Exercises

$T_{\text{sym}} = \frac{1}{\Delta f} = \frac{1}{15 \times 10^3} = 66.67\;\mu\text{s}$$

$B = N \cdot \Delta f = 1024 \times 15\;\text{kHz} = 15.36\;\text{MHz}$$

$B_c \approx \frac{1}{\tau_{\max}} = \frac{1}{5 \times 10^{-6}} = 200\;\text{kHz}$$Since$\Delta f = 15;\text{kHz} \ll B_c = 200;\text{kHz}$, each subcarrier experiences flat fading. The total bandwidth$B = 15.36;\text{MHz} \gg B_c$, confirming the channel is frequency-selective over the full band — precisely the scenario OFDM is designed for.$\blacksquare$

$I_{km} = \frac{1}{T_{\text{sym}}} \int_0^{T_{\text{sym}}} e^{j2\pi(k-m)\Delta f\, t}\, dtKATEXPLACEHOLDER0ENDI_{km} = \frac{1}{T_{\text{sym}}} \int_0^{T_{\text{sym}}} e^{j2\pi p\, t/T_{\text{sym}}}\, dt$$

$I_{kk} = \frac{1}{T_{\text{sym}}} \int_0^{T_{\text{sym}}} 1\, dt = 1$$

$I_{km} = \frac{1}{T_{\text{sym}}} \left[\frac{e^{j2\pi p\, t/T_{\text{sym}}}}{j2\pi p/T_{\text{sym}}}\right]_0^{T_{\text{sym}}} = \frac{e^{j2\pi p} - 1}{j2\pi p} = \frac{1 - 1}{j2\pi p} = 0$$Therefore$I_{km} = \delta_{km}$, confirming orthogonality.$\blacksquare$

$|x[n]| = \left|\frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} X[k]\, e^{j2\pi kn/N}\right| \leq \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} |X[k]| = \frac{N\sqrt{E_s}}{\sqrt{N}} = \sqrt{NE_s}$$So$|x[n]|^2 \leq NE_s$.

By Parseval's theorem for the DFT:

$$\frac{1}{N}\sum_{n=0}^{N-1} |x[n]|^2 = \frac{1}{N}\sum_{k=0}^{N-1} |X[k]|^2 = E_s$$

$\text{PAPR} = \frac{\max_n |x[n]|^2}{\frac{1}{N}\sum_n |x[n]|^2} \leq \frac{NE_s}{E_s} = N$$The bound is achieved when$X[k] = \sqrt{E_s}$for all$k$, giving$x[0] = \sqrt{NE_s}$and$x[n] = 0$for$n \neq 0$.$\blacksquare$

We need $N_{\text{cp}} \geq L - 1 = 10 - 1 = 9$. Since $N_{\text{cp}} = 16 \geq 9$, the CP is sufficient. There is a margin of $16 - 9 = 7$ samples for timing uncertainty.

$\eta_{\text{CP}} = \frac{N_{\text{cp}}}{N + N_{\text{cp}}} = \frac{16}{64 + 16} = \frac{16}{80} = 20\%$$

$T_{\text{sym}} = \frac{1}{\Delta f} = \frac{1}{312500} = 3.2\;\mu\text{s}KATEXPLACEHOLDER0ENDT_{\text{cp}} = \frac{N_{\text{cp}}}{N} \cdot T_{\text{sym}} = \frac{16}{64} \times 3.2 = 0.8\;\mu\text{s}KATEXPLACEHOLDER1ENDT_{\text{total}} = T_{\text{sym}} + T_{\text{cp}} = 3.2 + 0.8 = 4.0\;\mu\text{s}$$This matches the IEEE 802.11a/g standard.$\blacksquare$

With $W_4 = e^{-j2\pi/4} = e^{-j\pi/2} = -j$:

$$\mathbf{F}_4 = \frac{1}{2}\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -j & -1 & j \\ 1 & -1 & 1 & -1 \\ 1 & j & -1 & -j \end{bmatrix}$$

$\mathbf{F}_4^H = \frac{1}{2}\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & j & -1 & -j \\ 1 & -1 & 1 & -1 \\ 1 & -j & -1 & j \end{bmatrix}$

Computing $\mathbf{F}_4 \mathbf{F}_4^H$: the $(i,j)$ entry is $\frac{1}{4}\sum_{n=0}^{3} W_4^{in} W_4^{-jn} = \frac{1}{4}\sum_{n=0}^{3} W_4^{(i-j)n}$.

