Exercises

$\Delta\phi = 2\pi f_0 \tau_2 = 2k\pi$ $\Rightarrow \tau_2 = k/f_0 = k \times 0.556$ ns.

Smallest positive: $\tau_2 = 0.556$ ns (one period).

$\Delta\phi = (2k+1)\pi$ $\Rightarrow \tau_2 = (2k+1)/(2f_0) = 0.278$ ns (half period).

Max amplitude: $1 + 0.8 = 1.8$, power $= 3.24$. Min amplitude: $|1 - 0.8| = 0.2$, power $= 0.04$.

Ratio $= 10\log_{10}(3.24/0.04) = 10\log_{10}(81) = 19.1$ dB. $\blacksquare$

$\lambda = 2 \times 2.5 = 5$ cm $= 0.05$ m.

$f_0 = c/\lambda = 3 \times 10^8 / 0.05 = 6$ GHz.

This is in the 5G C-band (FR1). $\blacksquare$

$\gamma_{\text{threshold}} = \bar\gamma \times 10^{-10/10} = 1 \times 0.1 = 0.1$.

$P(\gamma < 0.1) = 1 - e^{-0.1/1} = 1 - 0.905 = 9.5$%.

$\gamma_{\text{threshold}} = 1 \times 10^{5/10} = 3.162$.

$P(\gamma > 3.162) = e^{-3.162/1} = e^{-3.162} = 4.2$%. $\blacksquare$

$K = 10$ (linear), so $A^2 = 10 \times 2\sigma^2$.

$\Omega = A^2 + 2\sigma^2 = 10 \times 2\sigma^2 + 2\sigma^2 = 11 \times 2\sigma^2 = 1$.

$2\sigma^2 = 1/11 = 0.0909$, $A^2 = 10/11 = 0.909$.

$A^2/\Omega = 10/11 = 90.9$% of total power is LOS.

New mean power $= 2\sigma^2 = 1/11 = 0.0909 = -10.4$ dBW.

Drop $= 10\log_{10}(1/0.0909) = 10\log_{10}(11) = 10.4$ dB.

Blocking the LOS causes a 10.4 dB drop — the channel reverts to Rayleigh fading with much lower mean power. $\blacksquare$

$K = 10^{6/10} = 3.981$ (linear).

$m = (3.981 + 1)^2 / (2 \times 3.981 + 1) = 24.81 / 8.962 = 2.77$.

For Nakagami-$m$ with $\Omega = 1$, the power CDF is the regularised incomplete gamma function:

$P(\gamma < x) = \frac{\gamma_{\text{inc}}(m, mx/\Omega)}{\Gamma(m)}$.

At $x = 0.1$ ($-10$ dB): $P(\gamma < 0.1) = \frac{\gamma_{\text{inc}}(2.77, 0.277)}{\Gamma(2.77)}$.

Using tables or computation: $\approx 0.8$%.

For exact Ricean: the CDF involves the Marcum Q-function, giving a similar result $\approx 0.9$%.

The Nakagami approximation is quite accurate. $\blacksquare$

$v = 80/3.6 = 22.22$ m/s. $f_D = 22.22 \times 3.5 \times 10^9 / 3 \times 10^8 = 259$ Hz.

$T_c = 0.423/259 = 1.63$ ms.

$0.5\text{ ms} / 1.63\text{ ms} = 0.31$.

Less than one coherence time fits in a slot, so the channel is approximately constant within each slot (slow fading). $\blacksquare$

$N_1 = \sqrt{2\pi} \times 100 \times 1 \times e^{-1} = 250.66 \times 0.368 = 92.2$ crossings/s.

$\rho = 0.1$: $N_{0.1} = \sqrt{2\pi} \times 100 \times 0.1 \times e^{-0.01} = 25.07 \times 0.990 = 24.8$ crossings/s.

$\bar\tau_{0.1} = \frac{1 - e^{-0.01}}{N_{0.1}} = \frac{0.00995}{24.8} = 0.401$ ms.

At 20 dB below the RMS level, the signal spends about 0.4 ms per fade event below the threshold. $\blacksquare$

$f = f_D\cos\theta$ maps angles to Doppler shifts. The density of angles mapping to a given $f$ is $|d\theta/df| = 1/\sqrt{f_D^2 - f^2}$.

Near $f = \pm f_D$: $\cos\theta \approx \pm 1$, so $\theta \approx 0, \pi$. Small angle changes produce small frequency changes — many angles pile up near $\pm f_D$.

Near $f = 0$: $\theta \approx \pi/2$. Here $\cos\theta$ changes rapidly — few angles map to $f \approx 0$.

A flat Doppler spectrum occurs when the scattering is not isotropic. For example, if scatterers are concentrated in a narrow angular range directly ahead of and behind the mobile, the spectrum would peak at a single Doppler shift rather than being U-shaped.

