Exercises

(a) $\Delta\phi = 2\pi/180 \approx 0.0349$ rad.

$\text{IRR} = 4/(0.05^2 + 0.0349^2) = 4/(0.0025 + 0.00122) = 4/0.00372 = 1075 \approx 30.3$ dB.

(b) 64-QAM requires EVM $\leq -25$ dB, i.e., IRR $\geq 25$ dB. With IRR $= 30.3$ dB $> 25$ dB, the receiver supports 64-QAM.

(c) For 256-QAM, EVM $\leq -30$ dB is needed (IRR $\geq 30$ dB). IRR $= 30.3$ dB just barely meets the 256-QAM requirement, but with negligible margin. Practically, 64-QAM is the safe maximum. $\blacksquare$

(a) $Y_k = \alpha_1 H_k X_k + \alpha_2 H_{N-k}^* X_{N-k}^* + W_k$

and symmetrically: $Y_{N-k} = \alpha_1 H_{N-k} X_{N-k} + \alpha_2 H_k^* X_k^* + W_{N-k}$

(b) Define $\mathbf{y} = [Y_k, Y_{N-k}^*]^T$ and $\mathbf{x} = [X_k, X_{N-k}^*]^T$. Then:

$$\mathbf{y} = \begin{bmatrix} \alpha_1 H_k & \alpha_2 H_{N-k}^* \\ \alpha_2^* H_k & \alpha_1^* H_{N-k}^* \end{bmatrix} \mathbf{x} + \mathbf{w} = \mathbf{G}\mathbf{x} + \mathbf{w}$$

The ZF equaliser is $\hat{\mathbf{x}} = \mathbf{G}^{-1}\mathbf{y}$.

(c) The equaliser fails when $\det(\mathbf{G}) = 0$, i.e., $|\alpha_1|^2|H_k||H_{N-k}| = |\alpha_2|^2|H_k||H_{N-k}|$, which occurs when $|\alpha_1| = |\alpha_2|$ (IRR $= 0$ dB). Also fails if either $H_k$ or $H_{N-k}$ is in a deep fade. $\blacksquare$

(a) $f_{\text{RF}} = f_{\text{LO}} - f_{\text{IF}} = 900 - 70 = 830$ MHz (low-side injection) or $f_{\text{RF}} = f_{\text{LO}} + f_{\text{IF}} = 970$ MHz (high-side). Assuming low-side: $f_{\text{RF}} = 830$ MHz.

(b) $f_{\text{img}} = f_{\text{LO}} + f_{\text{IF}} = 970$ MHz. The image is separated from the desired signal by $2f_{\text{IF}} = 140$ MHz.

(c) With 60 dB filter attenuation at 970 MHz, the SIR due to the image is 60 dB (assuming equal-power image signal). This is sufficient for all practical modulation orders. $\blacksquare$

(a) The I/Q distortion $\alpha_2 x^*$ and PA distortion $d_{\text{PA}}$ are applied at different stages and are approximately uncorrelated:

$$\text{EVM}_{\text{total}}^2 = \frac{|\alpha_2|^2}{|\alpha_1|^2} + \text{EVM}_{\text{PA}}^2 = \frac{1}{\text{IRR}} + \text{EVM}_{\text{PA}}^2$$

(b) $1/\text{IRR} = 10^{-30/10} = 0.001$. $\text{EVM}_{\text{PA}}^2 = 10^{-28/10} = 0.00158$.

$\text{EVM}_{\text{total}}^2 = 0.001 + 0.00158 = 0.00258$. $\text{EVM}_{\text{total}} = 10\log_{10}(0.00258) = -25.9$ dB.

(c) 256-QAM requires EVM $\leq -30$ dB. The combined EVM of $-25.9$ dB does not meet this requirement. Either the PA back-off must increase or the I/Q calibration must improve. $\blacksquare$

(a) $\text{IIP3} = \sqrt{4 \times 10/(3 \times 0.1)} = \sqrt{40/0.3} = \sqrt{133.3} \approx 11.5$ (voltage).

In dBm (assuming 50 $\Omega$): $P_{\text{IIP3}} = 10\log_{10}(133.3/2) + 30 \approx 48.2$ dBm... Let us work in relative terms.

$\text{IIP3}_{\text{dB}} = 10\log_{10}(4 \times 10/(3 \times 0.1)) /2 = 10\log_{10}(133.3)/2 = 10.6$ dB above 1 V peak.

(b) $\text{OIP3} = G \times \text{IIP3} = a_1^2 \times \text{IIP3}$ (power gain). $G = 20\log_{10}(10) = 20$ dB. $\text{OIP3} = \text{IIP3} + 20$ dB.

