Communication Questions in English

(B) The modulation index $m$ is defined as the ratio of the peak voltage of the message signal $(V_2)$ to the peak voltage of the carrier signal $(V_1)$,so $m = \frac{V_2}{V_1}$.
In amplitude modulation,the sideband frequencies are given by $v_{+} = v_1 + v_2$ and $v_{-} = v_1 - v_2$.
Adding these two equations: $v_{+} + v_{-} = (v_1 + v_2) + (v_1 - v_2) = 2v_1$.
Therefore,the carrier frequency is $v_1 = \frac{v_{+} + v_{-}}{2}$.
Thus,both option $B$ and the definition of $m$ are relevant,but based on standard physics problems of this type,$v_1 = \frac{v_{+} + v_{-}}{2}$ is a standard identity.

(D) The range $d$ of a transmitting antenna of height $h$ is given by the formula:
$d = \sqrt{2 R_e h}$
where $R_e$ is the radius of the Earth.
Given:
$d = 64 \ km = 64 \times 10^3 \ m$
$R_e = 6.4 \times 10^6 \ m$
Squaring both sides,we get:
$d^2 = 2 R_e h$
Rearranging for $h$:
$h = \frac{d^2}{2 R_e}$
Substituting the values:
$h = \frac{(64 \times 10^3)^2}{2 \times 6.4 \times 10^6} = \frac{4096 \times 10^6}{12.8 \times 10^6} = \frac{4096}{12.8} = 320 \ m$

(D) The modulation index $m$ is defined as the ratio of the peak voltage of the modulating signal $(A_m)$ to the peak voltage of the carrier wave $(A_c)$:
$m = \frac{A_m}{A_c}$
Given:
$A_c = 15 V$
$m = 80\% = 0.80$
Substituting the values into the formula:
$0.80 = \frac{A_m}{15 V}$
$A_m = 0.80 \times 15 V$
$A_m = 12 V$
Therefore, the peak voltage of the modulating signal should be $12 V$.