Exercises

$\lambda \approx 0.0857$ m. Side $a = 7\cdot 0.0857/2 = 0.30$ m. Diagonal $D = 0.30\sqrt{2} \approx 0.424$ m. $d_F = 2 D^2/\lambda = 2\cdot 0.180 / 0.0857 \approx 4.2$ m.

$\lambda \approx 0.0107$ m. $D = 127\cdot 0.00536 \approx 0.680$ m. $d_F = 2\cdot 0.463 / 0.0107 \approx 86.5$ m.

$\lambda \approx 0.00214$ m. $D = \sqrt{2} \approx 1.414$ m. $d_F = 2\cdot 2.0 / 0.00214 \approx 1869$ m. Nearly two kilometres of near-field — at this scale, "far field" is outside the cell. $\blacksquare$

$d_F = 2 D^2 / \lambda$, so doubling $D$ multiplies $d_F$ by $4$ — quadratic in aperture. Doubling $f_0$ halves $\lambda$, which only doubles $d_F$ — linear in carrier. The lesson is that aperture matters more than frequency for the near-field range. $\blacksquare$

$r_{k,n}^2 = (r_k\cos\theta_k - 0)^2 + (r_k\sin\theta_k - y_n)^2 = r_k^2 - 2 r_k y_n\sin\theta_k + y_n^2.$ Factor: $r_{k,n}^2 = r_k^2(1 + \epsilon_n)$ with $\epsilon_n = (-2 y_n\sin\theta_k + y_n^2/r_k)/r_k$.

$r_{k,n} = r_k\sqrt{1 + \epsilon_n} \approx r_k(1 + \epsilon_n/2 - \epsilon_n^2/8) = r_k + r_k\epsilon_n/2 - r_k\epsilon_n^2/8.$ Leading: $r_k\epsilon_n/2 = -y_n\sin\theta_k + y_n^2/(2 r_k)$. Quadratic: $r_k\epsilon_n^2/8 \approx r_k(2 y_n\sin\theta_k/r_k)^2/8 = (y_n\sin\theta_k)^2/(2 r_k)$ (to leading order in $y_n/r_k$).

$r_{k,n} \approx r_k - y_n\sin\theta_k + y_n^2/(2 r_k) - y_n^2\sin^2\theta_k/(2 r_k) = r_k - y_n\sin\theta_k + y_n^2(1 - \sin^2\theta_k)/(2 r_k) = r_k - y_n\sin\theta_k + y_n^2\cos^2\theta_k/(2 r_k).$ Validity: $|y_n|/r_k \ll 1$, i.e. aperture much smaller than the range — which is the Fresnel regime $d_R < r_k$ but weaker than $r_k \geq d_F$. $\blacksquare$

$\lambda \approx 1.07$ cm, $D = 63\cdot 0.00536 \approx 0.337$ m, $D/2 \approx 0.169$ m. $\kappa = 2\pi/\lambda \approx 587$ rad/m. Broadside: $\cos^2\theta = 1$.

$\Delta\phi = \kappa(D/2)^2/(2 d_k) = 587\cdot 0.0285/(2\cdot 5) \approx 1.67$ rad $\approx 0.53\pi$.

$\Delta\phi \approx \pi/2$ at the array edge is just beyond the $3$ dB threshold for Fresnel integrals. Numerically, the coherent gain drops by roughly $3$–$5$ dB compared to the ideal spherical- wave match. This is a large loss, and it is the direct consequence of using a plane-wave model for a user inside $d_F = 21.2$ m. $\blacksquare$

$d_k = 2 d_F d_k^2/(d_F^2 - d_k^2) \;\Longrightarrow\; d_F^2 - d_k^2 = 2 d_F d_k \;\Longrightarrow\; d_k^2 + 2 d_F d_k - d_F^2 = 0.$

Quadratic formula: $d_k^\star = (-2 d_F + \sqrt{4 d_F^2 + 4 d_F^2})/2 = d_F(-1 + \sqrt{2}) \approx 0.414\, d_F.$

Below $\approx 0.414 d_F$, the depth of focus is smaller than the focus range — i.e., the beam spot is tighter than the distance itself, and the range resolution of the array is comparable to where the user stands. Above this threshold, the depth of focus starts growing faster than the range itself, and by $d_k \to d_F$ it becomes infinite. The point $0.414 d_F$ marks the transition from "range-resolving near field" to "range- blurred near field." $\blacksquare$

$\lambda = 5$ mm. $D = 255\cdot 0.0025 = 0.638$ m. $d_F = 2\cdot 0.407 / 0.005 \approx 162.7$ m. $d_k = 3$ m is deep in the near field ($d_k/d_F \approx 0.018$).

$\Delta r = 2\cdot 162.7 \cdot 9/(162.7^2 - 9) = 2929/26468 \approx 0.111$ m $\approx 11$ cm.

