Question 1 of 27070KOjuL29v5aGZj1slBQwJ

What is the effective radiated power of a repeater with 50 watts transmitter power output, 4 dB feedline loss, 3 dB duplexer and circulator loss, and 6 dB antenna gain?

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Question 1 of 27070
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Q

What is the effective radiated power of a repeater with 50 watts transmitter power output, 4 dB feedline loss, 3 dB duplexer and circulator loss, and 6 dB antenna gain?

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🔍 Key Concepts

• dB (decibel) math: how to convert dB gains/losses into power ratios • The definition of effective radiated power (ERP): starting from transmitter power, subtract losses, add antenna gain (all in dB) • Relationship between 10·log10(P2/P1) and multiplying/dividing power in watts

💭 Think About

• First, combine all the dB values (feedline loss, duplexer/circulator loss, antenna gain) into a single net gain or loss in dB. Is the final result positive or negative? • Once you have the net dB value, how do you convert that dB value back into a power multiplier and apply it to the original 50 W? • Think about what happens to the power when you have more loss dB than gain dB versus more gain dB than loss dB. Should the ERP be higher or lower than 50 W in this case?

✅ Before You Answer

• Be sure you subtract losses and add gains when working in dB: losses are negative, gains are positive. • After finding the net dB, convert dB to a linear factor using factor=10(dB/10)\text{factor} = 10^{(\text{dB}/10)}factor=10(dB/10). • Finally, multiply the original 50 W by that factor and compare your result to the choices to see which is closest.