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Problem 9.22: Solution

A 150 MHz communication link uses two half-wave dipole antennas separated by 2 km. A transmitter power is required to achieve a signal-to-noise ratio of 17 dB for a receiver with a noise temperature of 600 K and bandwidth of 3 MHz. The Friis transmission formula is used to calculate that a transmitter power of 75 micro-watts is needed.

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121 views1 page

Problem 9.22: Solution

A 150 MHz communication link uses two half-wave dipole antennas separated by 2 km. A transmitter power is required to achieve a signal-to-noise ratio of 17 dB for a receiver with a noise temperature of 600 K and bandwidth of 3 MHz. The Friis transmission formula is used to calculate that a transmitter power of 75 micro-watts is needed.

Uploaded by

omairakhtar12345
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© © All Rights Reserved
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Problem 9.

22 A 150-MHz communication link consists of two vertical half-wave


dipole antennas separated by 2 km. The antennas are lossless, the signal occupies a
bandwidth of 3 MHz, the system noise temperature of the receiver is 600 K, and the
desired signal-to-noise ratio is 17 dB. What transmitter power is required?
Solution: From Eq. (9.77), the receiver noise power is
(9.71)
Pn = KTsys B = 1.38 1023 600 3 106 = 2.48 1014 W.

For a signal to noise ratio Sn = 17 dB = 50, the received power must be at least

Prec = Sn Pn = 50(2.48 1014 W) = 1.24 1012 W.


(9.49)
Since the two antennas are half-wave dipoles, Eq. (9.47) states Dt = Dr = 1.64, and
since the antennas are both lossless, Gt = Dt and Gr = Dr . Since the operating
frequency is f = 150 MHz, = c/ f = (3 108 m/s)/(150 106 Hz) = 2 m. Solving
the Friis transmission formula (Eq. (9.75)) for the transmitted power:
(9.69)
2
(4 )2 R2 (4 )2 (2 103 m)
Pt = Prec 2 = 1.24 1012 = 75 ( W).
Gr Gt (2 m)2 (1.64)(1.64)

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