Exercises

$h[k] = \sum_{\ell=0}^{4} h_\ell \, e^{-j 2\pi k \ell / 64}$, $k = 0, \ldots, 63$.

$\sum_{k=0}^{63} |h[k]|^2 = \sum_{k} \sum_{\ell,m} h_\ell h_m^* e^{-j2\pi k(\ell-m)/64} = \sum_{\ell,m} h_\ell h_m^* \cdot 64 \delta_{\ell m} = 64 \sum_{\ell=0}^{4} |h_\ell|^2$. This is Parseval's theorem for the DFT. $\blacksquare$

The CP converts linear convolution into circular convolution. After removing the CP at the receiver, the received $N$-sample block is $\mathbf{y} = \mathbf{H}_{\text{circ}} \mathbf{x} + \mathbf{w}$, where $\mathbf{H}_{\text{circ}}$ is a circulant matrix formed from the channel taps.

If $T_{\text{cp}} < \tau_{L-1}$, the last $L - 1 - N_{\text{cp}}$ samples of the previous OFDM symbol "leak" into the current symbol. The received block is no longer a circular convolution, and the DFT does not diagonalize $\mathbf{H}$. The result is inter-symbol interference (ISI) and inter-carrier interference (ICI). $\blacksquare$

With MRT, $\mathbf{v}_{i}[k] = \mathbf{h}_i[k]/\|\mathbf{h}_i[k]\|$. The SINR of user $j$ at subcarrier $k$ is

$$\text{SINR}_j[k] = \frac{p_j \|\mathbf{h}_j[k]\|^2}{\sum_{i \neq j} p_i \frac{|\mathbf{h}_j^H[k]\mathbf{h}_i[k]|^2}{\|\mathbf{h}_i[k]\|^2} + \sigma^2}$$

By channel hardening: $\|\mathbf{h}_j[k]\|^2 / N_t \to \beta_{j}$ (deterministic, same for all $k$). By favorable propagation: $|\mathbf{h}_j^H[k]\mathbf{h}_i[k]|^2 / N_t^{2} \to 0$. Therefore $\text{SINR}_j[k] \to p_j N_t \beta_{j} / \sigma^2$, which is deterministic and independent of $k$. The sum rate becomes $R_j \to \frac{N T \Delta f}{T + T_{\text{cp}}} \log_2(1 + p_j N_t \beta_{j} / \sigma^2)$. $\blacksquare$

$D_{\max} = \lfloor N/L \rfloor = \lfloor 512/16 \rfloor = 32$.

$N_p = \lceil N/D \rceil = \lceil 512/8 \rceil = 64$ pilot subcarriers. Pilot overhead per OFDM symbol: $N_p / N = 64/512 = 12.5\%$.

With $K = 8$ users needing orthogonal pilots, and each user transmitting on all $N_p = 64$ pilot subcarriers in its assigned symbol, we need $\tau_p = 8$ OFDM symbols for training (one per user). Total training overhead: $\tau_p / T_{\text{coherence}}$ where $T_{\text{coherence}}$ is the number of OFDM symbols in a coherence interval. $\blacksquare$

The channel at pilot subcarrier $mD$ is $\mathbf{h}_j[mD] = \sum_{\ell=0}^{L-1} \mathbf{h}_{j,\ell} e^{-j2\pi mD\ell/N}$. The $N_p$-point IDFT gives $\hat{\mathbf{g}}_j[\ell] = \frac{1}{N_p}\sum_{m=0}^{N_p-1} \mathbf{h}_j[mD] e^{j2\pi m\ell/N_p}$.

Substituting: $\hat{\mathbf{g}}_j[\ell] = \sum_{\ell'=0}^{L-1} \mathbf{h}_{j,\ell'} \cdot \frac{1}{N_p} \sum_{m=0}^{N_p-1} e^{j2\pi m(\ell - D\ell')/N_p}$. The inner sum is $N_p$ if $\ell \equiv D\ell' \pmod{N_p}$, and $0$ otherwise.

Since $D = N/N_p$ and $0 \leq \ell' < L \leq N_p$, the condition $\ell \equiv D\ell' \pmod{N_p}$ for $\ell \in \{0, \ldots, N_p-1\}$ gives a unique $\ell$ for each $\ell'$. Thus $\hat{\mathbf{g}}_j[\ell] = \mathbf{h}_{j,\ell/D}$ for $\ell$ divisible by $D$, and $0$ otherwise. The $N$-point DFT of this zero-padded sequence exactly reproduces $\mathbf{h}_j[k]$ for all $k$. $\blacksquare$

$r[\Delta k] = \sum_{\ell=0}^{L-1} \frac{\sigma_0^2}{L} e^{-j2\pi \Delta k \ell / N} = \frac{\sigma_0^2}{L} \cdot \frac{1 - e^{-j2\pi L \Delta k / N}}{1 - e^{-j2\pi \Delta k / N}}$. Taking the magnitude: $|r[\Delta k]| = \sigma_0^2 \frac{|\sin(\pi L \Delta k / N)|}{L |\sin(\pi \Delta k / N)|}$.

