NCERT Exemplar · Class 12 Physics Communication Systems

(i) \[x=\frac{\ln 4}{\alpha}\;\Rightarrow\; I=I_0\,e^{-\alpha x}=I_0\,e^{-\ln 4}=\frac{I_0}{4} \] \[\text{fall}=\frac{I_0-I_0/4}{I_0}\times100\,\%=75\,\% \] (ii) \[\frac{I}{I_0}=1-0.50=0.50 \] \[\mathrm{dB}=10\log_{10}(0.50)=-3.01\ \mathrm{dB}\quad\text{over }50\ \mathrm{km} \] \[\text{attenuation}=\frac{3.01\ \mathrm{dB}}{50\ \mathrm{km}}\approx 0.06\ \mathrm{dB\,km^{-1}} \] Answer: $\displaystyle 75$ % lost at \(\displaystyle x=\ln 4/\alpha\); attenuation \(\displaystyle \approx 0.06\ \mathrm{dB\,km^{-1}}\) (the minus sign only marks a loss).

The wave goes up to the satellite and comes down: the path is two slant lengths. \[2s=ct=(3\times10^{8}\ \mathrm{m\,s^{-1}})(4.04\times10^{-3}\ \mathrm{s})=1212\ \mathrm{km}\;\Rightarrow\; s=606\ \mathrm{km} \] Caution: the right triangle needs \(\displaystyle s=606\ \mathrm{km}\), not the height alone. \[x=\sqrt{s^{2}-h^{2}}=\sqrt{606^{2}-600^{2}}\ \mathrm{km}=\sqrt{7236}\ \mathrm{km}\approx85\ \mathrm{km} \] \[D=2x\approx170\ \mathrm{km} \] LOS antennas \[\lambda=\frac{c}{\nu}=\frac{3\times10^{8}}{50\times10^{6}}=6\ \mathrm{m}\;\Rightarrow\;\ell\gtrsim\frac{\lambda}{4}=1.5\ \mathrm{m} \] \[D=\sqrt{2Rh_T}+\sqrt{2Rh_R}\ \overset{h_T=h_R=h}{=}\ 2\sqrt{2Rh}\;\Rightarrow\;h=\frac{D^{2}}{8R} \] \[h=\frac{(170\ \mathrm{km})^{2}}{8\times6400\ \mathrm{km}}\approx0.56\ \mathrm{km} \] Answer: \(\displaystyle D\approx170\ \mathrm{km}\); LOS needs antennas \(\displaystyle \gtrsim1.5\ \mathrm{m}\) long, each on a tower about $\displaystyle 560$ m high.

From the envelope: largest peak-to-peak \(\displaystyle =100\ \mathrm V\), smallest peak-to-peak \(\displaystyle =20\ \mathrm V\). \[A_c+A_m=\frac{100\ \mathrm V}{2}=50\ \mathrm V \] \[A_c-A_m=\frac{20\ \mathrm V}{2}=10\ \mathrm V \] \[A_c=30\ \mathrm V,\qquad A_m=20\ \mathrm V \] \[\mu=\frac{A_m}{A_c}=\frac{20\ \mathrm V}{30\ \mathrm V}=0.67\;\Rightarrow\;66.7\,\% \] Answer: (i) $\displaystyle 66.7$ % (ii) \(\displaystyle A_c=30\ \mathrm V\) (iii) \(\displaystyle A_m=20\ \mathrm V\)

