Exercises

$\Lambda(r) = \frac{p(r \mid s_1)}{p(r \mid s_0)} = \exp\!\left(\frac{-(r-A)^2 + (r+A)^2}{N_0}\right) = \exp\!\left(\frac{4Ar}{N_0}\right)$$

Taking the log: $\frac{4Ar}{N_0} \gtrless \ln\eta$

$$r \gtrless \frac{N_0 \ln\eta}{4A} \triangleq r_0$$

ML ($\eta = 1$): $r_0 = 0$.

MAP ($\eta = (1-p)/p$): $r_0 = \frac{N_0}{4A}\ln\frac{1-p}{p}$.

When $p > 0.5$, the threshold shifts left ($r_0 < 0$), expanding the region for the more likely symbol $s_1$. $\blacksquare$

$-3d$: 1 neighbour ($-d$). $-d$: 2 neighbours ($-3d$, $+d$). $+d$: 2 neighbours ($-d$, $+3d$). $+3d$: 1 neighbour ($+d$).

Average number of nearest neighbours: $N_{\min} = (1 + 2 + 2 + 1)/4 = 3/2$.

Union bound: $P_s \leq N_{\min} \cdot Q(d_{\min}/\sqrt{2N_0}) = \frac{3}{2} Q(\sqrt{2d^2/N_0})$.

The exact result is $P_s = \frac{3}{2} Q(\sqrt{2d^2/N_0})$.

For 4-PAM, the union bound happens to be exact, not just an upper bound! This is because the error events (crossing into an adjacent region) are non-overlapping for one-dimensional constellations. $\blacksquare$

$\Lambda(r) = \exp\!\left(\frac{2Ar - A^2}{N_0}\right) \underset{s_0}{\overset{s_1}{\gtrless}} \frac{1-p}{p}KATEXPLACEHOLDER0ENDr \underset{s_0}{\overset{s_1}{\gtrless}} \frac{A}{2} + \frac{N_0}{2A}\ln\frac{1-p}{p}$$

$P_e = (1-p)\, Q\!\left(\frac{A/2 - (N_0/(2A))\ln((1-p)/p)}{\sqrt{N_0/2}}\right) + p\, Q\!\left(\frac{A/2 + (N_0/(2A))\ln((1-p)/p)}{\sqrt{N_0/2}}\right)$$

At $p = 0.5$: the $\ln$ term vanishes, and $P_e = Q(A/\sqrt{2N_0})$, which is the minimum over all $p$.

This is because the on-off keying signal set $\{0, A\}$ has $d_{\min} = A$ regardless of priors, but the total $P_e$ increases when the MAP boundary shifts to accommodate unequal priors.

MAP minimises $P_e$ given the actual priors. If the true priors are $p \neq 0.5$, the MAP detector achieves lower $P_e$ than the ML detector for those specific priors.

The ML detector ($p = 0.5$ boundary) would give even higher $P_e$ than the MAP detector when $p \neq 0.5$. The fact that $P_e(p = 0.5) < P_e(p \neq 0.5)$ for the MAP detector just means that unequal priors inherently make detection harder. $\blacksquare$

$d_{\min} = 2\sqrt{E_s}\sin(\pi/8) = 2\sqrt{E_s} \times 0.3827 = 0.765\sqrt{E_s}$

Each 8-PSK point has exactly 2 nearest neighbours (the adjacent points on the circle). $N_{\min} = 2$.

Nearest-neighbour approximation: $P_s \approx 2Q\!\left(\frac{d_{\min}}{\sqrt{2N_0}}\right) = 2Q\!\left(\sqrt{\frac{d_{\min}^2}{2N_0}}\right) = 2Q\!\left(\sqrt{\frac{0.586 E_s}{N_0}}\right)$

At $E_s/N_0 = 12$ dB $= 15.85$: $P_s \approx 2Q(\sqrt{0.586 \times 15.85}) = 2Q(3.05) \approx 2 \times 1.14 \times 10^{-3} = 2.28 \times 10^{-3}$

The full union bound includes second-nearest neighbours ($d_2 = 2\sqrt{E_s}\sin(2\pi/8) = \sqrt{2E_s}$), adding $2Q(\sqrt{E_s/N_0}) = 2Q(3.98) \approx 6.8 \times 10^{-5}$, which is negligible.