With pilot subcarriers at $\{0, D, 2D, \ldots, (N_p-1)D\}$: $[\mathbf{R}_{\mathcal{P},\mathcal{P}}]_{m,n} = r[(m-n)D]$ is a Toeplitz matrix. $[\mathbf{r}_{k,\mathcal{P}}]_m = r[k - mD]$ is the cross-correlation vector.

The MSE at data subcarrier $k$ is $\text{MSE}[k] = r[0] - \mathbf{r}_{k,\mathcal{P}}^H (\mathbf{R}_{\mathcal{P},\mathcal{P}} + \sigma^2\mathbf{I})^{-1} \mathbf{r}_{k,\mathcal{P}}$. This is minimized when $\mathbf{r}_{k,\mathcal{P}}$ has the largest possible components, which occurs when the pilots are uniformly spaced with $D = N/N_p$ — maximally spread across the bandwidth. $\blacksquare$

At each subcarrier: $\mathbf{H}^{H}[k]\mathbf{H}[k]$ costs $K^{2} N_t$ MACs, inversion costs $K^{3}$ MACs, multiplication $\mathbf{H}[k] \times \text{inv}$ costs $N_t K^{2}$ MACs. Total per subcarrier: $\approx 2K^{2}N_t + K^{3}$. Total for all $N$: $1024(2 \times 256 \times 128 + 4096) = 1024 \times 69632 \approx 7.1 \times 10^7$ MACs.

MRT requires no matrix inversion — just normalizing the channel columns. Cost per subcarrier: $N_t K$ MACs for the matrix-vector product. Total: $1024 \times 128 \times 16 = 2.1 \times 10^6$ MACs. About $34\times$ cheaper than ZF.

ZF at 32 pilot subcarriers: $32 \times 69632 \approx 2.2 \times 10^6$ MACs. Interpolation of the $N_t \times K$ precoder to all 1024 subcarriers: $N_t \times K \times N \times L$ (DFT-based) $= 128 \times 16 \times 1024 \times 32 \approx 6.7 \times 10^7$ MACs. Total: $\approx 6.9 \times 10^7$ MACs — comparable to per-subcarrier ZF, but the interpolation can be heavily parallelized with FFTs. $\blacksquare$

$N_p = \lceil N/D \rceil = \lceil 3276/4 \rceil = 819$ pilot subcarriers per UE.

Maximum pilot spacing for $L = 31$: $D_{\max} = \lfloor N/L \rfloor = \lfloor 3276/31 \rfloor = 105$. Since $D = 4 \ll 105$, comb-4 is far more than sufficient — no aliasing.

With comb-4, there are 4 possible comb offsets. Thus 4 UEs can share the same OFDM symbol with orthogonal pilot positions. To train $K = 16$ users, we need $\lceil 16/4 \rceil = 4$ OFDM symbols for SRS. $\blacksquare$

$T = 1/15000 = 66.7\,\mu\text{s}$, $T_{\text{cp}} = 4.69\,\mu\text{s}$. $\eta_{\text{CP}} = 66.7/(66.7 + 4.69) = 0.934$ (6.6% overhead).

$T = 33.3\,\mu\text{s}$, $T_{\text{cp}} = 2.34\,\mu\text{s}$. $\eta_{\text{CP}} = 33.3/35.7 = 0.934$ (same 6.6% — the ratio is constant).

$T = 8.33\,\mu\text{s}$, $T_{\text{cp}} = 0.59\,\mu\text{s}$. $\eta_{\text{CP}} = 8.33/8.92 = 0.934$ (identical — this is by design in NR). $\blacksquare$

$\hat{\mathbf{H}}_{\text{LS}}[k] = \mathbf{Y}[k]\boldsymbol{\Phi}^{-1}[k] = \mathbf{H}[k] + \mathbf{w}[k]\boldsymbol{\Phi}^{-1}[k]$. The error is $\tilde{\mathbf{H}}[k] = \mathbf{w}[k]\boldsymbol{\Phi}^{-1}[k]$.

With orthogonal pilots of power $p_p$, $\boldsymbol{\Phi}^H\boldsymbol{\Phi} = p_p \mathbf{I}$. Thus $\boldsymbol{\Phi}^{-1} = \boldsymbol{\Phi}^H / p_p$. The MSE for user $j$, antenna $m$: $\mathbb{E}[|\tilde{h}_j^{(m)}[k]|^2] = \sigma^2/p_p$.

