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An Amplitude Modulated (AM) wave is represented by the expression where a 10kHz sinusoidal message signal modulates a 1MHz carrier signal. The peak carrier amplitude is 10 V and the peak message amplitude is 6V.

Computer Science

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3 parts
i
0.6
ii

1.01 MHz1.01 \text{ MHz}

0.99 MHz0.99 \text{ MHz}

iii

1.18 W1.18 \text{ W}

Step-by-step solution

8 steps

Here's the solution for question 1a:

a) An Amplitude Modulated (AM) wave is represented by the expression where a 10kHz sinusoidal message signal modulates a 1MHz carrier signal. The peak carrier amplitude is 10 V and the peak message amplitude is 6V.

Part i· 2 steps

Calculate the Modulation Index (m) of the wave.

  1. Identify the given amplitudes.

    Peak carrier amplitude, Ac=10 VA_c = 10 \text{ V}.
    Peak message amplitude, Am=6 VA_m = 6 \text{ V}.

  2. Apply the formula for the modulation index.

    The modulation index mm is given by the ratio of the peak message amplitude to the peak carrier amplitude.

    m=AmAcm = \frac{A_m}{A_c} m=6 V10 Vm = \frac{6 \text{ V}}{10 \text{ V}} m=0.6m = 0.6

    The modulation index is 0.6\boxed{0.6}.

Part ii

1.01 MHz1.01 \text{ MHz}

Determine the Upper Sideband (USB) and Lower Sideband (LSB) frequencies.

  1. Identify the given frequencies.

    Message signal frequency, fm=10 kHz=10×103 Hzf_m = 10 \text{ kHz} = 10 \times 10^3 \text{ Hz}.
    Carrier signal frequency, fc=1 MHz=1×106 Hzf_c = 1 \text{ MHz} = 1 \times 10^6 \text{ Hz}.

  2. Calculate the Upper Sideband (USB) frequency.

    The USB frequency is the sum of the carrier frequency and the message frequency.

    fUSB=fc+fmf_{USB} = f_c + f_m fUSB=1×106 Hz+10×103 Hzf_{USB} = 1 \times 10^6 \text{ Hz} + 10 \times 10^3 \text{ Hz} fUSB=1,000,000 Hz+10,000 Hzf_{USB} = 1,000,000 \text{ Hz} + 10,000 \text{ Hz} fUSB=1,010,000 Hz=1.01 MHzf_{USB} = 1,010,000 \text{ Hz} = 1.01 \text{ MHz}
  3. Calculate the Lower Sideband (LSB) frequency.

    The LSB frequency is the difference between the carrier frequency and the message frequency.

    fLSB=fc−fmf_{LSB} = f_c - f_m fLSB=1×106 Hz−10×103 Hzf_{LSB} = 1 \times 10^6 \text{ Hz} - 10 \times 10^3 \text{ Hz} fLSB=1,000,000 Hz−10,000 Hzf_{LSB} = 1,000,000 \text{ Hz} - 10,000 \text{ Hz} fLSB=990,000 Hz=0.99 MHzf_{LSB} = 990,000 \text{ Hz} = 0.99 \text{ MHz}

    The Upper Sideband frequency is 1.01 MHz\boxed{1.01 \text{ MHz}} and the Lower Sideband frequency is 0.99 MHz\boxed{0.99 \text{ MHz}}.

Part iii

1.18 W1.18 \text{ W}

Calculate the total power transmitted if the antenna resistance is 50Ω.

  1. Identify the given values and the modulation index from part (i).

    Peak carrier amplitude, Ac=10 VA_c = 10 \text{ V}.
    Antenna resistance, R=50 ΩR = 50 \text{ } \Omega.
    Modulation index, m=0.6m = 0.6.

  2. Calculate the carrier power (PcP_c).

    The carrier power is given by Pc=Ac22RP_c = \frac{A_c^2}{2R}.

    Pc=(10 V)22×50 ΩP_c = \frac{(10 \text{ V})^2}{2 \times 50 \text{ } \Omega} Pc=100 V2100 ΩP_c = \frac{100 \text{ V}^2}{100 \text{ } \Omega} Pc=1 WP_c = 1 \text{ W}
  3. Calculate the total transmitted power (PtP_t).

    The total transmitted power for an AM wave is given by Pt=Pc(1+m22)P_t = P_c \left(1 + \frac{m^2}{2}\right).

    Pt=1 W(1+(0.6)22)P_t = 1 \text{ W} \left(1 + \frac{(0.6)^2}{2}\right) Pt=1 W(1+0.362)P_t = 1 \text{ W} \left(1 + \frac{0.36}{2}\right) Pt=1 W(1+0.18)P_t = 1 \text{ W} (1 + 0.18) Pt=1 W(1.18)P_t = 1 \text{ W} (1.18) Pt=1.18 WP_t = 1.18 \text{ W}

    The total power transmitted is 1.18 W\boxed{1.18 \text{ W}}.

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