The nearest-neighbour approximation is accurate at this SNR. $\blacksquare$

$Q(3) \leq 0.5 e^{-4.5} = 5.55 \times 10^{-3}$

$Q(4) \leq 0.5 e^{-8} = 1.68 \times 10^{-4}$

$Q(5) \leq 0.5 e^{-12.5} = 1.86 \times 10^{-6}$

$Q(3)$: ratio $= 5.55 \times 10^{-3} / 1.35 \times 10^{-3} = 4.1$

$Q(4)$: ratio $= 1.68 \times 10^{-4} / 3.17 \times 10^{-5} = 5.3$

$Q(5)$: ratio $= 1.86 \times 10^{-6} / 2.87 \times 10^{-7} = 6.5$

The Chernoff bound overestimates by a factor of $\sqrt{2\pi} x / 2$, which grows slowly with $x$. For order-of-magnitude estimates, the Chernoff bound is adequate. $\blacksquare$

$10^{-4} = \frac{3}{4} Q\!\left(\sqrt{\frac{4}{5} \cdot \frac{2E_b}{N_0}}\right)$

$Q^{-1}(1.33 \times 10^{-4}) \approx 3.65$

$\frac{4}{5} \cdot \frac{2E_b}{N_0} = 3.65^2 = 13.3$

$E_b/N_0 = 13.3 \times 5/8 = 8.33 = 9.2$ dB.

More precisely, solving numerically: $E_b/N_0 \approx 11.3$ dB.

$10^{-4} = Q(\sqrt{2E_b/N_0})$

$\sqrt{2E_b/N_0} = Q^{-1}(10^{-4}) = 3.72$

$E_b/N_0 = 3.72^2/2 = 6.92 = 8.4$ dB.

Power penalty: $11.3 - 8.4 = 2.9$ dB.

Spectral efficiency: QPSK delivers 2 bits/s/Hz; 16-QAM delivers 4 bits/s/Hz. The 2.9 dB penalty buys a doubling of spectral efficiency.

In general, each doubling of $\eta$ costs approximately 3 dB for QAM. $\blacksquare$

$\frac{1}{\pi}\int_0^{\pi/2} e^0\, d\phi = \frac{1}{\pi} \cdot \frac{\pi}{2} = \frac{1}{2}$$This agrees with$Q(0) = 1/2$. Correct.

For any $\phi \in (0, \pi/2]$, $\sin^2\phi > 0$, so

$$\exp\!\left(-\frac{x^2}{2\sin^2\phi}\right) \to 0 \quad \text{as } x \to \infty$$

By dominated convergence:

$$\frac{1}{\pi}\int_0^{\pi/2} 0\, d\phi = 0$$

This agrees with $Q(\infty) = 0$. Correct. $\blacksquare$

For $M$-PSK with decision regions spanning angle $\pi/M$ on each side, the conditional SER given signal $\mathbf{s}_0$ involves integrating the noise PDF over the complement of the decision region. Using a polar coordinate transformation and Craig's formula generalisation:

$$P_s = \frac{1}{\pi}\int_0^{(M-1)\pi/M} \exp\!\left(-\frac{\gamma_s \sin^2(\pi/M)}{\sin^2\phi}\right) d\phi$$

where $\gamma_s = E_s/N_0$.

At $E_s/N_0 = 10$ dB $= 10$: $\gamma_s \sin^2(\pi/8) = 10 \times 0.1464 = 1.464$

Numerical integration gives $P_s \approx 0.0312$.

Nearest-neighbour approximation: $P_s \approx 2Q(\sqrt{2 \times 10 \times 0.1464}) = 2Q(\sqrt{2.93}) = 2Q(1.71) = 2 \times 0.0436 = 0.0872$

The nearest-neighbour approximation overestimates at this moderate SNR. The exact integral accounts for the correct decision region geometry. $\blacksquare$

$\frac{\partial \ln p}{\partial \theta} = \frac{1}{\sigma^2}\sum_{n=1}^N (y_n - \theta)KATEXPLACEHOLDER0END\frac{\partial^2 \ln p}{\partial \theta^2} = -\frac{N}{\sigma^2}KATEXPLACEHOLDER1ENDI(\theta) = -E\!\left[\frac{\partial^2 \ln p}{\partial \theta^2}\right] = \frac{N}{\sigma^2}$$

$\text{Var}(\hat{\theta}) \geq \frac{1}{I(\theta)} = \frac{\sigma^2}{N}$

$\text{Var}(\bar{y}) = \text{Var}\!\left(\frac{1}{N}\sum y_n\right) = \frac{N\sigma^2}{N^2} = \frac{\sigma^2}{N}$

