Exercises

$\lambda = c/f = 3 \times 10^8 / 12 \times 10^9 = 0.025$ m = 2.5 cm.

$A_{\text{phys}} = \pi(0.6)^2 = 1.131$ m$^2$.

$A_e = 0.6 \times 1.131 = 0.679$ m$^2$.

$G = \frac{4\pi \times 0.679}{(0.025)^2} = \frac{8.539}{6.25 \times 10^{-4}} = 13{,}663$

$G_{\text{dBi}} = 10\log_{10}(13{,}663) = 41.4$ dBi. $\blacksquare$

$\Theta_E = 10 \times \pi/180 = 0.1745$ rad. $\Theta_H = 12 \times \pi/180 = 0.2094$ rad.

$D = \frac{4\pi}{0.1745 \times 0.2094} = \frac{12.566}{0.03654} = 344$.

$D_{\text{dBi}} = 10\log_{10}(344) = 25.4$ dBi.

$G = \eta D = 0.85 \times 344 = 292 = 24.7$ dBi. $\blacksquare$

$\text{PLF} = |\hat{\mathbf{z}} \cdot (\cos\gamma\,\hat{\mathbf{z}} + \sin\gamma\,\hat{\mathbf{y}})|^2 = |\cos\gamma|^2 = \cos^2\gamma$.

$\text{PLF} = \cos^2(45^\circ) = 0.5$.

Loss $= -10\log_{10}(0.5) = 3.01$ dB.

$\cos^2\gamma = 0.5 \Rightarrow \cos\gamma = 1/\sqrt{2} \Rightarrow \gamma = 45^\circ$.

The 3 dB polarization loss occurs at exactly $45^\circ$ tilt. $\blacksquare$

$\psi = kd\sin(0^\circ) = 0$.

$\mathbf{a}(0^\circ) = [1, 1, 1, 1]^T$.

$\psi = \pi\sin(30^\circ) = \pi/2$.

$\mathbf{a}(30^\circ) = [1, e^{j\pi/2}, e^{j\pi}, e^{j3\pi/2}]^T = [1, j, -1, -j]^T$.

$\psi = \pi\sin(90^\circ) = \pi$.

$\mathbf{a}(90^\circ) = [1, e^{j\pi}, e^{j2\pi}, e^{j3\pi}]^T = [1, -1, 1, -1]^T$.

Note: at broadside, all elements are in phase; at endfire, adjacent elements alternate in sign. $\blacksquare$

Nulls at $\psi = 2m\pi/N$ for $m = \pm 1, \pm 2, \ldots, \pm(N-1)$.

With $d = \lambda/2$: $\psi = \pi\sin\theta \in [-\pi, \pi]$.

Nulls exist for $|2m\pi/16| \leq \pi$, i.e., $|m| \leq 8$. Excluding $m = 0$ (main lobe): $m = \pm 1, \ldots, \pm 7, \pm 8$ gives $\mathbf{15}$ nulls on each side... but $m = 8$ gives $\psi = \pi$, so $\sin\theta = 1$, $\theta = 90^\circ$.

Total nulls in visible region: $2 \times 7 + 2 = 16$ (including $m = \pm 8$ at endfire). Actually, $m = 8$ and $m = -8$ both map to $\theta = \pm 90^\circ$, so there are $\mathbf{15}$ distinct null directions (7 on each side of broadside plus $\pm 90^\circ$, but $m = \pm 8$ each give one angle).

More precisely: 14 nulls at interior angles plus the 2 endfire nulls = 16 nulls total (if we count $\pm 90^\circ$), or 14 interior nulls.

$m = 1$: $\psi = 2\pi/16 = \pi/8$, $\sin\theta = 1/8$, $\theta_1 = 7.18^\circ$.

$m = 2$: $\sin\theta = 2/8 = 1/4$, $\theta_2 = 14.48^\circ$.

$m = 3$: $\sin\theta = 3/8$, $\theta_3 = 22.02^\circ$.

