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// Example no.5.7
// To find the peak current if (a) LO power = 10 dBm, (b) LO power = −10 dBm for the balanced receiver
// Page no.234
clc;
clear;
// Given data
L=100; // Length of fiber
loss=0.2*L; // Total fiber loss
PtdBm=12; // The peak power of the signal at the transmitter
R=0.9; // Responsivity in A/W
PrdBm=PtdBm-loss; // The power at the receiver
// (a) the peak current LO power = 10 dBm
PLO1dBm=10; // Power at local oscillator in dBm
PLO1=10^(0.1*PLO1dBm); // Power at local oscillator in mW
Pr=10^(0.1*PrdBm); // Power at receiver in mW
Id1=2*R*sqrt(Pr*PLO1); // The peak current LO power = 10 dBm
// Display result on command window
printf('\n The peak current for LO power 10 dBm = %0.4f mA',Id1)
// (b) the peak current LO power = -10 dBm
PLO2dBm=-10; // Power at local oscillator in dBm
PLO2=10^(0.1*PLO2dBm); // Power at local oscillator in mW
Id2=2*R*sqrt(Pr*PLO2); // The peak current LO power = -10 dBm
// Display result on command window
printf('\n The peak current for LO power -10 dBm = %0.4f mA',Id2)
// comment on the intermodulation cross-talk in a single-branch receiver and the balanced receiver
printf('\n A single-branch receiver would have a significant amount of cross-talk. In contrast, for a balanced receiver, intermodulation cross-talk is canceled out \n due to the balanced detection.')
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