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A 1-MHz, 40-V peak carrier is modulated by a 5-kHz intelligence signal so that m = 0.7. This AM signal is fed to a 50-???? antenna. Calculate the power of each spectral component fed to the antenna. (Pc = 16 W, Pusb = Plsb = 1.96 W)

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ProfLincoln

Answer:

16 W and 1.96 W

Explanation:

The spectral components in the AM wave are unmodulated carrier, USB & LSB.

Total Power of AM antenna = Power in the Unmodulated carrier (Pc) + Power in USB (Pusb) + Power in LSB (Plsb)

Given:

Carrier Frequency Fc=1 MHz,

Voltage of carrier Vc=40 V pk,

Modulated frequency Fm=5 KHz,

Modulation index m=0.7,

Load RL = 50 Ohms

Power in the Unmodulated carrier (Pc) = 16W

Pusb = Plsb = 1.96 W

Pc = Vc(sqr)/RL

= [40V/sqrt 2]/50ohms

Pc=16 W

Since Pusb = Plsb

Plsb = 1/2 × PC (m(sqr)/2)

= 1/2 × 16W × [0.7(sqr)/2]

= 1.96Watts (to confirm that which was already given)

Lanuel

The power of each spectral component fed to the antenna is 32 Watts and 7.84 Watts respectively.

Given the following data:

  • Carrier frequency = 1 MHz.
  • Voltage of carrier = 40 Volts.
  • Modulated frequency = 5 KHz.
  • Modulation index = 0.7.
  • Antenna resistance = 50 Ohms.
  • Power in unmodulated carrier = 16 Watts.
  • [tex]P_{usb} = P_{lsb}[/tex] = 1.96 Watt.

How to calculate the power.

The spectral components in the amplitude modulation (AM) wave include the following:

  • Unmodulated carrier
  • Upper sideband (USB)
  • Lower sideband (LSB).

Mathematically, the total power of amplitude modulation (AM) antenna is given by:

[tex]Total\;power =P_c+P_{usb} +P_{lsb}[/tex]

Now, we would calculate each of the powers:

[tex]P_c=\frac{V^2}{R} \\\\P_c=\frac{40^{2}}{50 } \\\\P_c= 32\;Watts[/tex]

[tex]P_{usb} =\frac{1}{2} P_c(m^2)\\\\P_{usb} =\frac{1}{2} \times 32 \times 0.7^2\\\\P_{usb} = P_{lsb}=7.84\;Watts[/tex]

Read more on amplitude modulation here: https://brainly.com/question/10690505

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