For uniform weights: the first sidelobe level of the Dirichlet kernel is $-13.3$ dB, independent of $N$. $\blacksquare$

Base beamwidth at broadside: $\text{HPBW}_0 = 0.886/(32 \times 0.5) = 0.0554$ rad $= 3.17^\circ$.

At $\theta_0 = 30^\circ$: $\text{HPBW} = 3.17/\cos(30^\circ) = 3.17/0.866 = 3.66^\circ$.

At $\theta_0 = 60^\circ$: $\text{HPBW} = 3.17/\cos(60^\circ) = 3.17/0.5 = 6.34^\circ$.

$\text{HPBW}(60^\circ)/\text{HPBW}(0^\circ) = 1/\cos(60^\circ) = 2.0$.

The beamwidth doubles at $60^\circ$ scan.

$1/\cos\theta_0 = 2 \Rightarrow \cos\theta_0 = 0.5 \Rightarrow \theta_0 = 60^\circ$.

This is exactly the $60^\circ$ case above. $\blacksquare$

$w_0 = 0.5(1 - \cos 0) = 0$ $w_1 = 0.5(1 - \cos(2\pi/3)) = 0.5(1 + 0.5) = 0.75$ $w_2 = 0.5(1 - \cos(4\pi/3)) = 0.5(1 + 0.5) = 0.75$ $w_3 = 0.5(1 - \cos(2\pi)) = 0$

$\mathbf{w} = [0, 0.75, 0.75, 0]^T$.

Normalised: $\mathbf{w} = [0, 0.5, 0.5, 0]^T$.

$\mathrm{AF}(\theta) = \sum_{n=0}^{3} w_n e^{jn\psi} = 0.5 e^{j\psi} + 0.5 e^{j2\psi}$

$= 0.5 e^{j3\psi/2}(e^{-j\psi/2} + e^{j\psi/2}) = e^{j3\psi/2}\cos(\psi/2)$

where $\psi = \pi\sin\theta$.

$|\mathrm{AF}(\theta)|^2 = \cos^2(\psi/2)$.

With only 2 active elements, $|\mathrm{AF}|^2 = \cos^2(\psi/2)$ has no sidelobes (the $\cos^2$ pattern has only a single lobe before the first null at $\psi = \pi$).

For uniform weights on 4 elements, the first sidelobe is at $-13.3$ dB. The Hanning weighting eliminates sidelobes entirely but at the cost of doubling the beamwidth (from $\sim 25^\circ$ to $\sim 50^\circ$ for $N = 4$). $\blacksquare$

$N_{\text{total}} = M \times N = 4 \times 8 = 32$.

$D \approx MN = 32 = 15.1$ dBi.

Elevation (4 elements): $\text{HPBW}_E = 0.886/(4 \times 0.5) = 0.443$ rad $= 25.4^\circ$.

Azimuth (8 elements): $\text{HPBW}_A = 0.886/(8 \times 0.5) = 0.221$ rad $= 12.7^\circ$.

The azimuth beam is narrower because there are more elements in that dimension. $\blacksquare$

$\gamma_n = 2\pi n/8 = n\pi/4$ for $n = 0, \ldots, 7$.

$(x_n, y_n) = \lambda(\cos(n\pi/4), \sin(n\pi/4))$.

Element 0: $(\lambda, 0)$; Element 1: $(\lambda/\sqrt{2}, \lambda/\sqrt{2})$; etc.

$kR = 2\pi$, $\sin(60^\circ) = \sqrt{3}/2$.

$[\mathbf{a}]_n = \exp\!\bigl(j\,2\pi \cdot \frac{\sqrt{3}}{2}\cos(45^\circ - n\pi/4)\bigr)$

$= \exp\!\bigl(j\pi\sqrt{3}\cos(\pi/4 - n\pi/4)\bigr)$.

Each entry has magnitude 1 since the argument of $\exp$ is purely imaginary.

$\|\mathbf{a}\|^2 = \sum_{n=0}^{7} |e^{j\phi_n}|^2 = \sum_{n=0}^{7} 1 = 8 = N$.

The steering vector always has $\|\mathbf{a}\| = \sqrt{N}$ since each entry has unit magnitude. $\blacksquare$

$\text{rank}(\mathbf{H}) \leq \min(L, N_r, N_t) = \min(3, 4, 4) = 3$.

