Exercises

$T = 1\cdot 0.8 + 2 \cdot 2.5 + (-1)(-0.7) + 1 \cdot 0.9 = 0.8 + 5.0 + 0.7 + 0.9 = 7.4$.

$E_s = \|\mathbf{s}\|^2 = 1+4+1+1 = 7$. With $N_0= 2\sigma^2 = 1$, $d = \sqrt{2 \cdot 7/1} = \sqrt{14} \approx 3.74$.

$\ell = 7.4/0.5 - 7/1 = 14.8 - 7 = 7.8$. A positive LLR favours $\mathcal{H}_1$.

Output $= \mathbf{h}^{\mathsf T}\mathbf{s} + \mathbf{h}^{\mathsf T}\mathbf{W}$. Signal component: $(\mathbf{h}^{\mathsf T}\mathbf{s})^2$. Noise variance: $\sigma^2\|\mathbf{h}\|^2 = \sigma^2$.

$(\mathbf{h}^{\mathsf T}\mathbf{s})^2 \leq \|\mathbf{h}\|^2 \|\mathbf{s}\|^2 = \|\mathbf{s}\|^2$, with equality iff $\mathbf{h} \propto \mathbf{s}$.

The maximum output SNR is $\|\mathbf{s}\|^2/\sigma^2 = 2E_s/N_0$, achieved by $\mathbf{h}^\star = \mathbf{s}/\|\mathbf{s}\|$.

$\mathbf{h}^{\mathsf T}\mathbf{s} = \rho \|\mathbf{s}\|$.

Output SNR = $\rho^2 \|\mathbf{s}\|^2/\sigma^2$; the matched filter gives $\|\mathbf{s}\|^2/\sigma^2$. Loss factor $\rho^2$, or $10\log_{10}(1/\rho^2)$ dB.

Template mismatch of correlation $\rho = 0.9$ costs $\approx 0.9$ dB; $\rho = 0.5$ costs $6$ dB. This quantifies the penalty of imperfect synchronisation/channel knowledge in practical receivers.

Under equal priors and known $E_s$, the ML rule compares the sign of the observation. The scalar statistic has mean $\pm\sqrt{E_s}$ and variance $\sigma^2 = N_0/2$.

$P_e = Q\!\big((2\sqrt{E_s})/(2\sqrt{\sigma^2})\big) = Q(\sqrt{2E_s/N_0})$.

At 10 dB: $P_e \approx 3.9\times 10^{-6}$; at 6 dB: $\approx 2.4\times 10^{-3}$.

$\|\mathbf{s}_1-\mathbf{s}_0\|^2 = 2E_s$, half that of antipodal signalling ($4E_s$).

$P_e^{\mathrm{orth}} = Q\!\big(\sqrt{E_s/N_0}\big)$.

BPSK gains $3$ dB: $P_e^{\mathrm{BPSK}} = Q(\sqrt{2E_s/N_0}) < P_e^{\mathrm{orth}}$ for all positive $E_s/N_0$.

$\hat{A} = (\mathbf{s}^{\mathsf T}\mathbf{y})/\|\mathbf{s}\|^2$ maximises the Gaussian likelihood under $\mathcal{H}_1$.

After substitution and cancellation, the log-GLRT becomes $\log L_G = \frac{(\mathbf{s}^{\mathsf T}\mathbf{y})^2}{2\sigma^2\|\mathbf{s}\|^2}$.

$(\mathbf{s}^{\mathsf T}\mathbf{y})^2 \gtrless \tau$: the test is a threshold on the squared matched-filter output, producing an F-distributed (or chi-squared) statistic under $\mathcal{H}_0$.

$\max_\phi \mathrm{Re}(e^{-j\phi}\mathbf{s}^H \mathbf{y}) = |\mathbf{s}^H\mathbf{y}|$.

$|\mathbf{s}^H \mathbf{y}|^2 \gtrless \tau$: the envelope of the complex matched filter. This is the standard noncoherent receiver.

Under $\mathcal{H}_0$, $|\mathbf{s}^H\mathbf{y}|^2$ is exponential; performance degrades by approximately $1$--$3$ dB versus coherent detection depending on SNR.

$\mathbf{C}^{-1} = \frac{1}{\sigma^2}\Big(\mathbf{I} - \frac{\alpha}{\sigma^2 + n\alpha}\mathbf{1}\mathbf{1}^{\mathsf T}\Big)$.

