NCERT Exemplar · Class 12 Physics Communication Systems

(a), (b), (d) — a $\displaystyle 15$ kHz signal needs \(\displaystyle l\approx\lambda/4\) (so (c) is wrong), lies below the sky-wave band, and a shorter antenna radiates very little power.\[\lambda=\frac{c}{f}=\frac{3\times10^{8}\ \mathrm{m\,s^{-1}}}{15\times10^{3}\ \mathrm{Hz}}=2\times10^{4}\ \mathrm{m}=20\ \mathrm{km} \]\[l\approx\frac{\lambda}{4}=\frac{20\ \mathrm{km}}{4}=5\ \mathrm{km} \]\[P_{\mathrm{rad}}\propto\left(\frac{l}{\lambda}\right)^{2}:\qquad l<5\ \mathrm{km}\Rightarrow P_{\mathrm{rad}}\downarrow \]\[f=15\ \mathrm{kHz}<3\ \mathrm{MHz}\quad(\text{ionosphere reflects }3\text{–}30\ \mathrm{MHz})\Rightarrow\text{no sky wave} \]

(b), (d) — the side bands lie at \(\displaystyle f_c\pm f_m\) and the bandwidth is \(\displaystyle 2f_m\).\[f_m=3\ \mathrm{kHz},\qquad f_c=1.5\ \mathrm{MHz}=1500\ \mathrm{kHz} \]\[f_c\pm f_m=(1500\pm3)\ \mathrm{kHz}=1503\ \mathrm{kHz},\ 1497\ \mathrm{kHz} \]\[\text{Bandwidth}=2f_m=2\times3\ \mathrm{kHz}=6\ \mathrm{kHz} \]

(b), (c), (d) — $\displaystyle 55$ km is the maximum LOS range, so every distance up to it is covered; $\displaystyle 100$ km lies beyond.\[d^{2}=(R+h)^{2}-R^{2}=2Rh+h^{2}\approx2Rh\qquad(h\ll R) \]\[d=\sqrt{2Rh}=\sqrt{2\times6.4\times10^{6}\ \mathrm{m}\times240\ \mathrm{m}}=\sqrt{3.07\times10^{9}}\ \mathrm{m}\approx5.5\times10^{4}\ \mathrm{m}\approx55\ \mathrm{km} \]\[24\ \mathrm{km},\ 50\ \mathrm{km},\ 55\ \mathrm{km}\le d\ (\text{received});\qquad100\ \mathrm{km}>d\ (\text{not received}) \]

(a), (b), (c) — the AM wave occupies \(\displaystyle \omega_c-\omega_m\) to \(\displaystyle \omega_c+\omega_m\), so a band-pass response is needed: (iii), or (i) and (ii) in cascade in either order. (iv) rejects this band.\[\text{AM wave}:\ \omega_c,\ \ \omega_c\pm\omega_m \]\[\text{(i) passes }\omega<\omega_c+\omega_m,\qquad\text{(ii) passes }\omega>\omega_c-\omega_m \]\[G(\omega)=G_{\mathrm{i}}(\omega)\,G_{\mathrm{ii}}(\omega)=G_{\mathrm{ii}}(\omega)\,G_{\mathrm{i}}(\omega)\ \Rightarrow\ \omega_c-\omega_m<\omega<\omega_c+\omega_m\ \equiv\ \text{(iii)} \]

(b), (d) — for \(\displaystyle m>1\) the message amplitude exceeds the carrier's, the envelope is distorted, and the extra side bands crowd and overlap, so information is lost.\[m=\frac{A_m}{A_c},\qquad y=(A_c+A_m\sin\omega_m t)\sin\omega_c t \]\[m\le1\Rightarrow A_c+A_m\sin\omega_m t\ge0\Rightarrow\text{envelope}=\text{message} \]\[m>1\Rightarrow A_m>A_c\Rightarrow A_c+A_m\sin\omega_m t<0\ \text{for part of each cycle} \]\[\text{envelope}\neq\text{message}\Rightarrow\text{distortion}\Rightarrow\text{extra side frequencies }\omega_c\pm n\omega_m\ (n=2,3,\dots)\Rightarrow\text{side bands overlap} \]