With well-separated AoA and AoD, the channel is generically rank-3.

$\mathbf{a}_r(10^\circ) = [1, e^{j\pi\sin 10^\circ}, e^{j2\pi\sin 10^\circ}, e^{j3\pi\sin 10^\circ}]^T$

$= [1, e^{j0.5453}, e^{j1.0906}, e^{j1.6359}]^T$

$\mathbf{a}_t(-20^\circ) = [1, e^{-j\pi\sin 20^\circ}, e^{-j2\pi\sin 20^\circ}, e^{-j3\pi\sin 20^\circ}]^T$

$= [1, e^{-j1.0746}, e^{-j2.1491}, e^{-j3.2237}]^T$

$\mathbf{H}_{1} = 1.0 \cdot \mathbf{a}_r(10^\circ)\,\mathbf{a}_t^H(-20^\circ) \in \mathbb{C}^{4 \times 4}$

This is a rank-1 matrix with $[\mathbf{H}_{1}]_{mn} = e^{j(m \times 0.5453 + n \times 1.0746)}$.

If clusters 1 and 2 share the same AoA ($\theta_1 = \theta_2$), then $\mathbf{a}_r(\theta_1) = \mathbf{a}_r(\theta_2)$, so

$\mathbf{H}_{1} + \mathbf{H}_{2} = \mathbf{a}_r(\theta_1)[\alpha_1\mathbf{a}_t^H(\phi_1) + \alpha_2\mathbf{a}_t^H(\phi_2)]$

This combined contribution is still rank-1 (outer product of one vector with another). The total rank would drop to $\min(2, 4, 4) = 2$. $\blacksquare$

For $d = \lambda/2$ and uniform distribution over $[-\Delta, \Delta]$ (with $\Delta$ in radians):

$\rho \approx \frac{\sin(\pi\Delta)}{\pi\Delta} = \mathrm{sinc}(\Delta)$ (using the normalised sinc).

$\Delta = 5^\circ = 0.0873$ rad: $\rho \approx \mathrm{sinc}(0.0873) = \frac{\sin(0.274)}{0.274} = 0.975$.

Very high correlation (low angular spread).

$\Delta = 30^\circ = 0.5236$ rad:

$\rho \approx \frac{\sin(\pi \times 0.5236)}{\pi \times 0.5236} = \frac{\sin(1.644)}{1.644} = \frac{0.998}{1.644} = 0.607$.

Solve $\mathrm{sinc}(\Delta) = 0.5$: $\sin(\pi\Delta)/(\pi\Delta) = 0.5$.

Numerically: $\pi\Delta \approx 1.895$, so $\Delta \approx 0.603$ rad $= 34.6^\circ$.

An angular spread of about $35^\circ$ is needed for $|\rho| < 0.5$ between adjacent half-wavelength-spaced elements. $\blacksquare$

$\|\mathbf{h}\|^2 = \sum_{n=1}^{N_r} |h_n|^2$ where each $|h_n|^2 \sim \text{Exp}(1)$.

$E[\|\mathbf{h}\|^2] = N_r$. $\text{Var}(\|\mathbf{h}\|^2) = N_r$ (sum of $N_r$ i.i.d. exponentials, each with variance 1).

$\frac{\|\mathbf{h}\|^2}{N_r} = \frac{1}{N_r}\sum_{n=1}^{N_r} |h_n|^2$

By the law of large numbers, this converges to $E[|h_n|^2] = 1$ as $N_r \to \infty$.

The variance of $\|\mathbf{h}\|^2/N_r$ is $N_r/N_r^2 = 1/N_r \to 0$.

$\text{CV} = \frac{\sqrt{N_r}}{N_r} = \frac{1}{\sqrt{N_r}} = \frac{1}{\sqrt{64}} = \frac{1}{8} = 12.5$%.

With 64 antennas, the channel gain fluctuates by only 12.5% around its mean — the fading is nearly eliminated. $\blacksquare$

Grating lobe at $\sin\theta_g = \sin(0) + \lambda/d = \lambda/d$. For $\theta_g$ to be invisible: $\lambda/d > 1$, so $d < \lambda$.