(c) The IM3-to-fundamental ratio follows the $2:1$ slope rule: at input $P_{\text{in}}$ dB below IIP3, the IM3 is $2(P_{\text{IIP3}} - P_{\text{in}})$ dB below the fundamental.

IM3 relative level $= -2(\text{IIP3}_{\text{dBm}} - P_{\text{in}})$. This gives the relative suppression. $\blacksquare$

(a) Let $g(r)$ be the PA AM/AM curve and $f(r)$ the pre-distorter output. Perfect linearisation requires $g(f(r)) = c \cdot r$ for all $r$, i.e., $f = g^{-1}(c \cdot (\cdot))$. Since $g$ is analytic and monotonically increasing for $r < A_{\text{sat}}$, $g^{-1}$ exists and can be expanded in a Taylor series. A polynomial of infinite order can represent $g^{-1}$ exactly.

(b) Define $\phi_q(n) = x[n]|x[n]|^{q-1}$ for $q = 1, 3, 5$. The basis matrix row for sample $n$ is: $[\phi_1(n), \phi_3(n), \phi_5(n)]$. The LS solution: $\mathbf{a} = (\mathbf{\Phi}^H\mathbf{\Phi})^{-1} \mathbf{\Phi}^H\mathbf{y}_{\text{desired}}$.

(c) With $Q = 5$, the first uncancelled nonlinear order is 7. For Rapp $p = 3$ at IBO = 3 dB ($\beta = 2$), the 7th-order residual contributes approximately $-35$ to $-40$ dB EVM, sufficient for 64-QAM but marginal for 256-QAM. $\blacksquare$

(a) $\text{SQNR} = 6.02 \times 12 + 1.76 = 74.0$ dB.

(b) $\text{SQNR}_{\text{Gauss}} = 74.0 - 4.0 = 70.0$ dB.

(c) The effective SNR is $(1/\text{SNR} + 1/\text{SQNR})^{-1}$. For 0.1 dB loss: $\text{SNR}_{\text{channel}} \leq \text{SQNR} - 16 \approx 54$ dB (from $10\log_{10}(1 + 10^{-1.6}) \approx 0.1$ dB). At any practical channel SNR ($< 40$ dB), 12-bit quantisation is transparent. $\blacksquare$

(a) $\alpha = \frac{1}{\sigma_x^2}\int_{-\infty}^{\infty} x\,\mathcal{Q}(x)\,\frac{1}{\sqrt{2\pi}\sigma_x} e^{-x^2/(2\sigma_x^2)}\,dx$

(b) $\mathbb{E}[x\,\text{sign}(x)] = \mathbb{E}[|x|] = \sigma_x\sqrt{2/\pi}$.

$\alpha = \sqrt{2/\pi}/\sigma_x \approx 0.798/\sigma_x$.

(c) A 2-bit midrise quantiser with thresholds $\{-\Delta, 0, \Delta\}$ and optimal $\Delta = 0.9816\sigma_x$ gives $\alpha \approx 0.8825/\sigma_x$, recovering about 88% of the signal power vs. 80% for 1-bit. $\blacksquare$

(a) $\rho = 10^{0.5} \approx 3.16$.

$\text{SINR}_k = \frac{(2/\pi) \times 128 \times 3.16} {(1-2/\pi) \times 128 \times 3.16 + (2/\pi) \times 7 \times 3.16 + 1}$

$= \frac{257.3}{146.2 + 14.1 + 1} = \frac{257.3}{161.3} = 1.595$

$R_k = \log_2(1 + 1.595) = \log_2(2.595) = 1.375$ bits/s/Hz.

(b) $\text{SINR}_k^{(\infty)} = 128 \times 3.16 / (8 \times 3.16 + 1) = 404.5 / 26.28 = 15.39$

$R_k^{(\infty)} = \log_2(16.39) = 4.03$ bits/s/Hz.

(c) Need $M'$ such that the 1-bit SINR equals 15.39. Solving the 1-bit SINR equation: this approaches the ceiling of $2/(\pi - 2) \approx 1.75$, so $\text{SINR} = 15.39$ is unreachable with 1-bit ADCs.

The 1-bit system fundamentally cannot match the 4.03 bits/s/Hz unquantised rate regardless of $M$. $\blacksquare$

(a) $-90 = 10\log_{10}(\beta_{3\text{dB}}/(\pi \times (10^5)^2))$

$10^{-9} = \beta_{3\text{dB}}/(3.14 \times 10^{10})$

$\beta_{3\text{dB}} = 31.4$ Hz.

(b) $\mathcal{L}(10^6) = 10\log_{10}(31.4/(\pi \times 10^{12})) = 10\log_{10}(10^{-10.99}) = -110$ dBc/Hz.

