8)Given a receive power of 80E-6 mW and a noise power in decibels of [N]= -125DBW, determine the decibel SNR value. a. [SNR] = -80e-6 dBm – 123.5 dBm = -173.5 dB b. [SNR] = -40.97BM – (123.5 dBm) = -165.97 dB c. [SNR] = -40.97 dBm – (-95 dBm) = 54.03 dB d. [SNR] = 40.97 dBm – (-125 dBm) = +84.03 dB
8)Given a receive power of 80E-6 mW and a noise power in decibels of [N]= -125DBW, determine the decibel SNR value. a. [SNR] = -80e-6 dBm – 123.5 dBm = -173.5 dB b. [SNR] = -40.97BM – (123.5 dBm) = -165.97 dB c. [SNR] = -40.97 dBm – (-95 dBm) = 54.03 dB d. [SNR] = 40.97 dBm – (-125 dBm) = +84.03 dB
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
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![8)Given a receive power of 80E-6 mW and a noise power in decibels of [N]= -125DBW, determine the
decibel SNR value.
a. [SNR] = -80e-6 dBm – 123.5 dBm = -173.5 dB
b. [SNR] = -40.97BM – (123.5 dBm) = -165.97 dB
c. [SNR] = -40.97 dBm – (-95 dBm) = 54.03 dB
d. [SNR] = 40.97 dBm – (-125 dBm) = +84.03 dB
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Transcribed Image Text:8)Given a receive power of 80E-6 mW and a noise power in decibels of [N]= -125DBW, determine the
decibel SNR value.
a. [SNR] = -80e-6 dBm – 123.5 dBm = -173.5 dB
b. [SNR] = -40.97BM – (123.5 dBm) = -165.97 dB
c. [SNR] = -40.97 dBm – (-95 dBm) = 54.03 dB
d. [SNR] = 40.97 dBm – (-125 dBm) = +84.03 dB
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%3D
%3D
%3D
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