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Unformatted text preview: ECE 3025 Homework 2: Reflections on Transmission Lines Solutions 1. The reflection coefficients for this transmission line are G =- . 818 and L = 0 . 818. Since the source is 2V DC, switched on at t = 0, the initial input voltage to the transmission line may be calculated using the voltage divider equation: V in = V S Z R G + Z = 1 . 82 Volts Now we can use our general form equation for the output of a transmission line as a function of time: V L ( t ) = (1 + L ) summationdisplay n =0 ( G L ) n V in u ( t- [2 n + 1] T ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright f ( t ) where u( ) is the unit step function. You dont have to write out this expression a simple graph constructed of step-by-step reflections suffices. However, this equation is useful if you want to produce a computer plot like the one below: 1 2 3 4 5 6 7 8 9 10 0.5 1 1.5 2 2.5 3 3.5 time (units of T) V L (t) (Volts) Output Voltage on Transmission Line (HW2, Prob 1) 3.31 V 1.09 V 2.57 V 1.58 V 2.25 V For those interested, the Matlab program used to construct this graph is listed below: % Solution to Problem 1 function VL = problem1 t = 0:.01:10; % time axis GS = (20-200)/(20+200); % source-side reflection coefficient GL = (2000-200)/(2000+200); % load-side reflection coefficient Vin = 200/(200+20)*2; % initial input voltage...
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- Spring '08