The two users are at $2$ m and $4$ m — separated by $2$ m, which is about $18\times$ the depth of focus. A beamformer focused at either user essentially nulls the other. The array can multiplex them using two polar codewords at the same angle. This is a spatial degree of freedom that the far-field model cannot see. $\blacksquare$

$\Delta r = 2 d_F d_k^2/(d_F^2(1 - (d_k/d_F)^2)) = (2 d_k^2/d_F)/(1 - (d_k/d_F)^2).$ For $d_k/d_F \to 0$, $1/(1 - (d_k/d_F)^2) \to 1$, so $\Delta r \to 2 d_k^2/d_F.$

At short ranges, the depth of focus grows quadratically in the focus range. This is the "microscope regime": near the array, the beam spots are very sharp; farther out, they smear. $\blacksquare$

$\Delta r \approx 2 d_k^2/d_F \Rightarrow d_k \approx \sqrt{d_F\Delta r/2} = \sqrt{86.5\cdot 2/2} = \sqrt{86.5} \approx 9.3$ m.

At $d_k = 9.3$ m from a $128$-element mmWave ULA, the depth of focus is on the order of $2$ m — two users spaced $2$ m apart along the same direction are separable. At closer ranges the minimum separation is smaller still, enabling very tight co-directional clustering. $\blacksquare$

$|\mathcal{V}_k| = \rho_kN_t = 0.25\cdot 1024 = 256$ elements.

Loss $= 10\log_{10}(1/\rho_k) = 10\log_{10}(4) = 6.02$ dB.

A full $6$ dB of array gain is left on the table if the receiver ignores non-stationarity. That is equivalent to pretending the array has $1024$ elements when, effectively, only $256$ are coherent with the user. $\blacksquare$

$\lambda = 3\times 10^8 / 10^{11} = 3$ mm. $L^2 = 4$ m$^2$. $L^2 / (\lambda d) = 4 / (0.003\cdot 10) = 4/0.03 \approx 133.3$.

$\eta_{\text{DoF}} \sim 133.3^2 \approx 17780$ streams.

The far-field LoS count is $1$ stream. The near-field count is $\sim 1.8\times 10^4$ — four orders of magnitude more. This is the quantitative case for holographic MIMO in sparse LoS environments: a single point-to-point LoS link can carry thousands of independent streams if the apertures sit in each other's near field. $\blacksquare$

$\mathbf{v}(\mathbf{p}_k) = (1/\sqrt{N_t}) [\ldots, e^{j\kappa r_{k,n}}, \ldots]^T$ and $\mathbf{a}_{\text{NF}}(\mathbf{p}_k)_n = e^{-j\kappa r_{k,n}}/\sqrt{N_t}$. So $\mathbf{v}^{H} \mathbf{a}_{\text{NF}}(\mathbf{p}_k) = \sum_n (1/\sqrt{N_t}) e^{-j\kappa r_{k,n}}\cdot (1/\sqrt{N_t}) e^{-j\kappa r_{k,n}}$ — wait, that's $\sum_n (1/N_t) = 1$. So the inner product magnitude is exactly $1$.

The unit-norm inner product equals $1$ by Cauchy–Schwarz at equality. Array gain is typically reported as the ratio of post-combining signal power to per-antenna signal power when weights and channel are aligned; here, the per-antenna channel has magnitude $|h_n| = \sqrt{\beta_{k}/N_t}$, so per-antenna received power is $\beta_{k}/N_t$. After combining with unit-norm $\mathbf{v}$, the signal power is $|\sum_n (1/\sqrt{N_t}) e^{j\kappa r_{k,n}}\cdot\sqrt{\beta_{k}/N_t} e^{-j\kappa r_{k,n}}|^2 = |\sqrt{\beta_{k}/N_t}\cdot N_t/\sqrt{N_t}|^2 = \beta_{k}.$ Divided by per-antenna power $\beta_{k}/N_t$, the ratio is $N_t$. $\blacksquare$

$\mathbf{v}(\mathbf{p}_k)_n = (1/\sqrt{N_t}) e^{j\kappa(r_k - y_n\sin\theta_k + y_n^2\cos^2\theta_k/(2 r_k))}.$ Absorb the $e^{j\kappa r_k}$ factor into a global phase.

As $r_k \to \infty$, the quadratic term $y_n^2\cos^2\theta_k/(2 r_k)$ vanishes uniformly over $y_n$ in the bounded aperture. The remaining factor is $(1/\sqrt{N_t}) e^{-j\kappa y_n\sin\theta_k\cdot(-1)} = (1/\sqrt{N_t}) e^{+j\kappa y_n\sin\theta_k} = \mathbf{a}(\theta_k)^*$.

The deviation is bounded by $\|\mathbf{v}(\mathbf{p}_k) - \mathbf{a}(\theta_k)^*\| \leq \kappa (D/2)^2\cos^2\theta_k / (2 r_k) = (\pi/8)(d_F/r_k)\cos^2\theta_k$. For $r_k = d_F$, this bound is $\pi/8 \approx 0.39$ rad per element — the conventional Fraunhofer threshold. Beyond $d_F$ the error is less than $\pi/8$ uniformly and the far-field approximation is accurate. $\blacksquare$

$N_\theta = 2/\Delta\sin\theta = N_t$.

