Exercises

$\text{FSPL}(28, 1\,\text{m}) = 32.4 + 20\log_{10}(28) = 32.4 + 28.94 = 61.3$ dB

$\text{FSPL}(39, 1\,\text{m}) = 32.4 + 20\log_{10}(39) = 32.4 + 31.82 = 64.2$ dB

$\text{FSPL}(73, 1\,\text{m}) = 32.4 + 20\log_{10}(73) = 32.4 + 37.27 = 69.7$ dB

Distance term: $10 \times 2.0 \times \log_{10}(100) = 40.0$ dB.

$PL^{CI}(28\,\text{GHz}, 100\,\text{m}) = 61.3 + 40.0 = 101.3$ dB

$PL^{CI}(39\,\text{GHz}, 100\,\text{m}) = 64.2 + 40.0 = 104.2$ dB

$PL^{CI}(73\,\text{GHz}, 100\,\text{m}) = 69.7 + 40.0 = 109.7$ dB

Additional loss at 73 GHz vs 28 GHz: $109.7 - 101.3 = 8.4$ dB.

This equals $20\log_{10}(73/28) = 8.3$ dB, confirming that the frequency dependence resides entirely in the FSPL anchor. $\blacksquare$

$\beta = \lambda_B \times w = 0.02 \times 0.4 = 0.008\;\text{m}^{-1}$$

At $d = 50$ m: $p_\text{LOS} = e^{-0.008 \times 50} = e^{-0.4} = 0.670$ (67.0%)

At $d = 150$ m: $p_\text{LOS} = e^{-0.008 \times 150} = e^{-1.2} = 0.301$ (30.1%)

$e^{-\beta d} = 0.5 \implies d = \frac{\ln 2}{\beta} = \frac{0.693}{0.008} = 86.6\;\text{m}$$Beyond 87 m, the link is more likely NLOS than LOS.$\blacksquare$

$\text{FSPL}(6, 1\,\text{m}) = 32.4 + 20\log_{10}(6) = 32.4 + 15.6 = 48.0$ dB

$\text{FSPL}(28, 1\,\text{m}) = 32.4 + 20\log_{10}(28) = 32.4 + 28.9 = 61.3$ dB

$\text{FSPL}(140, 1\,\text{m}) = 32.4 + 20\log_{10}(140) = 32.4 + 43.0 = 75.4$ dB

Increase from 6 to 140 GHz: $75.4 - 48.0 = 27.4$ dB.

This equals $20\log_{10}(140/6) = 27.4$ dB. This 27 dB penalty at 1 m must be compensated by antenna array gain at mmWave/sub-THz frequencies. $\blacksquare$

Indoor power: $-65 - 35 = -100$ dBm.

Since $-100 < -85$ dBm, the link fails with 15 dB margin deficit.

Indoor power: $-65 - 4 = -69$ dBm.

Since $-69 > -85$ dBm, the link works with 16 dB margin.

$L_\text{max} = -65 - (-85) = 20$ dB.

Any penetration loss below 20 dB maintains the link. $\blacksquare$

$\Delta PL = 101.3 - 81.9 = 19.4$ dB.

$10\log_{10}(N) = 19.4 \implies N = 10^{1.94} = 87.1$.

Rounding up: $N = 88$ elements (or $N = 64$ for a practical $8 \times 8$ UPA with 18 dBi gain, leaving a 1.4 dB gap).

In practice, both Tx and Rx arrays are used, so a $64$-element Tx array and a $16$-element Rx array together provide $18 + 12 = 30$ dBi of combined gain — more than enough. $\blacksquare$

$\mathbf{a}(30°) = \frac{1}{\sqrt{64}} \begin{bmatrix} 1 \\ e^{j\pi\sin(30°)} \\ e^{j2\pi\sin(30°)} \\ \vdots \\ e^{j63\pi\sin(30°)} \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 1 \\ e^{j\pi/2} \\ e^{j\pi} \\ \vdots \\ e^{j63\pi/2} \end{bmatrix}$$This vector has a uniform phase progression of$\pi/2$ per element.

After two OMP iterations selecting $\phi_1 = 30°$ and $\phi_2 = -15°$:

$$\mathbf{F}_\text{RF} = [\mathbf{a}(30°) \mid \mathbf{a}(-15°) \mid \mathbf{a}_3 \mid \mathbf{a}_4] \in \mathbb{C}^{64 \times 4}$$

where $\mathbf{a}_3$ and $\mathbf{a}_4$ are the remaining two columns selected by subsequent OMP iterations to minimise the residual $\|\mathbf{F}_\text{opt} - \mathbf{F}_\text{RF}\mathbf{F}_\text{BB}\|_F$.