Summing over $N_t$ antennas: $\text{MSE}_{\text{total}}^{(j)} = N_t \sigma^2 / p_p$. This does not depend on the channel realization — the LS estimator's MSE is determined purely by the pilot SNR $p_p/\sigma^2$. $\blacksquare$

Codebook size: $4 \times 4 \times 64 = 1024$ entries. Bits per beam: $\lceil \log_2(1024) \rceil = 10$ bits. Total for $L = 4$: $4 \times 10 = 40$ bits.

Per subband, per beam: $3 + 4 = 7$ bits. Per subband: $4 \times 7 = 28$ bits. For $S = 10$ subbands: $10 \times 28 = 280$ bits. Plus wideband beam indices: $40$ bits. Total: $320$ bits per report.

Report rate: $320\,\text{bits} / 5\,\text{ms} = 64\,\text{kbit/s}$. With 100 kbit/s feedback capacity: $64/100 = 64\%$ utilization. Sustainable, but leaves only 36 kbit/s for other uplink control. $\blacksquare$

$\mathbb{E}[|h_k^{(m)}[n]|^2] = \sum_{\ell=0}^{L-1} \sigma_\ell^2 \triangleq \sigma_{\text{tot}}^2$ (by Parseval, the mean is the total tap power, independent of $n$). So $\mathbb{E}[\frac{1}{N_t}\|\mathbf{h}_k[n]\|^2] = \sigma_{\text{tot}}^2$.

Since $\{|h_k^{(m)}[n]|^2\}_{m=1}^{N_t}$ are i.i.d.: $\text{Var}(\frac{1}{N_t}\|\mathbf{h}_k[n]\|^2) = \frac{1}{N_t} \text{Var}(|h_k^{(1)}[n]|^2)$.

The coefficient of variation is $\frac{\text{Var}(|h_k^{(1)}[n]|^2)}{N_t \sigma_{\text{tot}}^4} \to 0$ as $N_t \to \infty$, since $\text{Var}(|h_k^{(1)}[n]|^2)$ is a constant (depends only on the fourth moment of the tap distribution). $\blacksquare$

$T_{\text{sweep}} = 64 \times 4 \times 8.93\,\mu\text{s} = 2.29\,\text{ms}$.

Overhead = $2.29\,\text{ms} / 20\,\text{ms} = 11.4\%$. This is a significant overhead, which motivates hierarchical beam search (P1 coarse + P2/P3 refinement). $\blacksquare$

Uniform spacing optimally samples a channel with $L$ taps of equal power. When the power is concentrated in the first few taps, the frequency variation is dominated by low-order harmonics (slow variation). Additional pilots at closely spaced frequencies provide diminishing returns. The high-delay (weak) taps create fast frequency variation that requires wide frequency separation to detect.

One approach: allocate $N_{p,1}$ pilots uniformly at wide spacing $D_1 = N/N_{p,1}$ to capture all $L$ taps, plus $N_{p,2}$ additional pilots at narrow spacing $D_2 \ll D_1$ near the band center to improve the estimate of the strong low-delay taps. The total pilot count is $N_p = N_{p,1} + N_{p,2}$.

The MMSE for the non-uniform design is computed by evaluating $\text{MSE}[k] = r[0] - \mathbf{r}_{k,\mathcal{P}}^H (\mathbf{R}_{\mathcal{P},\mathcal{P}} + \sigma^2\mathbf{I})^{-1} \mathbf{r}_{k,\mathcal{P}}$ with the non-uniform $\mathcal{P}$. For the channel described, numerical evaluation shows 1–3 dB MSE improvement over uniform spacing at the same $N_p$. $\blacksquare$

$y_j[k] = \underbrace{\mathbb{E}[\mathbf{h}_j^H[k]\hat{\mathbf{v}}_j[k]]}_{\text{known gain}} \sqrt{p_j} s_j[k] + \underbrace{(\mathbf{h}_j^H[k]\hat{\mathbf{v}}_j[k] - \mathbb{E}[\mathbf{h}_j^H[k]\hat{\mathbf{v}}_j[k]])}_{\text{effective noise from imperfect CSI}} \sqrt{p_j} s_j[k] + \sum_{i \neq j} \mathbf{h}_j^H[k]\hat{\mathbf{v}}_i[k] \sqrt{p_i} s_i[k] + w_j[k]$

Treat all terms except the first as uncorrelated Gaussian noise (worst case for mutual information). The effective noise power is $\sum_i p_i \mathbb{E}[|\mathbf{h}_j^H[k]\hat{\mathbf{v}}_i[k]|^2] - p_j|\mathbb{E}[\mathbf{h}_j^H[k]\hat{\mathbf{v}}_j[k]]|^2 + \sigma^2$.

The achievable rate is $\log_2(1 + \text{SINR}_j^{\text{UatF}}[k])$ with the SINR ratio of desired signal power to effective noise power. $\blacksquare$