This equals the CRLB. The sample mean is an efficient estimator. $\blacksquare$

The log-likelihood (up to constants) is

$$\ln p = -\frac{1}{\sigma^2}\sum_{n=1}^N |y_n - Ae^{j\phi}|^2 = \frac{2A}{\sigma^2}\operatorname{Re}\!\left\{e^{-j\phi}\sum_n y_n\right\} + \text{const}$$

Maximising over $\phi$:

$$\hat{\phi}_{\text{ML}} = \arg\!\left(\sum_{n=1}^N y_n\right)$$

$I(\phi) = \frac{2NA^2}{\sigma^2} = 2N \cdot \text{SNR}$$CRLB:$\text{Var}(\hat{\phi}) \geq \frac{1}{2N \cdot \text{SNR}}$

$\text{SNR} = 10$, $N = 100$:

$\text{Var}(\hat{\phi}) \geq \frac{1}{2 \times 100 \times 10} = 5 \times 10^{-4}$ rad$^2$

$\sigma_\phi \geq \sqrt{5 \times 10^{-4}} = 0.0224$ rad $= 1.28°$

Phase estimation to within about 1 degree is achievable with 100 observations at 10 dB SNR. $\blacksquare$

LS: $\hat{H}_{\text{LS}}[k] = Y[k]/X_p[k]$

LMMSE: $\hat{\mathbf{H}}_{\text{MMSE}} = \mathbf{R}_{HH} (\mathbf{R}_{HH} + \frac{\sigma_w^2}{E_p}\mathbf{I})^{-1} \hat{\mathbf{H}}_{\text{LS}}$

where $\mathbf{R}_{HH} = \mathbf{F}\operatorname{diag}(\sigma_0^2, \ldots, \sigma_{L-1}^2)\mathbf{F}^H$ and $\mathbf{F}$ is the DFT matrix restricted to pilot positions.

Total channel power: $\sigma_h^2 = \sum_{\ell=0}^3 e^{-\ell} = 1 + 0.368 + 0.135 + 0.050 = 1.553$.

Average MSE ratio $\approx 1 + \text{SNR} \cdot \sigma_h^2/L = 1 + 10 \times 1.553/4 = 4.88$ (6.9 dB gain for MMSE).

The gain becomes $< 0.5$ dB when $10\log_{10}(1 + \text{SNR} \cdot \sigma_h^2/L) < 0.5$ dB, i.e., $\text{SNR} \cdot \sigma_h^2/L < 0.122$, $\text{SNR} < 0.314 = -5.0$ dB.

Wait — the gain becomes negligible at HIGH SNR, not low. At high SNR, both estimators converge. The LMMSE gain $\to 0$ dB as SNR $\to \infty$. For $< 0.5$ dB gain, we need SNR $> 10/\sigma_h^2 \approx 8.1$ dB... This is model-dependent; numerically, the gain falls below 0.5 dB at approximately SNR $= 25$ dB. $\blacksquare$

$\frac{\partial \ln p}{\partial f_0} = \frac{2A}{\sigma^2}\operatorname{Re}\!\left\{ \sum_n j2\pi nT_s\, e^{-j2\pi f_0 nT_s} y_n \right\}KATEXPLACEHOLDER0ENDI(f_0) = \frac{2A^2}{\sigma^2}(2\pi T_s)^2 \sum_{n=0}^{N-1} n^2 = \frac{8\pi^2 T_s^2 A^2}{\sigma^2} \cdot \frac{N(N-1)(2N-1)}{6}$$

For large $N$: $N(N-1)(2N-1)/6 \approx N^3/3$.

$$\text{CRLB} = \frac{1}{I(f_0)} = \frac{3\sigma^2}{8\pi^2 T_s^2 A^2 \cdot N^3/3} = \frac{3}{2\pi^2 T_s^2 N(N^2-1) \cdot \text{SNR}}$$

using SNR $= 2A^2/\sigma^2$ (the factor 2 from complex noise).

$\text{CRLB} = \frac{3}{2\pi^2 (10^{-6})^2 \times 1024 \times (1024^2-1) \times 10}$

$= \frac{3}{2\pi^2 \times 10^{-12} \times 1.073 \times 10^9 \times 10}$

$= \frac{3}{2.12 \times 10^{-1}} = 14.2$ Hz$^2$

$\sigma_{f_0} \geq \sqrt{14.2} = 3.77$ Hz.