The digital precoder is then: $\mathbf{F}_\text{BB} = (\mathbf{F}_\text{RF}^H\mathbf{F}_\text{RF})^{-1}\mathbf{F}_\text{RF}^H\mathbf{F}_\text{opt} \in \mathbb{C}^{4 \times 2}$.

Each data stream requires control of both magnitude and phase at the analog stage. Since each RF chain provides one complex degree of freedom (phase only for analog, but full complex for the combined $\mathbf{F}_\text{RF}\mathbf{F}_\text{BB}$), at least $2N_s$ RF chains are needed to independently control both the real and imaginary parts of the $N_s$-stream signal in the angular domain. With $L = 3$ paths and $N_\text{RF} = 4 \geq L$, all significant channel paths can be captured. $\blacksquare$

$N_t \times N_\text{RF} = 128 \times 8 = 1024$ phase shifters.

Quantisation step: $2\pi / 2^3 = \pi/4 = 45°$.

Maximum error: $\pi/2^3 = \pi/8 = 22.5°$.

This shows that 3-bit phase shifters incur $\sim$0.7 dB loss (acceptable), while 2-bit suffer 3 dB (significant). Most practical mmWave systems use 4–6 bit phase shifters. $\blacksquare$

Fully connected: $256 \times 16 = 4096$ phase shifters.

Sub-connected: $256$ phase shifters (each RF chain drives $256/16 = 16$ antennas).

RF chain power: $16 \times 250 = 4000$ mW = 4.0 W (same for both).

Fully connected PS power: $4096 \times 30 = 122{,}880$ mW = 122.9 W.

Sub-connected PS power: $256 \times 30 = 7680$ mW = 7.7 W.

Total: Fully connected = 126.9 W, Sub-connected = 11.7 W.

The sub-connected architecture reduces analog power by $\sim$10.8$\times$.

$\Delta R \approx N_s \log_2(N_\text{RF}) = 4 \times \log_2(16) = 4 \times 4 = 16\;\text{bps/Hz}$$This is an upper bound at high SNR; at moderate SNR (10–20 dB), the gap is typically 2–5 bps/Hz. Whether this performance loss justifies the$\sim$10$\times$power savings depends on the application.$\blacksquare$

$T_u = 1/\Delta f = 1/(120 \times 10^3) = 8.33\;\mu$s.

$T_\text{CP} \approx 0.57\;\mu$s (normal CP for $\mu = 3$).

$T_\text{sym} = 8.33 + 0.57 = 8.90\;\mu$s.

Each SSB occupies 4 symbols: $T_\text{SSB} = 4 \times 8.90 = 35.6\;\mu$s.

64 SSB blocks (assuming contiguous transmission): $T_\text{sweep} = 64 \times 35.6 = 2278\;\mu$s $= 2.28$ ms.

$\eta = 2.28 / 20 = 11.4\%$.

$T_\text{sweep} = 2.28$ ms $< T_c = 4.5$ ms.

Yes, the full sweep completes within one coherence interval, but it consumes $2.28/4.5 = 50.6\%$ of the coherence time. At higher speeds, the sweep may not complete within $T_c$. $\blacksquare$

$S = \log_2(256) = 8$ stages.

Exhaustive: $256 \times 0.25 = 64$ ms.

Hierarchical: $2 \times 8 \times 0.25 = 4$ ms.

Reduction factor: $64/4 = 16\times$.

Assuming 5% error probability per stage independently:

$$p_\text{correct} = (1 - 0.05)^8 = 0.95^8 = 0.663$$

There is a 33.7% probability of selecting the wrong beam! This motivates robust designs: multi-beam testing at each stage, wider margins at early stages, and fallback mechanisms. $\blacksquare$

$T_\text{interrupt} = T_\text{BFD} + T_\text{candidate} + T_\text{PRACH,max} + T_\text{gNB}$ $= 20 + 5 + 10 + 5 = 40$ ms.

40 ms $< 50$ ms requirement, so single-panel BFR barely meets the constraint. However, this is the worst case — if PRACH access is immediate, $T_\text{interrupt} = 30$ ms.

A dual-panel UE maintains beam pairs on both panels simultaneously. When panel 1 experiences blockage, panel 2 can immediately serve traffic without waiting for BFR. The effective interruption time is reduced to $\sim$0–5 ms (the time to switch data flow to the backup panel), well within the 50 ms requirement. $\blacksquare$

FSPL anchor: $32.4 + 20\log_{10}(10) = 32.4 + 20 = 52.4$ dB.