(i) Add both signals to the carrier and pass them through a square-law device, \(\displaystyle y=Bx+Cx^{2}\): \[x=A_1\sin\omega_1t+A_2\sin\omega_2t+A_c\sin\omega_ct \] Using \(\displaystyle \sin^{2}P=\tfrac12(1-\cos2P)\) and \(\displaystyle 2\sin P\sin Q=\cos(P-Q)-\cos(P+Q)\): \[y=B\,(A_1\sin\omega_1t+A_2\sin\omega_2t+A_c\sin\omega_ct)+\tfrac{C}{2}(A_1^{2}+A_2^{2}+A_c^{2})-\tfrac{C}{2}(A_1^{2}\cos2\omega_1t+A_2^{2}\cos2\omega_2t+A_c^{2}\cos2\omega_ct) \] \[+\,CA_1A_2\,[\cos(\omega_2-\omega_1)t-\cos(\omega_2+\omega_1)t] \] \[+\,CA_cA_1\,[\cos(\omega_c-\omega_1)t-\cos(\omega_c+\omega_1)t]+CA_cA_2\,[\cos(\omega_c-\omega_2)t-\cos(\omega_c+\omega_2)t] \] \[\underbrace{0,\ \omega_1,\ \omega_2,\ \omega_2-\omega_1,\ \omega_1+\omega_2,\ 2\omega_1,\ 2\omega_2}_{\text{low group}}\ ;\qquad \underbrace{\omega_c,\ \omega_c\pm\omega_1,\ \omega_c\pm\omega_2}_{\text{carrier group}},\ 2\omega_c \] (ii) No. The pairs \(\displaystyle \omega_c\pm\omega_1\) and \(\displaystyle \omega_c\pm\omega_2\) are symmetric about \(\displaystyle \omega_c\), but \(\displaystyle \omega_1,\ \omega_2,\ \omega_2\pm\omega_1,\ 2\omega_1,\ 2\omega_2\) have no counterpart above \(\displaystyle \omega_c\): \[\omega<\omega_c:\ 0,\ \omega_1,\ \omega_2,\ \omega_2\pm\omega_1,\ 2\omega_1,\ 2\omega_2,\ \omega_c-\omega_2,\ \omega_c-\omega_1\quad(9\ \text{lines}) \] \[\omega>\omega_c:\ \omega_c+\omega_1,\ \omega_c+\omega_2,\ 2\omega_c\quad(3\ \text{lines}) \] The region \(\displaystyle \omega<\omega_c\) is crowded. (iii) More signals \(\displaystyle \Rightarrow\) more cross-terms: \[n\ \text{signals}\ \Rightarrow\ 2\binom{n}{2}=n(n-1)\ \text{sum and difference lines below}\ \omega_c \;\Rightarrow\; \text{overlap, distortion} \] (iv) Keep the low group clear of the carrier group, then filter around \(\displaystyle \omega_c\): \[2\omega_2<\omega_c-\omega_2\;\Rightarrow\;\omega_c>3\omega_2 \] \[\Delta\omega=(\omega_c+\omega_2)-(\omega_c-\omega_2)=2\omega_2 \] With \(\displaystyle \omega_c\gg\omega_2\), one channel (one carrier's band, carrying both \(\displaystyle \omega_1\) and \(\displaystyle \omega_2\)) occupies only \(\displaystyle 2\omega_2\ll\omega_c\), so further channels get their own carriers at least \(\displaystyle 2\omega_2\) apart. This room for many channels at high frequency is the bandwidth advantage. Answer: Not symmetric about \(\displaystyle \omega_c\): the low group crowds \(\displaystyle \omega<\omega_c\); more signals add overlapping lines; use \(\displaystyle \omega_c>3\omega_2\) (in practice \(\displaystyle \omega_c\gg\omega_2\)), filter, and space carriers at least \(\displaystyle 2\omega_2\) apart.

\[\mathrm{BW}=2f_m=3\ \mathrm{kHz}\;\Rightarrow\;f_m=1.5\ \mathrm{kHz} \] Caution: \(\displaystyle f_m\) is half the band width. \[\frac1{f_c}=\frac{1}{20\times10^{6}\ \mathrm{Hz}}=0.05\ \mu\mathrm s,\qquad \frac1{f_m}=\frac{1}{1.5\times10^{3}\ \mathrm{Hz}}\approx667\ \mu\mathrm s \] \[\text{need}\quad \frac1{f_c}\ll RC\ll\frac1{f_m}:\qquad 0.05\ \mu\mathrm s\ll RC\ll667\ \mu\mathrm s \] (i) \[RC=(10^{3}\ \Omega)(10^{-8}\ \mathrm F)=10\ \mu\mathrm s \] Inside the range: yes. (ii) \[RC=(10^{4}\ \Omega)(10^{-8}\ \mathrm F)=100\ \mu\mathrm s \] Inside the range: yes, with less margin than (i). (iii) \[RC=(10^{4}\ \Omega)(10^{-7}\ \mathrm F)=1000\ \mu\mathrm s>667\ \mu\mathrm s \] Capacitor discharges too slowly to follow the envelope: no. Answer: (i) and (ii) can demodulate; (iii) cannot.