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  2. Help me understand! A message signal of frequency 100 MHz and peak voltage 100V is used to execute amplitudemodulation on a carrier wave of frequency 300 GHz and peak voltage 400V. The modulationindex and difference between the two side band frequenc
# Engineering

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)
P Plabita

 

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, fm

- 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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