A truly flat spectrum requires a specific non-uniform angular distribution — the 3D isotropic case (scattering from all elevations as well) gives a flat spectrum within $[-f_D, f_D]$. $\blacksquare$

$\sum P_l = 1 + 0.5 + 0.1 = 1.6$.

$\bar\tau = (1 \times 0 + 0.5 \times 1 + 0.1 \times 3) / 1.6 = 0.8/1.6 = 0.5\,\mu$s.

$\overline{\tau^2} = (1 \times 0 + 0.5 \times 1 + 0.1 \times 9)/1.6 = 1.4/1.6 = 0.875\,\mu$s$^2$.

$\sigma_\tau = \sqrt{0.875 - 0.25} = \sqrt{0.625} = 0.791\,\mu$s.

$B_c \approx 1/(5\sigma_\tau) = 1/(5 \times 0.791 \times 10^{-6}) = 253$ kHz. $\blacksquare$

$B_c = 1/(5 \times 10^{-6}) = 200$ kHz. $W = 200$ kHz $= B_c$. Borderline — typically classified as frequency-selective since $W \not\ll B_c$.

$B_c = 1/(5 \times 50 \times 10^{-9}) = 4$ MHz. $W = 20$ MHz $\gg B_c$. Frequency-selective.

$B_c = 1/(5 \times 100 \times 10^{-9}) = 2$ MHz. $W = 5$ MHz $> B_c$. Frequency-selective. $\blacksquare$

$B_c \approx 1/(5\sigma_\tau) = 1/(5 \times 2 \times 10^{-6}) = 100$ kHz.

For flat fading: $\Delta f \ll B_c$. A practical rule is $\Delta f \leq B_c/5 = 20$ kHz, giving symbol duration $T_u = 1/\Delta f = 50\,\mu$s.

$\text{CP} \geq 4\sigma_\tau = 8\,\mu$s. Total symbol: $T_s = T_u + \text{CP} = 50 + 8 = 58\,\mu$s.

$\Delta f_{\max} \approx B_c/5 = 20$ kHz (for $\leq -20$ dB variation across a subcarrier).

$T_c = 0.423/200 = 2.12$ ms. $T_s = 58\,\mu$s $\ll T_c = 2.12$ ms.

The channel is slow fading — constant over each OFDM symbol. $\blacksquare$

$H(f; t) = \int h(\tau; t)\,e^{-j2\pi f\tau}\,d\tau$

$= e^{j2\pi f_1 t} + 0.5\,e^{j2\pi f_2 t}\,e^{-j2\pi f\tau_1}$

$= e^{j2\pi(50)t} + 0.5\,e^{j2\pi(-30)t}\,e^{-j2\pi f \times 10^{-6}}$.

$S(\tau; \nu) = \int h(\tau; t)\,e^{-j2\pi\nu t}\,dt$

$= \delta(\nu - f_1)\,\delta(\tau) + 0.5\,\delta(\nu - f_2)\,\delta(\tau - \tau_1)$

$= \delta(\nu - 50)\,\delta(\tau) + 0.5\,\delta(\nu + 30)\,\delta(\tau - 10^{-6})$.

The channel has two points in the delay-Doppler plane: $(0, 50\text{ Hz})$ and $(1\,\mu\text{s}, -30\text{ Hz})$. $\blacksquare$

$\int_{-f_D}^{f_D} C_S(\tau,\nu)\,d\nu = \frac{1}{\sigma_\tau}e^{-\tau/\sigma_\tau} \underbrace{\int_{-f_D}^{f_D}\frac{1}{\pi f_D\sqrt{1-(\nu/f_D)^2}}\,d\nu}_{= 1}$

$= \frac{1}{\sigma_\tau}e^{-\tau/\sigma_\tau} = P(\tau)$. ✓

$\int_0^\infty C_S(\tau,\nu)\,d\tau = \frac{1}{\pi f_D\sqrt{1-(\nu/f_D)^2}} \underbrace{\int_0^\infty \frac{1}{\sigma_\tau}e^{-\tau/\sigma_\tau}\,d\tau}_{= 1}$

$= S_H(\nu)$. ✓

$\int\int C_S\,d\tau\,d\nu = 1 \times 1 = 1$ (normalised). $\blacksquare$

$T_s = 1/5 \times 10^6 = 200$ ns. $\tau_{\max} = 700$ ns. $L = \lceil 700/200 \rceil + 1 = 4$.

Taps: $l = 0, 1, 2, 3$ (delays 0, 200, 400, 600 ns).

• Tap 0 ($\tau = 0$): receives path at 0 ns → power 0 dB
• Tap 1 ($\tau = 200$ ns): receives path at 200 ns → power $-3$ dB
• Tap 2 ($\tau = 400$ ns): path at 500 ns rounds to tap 2.5, split between taps 2 and 3
• Tap 3 ($\tau = 600$ ns): path at 700 ns → closest to tap 3.5

In practice, the 500 ns and 700 ns paths would be assigned to the nearest tap indices (2 and 3 respectively), with adjusted powers. All 4 taps are active. $\blacksquare$

$\bar{C} = \int_0^\infty \log_2(1+x) \cdot \frac{1}{\bar\gamma}e^{-x/\bar\gamma}\,dx$

With $\bar\gamma = 100$ (linear): $\bar{C} \approx 5.2$ bits/s/Hz.

