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klasskru [66]
2 years ago
13

A transmission channel between two communicating DTEs is made up of three sections. The first introduces an attenuation of 16 dB

, the second introduces an amplification of 20 dB, and the third introduces an attenuation of 10 dB. Assuming a mean transmitted power level of 400 mW, determine the mean output power level of the channel. (3 points)
Engineering
1 answer:
murzikaleks [220]2 years ago
7 0

Answer:

the mean output power level of the channel is 100.475 mW

Explanation:

Given the data in the question;

for the first section; attenuation

16 = 10log₁₀  400/P₂

P₂ = 10.0475 mW

For the second section;  amplification

20 = 10log₁₀  p₂/10.0475

p₂ = 1004.75 mW

For the third section; attenuation

10 = 10log₁₀  p₂/1004.75  

p₂ = 100.475 mW

i.e the mean output power level is 100.475 mW

or better still overall attenuation channel = ( 16 - 20) + 10 = 6dB

so

6 = 10log₁₀  400/P₂

p₂ = 100.475 mV

the mean output power level of the channel is 100.475 mW

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air at 600 kPa, 330 K enters a well-insulated, horizontal pipe having a diameter of 1.2 cm and exits at 120 kPa, 300 K. Applying
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h_{i}+\frac{V_{i}^{2} }{2} =h_{2}+\frac{V_{o}^{2} }{2} \\h_{i}-h_{o}+(\frac{V_{i}^{2}-V_{o}^{2}  }{2} )=0\\V_{o}=\sqrt{\frac{2(h_{i}-h_{o})}{0.9516} } =\sqrt{\frac{2*(330.24-300.19)x10^{3} }{0.9516} } =251.31m/s

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b) The mass flow rate is:

m=\frac{A_{o}V_{o}}{v_{o}} =\frac{\pi d^{2}V_{o}P_{o} }{4RT_{o}} =\frac{\pi *(0.012^{2})*251.31*120x10^{3}  }{4*287*300} =0.0396kg/s

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See explanation for step by step procedure to get answer.

Explanation:

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