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A message signal of frequency 100 MHz and peak voltage 100V is used to execute amplitude

modulation on a carrier wave of frequency 300 GHz and peak voltage 400V. The modulation

index and difference between the two side band frequencies are:
Option 1)
$4;1\times 10^{8}\: Hz$
Option 2)
$4;2\times 10^{8}\: Hz$
Option 3)
$0.25;2\times 10^{8}\: Hz$
Option 4)
$0.25;1\times 10^{8}\: Hz$

Answers (1)

best_answer

Modulation Index -

The ratio of change of amplitude of carrier wave to the amplitude of original carrier wave.

- wherein

$m_{a}= \frac{E_{m}}{E_{c}}$

Side band frequency -

AM wave contains three frequency $f_{c},\left ( f_{c}+f_{m} \right )\: and \: \left ( f_{c}-f_{m} \right )$

- wherein

$f_{c}$ is carrier frequency, $f_{c} -f_{m} \:and\: f_{c}+f_{m}$ are side band frequency.

Band Width -

The two side band lie on either side of the carrier frequency at equal frequency intravel, f_m

- wherein

$Band \: W\! idth= \left ( f_{c}\: + f_{m}\right )- \left ( f_{c}\: - f_{m}\right )$

$= 2 f_{m}$

$f_m=100MHz$

$E_m=100V$

$f_c=300GHz$

$E_c=400V$

Modulation index $=m=\frac{E_m}{E_c}=\frac{100}{400}=0.25$

and

$\Delta f=(f_c+f_m)-(f_c-f_m)=2f_m$

$\Delta f=2\times 10^{2}\times 10^{6}Hz$

$\Delta f=2\times 10^{8}Hz$

Option 1)

$4;1\times 10^{8}\: Hz$

Option 2)

$4;2\times 10^{8}\: Hz$

Option 3)
$0.25;2\times 10^{8}\: Hz$
Option 4)

$0.25;1\times 10^{8}\: Hz$

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