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Unformatted text preview: 50GHz apart at approximately 1550nm? df c d 2 = = ((1550x10-9 ) 2 /3x10 8 ) x 50x10 9 = 0.4nm (n.b. this formula can be found be differentiating f=c/ with respect to and re-arranging for d , it can only be used for small d or df (i.e <10nm or 1250GHz) 5. What is the peak-to-peak voltage of a 10dBm electrical signal into a 50 load? P = 10 (10dBm/10) = 10mW Vrms = (P x R) 1/2 = (0.01 x 50) 1/2 = 0.707V Vpp = Vrms x 2 1/2 x 2 = 2V 6. A -13dBm optical signal is incident on a photodiode with a response of 0.6A/W, this is followed by a transimpedance amplifier with a transimpedance of 11k , what is the output voltage? P = 10 (-13/10) = 50uW I = P x R = 50uW x 0.6A/W = 30uA V = R x I = 11k x 30uA = 0.33V...
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- Spring '11