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lecture24

# lecture24 - Lecture 24 Time Domain Analysis of Transmission...

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1 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Lecture 24 Time Domain Analysis of Transmission Lines In this lecture you will learn: • Time domain analysis of transmission lines • Transients in transmission lines ECE 303 – Fall 2007 – Farhan Rana – Cornell University o Z 0 = z ( ) t V s l = z Time Domain Analysis - Basics Question: How does one handle transmission lines for signals that are NOT time harmonic and when one is NOT dealing with the sinusoidal steady state? a) First thing to realize is that the notion of complex impedance has meaning only for the sinusoidal steady state b) For an arbitrary source voltage V s ( t ), one needs to work in the time domain and start from the basic time-domain equations: s R L R ( ) ( ) t t z I L z t z V = , , ( ) ( ) t t z V C z t z I = , , ( ) ( ) 2 2 2 2 2 , 1 , t t z V v z t z V = ( ) ( ) 2 2 2 2 2 , 1 , t t z I v z t z I = LC v 1 =

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2 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Time Domain Analysis - Basics o Z The equation: ( ) ( ) 2 2 2 2 2 , 1 , t t z V v z t z V = Has forward moving solutions of the form: ( ) ( ) vt z V t z V = + , And backward moving solutions of the form: ( ) ( ) vt z V t z V + = , Examples: ( ) vt z V + z v ( ) vt z V + z v LC v 1 = ECE 303 – Fall 2007 – Farhan Rana – Cornell University Voltages and Currents o Z ( ) vt z V + z v Voltage: ( ) vt z I + z v Corresponding Current: ( ) ( ) t t z I L z t z V = , , The current is related to the voltage and satisfies: And this: ( ) ( ) t t z V C z t z I = , , ( ) ( ) ( ) ( ) o o Z vt z V vt z I Z vt z V vt z I + = + = + + and The solution is: + + + + + + - - - - - - Current is proportional to voltage since higher voltage means more surface charges and more surface charges mean more current flow v
3 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Visualizing Propagation o Z 0 = z l = z t 1 Forward moving solutions are of the form: ( ) vt z V + And backward moving solutions are of the form: ( ) vt z V + So suppose somebody tells you that at z = - : ( ) = + t V , l Then what is the forward moving voltage on the line at t = / 2 v ? T 0 = z l = z ( ) v t z V 2 , l = + v t 2 l = vT ECE 303 – Fall 2007 – Farhan Rana – Cornell University Load End Boundary Condition o Z 0 = z ( ) t V s l = z s R L R Load end boundary condition: ( ) ( ) L R t z I t z V , 0 , 0 = = = ( ) ( ) ( ) t z V t z V t z V , 0 , 0 , 0 = + = = = + ( ) ( ) ( ) ( ) ( ) o o Z t z V Z t z V t z I t z I t z I , 0 , 0 , 0 , 0 , 0 = = = = + = = = + + ( ) ( ) L t z V t z V Γ = = = + , 0 , 0 1 1 + = Γ o L o L L Z R Z R + - ( ) t z V , 0 = ( ) t z I , 0 = For all time t we must have: Substitute these in this to get:

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4 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Source End Boundary Condition o Z 0 = z ( ) t V s l = z s R L R + - Source end boundary condition: ( ) ( ) ( ) t z V R t z I t V s s , , l l = + = = ( ) ( ) ( ) t z V t z V t z V , , , l l l = + = = = + ( ) ( ) ( ) ( ) ( ) o o Z t z V Z t z V t z I t z I t z
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