Exercises

$E_x = \int_0^\infty e^{-4t}\,dt = 1/4 < \infty$. Energy signal.

$P_x = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T 9\cos^2(10\pi t)\,dt = 9/2$. Infinite energy but finite power. Power signal.

$E_x = \int_0^\infty t^2\,dt = \infty$. $P_x = \lim_{T\to\infty}\frac{1}{2T}\int_0^T t^2\,dt = \lim_{T\to\infty} T^2/6 = \infty$. Neither.

$E_x = \int_{-\infty}^\infty e^{-2|t|}\,dt = 2\int_0^\infty e^{-2t}\,dt = 1$. Energy signal. $\blacksquare$

Both $x(\tau)$ and $h(t-\tau)$ have no overlap. $y(t) = 0$.

$y(t) = \int_0^t e^{-(t-\tau)}\,d\tau = 1 - e^{-t}$.

$y(t) = \int_0^2 e^{-(t-\tau)}\,d\tau = e^{-t}(e^2 - 1) = (1-e^{-2})e^{-(t-2)}$.

$y(t) = \begin{cases} 0 & t < 0 \\ 1 - e^{-t} & 0 \le t < 2 \\ (1-e^{-2})e^{-(t-2)} & t \ge 2 \end{cases}$$$\blacksquare$

$H_1(f) = \frac{1}{1 + j2\pi f}$, $H_2(f) = \frac{1}{2 + j2\pi f}$.

$H(f) = H_1(f) H_2(f) = \frac{1}{(1+j2\pi f)(2+j2\pi f)} = \frac{1}{1+j2\pi f} - \frac{1}{2+j2\pi f}$ (partial fractions).

Inverse transforming: $h(t) = (e^{-t} - e^{-2t})u(t)$. $\blacksquare$

$\mathcal{F}\{\Lambda(t)\} = \mathcal{F}\{\mathrm{rect}(t)\}^2 = \mathrm{sinc}^2(f)$.

Verification: $\Lambda(0) = 1$ and $\int \mathrm{sinc}^2(f)\,df = 1$. $\checkmark$ $\blacksquare$

Since $f_0 = 150 > f_s/2 = 100$, aliasing occurs.

$f_{\text{alias}} = f_s - f_0 = 200 - 150 = 50$ Hz.

The reconstructed signal looks like $\cos(2\pi \cdot 50 t)$ — a 50 Hz sinusoid instead of the original 150 Hz. $\blacksquare$

$y(t) = \frac{1}{2}x(t)e^{j2\pi f_c t} + \frac{1}{2}x(t)e^{-j2\pi f_c t}$.

$Y(f) = \frac{1}{2}X(f - f_c) + \frac{1}{2}X(f + f_c) = \frac{1}{2}\mathrm{rect}\!\left(\frac{f - f_c}{2W}\right) + \frac{1}{2}\mathrm{rect}\!\left(\frac{f + f_c}{2W}\right)$.

The baseband signal is shifted to two sidebands centred at $\pm f_c$, each with amplitude $1/2$. $\blacksquare$

$E_x = \int_{-\infty}^{\infty} e^{-2a|t|}\,dt = 2\int_0^\infty e^{-2at}\,dt = \frac{1}{a}$.

$E_x = \int_{-\infty}^{\infty} \frac{4a^2}{(a^2 + 4\pi^2 f^2)^2}\,df$.

Using $\int_{-\infty}^{\infty}\frac{d\xi}{(1+\xi^2)^2} = \frac{\pi}{2}$ with substitution $\xi = 2\pi f/a$:

$E_x = \frac{4a^2}{(2\pi/a)} \cdot \frac{\pi}{2} \cdot \frac{1}{a^2} = \frac{1}{a}$. $\checkmark$ $\blacksquare$

$x[n] = \cos(2\pi n/8) = \frac{1}{2}e^{j2\pi n/8} + \frac{1}{2}e^{-j2\pi n/8}$.

The second term equals $e^{j2\pi \cdot 7n/8}$, so this is a sum of DFT basis functions at $k = 1$ and $k = 7$.