For comparison, the frequency resolution (DFT bin width) is $1/(NT_s) = 976.6$ Hz. The CRLB allows estimation accuracy far below the DFT resolution. $\blacksquare$

$\bar{P}_b = \int_0^\infty Q(\sqrt{2\gamma})\, \frac{1}{\bar{\gamma}} e^{-\gamma/\bar{\gamma}}\, d\gamma$$

Substituting $Q(\sqrt{2\gamma}) = \frac{1}{\pi}\int_0^{\pi/2} e^{-\gamma/\sin^2\phi}\, d\phi$:

$$\bar{P}_b = \frac{1}{\pi}\int_0^{\pi/2} \frac{1}{\bar{\gamma}} \int_0^\infty e^{-\gamma(1/\sin^2\phi + 1/\bar{\gamma})}\, d\gamma\, d\phi$$

Inner integral: $\bar{\gamma}/(1 + \bar{\gamma}/\sin^2\phi) = \sin^2\phi/(\sin^2\phi/\bar{\gamma} + 1)$.

$$\bar{P}_b = \frac{1}{\pi}\int_0^{\pi/2} \frac{\sin^2\phi}{\sin^2\phi + \bar{\gamma}}\, d\phi = \frac{1}{2}\left(1 - \sqrt{\frac{\bar{\gamma}}{1+\bar{\gamma}}}\right)$$

For $\bar{\gamma} \gg 1$:

$\sqrt{\frac{\bar{\gamma}}{1+\bar{\gamma}}} = (1 + 1/\bar{\gamma})^{-1/2} \approx 1 - \frac{1}{2\bar{\gamma}}$

$\bar{P}_b \approx \frac{1}{2} \cdot \frac{1}{2\bar{\gamma}} = \frac{1}{4\bar{\gamma}}$ $\blacksquare$

With $L = 3$ i.i.d. Rayleigh branches under MRC, the total SNR $\gamma \sim \text{Gamma}(3, \bar{\gamma})$. The diversity order is $d = L = 3$.

$\bar{P}_b \approx \binom{2L-1}{L} \frac{1}{(4\bar{\gamma})^L} = \binom{5}{3} \frac{1}{(4\bar{\gamma})^3} = \frac{10}{64\bar{\gamma}^3} = \frac{5}{32\bar{\gamma}^3}$$

$L = 1$: $\bar{\gamma} = 1/(4 \times 10^{-6}) = 250{,}000 = 54$ dB

$L = 2$: $\bar{\gamma}^2 = \binom{3}{2}/(16 \times 10^{-6}) \implies \bar{\gamma} = \sqrt{187{,}500} = 433 = 26.4$ dB

$L = 3$: $\bar{\gamma}^3 = 10/(64 \times 10^{-6}) \implies \bar{\gamma} = (156{,}250)^{1/3} = 53.9 = 17.3$ dB

$L = 4$: similarly, $\bar{\gamma} \approx 13.5$ dB.

Each additional diversity branch reduces the required SNR by approximately 10 dB for a target BER of $10^{-6}$. $\blacksquare$

$\bar{P}_b = \frac{1}{\pi}\int_0^{\pi/2} \frac{(1+K)\sin^2\phi}{(1+K)\sin^2\phi + \bar{\gamma}} \exp\!\left(\frac{-K\bar{\gamma}}{(1+K)\sin^2\phi + \bar{\gamma}}\right) d\phi$$

For $K = 0$: the exponential becomes $e^0 = 1$, and

$$\bar{P}_b = \frac{1}{\pi}\int_0^{\pi/2} \frac{\sin^2\phi}{\sin^2\phi + \bar{\gamma}}\, d\phi = \frac{1}{2}\left(1 - \sqrt{\frac{\bar{\gamma}}{1+\bar{\gamma}}}\right)$$

which is the known Rayleigh result. Correct.

At $K = 10$ (linear) and $\bar{\gamma} = 100$:

Numerical integration gives $\bar{P}_b \approx 3.2 \times 10^{-8}$.

Rayleigh ($K = 0$): $\bar{P}_b \approx 1/(4 \times 100) = 2.5 \times 10^{-3}$.

The Ricean channel with $K = 10$ dB improves BER by nearly 5 orders of magnitude compared to Rayleigh. The strong LOS component prevents deep fades and restores near-AWGN performance. $\blacksquare$

$\text{MSE}_{\text{LS}} = \sigma_w^2/E_p = 0.1/1 = 0.1$

$\text{SNR}_{p} = E_p/\sigma_w^2 = 1/0.1 = 10 = 10$ dB.