For $i = j$: $\frac{1}{4}(1+1+1+1) = 1$.

For $i \neq j$: geometric series $\frac{1}{4} \cdot \frac{1 - W_4^{4(i-j)}}{1 - W_4^{(i-j)}} = 0$ since $W_4^4 = 1$.

Therefore $\mathbf{F}_4 \mathbf{F}_4^H = \mathbf{I}_4$. $\blacksquare$

Let $\mathbf{f}_k = \frac{1}{\sqrt{N}}[1, e^{j2\pi k/N}, \ldots, e^{j2\pi k(N-1)/N}]^T$ be the $k$-th column of $\mathbf{F}^H$. The $n$-th element of $\mathbf{C}\mathbf{f}_k$ is:

$$[\mathbf{C}\mathbf{f}_k]_n = \sum_{m=0}^{N-1} c[(n-m) \bmod N]\, \frac{1}{\sqrt{N}} e^{j2\pi km/N}$$

Substituting $l = n - m$:

$$= \frac{e^{j2\pi kn/N}}{\sqrt{N}} \sum_{l=0}^{N-1} c[l]\, e^{-j2\pi kl/N} = \frac{e^{j2\pi kn/N}}{\sqrt{N}} \cdot \lambda_k$$

where $\lambda_k = \sum_{l=0}^{N-1} c[l] e^{-j2\pi kl/N} = \sqrt{N}\, [\mathbf{F}\mathbf{c}]_k$.

Since $\mathbf{C}\mathbf{f}_k = \lambda_k \mathbf{f}_k$ for all $k$:

$$\mathbf{C} \mathbf{F}^H = \mathbf{F}^H \boldsymbol{\Lambda}$$

Multiplying by $\mathbf{F}$ from the right:

$$\mathbf{C} = \mathbf{F}^H \boldsymbol{\Lambda} \mathbf{F}$$

This is the fundamental result enabling OFDM: the channel circulant matrix is diagonalised by the DFT. $\blacksquare$

The maximum delay index is 7, so $L - 1 = 7$. We need $N_{\text{cp}} \geq 7$. Minimum $N_{\text{cp}} = 7$.

(In practice, we would choose $N_{\text{cp}} = 8$ or larger for margin.)

$\eta_{\text{CP}} = \frac{7}{256 + 7} = \frac{7}{263} = 2.66\%$$

$\eta_{\text{CP}} = \frac{7}{1024 + 7} = \frac{7}{1031} = 0.68\%$$Quadrupling$N$reduces the CP overhead by approximately a factor of 4. This illustrates the trade-off: larger$N$reduces CP overhead but increases sensitivity to Doppler spread and PAPR.$\blacksquare$

SNR $= 20$ dB means $|X_p|^2/\sigma^2 = 100$, so $\sigma^2 = 0.01$.

$$\text{MSE}_{\text{LS}} = \frac{\sigma^2}{|X_p|^2} = \frac{0.01}{1} = 0.01 = -20\;\text{dB}$$

By the sampling theorem, pilot spacing $D_f = 4$ can resolve channel responses with up to $N/D_f = 64/4 = 16$ taps.

The channel has $L = 20 > 16$ taps, so the pilot spacing is not adequate. The channel estimate will suffer from aliasing. We need $D_f \leq N/L = 64/20 = 3.2$, so $D_f \leq 3$ is required. $\blacksquare$

After channel convolution and CP removal, the received sample is:

$$y[n] = e^{j2\pi \epsilon_F n/N} \sum_{k=0}^{N-1} H[k] X[k] \frac{1}{\sqrt{N}} e^{j2\pi kn/N} + w[n]$$

The CFO introduces a phase rotation $e^{j2\pi \epsilon_F n/N}$ that varies across samples within the DFT window.

$Y[l] = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} y[n] e^{-j2\pi ln/N}KATEXPLACEHOLDER0END= \sum_{k=0}^{N-1} H[k] X[k] \underbrace{\frac{1}{N}\sum_{n=0}^{N-1} e^{j2\pi(k-l+\epsilon_F)n/N}}_{S(k-l+\epsilon_F)} + W[l]$$

$S(p) = \frac{1}{N}\sum_{n=0}^{N-1} e^{j2\pi pn/N} = \frac{1}{N} \cdot \frac{1 - e^{j2\pi p}}{1 - e^{j2\pi p/N}}KATEXPLACEHOLDER0END= \frac{\sin(\pi p)}{N\sin(\pi p/N)} \cdot e^{j\pi p(N-1)/N}$$For$\epsilon_F = 0$:$S(0) = 1$and$S(k-l) = 0$for$k \neq l$ (perfect orthogonality).