Maximum spacing: $d_{\max} = \lambda$.

$\sin\theta_g = \sin(45^\circ) + \lambda/d = 0.707 + \lambda/d$. For invisible: $0.707 + \lambda/d > 1$, so $\lambda/d > 0.293$, $d < 3.41\lambda$.

But also check the other grating lobe: $\sin\theta_g = 0.707 - \lambda/d$. For this to not be $< -1$: $0.707 - \lambda/d < -1$, so $\lambda/d > 1.707$, $d < 0.586\lambda$.

Maximum spacing: $d_{\max} = \lambda/(1 + \sin 45^\circ) = 0.586\lambda$.

$d_{\max} = \lambda/(1 + \sin 60^\circ) = \lambda/(1 + 0.866) = 0.536\lambda$.

For wide-scan arrays, spacing must be less than $\lambda/2$ to completely avoid grating lobes at all scan angles. $\blacksquare$

$|Z_{11}| = \sqrt{73^2 + 42.5^2} = 84.5$ $\Omega$.

$d = \lambda/4$: $Z_{12} = (73 + j42)\,\mathrm{sinc}(0.5) = (73 + j42) \times 0.637$. $|Z_{12}| = 53.8$ $\Omega$. $|Z_{12}/Z_{11}| = 0.637 = -3.9$ dB.

$d = \lambda/2$: $\mathrm{sinc}(1) = 0$. $|Z_{12}/Z_{11}| = 0 = -\infty$ dB (no coupling at this simplified model level).

$d = \lambda$: $\mathrm{sinc}(2) = 0$. Again zero in this approximation.

(Note: the actual mutual impedance has a more complex dependence; this sinc model is a first-order approximation.)

$|Z_{12}/Z_{11}| < 0.1 \Rightarrow \mathrm{sinc}(2d/\lambda) < 0.1$.

$\sin(\pi \times 2d/\lambda)/(\pi \times 2d/\lambda) < 0.1$.

Numerically: $2d/\lambda > 3.3$, so $d > 1.65\lambda$.

Mutual coupling modifies the effective excitation currents, causing: (1) main beam pointing error, (2) increased sidelobe levels, (3) impedance mismatch losses, (4) degradation of null depths. These effects are most severe for $d < \lambda/2$. $\blacksquare$

Analog: 1 beam (single RF chain). Digital: up to 64 simultaneous beams (64 RF chains).

6-bit shifters: $2^6 = 64$ phase states. Resolution: $360^\circ/64 = 5.625^\circ$. Maximum error: $\Delta\phi_{\max} = 5.625^\circ/2 = 2.8^\circ = 0.049$ rad.

Phase error variance: $\sigma_\phi^2 = (\pi/2^6)^2/3 = 8.0 \times 10^{-4}$ rad$^2$.

Array gain loss: $\Delta G \approx -10\log_{10}(1 - 8.0 \times 10^{-4}) \approx 0.003$ dB.

With 6-bit phase shifters, the quantization loss is negligible.

Digital: $64 \times \$50 = \$3{,}200$ (RF chains only).

Analog: $1 \times \$50 + 64 \times \$5 = \$370$.

Digital is $8.6\times$ more expensive in RF hardware alone, not counting the 64 ADCs/DACs needed. This cost gap motivates hybrid architectures at mmWave. $\blacksquare$

Each subarray of 16 elements (half-wavelength spacing): $G_{\text{sub}} \approx 16 = 12.0$ dBi.

With hybrid beamforming, the analog stage provides 12 dBi per subarray, and the digital stage provides coherent combining across 4 RF chains for each user's direction (if well-separated):

Effective gain $\approx N_{\text{RF}} \times G_{\text{sub}} = 4 \times 16 = 64 = 18.1$ dBi (same as the full array).

However, in multi-user mode, the digital precoder must balance beam gain with inter-user interference suppression, so the effective per-user gain is typically 3-6 dB less than the single-user maximum.

