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What is time division switching? With the help of block diagram, explain basic time division time switching method.

Computer Science

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Answer

19.00 W\mathbf{19.00 \text{ W}}

Step-by-step solution

4 steps

here's how to solve question 4b:

  1. List the given parameters and constants.

    • Transmitted power, Pt=500 WP_t = 500 \text{ W}
    • Antenna gain (each), GdB=70 dBG_{dB} = 70 \text{ dB}
    • Signal frequency, f=12 GHz=12×109 Hzf = 12 \text{ GHz} = 12 \times 10^9 \text{ Hz}
    • Transmission path distance, d=102 km=102×103 md = 102 \text{ km} = 102 \times 10^3 \text{ m}
    • Speed of light, c=3×108 m/sc = 3 \times 10^8 \text{ m/s}

  2. Calculate the wavelength (λ\lambda).

    The wavelength is given by the formula λ=cf\lambda = \frac{c}{f}.

    λ=3×108 m/s12×109 Hz\lambda = \frac{3 \times 10^8 \text{ m/s}}{12 \times 10^9 \text{ Hz}} λ=0.025 m\lambda = 0.025 \text{ m}

  3. Convert the antenna gains from decibels (dB) to a linear ratio.

    The linear gain GG is related to the dB gain GdBG_{dB} by the formula G=10(GdB/10)G = 10^{(G_{dB}/10)}.

    Gt=Gr=10(70/10)=107G_t = G_r = 10^{(70/10)} = 10^7

  4. Apply the Friis transmission equation to calculate the received power (PrP_r).

    The Friis transmission equation is given by:

    Pr=Pt⋅Gt⋅Gr⋅(λ4πd)2P_r = P_t \cdot G_t \cdot G_r \cdot \left(\frac{\lambda}{4 \pi d}\right)^2

    Substitute the values:

    Pr=500 W⋅(107)⋅(107)⋅(0.025 m4π⋅102×103 m)2P_r = 500 \text{ W} \cdot (10^7) \cdot (10^7) \cdot \left(\frac{0.025 \text{ m}}{4 \pi \cdot 102 \times 10^3 \text{ m}}\right)^2 Pr=500⋅1014⋅(0.0251282566.08)2P_r = 500 \cdot 10^{14} \cdot \left(\frac{0.025}{1282566.08}\right)^2 Pr=500⋅1014⋅(1.9492×10−8)2P_r = 500 \cdot 10^{14} \cdot (1.9492 \times 10^{-8})^2 Pr=500⋅1014⋅(3.7994×10−16)P_r = 500 \cdot 10^{14} \cdot (3.7994 \times 10^{-16}) Pr=1899.7×10−2 WP_r = 1899.7 \times 10^{-2} \text{ W} Pr=18.997 WP_r = 18.997 \text{ W}

    The power received is 19.00 W\boxed{\mathbf{19.00 \text{ W}}} (rounded to two decimal places).

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