$T(\mathbf{y}) = \mathbf{s}^{\mathsf T}\mathbf{C}^{-1}\mathbf{y} = \frac{1}{\sigma^2}\big(\mathbf{s}^{\mathsf T}\mathbf{y} - \frac{\alpha}{\sigma^2 + n\alpha}(\mathbf{s}^{\mathsf T}\mathbf{1})(\mathbf{1}^{\mathsf T}\mathbf{y})\big)$.

The detector subtracts a common-mode contamination term proportional to the mean of $\mathbf{y}$ and the DC component of $\mathbf{s}$ --- which is exactly the optimal way to suppress spatially correlated noise.

Let $\mathbf{L}\mathbf{L}^{\mathsf T} = \mathbf{C}$. Transform $\tilde{\mathbf{y}} = \mathbf{L}^{-1}\mathbf{y}$, $\tilde{\mathbf{s}} = \mathbf{L}^{-1}\mathbf{s}$.

$d^2 = \|\tilde{\mathbf{s}}\|^2 = \mathbf{s}^{\mathsf T}\mathbf{L}^{-\mathsf T}\mathbf{L}^{-1}\mathbf{s} = \mathbf{s}^{\mathsf T}\mathbf{C}^{-1}\mathbf{s}$.

$h(t) = s(T-t) = \sqrt{2/T}\cos(2\pi f_c(T-t))\cdot\mathbf{1}_{[0,T]}(t)$.

For unit-energy $s$, peak $= \int_0^T s^2(t) dt = 1$; in general $= E_s$.

$\mathcal{F}\{s(-t)\}(f) = S^*(f)$ (for real $s$).

$\mathcal{F}\{s(T-t)\}(f) = S^*(f)\cdot e^{-j 2\pi f T}$.

$\phi_1 = s_0/\|s_0\|$; $\psi_2 = s_1 - \langle s_1,\phi_1\rangle\phi_1$, $\phi_2 = \psi_2/\|\psi_2\|$; $\psi_3 = s_2 - \langle s_2,\phi_1\rangle\phi_1 - \langle s_2,\phi_2\rangle\phi_2$, $\phi_3 = \psi_3/\|\psi_3\|$.

Max dimension is $N \leq M = 3$; strictly less if any $s_k$ is a linear combination of the previous signals.

$f(\mathbf{y};\mu) \propto \exp(-\tfrac{1}{2\sigma^2}\|\mathbf{y}-\mu\mathbf{1}\|^2)$.

$\|\mathbf{y}-\mu\mathbf{1}\|^2 = n(\bar{y}-\mu)^2 + \|\mathbf{y}-\bar{y}\mathbf{1}\|^2$.

Set $g(\bar{y},\mu) = \exp(-n(\bar{y}-\mu)^2/(2\sigma^2))$ and $h(\mathbf{y}) = \exp(-\|\mathbf{y}-\bar{y}\mathbf{1}\|^2/(2\sigma^2))$; $\bar{Y}$ is sufficient.

$\int_0^{2\pi}\exp(2\mathrm{Re}(e^{-j\phi}\mathbf{s}_m^H\mathbf{y})/\sigma^2)\,d\phi/(2\pi) = I_0(2|\mathbf{s}_m^H\mathbf{y}|/\sigma^2)$.

$I_0$ is monotone in its argument, so the decision reduces to comparing $|\mathbf{s}_m^H\mathbf{y}|^2$.

Compute both matched filters, square the envelopes, pick the largest: the canonical noncoherent FSK receiver.

Let $\tilde{\mathbf{s}} = \mathbf{U}^{\mathsf T}\mathbf{s}$. Then $d^2 = \sum_k \tilde{s}_k^2/\lambda_k$.

Place all energy on the smallest eigenvalue $\lambda_{\min}$: $d^2_{\max} = 1/\lambda_{\min}$.

Waveform design for detection in colored noise puts signal energy where noise is weakest --- the dual of water-filling for capacity.

$P_d = Q(3.09 - 3) = Q(0.09) \approx 0.464$.

$P_d = Q(3.09 - 5) = Q(-1.91) \approx 0.972$.

$Q^{-1}(0.9) = Q^{-1}(10^{-4}) - d$. $-1.28 = 3.72 - d \Rightarrow d = 5.00$.

$d^2 = 2E_s/N_0 = 25 \Rightarrow E_s/N_0 = 12.5$, i.e. $10\log_{10}(12.5) \approx 11.0$ dB.