CI path loss: $52.4 + 10 \times 2.1 \times \log_{10}(500) = 52.4 + 21 \times 2.699 = 52.4 + 56.7 = 109.1$ dB.

$N_0 = -174 + 10\log_{10}(8 \times 10^8) + 6 = -174 + 89.0 + 6 = -79.0$ dBm.

$\text{SNR} = 25 + 18 + 3 - 109.1 + 79.0 = 15.9$ dB.

$C = 800 \times 10^6 \times \log_2(1 + 10^{1.59}) = 800 \times \log_2(39.9) = 800 \times 5.32 = 4.25$ Gbps.

At $P_t = 15$ dBm: $\text{SNR} = 15.9 - 10 = 5.9$ dB.

$C' = 800 \times \log_2(1 + 10^{0.59}) = 800 \times \log_2(4.89) = 800 \times 2.29 = 1.83$ Gbps.

Capacity loss: $4.25 - 1.83 = 2.42$ Gbps (57% reduction). A 10 dB power reduction more than halves the capacity. $\blacksquare$

Case 1: $f = 28$ GHz, $\lambda = 10.7$ mm, $N = 64$. $D = 64 \times 10.7/2 = 342.9$ mm $= 0.343$ m. $d_F = 2 \times 0.343^2 / 0.0107 = 0.235 / 0.0107 = 22.0$ m.

Case 2: $f = 140$ GHz, $\lambda = 2.14$ mm, $N = 256$. $D = 256 \times 2.14/2 = 274.0$ mm $= 0.274$ m. $d_F = 2 \times 0.274^2 / 0.00214 = 0.150 / 0.00214 = 70.1$ m.

Case 3: $f = 300$ GHz, $\lambda = 1.00$ mm, $N = 1024$. $D = 1024 \times 1.00/2 = 512.0$ mm $= 0.512$ m. $d_F = 2 \times 0.512^2 / 0.001 = 0.524 / 0.001 = 524.3$ m.

At 300 GHz with 1024 elements, even a 100 m link is in the near field.

With $d_s = \lambda/2$ and $D \approx N\lambda/2$:

$$d_F = \frac{2D^2}{\lambda} = \frac{2(N\lambda/2)^2}{\lambda} = \frac{N^2\lambda}{2} = \frac{N^2 c}{2f}$$

So $d_F \propto N^2/f$. Doubling the array size quadruples $d_F$; doubling the frequency halves it. However, since practical array sizes tend to increase faster with frequency (to maintain aperture), the net effect is that $d_F$ increases dramatically at sub-THz frequencies. $\blacksquare$

FSPL anchor: $32.4 + 20\log_{10}(140) = 32.4 + 42.9 = 75.3$ dB.

CI spreading loss: $75.3 + 20\log_{10}(200) = 75.3 + 46.0 = 121.3$ dB.

Atmospheric absorption: $1.5 \times 200/1000 = 0.3$ dB.

Total: $PL_\text{total} = 121.3 + 0.3 = 121.6$ dB.

$N_0 = -174 + 10\log_{10}(20 \times 10^9) + 8 = -174 + 103.0 + 8 = -63.0$ dBm.

$\text{SNR} = 20 + 30 + 30 - 121.6 + 63.0 = 21.4$ dB.

At $d = 200$ m: absorption = 0.3 dB out of 121.6 dB total (0.25%).

For 10% contribution: $\alpha d/1000 = 0.1 \times PL_\text{total}$.

Since $PL_\text{total} \approx 75.3 + 20\log_{10}(d) + 1.5d/1000$, this must be solved numerically. For large $d$: $0.0015d \approx 0.1 \times (75.3 + 20\log_{10}(d))$, giving $d \approx 6$–$8$ km — well beyond typical sub-THz link ranges.

Conclusion: at 140 GHz, atmospheric absorption is negligible for links below $\sim$500 m.

FSPL anchor: $32.4 + 20\log_{10}(183) = 32.4 + 45.3 = 77.7$ dB.

CI spreading loss: $77.7 + 46.0 = 123.7$ dB.

Atmospheric absorption: $30 \times 200/1000 = 6.0$ dB.

Total: $PL_\text{total} = 123.7 + 6.0 = 129.7$ dB.

SNR $= 20 + 30 + 30 - 129.7 + 63.0 = 13.3$ dB.

The 183 GHz water vapour resonance adds 6 dB of loss at 200 m — this frequency should be avoided for outdoor links. $\blacksquare$

$\lambda = 3 \times 10^8 / 150 \times 10^9 = 2.0$ mm.