The desired signal power on subcarrier $l$ is $|S(\epsilon_F)|^2 |H[l]|^2 E_s$.

The ICI power (averaging over data and channel) is $(1 - |S(\epsilon_F)|^2) E_s \bar{|H|^2}$.

For equal average channel gains $\bar{|H|^2}$:

$$\text{SINR} \approx \frac{|S(\epsilon_F)|^2 \cdot \text{SNR}}{(1 - |S(\epsilon_F)|^2) \cdot \text{SNR} + 1}$$

$\blacksquare$

$|S(\epsilon_F)|^2 \approx 1 - \frac{\pi^2 (0.03)^2}{3} = 1 - \frac{0.00888}{3} = 1 - 0.00296 = 0.99704$$

$\text{SINR}_{\max} = \frac{|S|^2}{1 - |S|^2} = \frac{0.99704}{0.00296} = 336.8KATEXPLACEHOLDER0END\text{SINR}_{\max,\text{dB}} = 10\log_{10}(336.8) = 25.3\;\text{dB}$$

With SNR $= 10^{2.5} = 316.2$:

$$\text{SINR} = \frac{0.99704 \times 316.2}{0.00296 \times 316.2 + 1} = \frac{315.3}{1.936} = 162.8$$

$$\text{SINR}_{\text{dB}} = 10\log_{10}(162.8) = 22.1\;\text{dB}$$

Degradation: $25.0 - 22.1 = 2.9$ dB. Even this small CFO causes nearly 3 dB degradation at 25 dB SNR. $\blacksquare$

By the CLT, $x[n] \sim \mathcal{CN}(0, \sigma_x^2)$ for large $N$. Therefore $|x[n]|^2 / \sigma_x^2 \sim \text{Exp}(1)$:

$$\Pr\!\left(\frac{|x[n]|^2}{\sigma_x^2} \leq \gamma\right) = 1 - e^{-\gamma}$$

PAPR $= \max_n |x[n]|^2 / \sigma_x^2$. Assuming approximate independence of the $N$ samples:

$$\Pr(\text{PAPR} \leq \gamma) = \prod_{n=0}^{N-1} \Pr\!\left(\frac{|x[n]|^2}{\sigma_x^2} \leq \gamma\right) = (1 - e^{-\gamma})^N$$

$$\text{CCDF}(\gamma) = 1 - (1 - e^{-\gamma})^N$$

Solve $1 - (1 - e^{-\gamma})^{128} = 10^{-3}$:

$(1 - e^{-\gamma})^{128} = 0.999$

$1 - e^{-\gamma} = 0.999^{1/128} = e^{\ln(0.999)/128} = e^{-7.81 \times 10^{-6}} \approx 0.99999$

$e^{-\gamma} \approx 7.82 \times 10^{-6}$

$\gamma = -\ln(7.82 \times 10^{-6}) = 11.76$

$$\text{PAPR}_{0.1\%} = 11.76 \approx 10.7\;\text{dB}$$

$\blacksquare$

$\text{bits} = \lceil\log_2 4\rceil = 2\;\text{bits}$$

If each candidate has independent PAPR with CCDF $\text{CCDF}_1(\gamma) = 1 - (1 - e^{-\gamma})^N$, then the probability that all $U$ candidates exceed $\gamma$ is:

$$\text{CCDF}_{\text{SLM}}(\gamma) = [\text{CCDF}_1(\gamma)]^U = [1 - (1-e^{-\gamma})^N]^U$$

Without SLM ($U = 1$): solve $1 - (1-e^{-\gamma})^{64} = 10^{-3}$, giving $\gamma_1 \approx 10.05$ ($10.0$ dB).

With SLM ($U = 4$): solve $[1 - (1-e^{-\gamma})^{64}]^4 = 10^{-3}$, i.e., $1 - (1-e^{-\gamma})^{64} = 10^{-3/4} = 0.1778$.

$(1-e^{-\gamma})^{64} = 0.8222$, so $e^{-\gamma} = 1 - 0.8222^{1/64} \approx 0.003063$, $\gamma \approx 5.79$ ($7.6$ dB).

