Transmission Lines (Bounce Diagram) 3

Capacitive Load (cont.) t=0 V0 [ V ] Rg = Z 0 + v (t ) g - CL Z0 z=L z=0 −1 At t = T : The capacitor acts as a short circuit: Γ L (T ) = At t = ∞: The capacitor acts as an open circuit: Γ L ( ∞ ) =1 Between t = T and t = ∞, there is an exponential time-constant behavior. General time-constant formula: F ( t )= F ( ∞ ) +  F (T ) − F ( ∞ )  e −( t −T ) /τ t ≥T Hence, we have: ΓL (t ) = 1 − 2 e −( t −T ) /τ , t ≥ T τ = Z 0CL 27 Capacitive Load (cont.) t=0 Rg = Z 0 + + v (t ) g - V0 [ V ] V + = V0 / 2 Assume z = 0 - v ( 0, t ) Z0 z=L z=0 ΓL (t ) Γ g =0 t =0 t T V+ 0 V+ CL V0 Steady-state ( z V0 1 − e −( t − 2T ) /τ v ( 0, t ) Γ L ( t − T )V + ) V0 / 2 2T V + (1 + Γ L ( t − T ) ) 3T z=0 Γ L ( t ) = 1 − 2 e −( t −T ) /τ , T ( ) 1 + Γ L ( t − T ) = 2 1 − e −( t − 2T ) /τ , t 2T ( V + (1 + Γ L ( t − T ) ) = V0 1 − e −( t − 2T ) /τ ) 28 Inductive Load Rg = Z 0 t=0 V0 [ V ] + - vg ( t ) LL Z0 z=L z=0 1 At t = T: inductor acts as an open circuit: Γ L (T ) = At t = ∞: inductor acts as a short circuit: Γ L ( ∞ ) =−1 Between t = T and t = ∞, there is an exponential time-constant behavior. Γ L ( t ) =−1 + 2 e −( t −T ) /τ τ = LL / Z 0 , t ≥T Recall: F ( t )= F ( ∞ ) +  F (T ) − F ( ∞ )  e −( t −T ) /τ t ≥T 29 Inductive Load (cont.) Assume z = 0 Rg = Z 0 t=0 + V0 [ V ] - V + = V0 / 2 + vg ( t ) v ( 0, t ) - LL Z0 z=0 t =0 t V0 ΓL (t ) Γ g =0 T z=L V + 2T V 0 + Γ L ( t − T )V L v ( 0, t ) + (1 + Γ ( t − T ) ) V 3T z V0 e −( t − 2T ) /τ V0 / 2 Steady-state + z=0 Γ L ( t ) = −1 + 2 e −( t −T ) /τ , 1 + Γ L ( t − T ) = 2 e −( t − 2T ) /τ , T 2T t V + (1 + Γ L ( t − T ) ) = V0 e −( t − 2T ) /τ 30 Time-Domain Reflectometer (TDR) This is a device that is used to look at reflections on a line, to look for potential problems such as breaks on the line. “The 20/20 Step Time Domain Reflectometer (TDR) was designed to provide the clearest picture of coaxial or twisted pair cable lengths and to pin-point cable faults.” AEA Technology, Inc. 31 Time-Domain Reflectometer (cont.) Rg = Z 0 t=0 + V0 [ V ] td (round - trip time down to fault) - z = zF vg ( t ) Z0 td = 2 z F / cd RL = Z 0 Z0 z=L z=0 Fault The fault is modeled as a load resistor RF at z = z F : The fault might be acting as a parallel resistor or a series resistor. RF < Z 0 or RF > Z 0 Rg = Z 0 t=0 V0 [ V ] td = 2 z F / cd + - z = zF vg ( t ) Z0 z=0 RF 32 Time-Domain Reflectometer (cont.) Rg = Z 0 t=0 V0 [ V ] td (round - trip time down to fault) + - z = zF vg ( t ) Z0 z=0 td = 2 z F / cd RL = Z 0 Z0 z=L Fault The fault is modeled as a load resistor RF at z = z F . v ( 0, t ) The time td indicates where the fault is. v ( 0, t ) t = td t = td Resistive load, RF > Z0 t t Resistive load, RF < Z0 33 Time-Domain Reflectometer (cont.) The TDR can also tell us what kind of a load we have at the end of a line. Rg = Z 0 t=0 V0 [ V ] + - vg ( t ) Load Z0 z=L z=0 v ( 0, t ) v ( 0, t ) t Capacitive load t Inductive load 34 Appendix: Steady-State Solution Adding all infinite number of bounces (t = ∞) for the step response, we have: V ( z,= ∞) V + (1 + Γ g Γ L + Γ 2g Γ 2L + Γ3g Γ3L + ) + V + Γ L (1 + Γ g Γ L + Γ 2g Γ 2L + Γ3g Γ3L + )     Sum of all right-traveling waves Sum of all left-traveling waves (1 + Γ L ) + V+ V + ΓL V = + = 1 − Γg ΓL 1 − Γg ΓL 1 − Γg ΓL = =  RL − Z 0  1 +   R + Z L 0    Rg − Z 0   RL − Z 0  1−   R + Z   R + Z  0  L 0   g Note: We have used ∞  Z0  V0     Rg + Z 0   RL − Z 0  Rg + Z 0 ) ( RL + Z 0 ) 1 + (   RL + Z 0   ( Rg + Z 0 ) ( RL + Z 0 ) − ( Rg − Z 0 ) ( RL − Z 0 ) ∑ zn = n =0 1 1− z ( z < 1)  Z0  V0     Rg + Z 0  35 Appendix: Steady-State Solution (cont.) Simplifying, we have:  RL − Z 0  1 +  ( Rg + Z 0 ) ( RL + Z 0 ) RL + Z 0  V ( z , ∞) =  ( Rg + Z0 ) ( RL + Z0 ) − ( Rg − Z0 ) ( RL − Z0 )  Z0  V0     Rg + Z 0   2 RL    ( Rg + Z 0 ) ( RL + Z 0 )  RL + Z 0  = ( Rg + Z0 ) ( RL + Z0 ) − ( Rg − Z0 ) ( RL − Z0 )  Z0  V0    + R Z 0  g  = (R g 2 RL ( Rg + Z 0 ) + Z 0 ) ( RL + Z 0 ) − ( Rg − Z 0 ) ( RL − Z 0 )  Z0  V0    R Z + 0  g  = 2 RL Z 0V0 ( Rg + Z0 ) ( RL + Z 0 ) − ( Rg − Z0 ) ( RL − Z0 ) = 2 RL Z 0V0 Rg RL + Z 02 + RL Z 0 + Rg Z 0 − Rg RL − Z 02 + RL Z 0 + Rg Z 0 = 2 RL Z 0V0 + RL Z 0 + Rg Z 0 + RL Z 0 + Rg Z 0 36 Appendix: Steady-State Solution (cont.) Continuing with the simplification: 2 RL Z 0V0 V ( z, ∞) = RL Z 0 + Rg Z 0 + RL Z 0 + Rg Z 0 = 2 RL Z 0V0 2 ( RL Z 0 + Rg Z 0 ) RL Z 0V0 = ( RL Z0 + Rg Z0 ) = RLV0 ( RL + Rg ) Hence we finally have:  RL V ( z , ∞) =   RL + Rg   V0  37