$.

For $r \in [0, 2]$: same as full Zheng-Tse. Corners $(0, 9)$, $(1, 4)$, $(2, 1)$ — note $d^*(2) = 1 \ne 0$ here (the truncation stops the curve, it doesn't force $d^* = 0$ at the cutoff!).

For $r > 2$: $d^*(r) = 0$ (no coding scheme can support multiplexing above block length).

Piecewise-linear from $(0, 9)$ to $(1, 4)$ to $(2, 1)$; then a jump to $d^*(2^+) = 0$ (discontinuity at $r = L$); then flat $d^* = 0$ for $r > 2$. The curve is NOT continuous at $r = L$ in general — it drops from $1$ to $0$.

Thm. TDMT Invariance under Full-Rank Tx Correlation: $d^*(1)$ is the same as i.i.d. Rayleigh on $2 \times 2$: $d^*(1) = (2-1)(2-1) = 1$.

$\det \mathbf{R}_t = 1 - 0.25 = 0.75$. Factor $(\det \mathbf{R}_t)^{-1/n_t} = 0.75^{-1/2} \approx 1.155$.

In dB, the loss is $10 \log_{10} 1.155 \approx 0.63$ dB — a modest coding-gain penalty that a well-designed STC will absorb. The DMT slope of $-1$ at high SNR is unchanged, so reliability scaling is not affected.

$k = 0$: slope $= -(8 + 8 - 1) = -15$. First unit of $r$ costs $15$ units of $d^*$.

$k = 7$: slope $= -(8 + 8 - 15) = -1$. Last unit of $r$ costs $1$ unit of $d^*$.

Solve $-(8 + 8 - 2k - 1) = -8 \Rightarrow 2k + 1 = 7 \Rightarrow k = 3$. On segment $[3, 4]$ the DMT slope is $-(16 - 7) = -9$... wait, recompute: $-(n_t + n_r - 2k - 1) = -(16 - 2 \cdot 3 - 1) = -9$. Adjust: solve $-(15 - 2k) = -8 \Rightarrow k = 3.5$ — not integer. The slope $-8$ is never exactly achieved; segments change in increments of $2$, from $-15, -13, -11, -9, -7, -5, -3, -1$. The slope $-n_r = -8$ falls between the $k=3$ ($-9$) and $k=4$ ($-7$) segments — "crossing over" receive diversity in the middle of the curve.

With $n_r - n_t = 0$, coefficients are $(2i - 1)$ = $1, 3$. Objective: $\alpha_1 + 3 \alpha_2$.

Constraint: $(1 - \alpha_1)^+ + (1 - \alpha_2)^+ < 0.5$, ordering $\alpha_1 \le \alpha_2$, non-negativity $\alpha_1, \alpha_2 \ge 0$.

Boundary: $(1 - \alpha_1)^+ + (1 - \alpha_2)^+ = 0.5$. With $\alpha_1 = 0$, we need $(1 - \alpha_2)^+ = -0.5$ — infeasible with $\alpha_1 = 0$ since $(1 - 0)^+ = 1$. So actually we need $\alpha_1 > 0$ such that $1 - \alpha_1 + (1 - \alpha_2)^+ = 0.5$. Taking $\alpha_2 = 1$ (second eigenvalue at exact fade threshold) gives $1 - \alpha_1 = 0.5$, i.e. $\alpha_1 = 0.5$. Objective: $0.5 + 3 \cdot 1 = 3.5$...

Reconciliation: the Zheng-Tse formula $(n_t - k)(n_r - k) = (2-k)(2-k)$ at $k = 0$ gives $4$; at $k = 1$ gives $1$. Interpolated at $r = 0.5$: $d^*(0.5) = 4 + (1 - 4) \cdot 0.5 = 2.5$. The LP should give the same value. The issue above was the ordering convention; the correct LP (with $\alpha_1 \le \alpha_2$ and coefficients $1, 3$) at $r = 0.5$ gives $\alpha_1^\star = 0.5, \alpha_2^\star = 1$ and objective $0.5 + 3 = 3.5$ — which is NOT $2.5$. The discrepancy is a reordering issue: in Zheng-Tse's paper, the coefficients are $(2i - 1 + n_r - n_t)$ under the convention that $\alpha_1$ is the largest (worst-fading) eigenvalue, not the smallest. Reversing: coefficients $3, 1$ (not $1, 3$), giving $3 \cdot 0.5 + 1 \cdot 1 = 2.5$. $\checkmark$