$\Delta(1/r) = \lambda/D^2$. The total range of $1/r$ is $1/r_{\min} - 1/d_F \approx 1/r_{\min}$ for $r_{\min} \ll d_F$. So $N_r \approx (1/r_{\min})/(\lambda/D^2) = D^2/(r_{\min}\lambda)$. Using $D \approx N_t\lambda/2$: $N_r \approx N_t^{2}\lambda^{2}/(4 r_{\min}\lambda) = N_t^{2}\lambda/(4 r_{\min})$.

$N_\theta\cdot N_r \approx N_t\cdot N_t^{2}\lambda/(4 r_{\min}) = N_t^{3}\lambda/(4 r_{\min})$. The $\mathcal{O}(N_t^{3/2})$ bound in the literature comes from restricting to one range bin per depth of focus and observing that the depth of focus itself shrinks with $N_t$; when $r_{\min}$ is chosen as the reactive near-field boundary $\propto \sqrt{D^3/\lambda} \propto N_t^{3/2}$, the product collapses to $\mathcal{O}(N_t^{3/2})$. $\blacksquare$

$\lambda = 3$ mm. $d_F = 2\cdot 1/0.003 \approx 667$ m. $\Delta r \approx 2 d_k^2/d_F = 2\cdot 25/667 \approx 0.075$ m $= 7.5$ cm.

$\sigma = 7.5/(2\sqrt{2\ln 2}) \approx 3.2$ cm. At $\Delta = 2$ m, $\Delta/\sigma \approx 62.5$, so the Gaussian gain is $\exp(-0.5\cdot 62.5^2) \approx 10^{-848}$ — essentially zero.

At such a short range with a $1$ m aperture, the depth of focus is a few centimetres and the fall-off is extremely steep. A scatterer two metres deeper is completely off the focused beam. This is the level of range resolution that holographic apertures can in principle offer. Practically, the Gaussian profile is an approximation and the true Fresnel-integral profile has sidelobes rather than a clean Gaussian tail — but the order of magnitude is correct. $\blacksquare$

Under LoS, $\mathbf{h}_k$ is a known non-linear function of $(r_k, \theta_k)$ (and an overall phase and path-loss, which we subsume into $\beta_{k}$). Two real parameters are enough to pin down the channel — a massive dimensionality reduction compared to $N_t$ free complex numbers.

With the Fresnel expansion, the phase at element $n$ is $\phi_n = \kappa(r_k - y_n\sin\theta_k + y_n^2\cos^2\theta_k/(2 r_k))$. The Fisher information for $r_k$ is $F_{rr} = 2\text{SNR}\cdotN_t\sum_n(\partial\phi_n/\partial r_k)^2/N_t$ $\approx 2\text{SNR}\cdotN_t\cdot \mathbb{E}[(\kappa y_n^2\cos^2\theta_k/(2 r_k^2))^2]$. For a ULA with $y_n \in [-D/2, D/2]$, $\mathbb{E}[y_n^4] \propto D^4$.

$F_{rr} \propto \text{SNR}\cdotN_t\cdot\kappa^2 D^4/r_k^4$. RMS range error $\sigma_r \propto 1/\sqrt{F_{rr}} \propto r_k^2/(\kappa D^2\sqrt{\text{SNR}\cdotN_t}) = d_k^2/(\kappa D^2\sqrt{\text{SNR}\cdotN_t})$. With $\kappa = 2\pi/\lambda$ absorbed into a constant, the scaling is exactly $d_k^2/(D^2\sqrt{\text{SNR}\cdotN_t})$. Range resolution is better (smaller) for larger apertures, shorter ranges, higher SNR, more antennas — exactly the near-field beam-focusing formula in disguise. $\blacksquare$

Write the visibility-restricted channel matrix \tilde\mathbf{H} = [\mathbf{h}_1, \mathbf{h}_2] restricted to $\mathcal{V}_1\cup\mathcal{V}_2$. The Gram matrix is \tilde\mathbf{H}^{H}\tilde\mathbf{H} = m\begin{bmatrix}1 & \alpha \\ \alpha^* & 1\end{bmatrix} (to leading order, after normalising per-antenna energies).

ZF inverts this Gram: (\tilde\mathbf{H}^{H}\tilde\mathbf{H})^{-1} = (1/(m(1-|\alpha|^2)))\begin{bmatrix}1 & -\alpha \\ -\alpha^* & 1\end{bmatrix}. The per-user noise amplification factor is $1/(1-|\alpha|^2)$.

ZF becomes ill-conditioned as $|\alpha| \to 1$, i.e. when the two visibility regions completely overlap. At that point the two channels are linearly dependent (both users are in the same sub-array and see the same geometry at the same angle). The remedy is either to separate the users into different time- frequency slots or to exploit range-axis differences by using a polar (range-aware) codebook, which restores the rank even when the angular channels coincide. $\blacksquare$