Hybrid: 4 streams, each with $\sim 15$ dBi effective gain. $\text{SE}_{\text{hybrid}} \approx 4 \times \log_2(1 + \text{SNR}_{15\text{dBi}})$.

Fully digital: 4 streams with true zero-forcing, each with $\sim 18$ dBi gain. $\text{SE}_{\text{digital}} \approx 4 \times \log_2(1 + \text{SNR}_{18\text{dBi}})$.

The fully digital architecture achieves 3-6 dB higher per-stream SNR due to perfect interference nulling, translating to $\sim 4$-$8$ bits/s/Hz higher total spectral efficiency. $\blacksquare$

$\mathbf{H} = \mathbf{A}_r \,\mathrm{diag}(\boldsymbol{\alpha})\, \mathbf{A}_t^H$

where $\mathbf{A}_r = [\mathbf{a}_r(10^\circ), \mathbf{a}_r(50^\circ)] \in \mathbb{C}^{8 \times 2}$ and $\mathbf{A}_t = [\mathbf{a}_t(-15^\circ), \mathbf{a}_t(30^\circ)] \in \mathbb{C}^{4 \times 2}$.

$\text{rank}(\mathbf{H}) = \text{rank}(\mathbf{A}_r \,\mathrm{diag}(\boldsymbol{\alpha})\, \mathbf{A}_t^H) \leq \min(\text{rank}(\mathbf{A}_r), 2, \text{rank}(\mathbf{A}_t)) = 2$.

Since $\theta_1 \neq \theta_2$ and $\phi_1 \neq \phi_2$, both $\mathbf{A}_r$ and $\mathbf{A}_t$ have rank 2. So $\text{rank}(\mathbf{H}) = 2$.

The Gram matrix is $\mathbf{G} = \mathbf{H}^{H}\mathbf{H} \in \mathbb{C}^{4 \times 4}$.

For well-separated angles (near-orthogonal steering vectors), $\mathbf{A}_r^H\mathbf{A}_r \approx N_r \mathbf{I}_2$ and $\mathbf{A}_t^H\mathbf{A}_t \approx N_t \mathbf{I}_2$.

In this regime: $\sigma_1^2 \approx N_r N_t |\alpha_1|^2 = 8 \times 4 \times 1 = 32$, $\sigma_2^2 \approx 32 \times 0.49 = 15.7$.

$\sigma_1 \approx 5.66$, $\sigma_2 \approx 3.96$.

$\kappa = \sigma_1/\sigma_2 \approx 5.66/3.96 = 1.43$.

The condition number is low (close to 1), indicating that both spatial streams have similar quality. This is because the well-separated angles create nearly orthogonal steering vectors. $\blacksquare$

$N_r - K = 128 - 8 = 120$.

$\sum_{j=1}^{8} \beta_j = 1 + 1/2 + 1/3 + \ldots + 1/8 = 2.717$.

For user $k$: $\text{SINR}_k = \frac{120 \times \beta_k}{2.717 - \beta_k}$.

$\text{SE} = \sum_{k=1}^{8} \log_2(1 + \text{SINR}_k)$

$\approx 6.15 + 4.81 + 4.15 + 3.72 + 3.39 + 3.14 + 2.95 + 2.77$

$= 31.1$ bits/s/Hz.

$\text{SINR}_k \propto (N_r - K)$ for fixed $K$ and $\beta_k$.

As $N_r \to \infty$: $\text{SINR}_k \to \infty$, so $\text{SE} \approx K\log_2(N_r - K)$, which grows logarithmically with $N_r$.

More importantly, the interference becomes negligible relative to the desired signal, so even simple MF precoding becomes near-optimal.

ZF: $\text{SINR}_k^{\text{ZF}} = (N_r - K)\beta_k = 120\beta_k$.

$\text{SE}_{\text{ZF}} \approx 6.92 + 5.93 + 5.36 + 4.95 + 4.64 + 4.39 + 4.19 + 4.01 = 40.4$ bits/s/Hz.

ZF gains $\sim 9$ bits/s/Hz over MF by eliminating inter-user interference, at the cost of noise enhancement and requiring full CSI at the transmitter. $\blacksquare$