This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 5. The power into an optical amplifier is 5mW. The gain of the amplifier is 23dB and the saturation output power is 33dBm. What would the amplifiers output power be in mW? Pin = 10log10(5) = 7dBm Pout = Pin + gain = 7 + 23 = 30dBm Pout (mW) = 10 (30/10) = 1000mW = 1W 6. The input power is increased to 20mW what would the new output power be? Pin = 10log10(20) = 13dBm Pout = Pin + gain = 13 + 23 = 36dBm But because this is above the amplifiers saturation output power the amplifier will be in saturation. Therefore the output power will be limited to the saturation output power of 33dBm or 2W....
View Full Document
This note was uploaded on 03/23/2011 for the course EEE 174 taught by Professor Kasab during the Spring '11 term at ASU.
- Spring '11