(20 dB lower than at 100 kHz, as expected from $1/f^2$ rolloff.)

(c) $\text{SIR} = \Delta f_{\text{SCS}}/(2\pi\beta_{3\text{dB}}) = 30000/(2\pi \times 31.4) = 152 = 21.8$ dB. $\blacksquare$

(a) $10^{2.5} = 316 = 120000/(2\pi\beta_{3\text{dB}})$

$\beta_{3\text{dB}} \leq 120000/(2\pi \times 316) = 60.4$ Hz.

(b) $\mathcal{L}(10^6) = 10\log_{10}(60.4/(\pi \times 10^{12})) = 10\log_{10}(1.92 \times 10^{-11}) = -107.2$ dBc/Hz.

(c) With $\mathcal{L}(10^6) = -105$ dBc/Hz, the achieved linewidth is: $\beta = \pi \times 10^{12} \times 10^{-10.5} = 99.3$ Hz.

$\text{SIR} = 120000/(2\pi \times 99.3) = 192 = 22.8$ dB. Margin: $22.8 - 25 = -2.2$ dB (insufficient!).

This shows that 64-QAM at 28 GHz with 120 kHz SCS is challenging with current oscillator technology. $\blacksquare$

(a) RF chains: 32. ADCs: 64 (2 per chain). Phase shifters: 0. Total RF components: 96.

(b) RF chains: 1. ADCs: 2. Phase shifters: 32. Total: 35.

(c) RF chains: 4. ADCs: 8. Phase shifters: $32 \times 4 = 128$ (fully connected). Total: 140.

Sub-connected variant: $32$ phase shifters, 4 RF chains, 8 ADCs. Total: 44. $\blacksquare$

(a) After analog combining: $r = \mathbf{f}^H\mathbf{y} = \mathbf{f}^H\mathbf{h}\,s + \mathbf{f}^H\mathbf{n}$.

$\text{SNR}_{\text{out}} = |\mathbf{f}^H\mathbf{h}|^2 \cdot \text{SNR}$.

(b) For a ULA with $\mathbf{h} = \sqrt{N}\,\mathbf{a}(\theta_0)$:

$|\mathbf{f}^H\mathbf{h}|^2 = N \cdot |\mathbf{a}^H(\theta_0) \mathbf{a}(\theta_0)|^2/N = N$.

Array gain = $N$ (10 dB for $N = 10$, 18 dB for $N = 64$).

(c) With one RF chain, only one baseband signal exists. The analog combiner produces a single scalar $r = \mathbf{f}^H\mathbf{y}$. Two independent data streams would require two independent observations, i.e., two RF chains. Spatial multiplexing rank $\leq N_{\text{RF}} = 1$. $\blacksquare$

(a) With 32 uniform angles over $[-90^{\circ}, 90^{\circ}]$, the spacing is $180/32 = 5.625^{\circ}$. The closest atoms are at $11.25^{\circ}$ (index 18) and $50.625^{\circ}$ (index 25). OMP selects these two atoms first (highest correlation with the channel).

(b) With $L = 2$ paths and 2 well-matched dictionary atoms, the residual is primarily due to the angular mismatch:

$\|\mathbf{R}\|_F^2 \leq |\sigma_1|^2(1 - |\mathbf{a}^H(\theta_1) \mathbf{a}(\hat{\theta}_1)|^2/N) + |\sigma_2|^2(\ldots)$

For $\Delta\theta \approx 1.25^{\circ}$ and $N = 16$: $|\mathbf{a}^H(\theta)\mathbf{a}(\hat{\theta})|^2/N \approx 0.97$. Residual $\approx 3$% of channel energy.

(c) The 3rd and 4th OMP iterations select atoms adjacent to the first two, refining the beam directions. This reduces the residual to $< 0.5$% — the extra RF chains provide angular interpolation capability beyond the dictionary resolution. $\blacksquare$

(a) $P = N_{\text{RF}} \times 250\;\text{mW} + N_{\text{ant}} \times N_{\text{RF}} \times 30\;\text{mW} + 100 + 10 \times N_{\text{RF}}^2\;\text{mW}$ (baseband scales quadratically with streams).

(b) For sparse mmWave with $L = 3$ paths, $K = 4$ users: $R(N_{\text{RF}}) \approx R_{\text{digital}} \cdot \min(1, N_{\text{RF}}/(2K))$ for $N_{\text{RF}} \leq 2K$, saturating at $R_{\text{digital}}$ for $N_{\text{RF}} \geq 8$.

