Would sky waves be suitable for transmission of TV signals of 60 MHz frequency?
A signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz . But, here the frequency of TV signals are 60 MHz which is beyond the required range. So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.
Two waves $A$ and $B$ of frequencies 2 MHz and 3 MHz , respectively are beamed in the same direction for communication via sky wave. Which one of these is likely to travel longer distance in the ionosphere before suffering total internal reflection?
As the frequency of wave $B$ is more than wave $A$, it means the refractive index of wave $B$ is more than refractive index of wave $A$ (as refractive index increases with frequency increases).
For higher frequency wave (i.e., higher refractive index) the angle of refraction is less i.e., bending is less. So, wave $B$ travel longer distance in the ionosphere before suffering total internal reflection.
The maximum amplitude of an AM wave is found to be 15 V while its minimum amplitude is found to be 3 V . What is the modulation index?
Let $A_c$ and $A_m$ be the amplitudes of carrier wave and modulating wave respectively. So,
Maximum amplitude $\longrightarrow A_{\text {max }}=A_c+A_m=15 \mathrm{~V}\quad\text{.... (i)}$
Minimum amplitude $\longrightarrow A_{\min }=A_c-A_m=3 \mathrm{~V}\quad\text{.... (ii)}$
Adding Eqs. (i) and (ii), we get
$$2 A_C=18$$
$$\begin{aligned} \text{or}\quad & A_c=9 \mathrm{~V} \\ \text{and}\quad & A_m=15-9=6 \mathrm{~V} \end{aligned}$$
Modulating index of wave $\propto=\frac{A_m}{A_c}=\frac{6}{9}=\frac{2}{3}$
Compute the $L C$ product of a tuned amplifier circuit required to generate a carrier wave of 1 MHz for amplitude modulation.
Given, the frequency of carrier wave is 1 MHz .
Formula for the frequency of tuned amplifier,
$$\begin{aligned} \frac{1}{2 \pi \sqrt{L C}} & =1 \mathrm{MHz} \\ \sqrt{L C} & =\frac{1}{2 \pi \times 10^6} \\ L C & =\frac{1}{\left(2 \pi \times 10^6\right)^2}=2.54 \times 10^{-14} \mathrm{~s} \end{aligned}$$
Thus, the product of $L C$ is $2.54 \times 10^{-14} \mathrm{~s}$.
Why is a AM signal likely to be more noisy than a FM signal upon transmission through a channel?
In case of AM, the instantaneous voltage of carrier waves is varied by the modulating wave voltage. So, during the transmission, nosie signals can also be added and receiver assumes noise a part of the modulating signal. In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal is transmitting in channel. So, noise does not affect FM signal.