PAPR reduction $\approx 10.0 - 7.6 = 2.4$ dB. $\blacksquare$

$\eta = \frac{144}{2048 + 144} = \frac{144}{2192} = 6.57\%$$

$\eta = \frac{512}{2048 + 512} = \frac{512}{2560} = 20.0\%$$

The extended CP is designed for large delay spread environments (e.g., MBSFN) at the cost of significant overhead.

With $\Delta f = 30$ kHz (double the LTE spacing), the symbol duration is halved but the FFT size is the same:

$$\eta = \frac{144}{2048 + 144} = 6.57\%$$

The CP overhead fraction is the same, but the actual CP duration is halved to $T_{\text{cp}} \approx 2.3\;\mu$s, protecting against half the delay spread. $\blacksquare$

$H[k] = 1 + 0.5 e^{-j2\pi k/8} + 0.3 e^{-j2\pi \cdot 2k/8} = 1 + 0.5 e^{-j\pi k/4} + 0.3 e^{-j\pi k/2}$$Computing for each$k$:$H[0] = 1 + 0.5 + 0.3 = 1.8$$H[1] = 1 + 0.5 e^{-j\pi/4} + 0.3 e^{-j\pi/2} = 1 + 0.354 - j0.354 - j0.3$$\quad= 1.354 - j0.654$$H[2] = 1 + 0.5 e^{-j\pi/2} + 0.3 e^{-j\pi} = 1 - j0.5 - 0.3 = 0.7 - j0.5$$H[3] = 1 + 0.5 e^{-j3\pi/4} + 0.3 e^{-j3\pi/2} = 1 - 0.354 - j0.354 + j0.3$$\quad= 0.646 - j0.054$$H[4] = 1 - 0.5 + 0.3 = 0.8$$H[5] = 1 + 0.5 e^{-j5\pi/4} + 0.3 e^{-j5\pi/2} = 1 - 0.354 + j0.354 - j0.3$$\quad= 0.646 + j0.054$$H[6] = 1 + j0.5 - 0.3 = 0.7 + j0.5$$H[7] = 1 + 0.354 + j0.354 + j0.3 = 1.354 + j0.654$

Since $X[k] = 1$ for all $k$: $Y[k] = H[k] \cdot 1 = H[k]$.

$\hat{X}[k] = \frac{Y[k]}{H[k]} = \frac{H[k]}{H[k]} = 1$$for all$k$. The ZF equaliser perfectly recovers all data symbols in the absence of noise.$\blacksquare$

The received samples during symbol $m$ are:

$y_m[0] = h_0 x_m[0] + h_1 x_{m-1}[3]$ (ISI!)

$y_m[1] = h_0 x_m[1] + h_1 x_m[0]$

$y_m[2] = h_0 x_m[2] + h_1 x_m[1]$

$y_m[3] = h_0 x_m[3] + h_1 x_m[2]$

$\mathbf{y}_m = \underbrace{\begin{bmatrix} h_0 & 0 & 0 & 0 \\ h_1 & h_0 & 0 & 0 \\ 0 & h_1 & h_0 & 0 \\ 0 & 0 & h_1 & h_0 \end{bmatrix}}_{\mathbf{H}_{\text{lin}}} \mathbf{x}_m + \underbrace{\begin{bmatrix} 0 & 0 & 0 & h_1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}}_{\mathbf{H}_{\text{ISI}}} \mathbf{x}_{m-1}$$

$\mathbf{H}_{\text{lin}}$ is Toeplitz but not circulant: the $(0,3)$ entry is 0 instead of $h_1$. In a circulant matrix, row 0 would have $h_1$ in position $(0,3)$.

Since $\mathbf{H}_{\text{lin}}$ is not circulant, it is not diagonalised by the DFT. After applying the DFT: $\mathbf{Y}_m = \mathbf{F}\mathbf{H}_{\text{lin}}\mathbf{F}^H \mathbf{X}_m + \ldots$ and $\mathbf{F}\mathbf{H}_{\text{lin}}\mathbf{F}^H$ is not diagonal, causing ICI.