$D = 64 \times 2.0/2 = 64.0$ mm $= 0.064$ m.

$d_F = 2 \times 0.064^2 / 0.002 = 2 \times 0.004096 / 0.002 = 4.10$ m.

So $d_0 = 10$ m is in the far field for this array (barely). Let us use the larger diagonal aperture: $D_\text{diag} = 0.064\sqrt{2} = 0.0905$ m.

$d_F = 2 \times 0.0905^2 / 0.002 = 8.19$ m.

With the diagonal aperture, $d_0 = 10$ m is marginally in the transition region.

$\Delta r = \frac{8\lambda d_0^2}{\pi D^2} = \frac{8 \times 0.002 \times 100}{\pi \times 0.064^2} = \frac{1.6}{\pi \times 0.004096} = \frac{1.6}{0.01287} = 124.3\;\text{m}$$This enormous depth of focus means that at$d_0 = 10$ m (which is near the Fraunhofer boundary), range resolution is very poor.

$|d_1 - d_2| = 4$ m $\ll \Delta r = 124$ m.

The two users cannot be separated by range at this distance and array size. The depth of focus is too large.

For effective near-field spatial multiplexing by range, we need $d_0 \ll d_F$ (deep near field). At $d_0 = 2$ m: $\Delta r = 8 \times 0.002 \times 4 / (\pi \times 0.004096) = 4.97$ m, which still does not resolve a 4 m separation.

This illustrates that near-field range multiplexing requires either much larger arrays or much shorter distances.

Angular resolution: $\theta_{3\text{dB}} \approx 0.886\lambda/D = 0.886 \times 0.002/0.064 = 0.028$ rad $= 1.6°$.

At $d = 10$ m, this corresponds to a lateral resolution of $d \times \theta_{3\text{dB}} = 10 \times 0.028 = 0.28$ m.

Range resolution: $\Delta r = 124$ m.

The angular resolution is far superior to the range resolution at this operating point. Near-field range multiplexing becomes advantageous only for very large arrays (hundreds of wavelengths aperture) at close range. $\blacksquare$

Per-stream SNR: $\rho/N_s = 100/4 = 25$ (linear).

$$R_\text{digital} = \sum_{i=1}^{4} \log_2(1 + 25\sigma_i^2)$$

$= \log_2(1 + 25 \times 10.24) + \log_2(1 + 25 \times 3.24) + \log_2(1 + 25 \times 0.81) + \log_2(1 + 25 \times 0.16)$

$= \log_2(257) + \log_2(82) + \log_2(21.25) + \log_2(5)$

$= 8.01 + 6.36 + 4.41 + 2.32 = 21.10$ bps/Hz.

$\tilde{\sigma}_i = 0.95\sigma_i$, so $\tilde{\sigma}_i^2 = 0.9025\sigma_i^2$.

$$R_\text{hybrid,8} = \log_2(1 + 25 \times 0.9025 \times 10.24) + \cdots$$

$= \log_2(232.0) + \log_2(74.1) + \log_2(19.27) + \log_2(4.61)$

$= 7.86 + 6.21 + 4.27 + 2.21 = 20.55$ bps/Hz.

Gap: $21.10 - 20.55 = 0.55$ bps/Hz.

$\tilde{\sigma}_i^2 = 0.7225\sigma_i^2$.

$$R_\text{hybrid,4} = \log_2(1 + 25 \times 0.7225 \times 10.24) + \cdots$$

$= \log_2(186.0) + \log_2(59.5) + \log_2(15.63) + \log_2(3.89)$

$= 7.54 + 5.89 + 3.97 + 1.96 = 19.36$ bps/Hz.

Gap: $21.10 - 19.36 = 1.74$ bps/Hz.

From the above, $N_\text{RF} = 8$ (95% capture) gives a 0.55 bps/Hz gap. The rule of thumb $N_\text{RF} \geq 2N_s = 8$ achieves less than 1 bps/Hz gap, consistent with theory. $\blacksquare$

The exact distance between Tx element $m$ and Rx element $n$ is:

$$d_{n,m} = \sqrt{d_0^2 + (p_n^r - p_m^t)^2}$$

For $|p_n^r - p_m^t| \ll d_0$ (paraxial regime):

$$d_{n,m} \approx d_0 + \frac{(p_n^r - p_m^t)^2}{2d_0}$$

The channel entry is $h_{n,m} \propto e^{-j2\pi d_{n,m}/\lambda}$, and ignoring the common phase $e^{-j2\pi d_0/\lambda}$:

$$h_{n,m} \propto \exp\!\left(-j\frac{\pi(p_n^r - p_m^t)^2}{\lambda d_0}\right)$$

Expanding the square:

$$h_{n,m} \propto e^{-j\frac{\pi}{\lambda d_0}(p_n^r)^2} \cdot e^{+j\frac{2\pi}{\lambda d_0}p_n^r p_m^t} \cdot e^{-j\frac{\pi}{\lambda d_0}(p_m^t)^2}$$

The middle term $e^{j2\pi p_n^r p_m^t/(\lambda d_0)}$ has the structure of a DFT matrix with "frequency spacing" $\Delta = \lambda/(2\lambda d_0) = 1/(2d_0)$ and the outer terms are diagonal phase matrices.