(Exact: $\bar{C} = e^{1/\bar\gamma} E_1(1/\bar\gamma)/\ln 2 \approx 5.2$ bits/s/Hz.)

$P(\gamma < 2^R - 1) = 1 - e^{-(2^R-1)/100} = 0.01$

$e^{-(2^R-1)/100} = 0.99$

$2^R - 1 = -100\ln(0.99) = 1.005$

$R = \log_2(2.005) = 1.00$ bit/s/Hz.

The 1% outage capacity is only 1.0 bit/s/Hz, much less than the ergodic capacity of 5.2 bits/s/Hz — illustrating how fading creates a severe penalty for delay-sensitive traffic. $\blacksquare$

Actual delay $=$ normalised delay $\times r_\tau \times \sigma_\tau$, where $r_\tau$ is the delay scaling factor. For simplicity, taking the normalised delays as multiples of $\sigma_\tau$:

Tap 2 delay $= 114.6$ ns. Sample period $= 1/30.72 \times 10^6 = 32.55$ ns. Samples $= 114.6/32.55 = 3.52$ samples.

The tap would be placed between sample indices 3 and 4, requiring fractional delay interpolation. $\blacksquare$

$\lambda = 3 \times 10^8 / 28 \times 10^9 = 0.0107$ m. $v = 100/3.6 = 27.78$ m/s. $f_D = v/\lambda = 27.78/0.0107 = 2{,}596$ Hz. $T_c = 0.423/2596 = 163\,\mu$s. $B_c = 1/(5 \times 50 \times 10^{-9}) = 4$ MHz.

Full band (100 MHz): $W = 100$ MHz $\gg B_c = 4$ MHz → frequency-selective.

Single subcarrier (120 kHz): $\Delta f = 120$ kHz $\ll B_c = 4$ MHz → flat fading per subcarrier.

Single OFDM symbol: $T_s \approx 1/120000 + 0.59\,\mu$s $= 8.33 + 0.59 = 8.92\,\mu$s $\ll T_c = 163\,\mu$s → slow fading per symbol.

Maximum excess delay $\approx 4\sigma_\tau = 200$ ns. CP $= 0.59\,\mu$s $= 590$ ns $> 200$ ns.

The CP is sufficient. At mmWave, delay spreads are typically shorter than at sub-6 GHz. $\blacksquare$

$P_{\text{out}} = 1 - e^{-\gamma_0/\bar\gamma}$.

For $\bar\gamma \gg \gamma_0$: $e^{-\gamma_0/\bar\gamma} \approx 1 - \gamma_0/\bar\gamma$,

so $P_{\text{out}} \approx \gamma_0/\bar\gamma$.

$\log P_{\text{out}} \approx \log\gamma_0 - \log\bar\gamma$.

The slope of $\log P_{\text{out}}$ vs $\log\bar\gamma$ is $-1$, giving diversity order $d = 1$.

With $L$ independent diversity branches (MRC combining): $P_{\text{out}} \propto (1/\bar\gamma)^L$, so diversity order $= L$. This motivates MIMO and diversity techniques (Chapter 11). $\blacksquare$

Set $N_0 = (M-1)/2 = 3$ (rounding), $\beta_k = \pi k/4$. Frequencies: $f_k = f_D\cos(2\pi k/8)$ for $k = 1, 2, 3$.

Generate $h_I[n]$, $h_Q[n]$ using the sum-of-sinusoids formula, normalise, and form $h[n] = (h_I + jh_Q)/\sqrt{2(N_0+1)}$.

The envelope histogram should match $f_r(r) = (r/\sigma^2)e^{-r^2/(2\sigma^2)}$ with $\sigma^2 = 0.5$.

The autocorrelation $R_h[\Delta n]$ should approximate $J_0(2\pi \times 50 \times 0.0001 \times \Delta n) = J_0(0.0314\,\Delta n)$.

With only 8 oscillators, there will be some deviation from the ideal Bessel function; more oscillators improve accuracy. $\blacksquare$

Using the formula or numerical integration:

$C_{\text{total}} = B \sum_l C_l = 1 \times (5.21 + 4.43 + 3.66 + 2.65) = 15.95$ Mbits/s.

Average SNR $= 15.25$ dB $= 33.5$ (linear). Capacity at average SNR: $C = 4 \times E[\log_2(1 + \gamma)]$ with $\bar\gamma = 33.5$: $C_{\text{avg}} \approx 4 \times 3.97 = 15.88$ Mbits/s.

The difference is small ($0.07$ Mbits/s), showing that the ergodic capacity is nearly achieved by the average SNR — a consequence of Jensen's inequality being nearly tight for $\log(1+x)$ at high SNR. At low SNR the gap would be larger. $\blacksquare$