The ISI term $\mathbf{H}_{\text{ISI}}\mathbf{x}_{m-1}$ further corrupts the signal. $\blacksquare$

The pilot spacing in frequency samples the channel in the delay domain. With $D_f = 6$, the maximum resolvable delay spread (in samples) is $N/D_f$. In time:

$$\tau_{\max} = \frac{1}{D_f \cdot \Delta f} = \frac{1}{6 \times 15000} = 11.1\;\mu\text{s}$$

$f_{D,\max} = \frac{1}{2 D_t T_{\text{total}}} = \frac{1}{2 \times 4 \times 71.4 \times 10^{-6}} = \frac{1}{5.71 \times 10^{-4}} = 1751\;\text{Hz}$$

$v_{\max} = \frac{f_{D,\max} \cdot c}{f_c} = \frac{1751 \times 3 \times 10^8}{2 \times 10^9} = 262.7\;\text{m/s} \approx 945\;\text{km/h}$$This is sufficient for high-speed trains.$\blacksquare$

The "water levels" for each subcarrier are:

$\sigma^2/|H[0]|^2 = 0.1/4 = 0.025$

$\sigma^2/|H[1]|^2 = 0.1/1 = 0.1$

$\sigma^2/|H[2]|^2 = 0.1/0.25 = 0.4$

$\sigma^2/|H[3]|^2 = 0.1/0.01 = 10$

Power allocation: $P_k = (\mu - \sigma^2/|H[k]|^2)^+$

Try all 4 subcarriers active: $\sum P_k = 4\mu - (0.025 + 0.1 + 0.4 + 10) = 4$. $4\mu = 14.525$, $\mu = 3.631$. But $\mu < 10$, so subcarrier 3 gets negative power — exclude it.

Try 3 subcarriers: $\sum P_k = 3\mu - (0.025 + 0.1 + 0.4) = 4$. $3\mu = 4.525$, $\mu = 1.508$. Check: $\mu > 0.4$ (OK).

$P_0 = 1.508 - 0.025 = 1.483$, $P_1 = 1.508 - 0.1 = 1.408$, $P_2 = 1.508 - 0.4 = 1.108$, $P_3 = 0$.

Total: $1.483 + 1.408 + 1.108 + 0 = 3.999 \approx 4$. Correct.

Water-filling capacity:

$C_{\text{WF}} = \sum_{k:P_k>0} \log_2(1 + P_k |H[k]|^2/\sigma^2)$

$= \log_2(1 + 1.483 \times 4/0.1) + \log_2(1 + 1.408 \times 1/0.1) + \log_2(1 + 1.108 \times 0.25/0.1)$

$= \log_2(60.32) + \log_2(15.08) + \log_2(3.77)$

$= 5.91 + 3.91 + 1.91 = 11.74$ bits/symbol

Equal power ($P_k = 1$ for all $k$):

$C_{\text{eq}} = \log_2(1+40) + \log_2(1+10) + \log_2(1+2.5) + \log_2(1+0.1)$

$= 5.36 + 3.46 + 1.81 + 0.14 = 10.77$ bits/symbol

Water-filling gain: $11.74 - 10.77 = 0.97$ bits/symbol. $\blacksquare$

The $M = 4$ DFT outputs are mapped to 4 consecutive subcarriers, e.g., subcarriers $\{4, 5, 6, 7\}$:

Allocation: $[-, -, -, -, \mathbf{D}, \mathbf{D}, \mathbf{D}, \mathbf{D}, -, -, -, -, -, -, -, -]$

Advantage: Simpler scheduling, enables frequency-selective scheduling.

Disadvantage: No frequency diversity within the allocation.

The $M = 4$ DFT outputs are mapped to 4 evenly spaced subcarriers with spacing $N/M = 4$:

Allocation: $[\mathbf{D}, -, -, -, \mathbf{D}, -, -, -, \mathbf{D}, -, -, -, \mathbf{D}, -, -, -]$

Advantage: Full frequency diversity across the band.

Disadvantage: More complex scheduling; less flexibility for frequency-selective scheduling.

LTE uses localised SC-FDMA for the uplink (PUSCH), as frequency-selective scheduling provides larger gains in practice than distributed diversity. The term "LFDMA" and "IFDMA" (interleaved FDMA) are also used for these modes. $\blacksquare$

Constraints on $\Delta f$:

1. $\Delta f \gg f_D = 200$ Hz (to limit ICI from Doppler). Rule of thumb: $\Delta f \geq 10 f_D = 2$ kHz.

2. $\Delta f \ll B_c = 1/\tau_{\max} = 100$ kHz (flat fading per subcarrier). Rule of thumb: $\Delta f \leq B_c/10 = 10$ kHz.