The LP setup requires careful attention to eigenvalue ordering. Zheng-Tse 2003 uses $\alpha_1 \ge \alpha_2 \ge \cdots$ (first $\alpha$ = largest = worst fade). With this convention the coefficient of $\alpha_i$ is $(2i - 1 + n_r - n_t)$, and the LP optimum matches the Zheng-Tse corner formula. $\blacksquare$

$r = \lim_{\text{SNR}\to\infty} 4 / \log_2 \text{SNR} = 0$. Fixed rate corresponds to $r = 0$.

$d^*(0) = n_t n_r = 4$. At fixed rate $4$ bits/use, the Alamouti scheme already achieves this (full-diversity, fixed rate, DMT-optimal at $r = 0$).

$d^*(0) = n_t n_r$. Maximum diversity; fixed rate. Alamouti sits here on $2 \times n_r$ channels.

$d^*(\min) = 0$. Maximum multiplexing; zero diversity margin. V-BLAST-ML and V-BLAST-ZF both reach this corner (but only this corner, not the intermediate points).

The two corners are the "extremes" of the DMT curve; Alamouti and V-BLAST are canonical "extremists". DMT-optimal codes (CDA / Golden) are "centrists" that achieve every interior point as well.

On segment $[k, k+1]$, slope $= (n_t - k - 1)(n_r - k - 1) - (n_t - k)(n_r - k)$. Expanding: $n_t n_r - k(n_t + n_r) + k^2 - (n_t + n_r - 2k - 1) - [n_t n_r - k(n_t + n_r) + k^2] = -(n_t + n_r - 2k - 1)$.

Let $s_k = -(n_t + n_r - 2k - 1)$. Then $s_{k+1} - s_k = 2 > 0$: slopes increase with $k$ (become less negative).

A piecewise-linear function with monotone non-decreasing slopes is concave. Therefore $d^*(r)$ is concave on $r \in [0, \min(n_t, n_r)]$. $\blacksquare$

Geometric intuition. Concavity of $d^*(r)$ means that time- sharing between two operating points $(r_1, d_1)$ and $(r_2, d_2)$ on the DMT curve yields an achievable point strictly above the chord between them — which is a point below the DMT curve (the chord IS the concave hull). So pure time- sharing is DMT-sub-optimal; achieving the full curve requires codes that naturally sit on the piecewise-linear corners.

Corners: $(0, 2), (1, 0)$. $d^*(1) = 0$.

By symmetry same as (a): Corners $(0, 2), (1, 0)$. $d^*(1) = 0$.

Corners: $(0, 4), (1, 1), (2, 0)$. $d^*(1) = 1$.

At $r = 1$, both the SIMO and the MISO achieve $d^* = 0$ — no reliability at maximum multiplexing. The $2 \times 2$ gives $d^* = 1$, a strictly positive exponent.

Message. Adding antennas on both sides multiplies the $(r, d)$ operating point in a non-linear way — you cannot substitute a single Tx-antenna addition for a single Rx-antenna addition. Each side independently contributes to both the multiplexing cap $r_{\max} = \min(n_t, n_r)$ and the corner diversity $(n_t - k)(n_r - k)$.

$r = \lim 2 \log_2 \text{SNR} / \log_2 \text{SNR} = 2$. Golden code at this rate has $r = 2 = r_{\max}$.

$d^*(2) = (2-2)(2-2) = 0$. So Golden at $r = 2$ has $d = 0$.

Golden code at the right endpoint of the DMT curve has $d = 0$ — no reliability margin. This is OK; at $r = r_{\max}$ no code can have positive diversity (the channel is outage-limited at the ergodic-capacity slope). The Golden code's strength is achieving $d^*(r)$ for every $r \in [0, 2]$, not a spectacular value at $r = 2$ specifically.

The DMT is asymptotic; at finite SNR the two channels have different coding gains. On $4 \times 2$ with rank-$2$ codes, the transmitter has $2$ "extra" antennas that can be used for precoding but not for additional rank — a coding-gain benefit (up to $3$ dB from optimal precoding) not captured by DMT.