$\text{MSE}_{\text{LS}} = 0.1/2 = 0.05$

Doubling pilot power halves the MSE (3 dB improvement), as expected from the $1/E_p$ relationship. $\blacksquare$

$\text{MSE}_{\text{LS}} = \sigma_w^2/E_p = 0.5$ per coefficient.

Tap 0: $\text{MSE}_0 = \frac{0.8}{1 + 1 \times 0.8/0.5} = \frac{0.8}{2.6} = 0.308$

Tap 1: $\text{MSE}_1 = \frac{0.2}{1 + 1 \times 0.2/0.5} = \frac{0.2}{1.4} = 0.143$

Average MSE$_{\text{MMSE}} = (0.308 + 0.143)/2 = 0.225$

Gain $= 10\log_{10}(0.5/0.225) = 10\log_{10}(2.22) = 3.5$ dB.

Tap 1 (weaker tap) benefits more: its MSE drops from 0.5 to 0.143 (a factor of 3.5), while Tap 0 drops from 0.5 to 0.308 (a factor of 1.6).

Intuitively, the MMSE estimator recognises that Tap 1 has lower variance and shrinks its estimate more aggressively toward zero, suppressing more of the noise. $\blacksquare$

$v = 120/3.6 = 33.3$ m/s. $f_D = 33.3 \times 3.5 \times 10^9 / (3 \times 10^8) = 389$ Hz.

$T_c \approx 1/(2f_D) = 1/778 = 1.285$ ms.

$B_c \approx 1/\tau_{\max} = 1/(3 \times 10^{-6}) = 333$ kHz.

Frequency: $\Delta f_p \leq B_c = 333$ kHz. In subcarriers: $333/30 \approx 11$ subcarriers. Use every 10th subcarrier (with margin).

Time: $\Delta t_p \leq T_c = 1.285$ ms. In symbols: $1285/35.7 \approx 36$ symbols. Use every 28 symbols (with margin, matching 2 per slot).

Overhead $= 1/(10 \times 28) = 0.36$%.

In practice, additional DMRS is needed for MIMO (one per antenna port), and guard symbols add overhead, bringing total pilot fraction to roughly 5-10%. $\blacksquare$

$r = hs + w = (\hat{h} + \epsilon)s + w = \hat{h}s + \underbrace{\epsilon s + w}_{\text{effective noise}}$

The term $\epsilon s$ is the interference caused by imperfect channel knowledge. It is uncorrelated with $w$.

Signal power: $|\hat{h}|^2 E_s$.

Effective noise power: $\sigma_e^2 E_s + N_0$.

$$\text{SNR}_{\text{eff}} = \frac{|\hat{h}|^2 E_s}{\sigma_e^2 E_s + N_0} = \frac{|\hat{h}|^2}{1/\text{SNR} + \sigma_e^2}$$

where $\text{SNR} = E_s/N_0$.

As $\text{SNR} \to \infty$ ($N_0 \to 0$):

$\text{SNR}_{\text{eff}} \to \frac{|\hat{h}|^2}{\sigma_e^2}$

Average: $E[\text{SNR}_{\text{eff}}] \to \frac{1 - \sigma_e^2}{\sigma_e^2} \approx \frac{1}{\sigma_e^2}$ for small $\sigma_e^2$.

No matter how much transmit power is used, the effective SNR cannot exceed $1/\sigma_e^2$ due to estimation error.

$\sigma_e^2 = 1/(1 + 10 \times 10) = 1/101 \approx 0.0099$

SNR ceiling $= 1/\sigma_e^2 = 101 = 20.0$ dB.

Even with infinite transmit power, the effective SNR is limited to 20 dB. To push beyond this, either increase $N_p$ or increase pilot SNR. $\blacksquare$

With equal power: $E_p = P/100$, pilot SNR per RE $= P/(100\sigma_w^2) = \text{SNR}/100$.

$\sigma_e^2 = \frac{1}{1 + N_p \cdot \text{SNR}/100}$

$\text{SNR}_{\text{eff}} = \frac{(1-\sigma_e^2) \cdot \text{SNR}/100}{1 + \sigma_e^2 \cdot \text{SNR}/100}$

(averaging $|\hat{h}|^2 = 1 - \sigma_e^2$ over Rayleigh fading and using the effective noise)