(c) Since $R$ saturates at $N_{\text{RF}} = 8$ but $P$ keeps growing, $\text{EE}$ peaks near $N_{\text{RF}} = 2K = 8$ and decreases for $N_{\text{RF}} > 8$.

For $N_{\text{ant}} = 64$: $P(8) = 2 + 15.4 + 0.74 = 18.1$ W. $P(64) = 16 + 122.9 + 41.1 = 180$ W.

EE improves by $\sim 10\times$ with hybrid ($N_{\text{RF}} = 8$) vs. digital ($N_{\text{RF}} = 64$) at nearly the same rate. $\blacksquare$

(a) Converting to linear EVM$^2$: $10^{-3.5} + 10^{-3.0} + 10^{-2.8} + 10^{-4.0}$ $= 0.000316 + 0.001 + 0.00158 + 0.0001 = 0.003$

$\text{EVM}_{\text{total}} = 10\log_{10}(0.003) = -25.2$ dB.

(b) Requirements: QPSK $-15$ dB, 16-QAM $-20$ dB, 64-QAM $-25$ dB, 256-QAM $-30$ dB.

Total EVM = $-25.2$ dB: supports 64-QAM (barely), not 256-QAM.

(c) Phase noise ICI ($-28$ dB) is the largest contributor (53% of total EVM$^2$). Improving it by 3 dB (to $-31$ dB) would reduce total EVM$^2$ to $0.00222$, giving $-26.5$ dB. This is the most impactful single improvement. $\blacksquare$

(a) For 1-bit ADC with input power $\sigma_x^2 + \sigma_n^2$: $A = \sqrt{2/(\pi \cdot (1 + 1/\text{SNR}))} = \sqrt{2/(\pi \cdot 1.1)} = 0.763$.

The effective model is $\mathbf{r} = A\mathbf{Hx} + A\mathbf{n} + \mathbf{q}$, where $\mathbf{q}$ has covariance $\mathbf{C}_q = \mathbf{I} - A^2(\sigma_x^2 + \sigma_n^2)\mathbf{I}$.

(b) With MRC and large $M$, the per-user SINR is approximately:

$\text{SINR}_k \approx \frac{A^2 M \text{SNR}}{A^2(K-1)\text{SNR} + A^2 + (1-A^2(1+\text{SNR}^{-1}))}$

$= \frac{0.582 \times 128 \times 10}{0.582 \times 7 \times 10 + 0.582 + 0.418}$ $= \frac{745}{40.7 + 1} = 17.9$ (12.5 dB).

Rate $= \log_2(1 + 17.9) = 4.24$ bits/s/Hz per user.

(c) With infinite resolution: $\text{SINR}_{\infty} = M \cdot \text{SNR} / (K-1)\text{SNR} + 1) = 1280/71 = 18.0$ (12.6 dB).

Rate$_{\infty} = \log_2(1+18.0) = 4.25$ bits/s/Hz.

Rate loss $\approx 0.01$ bits/s/Hz ($< 1$%). At $M/K = 16$, the massive MIMO array gain largely compensates for 1-bit quantisation distortion. $\blacksquare$

(a) $\mathbf{F}_{\text{RF}} = \text{blkdiag}(\mathbf{f}_1, \ldots, \mathbf{f}_{N_{\text{RF}}})$ where $\mathbf{f}_i \in \mathbb{C}^{N_{\text{sub}} \times 1}$ with unit-modulus entries.

Phase shifters: $N_{\text{RF}} \times N_{\text{sub}} = N = 64$ total (vs. $N \times N_{\text{RF}} = 256$ for fully connected).

(b) Each subarray has aperture $N_{\text{sub}} = 16$. The array gain per RF chain is $10\log_{10}(16) = 12$ dB vs. $10\log_{10}(64) = 18$ dB for fully connected (6 dB loss per chain).

However, the digital precoder across $N_{\text{RF}} = 4$ chains recovers some multiplexing gain. For $K = 2$, $L = 3$:

$R_{\text{sub}} \approx R_{\text{full}} - K\log_2(N/N_{\text{sub}}) \approx R_{\text{full}} - 2\log_2(4) = R_{\text{full}} - 4$ bits/s/Hz.

Typical loss is 15--25% depending on angular spread.

(c) For clustered angles, interleaved subarray allocation (antennas $\{1, 5, 9, \ldots\}$ to RF chain 1, $\{2, 6, 10, \ldots\}$ to RF chain 2, etc.) ensures each subarray spans the full physical aperture, maintaining angular resolution.

This gives each subarray the full $N$-element beamwidth (though with grating lobes spaced at $N_{\text{RF}} \cdot \lambda/2$). The digital precoder can suppress grating lobes, recovering $\sim 80$% of the fully-connected performance. $\blacksquare$