$X[1] = 4$, $X[7] = 4$, and $X[k] = 0$ for all other $k$.

(The factor $N/2 = 4$ comes from the orthogonality of the DFT basis.) $\blacksquare$

$x_I = 2$, $x_Q = 3$.

$\tilde{x} = x_I + jx_Q = 2 + j3$.

$|\tilde{x}| = \sqrt{4 + 9} = \sqrt{13} \approx 3.606$.

$\angle\tilde{x} = \arctan(3/2) \approx 56.3°$. $\blacksquare$

$v = 100/3.6 = 27.78$ m/s.

$f_D = \frac{v}{c} f_c = \frac{27.78}{3 \times 10^8} \times 1.8 \times 10^9 = 166.7$ Hz.

Since the car moves toward the base station, the received frequency increases by $f_D$: $f_{\text{received}} = f_c + f_D = 1800.000167$ MHz. $\blacksquare$

$v = 300/3.6 = 83.33$ m/s.

$f_D = (83.33/3\times10^8) \times 3.5\times10^9 = 972$ Hz.

$T_c \approx 1/972 = 1.03$ ms.

$T_{\text{sym}}/T_c = 35.7\times10^{-6}/1.03\times10^{-3} = 0.035 \ll 1$.

The ratio is small, so the quasi-static assumption is still valid even at 300 km/h. However, the channel changes significantly over a subframe (14 symbols $\approx 0.5$ ms), requiring frequent channel estimation. $\blacksquare$

$S_X(f) = 4 \cdot \frac{2 \cdot 2}{4 + (2\pi f)^2} = \frac{16}{4 + 4\pi^2 f^2}$.

$P_X = R_X(0) = 4$ W.

Verification: $\int_{-\infty}^{\infty} \frac{16}{4 + 4\pi^2 f^2}\,df = \frac{16}{2\pi} \cdot \frac{\pi}{1} = 4$. $\checkmark$ $\blacksquare$

$|H(f)|^2 = \frac{1}{1 + (2\pi f RC)^2}$.

$P_n = \frac{N_0}{2}\int_{-\infty}^{\infty}\frac{df}{1 + (2\pi f RC)^2}$.

Let $u = 2\pi f RC$: $df = du/(2\pi RC)$.

$P_n = \frac{N_0}{2} \cdot \frac{1}{2\pi RC} \cdot \pi = \frac{N_0}{4RC}$.

Equivalently, the noise bandwidth is $B_n = 1/(4RC)$ and $P_n = N_0 B_n$. $\blacksquare$

$|x(t)|^2 = e^{-2\pi t^2}$ is a Gaussian with variance $\sigma_t^2 = 1/(4\pi)$, so $\Delta t = 1/(2\sqrt{\pi})$.

$|X(f)|^2 = e^{-2\pi f^2}$, so $\Delta f = 1/(2\sqrt{\pi})$.

$\Delta t \cdot \Delta f = \frac{1}{4\pi}$.

This equals the lower bound, confirming the Gaussian achieves equality. $\blacksquare$

$T_s = 1/f_s = 1/1200$ s.

$x(t) = \sum_n x[n]\,\mathrm{sinc}\!\left(\frac{t - nT_s}{T_s}\right)$.

Since $f_s > 2W = 1000$, this recovers $x(t)$ perfectly.

Zeros of $\mathrm{sinc}(t/T_s)$: $t = kT_s = k/1200$ for $k = \pm 1, \pm 2, \pm 3, \ldots$

First three positive zeros: $t = 1/1200, 2/1200, 3/1200$ s $\approx 0.833, 1.667, 2.500$ ms. $\blacksquare$

$\tilde{H}(f) = 2H(f + f_c)\big|_{f \in [-W, W]} = 2\,\mathrm{rect}(f/(2W))$.

$\tilde{h}(t) = \mathcal{F}^{-1}\{2\,\mathrm{rect}(f/(2W))\} = 4W\,\mathrm{sinc}(2Wt)$. $\blacksquare$

$P_1 = 1$, $P_2 = 0.5$, $P_3 = 0.25$. Total: $P = 1.75$.