Since diagonal matrices are full rank, the rank of $\mathbf{H}$ equals the rank of the DFT-like core. The effective rank is determined by how many distinct "frequency bins" the product $p_n^r p_m^t$ can resolve:

$$\text{rank}(\mathbf{H}) \approx \min\!\left(N_t,\, N_r,\, \frac{D_t D_r}{4\lambda d_0} + 1\right)$$

where $D_t = N_t\lambda/2$ and $D_r = N_r\lambda/2$.

$\lambda = 2.14$ mm, $D_t = D_r = 128 \times 2.14/2 = 137.0$ mm $= 0.137$ m.

Rank bound: $D_t D_r/(4\lambda d_0) = 0.137^2/(4 \times 0.00214 \times d_0) = 0.01877/(0.00856 \times d_0) = 2.192/d_0$.

At $d_0 = 1$ m: rank $\approx \min(128, 128, 2.19 + 1) \approx 3$.

At $d_0 = 5$ m: rank $\approx \min(128, 128, 0.44 + 1) \approx 1$.

At $d_0 = 20$ m: rank $\approx 1$ (far field, LOS = rank 1).

Even at $d_0 = 1$ m, the LOS near-field channel only supports $\sim$3 spatial streams for this array size. Larger arrays ($N \geq 512$) are needed for significant near-field multiplexing gains. $\blacksquare$

$\eta = \frac{R_\text{min}}{B} = \frac{500 \times 10^6}{400 \times 10^6} = 1.25\;\text{bps/Hz}$$With practical coding efficiency$\sim$80efficiency needed is$\eta_\text{raw} \approx 1.25/0.8 = 1.56$ bps/Hz.

$p_\text{LOS}(200) = e^{-0.015 \times 200} = e^{-3.0} = 0.050$ (5.0%).

FSPL anchor at 39 GHz: $32.4 + 20\log_{10}(39) = 64.2$ dB.

$PL_\text{LOS} = 64.2 + 20 \times \log_{10}(200) = 64.2 + 46.0 = 110.2$ dB.

$PL_\text{NLOS} = 64.2 + 33 \times \log_{10}(200) = 64.2 + 75.9 = 140.1$ dB.

$\overline{PL} = 0.05 \times 110.2 + 0.95 \times 140.1 = 5.5 + 133.1 = 138.6$ dB.

The cell edge is almost certainly NLOS at 39 GHz.

Required SNR: $\text{SNR}_\text{req} = 2^{1.56} - 1 = 1.95$ (linear) = 2.9 dB.

With 10 dB margin: SNR target = 12.9 dB.

Noise: $N_0 = -174 + 86.0 + 7 = -81.0$ dBm.

Link budget: $G_\text{total} = \text{SNR}_\text{target} + PL - P_t + N_0$ $= 12.9 + 138.6 - 33 + (-81.0) = 37.5$ dBi.

So Tx + Rx combined gain must be at least 37.5 dBi.

With $G_t + G_r = 37.5$ dBi, a practical split:

Tx: $N_t = 256$ elements ($16 \times 16$ UPA), gain $\approx 24$ dBi.

Rx: $N_r = 16$ elements ($4 \times 4$ UPA), gain $\approx 12$ dBi.

Combined: $24 + 12 = 36$ dBi (1.5 dB short; acceptable with 4-stream MU-MIMO gain).

Hybrid BF: With $K = 8$ users and $N_s = 1$ stream per user, need $N_\text{RF} \geq 2 \times 8 = 16$ RF chains.

Proposed: $N_t = 256$, $N_\text{RF} = 16$, sub-connected architecture (256 phase shifters, manageable power). Each RF chain drives a $4 \times 4$ sub-array of 16 elements.

Phase shifters: $256$ (sub-connected) at 30 mW each = 7.7 W. RF chains: 16 at 250 mW each = 4.0 W. Total analog: 11.7 W — feasible for a small cell. $\blacksquare$