Physics Part Ii Chapter 15 Communication Systems

Skywave is used to propagate shortwave broadcast services.

It is given that peak voltage of carrier wave, A_c = 15V

and, Modulation index, m_u = 60% = 60/100

Then, the peak voltage of the modulating signal (A_m) will be given by the relation,

$$\begin{gathered} \mathrm{\mu } = \frac{{\mathrm{A}}_{\mathrm{m}}}{{\mathrm{A}}_{\mathrm{c}}} \\ 60\% = \frac{15 \mathrm{V}}{{\mathrm{A}}_{\mathrm{c}}} \\ {\mathrm{A}}_{\mathrm{c}} =\frac{15}{60} \mathrm{x} 100 = 25 \mathrm{V} \end{gathered}$$

Reasons:

1. Height of antenna
2. Power of signal
3. Utilization of frequency

Audio Signal:

D.

100

C.

100 Hz

frequency = 2 (frequency of input signal).

D.

2 x 10⁵

If n = no. of channels

10% of 10 GHz = n x 5 KHz

or $$\begin{gathered} \frac{10}{1000}x 10 x {10}^9 \\ = n x 5 x {10}^3 \\ \Rightarrow n = 2 x {10}^5 \end{gathered}$$

D.

(9/4) fold

We know that a phase change of π occurs when the reflection takes place at the boundary of the denser medium. This is equivalent to a path difference of λ/2.

Therefore, Total phase difference = π -π = 0

Thus, the two waves superimpose in phase.

Resultant,

$$\begin{gathered} \mathrm{Resultant} \mathrm{amplitude} = \sqrt{\mathrm{I}} + \sqrt{\left(\frac{\mathrm{I}}{4}\right)} = \frac{3}{2}\sqrt{\mathrm{I}} \\ \mathrm{Resultant} \mathrm{amplitude} = {\left(\frac{3}{2}\sqrt{\mathrm{I}}\right)}^2 = \frac{9}{4} \mathrm{I} = \frac{9}{4} \mathrm{fold} \end{gathered}$$ 

D.

100

Let intensity of sound be I and I'. Loudness of sound initially

$$\begin{gathered} {\mathrm{\beta }}_1 = 10 \log \left(\frac{\mathrm{I}}{{\mathrm{I}}_{\mathrm{o}}}\right) \\ {\mathrm{\beta }}_2 = 10 \log \left(\frac{\mathrm{I}\text{'}}{{\mathrm{I}}_{\mathrm{o}}}\right) \end{gathered}$$

β₂ - β₁ = 20

$\therefore 20 = 20 \log \left(\frac{\mathrm{I}\text{'}}{\mathrm{I}}\right)$

∴ I' = 100 I

A.

5 MHz

The ionosphere extends from a height of 80 km to 300 km. The refractive index of ionosphere is less than its free space value. That is, it behaves as a rarer medium. As we go deep into the ionosphere, the refractive index keeps on decreasing. The bending of beam (away from the normal) will continue till it reaches critical angle after which it will be reflected back. The different points on earth receive signals reflected from different depths of the ionosphere. There is a critical frequency f, (S to 100 MHz) beyond which the waves cross the ionosphere and do not return back to earth.

A.

perpendicular to the diameter

When the point is on the diameter and away from the centre of hemisphere which is charged uniformly and positively, the component of electric field intensity parallel to the diameter cancel out. So the electric field is perpendicular to the diameter.