$\bar{\tau} = (0 + 0.5 + 0.75)/1.75 = 0.714\;\mu$s.

$\overline{\tau^2} = (0 + 0.5 + 2.25)/1.75 = 1.571\;\mu\text{s}^2$.

$\sigma_\tau = \sqrt{1.571 - 0.714^2} = \sqrt{1.061} = 1.03\;\mu$s.

$B_c \approx 1/\sigma_\tau \approx 970$ kHz.

Signals with bandwidth $\ll 970$ kHz experience flat fading; wider-band signals experience frequency-selective fading. $\blacksquare$

$h(t) = s(T - t) = \frac{1}{2}\left[1 - \cos\!\left(\frac{2\pi(T-t)}{T}\right)\right] = \frac{1}{2}\left[1 - \cos\!\left(\frac{2\pi t}{T}\right)\right] = s(t)$.

The raised-cosine pulse is symmetric, so the matched filter equals the signal itself.

$E_s = \int_0^T \frac{1}{4}\left[1 - \cos\!\left(\frac{2\pi t}{T}\right)\right]^2 dt$.

Expanding: $[1-\cos\theta]^2 = 1 - 2\cos\theta + \cos^2\theta = 3/2 - 2\cos\theta + \cos(2\theta)/2$.

$E_s = \frac{1}{4} \cdot \frac{3T}{2} = \frac{3T}{8}$.

$\text{SNR} = \frac{2E_s}{N_0} = \frac{3T}{4N_0}$. $\blacksquare$

Let $r(t) = s_i(t) + n(t)$. Correlate with $s_1(t) = A$:

$z = \int_0^T r(t) \cdot A\,dt = A \int_0^T s_i(t)\,dt + A\int_0^T n(t)\,dt$.

If $s_1$ sent: $\mathbb{E}[z] = A \cdot AT = A^2T$. If $s_2$ sent: $\mathbb{E}[z] = A \cdot (-A)T = -A^2T$.

Noise component: $n_z = A\int_0^T n(t)\,dt$. $\mathrm{Var}(n_z) = A^2 \cdot (N_0/2) \cdot T = A^2 T N_0/2$.

The decision variable under the two hypotheses is Gaussian with means $\pm A^2T$ and equal variance. The distance is $2A^2T$:

$P_e = Q\!\left(\frac{A^2T}{\sqrt{A^2TN_0/2}}\right) = Q\!\left(\sqrt{\frac{2A^2T}{N_0}}\right) = Q\!\left(\sqrt{\frac{2E_b}{N_0}}\right)$

where $E_b = A^2T$ is the energy per bit. $\blacksquare$

$X(f) = \frac{1}{1+j2\pi f} \cdot \frac{1}{2+j2\pi f}$ (product of individual transforms).

$\tilde{H}_c(f) = H_c(f + f_c)$ for $|f| \ll f_c$. For narrowband signals, $|f+f_c| \approx f_c + f$:

$\tilde{H}_c(f) \approx e^{-\alpha(f_c + f)} e^{-j\beta(f + f_c)} = e^{-\alpha f_c} e^{-\alpha f} e^{-j\beta f_c} e^{-j\beta f}$.

The constant factors are an overall attenuation and phase; the frequency-dependent part introduces slight distortion and delay.

The received baseband signal is $\tilde{S}(f) = \tilde{X}(f)\tilde{H}_c(f)$. The matched filter has $H_{\text{MF}}(f) = \tilde{S}^*(f) e^{-j2\pi fT}$.

$\text{SNR} = \frac{2E_s}{N_0}$ where $E_s = \int |\tilde{S}(f)|^2\,df = \int |X(f)|^2 |H_c(f+f_c)|^2\,df$.

Since $|H_c|$ introduces negligible attenuation for $|f| \ll f_c$:

$E_x = \int_0^\infty (e^{-t} - e^{-2t})^2\,dt = 1/2 - 2/3 + 1/4 = 1/12$.

$\text{SNR} = \frac{2 \times (1/12)}{2\times10^{-10}} \approx 8.33\times10^8$ (89.2 dB). $\blacksquare$