On $2 \times 4$ (more receive) V-BLAST-ZF is feasible ($n_r = 4 \ge n_t = 2$) with moderate complexity per stream. On $4 \times 2$ V-BLAST-ZF is infeasible ($n_r = 2 < n_t = 4$); the receiver cannot linearly separate $4$ streams with only $2$ receive antennas. ML detection is still possible but complexity-exponential.

On $4 \times 2$ (downlink, base $\to$ UE), the precoder has $4$ antennas to configure based on $2$ feedback streams — the CSI feedback is concise. On $2 \times 4$ (uplink), the receiver can learn $4 \cdot 2 = 8$ channel coefficients without UE-side feedback at all. This complexity asymmetry matters in scheduling and power allocation, but not in DMT.

The DMT is a performance-limit statement. When it gives the same number for two different channels, that means the best-case asymptotic performance is the same — not that the implementations, costs, and finite-SNR gaps are the same.

$d(0) = \min(n_t n_r, (n_r - n_t + 1) n_t) = \min(16, 4) = 4$ on $4 \times 4$. Since $d^*(0) = 16$, this code has $d(0) = 4 \ne 16$ — it is not DMT-optimal at $r = 0$. Sub-optimal by a factor of $4$ in the SNR exponent.

The code's trace is a linear interpolation between the two corners $(0, 4)$ and either $(1, 0)$ (if $n_t n_r$ dominates) or $(n_t, 0)$ (if $(n_r - n_t + 1)$ dominates). Let's take the latter: chord from $(0, 4)$ to $(4, 0)$, slope $-1$.

$d^*$: $(0, 16), (1, 9), (2, 4), (3, 1), (4, 0)$, slope $-7, -5, -3, -1$. The code's trace is below $d^*$ for all $r \in [0, 4)$: even at $r = 2$, the code gives $d = 2$ but $d^* = 4$. Nowhere DMT-optimal except at $r = r_{\max} = 4$.

This operating trace matches V-BLAST-ML's: $d_{\rm ML}(r) = n_r(1 - r/n_t)^+ = 4(1 - r/4)^+$ — a chord from $(0, 4)$ to $(4, 0)$. V-BLAST-ML is DMT-optimal at only the right endpoint $r = r_{\max}$ and is elsewhere strictly sub-optimal.

At $r = k$: $\alpha_i^\star = 0$ for $i \le k$, $\alpha_i^\star = 1$ for $i > k$. Value: $\sum_{i > k} (2i - 1 + n_r - n_t) \cdot 1 = (n_t - k)(n_r - k)$ (after the corner-point algebra).

At $r = k + \epsilon$: the LP budget grows, so we can relax $\alpha_{k+1}$ from $1$ to $1 - \epsilon$ (just-satisfy the constraint $\sum (1 - \alpha_i)^+ = r$). $\alpha_i^\star = 0$ for $i \le k$, $\alpha_{k+1}^\star = 1 - \epsilon$, and $\alpha_i^\star = 1$ for $i > k + 1$.

$\alpha_{k+1}^\star$ is a continuous function of $r$ on $[k, k+1]$ (linear from $1$ at $r = k$ to $0$ at $r = k + 1$). The LP optimum value $(n_t - k - 1)(n_r - k - 1) + (2(k+1) - 1 + n_r - n_t)(1 - \epsilon) = (n_t - k)(n_r - k) - \epsilon (n_t + n_r - 2k - 1) + O(\epsilon^2)$ — continuous in $r$, but the slope changes at integer $r$ (the active coefficient switches from index $k + 1$ to $k + 2$).

The piecewise-linear structure of $d^*(r)$ arises from the active-set changes in the LP at integer $r$, not from discontinuities in the optimiser. The curve $d^*(r)$ is continuous (even at corners), but its derivative is discontinuous — each corner is a slope change, no value jump. $\blacksquare$

$d^*(0) = n_t n_r$, $d^*(\min(n_t, n_r)) = 0$.

Non-increasing in $r$: more multiplexing never raises diversity.

Piecewise-linear with non-decreasing slopes $\Rightarrow$ concave (see Ex. Eex-ch12-15).

$d^*(r)$ is identical under $(n_t, n_r) \leftrightarrow (n_r, n_t)$ (Thm. $(n_t, n_r)$" data-ref-type="theorem">TDMT Symmetry in $(n_t, n_r)$).