Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation-equation of modulating wave is, m (t)= A_msinw_mt

Carrier wave equation is C_m (t)= (A_c+A_msinw_mt)sinw_ct

 = A_c [1+ $\frac{A_m}{A_c}sin{w}_{m}t$ ]sinw_ct

Also $\frac{A_m}{A_c}$ = M

C_mt= (A_c+A_{c $*\propto sin{w}_{m}t$} )sinw_ct

Ac $*\propto$ =K

Sinw_mt= V_m

equation having phase $\varnothing$

so equation is A_c+K₂V_mt}+sinw_ct+ $?$ ]

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation- it is represented by a diagram given below

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation- Here, in this question, the frequency of modulated signal received becomes more, which is possible with the poor bandwidth selection of amplifiers.

This happens because bandwidth in amplitude modulation is equal to twice the

Frequency of modulating signal. But, the frequency of male voice is less than that of a female.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation- The device which follows square law is used for modulation purpose. Characteristics shown by (i) and (iii) corresponds to linear devices.

Characteristics shown by (ii) corresponds to square law device. Some part of (i) also

Follow square law.

Hence, (ii)and (iv) can be used for modulation.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation- frequency of career wave and frequency of amplitude modulated wave is same which is w_c .

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a

Frequency of career signal = 1MHz and frequency = 3KHz= 0.003MHz

So frequency of side bands = 1 $?$ 0.003

So 1.003MHz and 0.997MHz

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b

p = 1kW= 1000W

Attenuation of signal = -2dB/km and total path length = 5km

Gain in dB= 5 $*-2=-10dB$

Gain in dB= 10 log (p₀/pi)……. (i)

-10=10 log (p₀/pi)= -10log (p_i/p_o)

logP_i/p_o=1 = log (p_i/p_o)= log10

pi/p₀=10= 1000W= 10p₀

p₀=100W

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a

length of building is given by l= 500m

And $\lambda ~$ 4l= 4 $*$  100 =400mf

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Ground wave propagation – 530KHz to 1710KHz

Sky wave propagation- 1710KHz- 40MHz

Space wave propagation- 54MHz to 42GHz 

This is a  Short Answer Type Questions as classified in NCERT Exemplar

In amplitude modulated wave side frequency contain the information and from the question the power will be ω_c, (ω_c – ω_m) and (ω_c + ω_m)

So to minimise cost of radiation we use both (ω_c – ω_m) and (ω_c + ω_m)