1

### JEE Main 2019 (Online) 11th January Morning Slot

An amplitude modulated signal is given by V(t) = 10[1 + 0.3cos(2.2 $\times$ 104 t)] sin(5.5 $\times$ 105 t). Here t is in seconds. The sideband frequencies (in kHz) are, [Given $\pi$ = 22/7]
A
892.5 and 857.5
B
89.25 and 85.75
C
1785 and 1715
D
178.5 and 171.5

## Explanation

V(t) = 10 + ${3 \over 2}$ [2cos A sinB]

= 10 + ${3 \over 2}$ [sin(A+B) $-$ sin(A $-$ B)]

= 10+${3 \over 2}$[sin (57.2 $\times$ 104 t) $-$ sin(52.8 $\times$ 104 t)]

$\omega$1 = 57.2 $\times$ 104 = 2$\pi$f1

f1 = ${{57.2 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}} = 9.1 \times {10^4}$

$\simeq$ 91KHz

f2 = ${{52.8 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}}$

$\simeq$ 84 KHz Side band frequency are

f1 = fc $-$ fw = ${{52.8 \times {{10}^4}} \over {2\pi }}$ $\simeq$ 85.00 kHz

f2 = fc + fw = ${{57.2 \times {{10}^4}} \over {2\pi }}$ $\simeq$ 90.00 kHz
2

### JEE Main 2019 (Online) 11th January Evening Slot

An amplitude modulated signal is plotted below : Which one of the following best describes the above signal ?
A
(1 + 9sin (2$\pi$ $\times$ 104 t)) sin(2.5$\pi$ $\times$ 105t) V
B
(9 + sin (4$\pi$ $\times$ 104 t)) sin(5$\pi$ $\times$ 105t) V
C
(9 + sin (2$\pi$ $\times$ 104 t)) sin(2.5$\pi$ $\times$ 105t) V
D
(9 + sin (2.5$\pi$ $\times$ 104 t)) sin(2$\pi$ $\times$ 104t) V

## Explanation

Analysis of graph says

(1)   Amplitude varies as 8 $-$ 10 V or 9 $\pm$ 1

(2)  Two time period as

100 $\mu$s (signal wave) & 8 $\mu$s (carrier wave)

Hence signal is $\left[ {9 \pm 1\sin \left( {{{2\pi t} \over {{T_1}}}} \right)} \right]\sin \left( {{{2\pi t} \over {{T_2}}}} \right)$

= 9 $\pm$ 1sin (2$\pi$ $\times$ 104t) sin 2.5$\pi$ $\times$ 105 t
3

### JEE Main 2019 (Online) 12th January Morning Slot

A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index ?
A
0.5
B
0.6
C
0.4
D
0.3

## Explanation

Em + Ec = 160

Em + 100 = 160

Em = 60

$\mu = {{{E_m}} \over {{E_C}}} = {{60} \over {100}}$

$\mu$ = 0.6
4

### JEE Main 2019 (Online) 12th January Evening Slot

To double the covering range of a TV transmittion tower, its height should be multiplied by :
A
4
B
2
C
$\sqrt 2$
D
${1 \over {\sqrt 2 }}$