Transmission Lines (Bounce Diagram) 2

Two Transmission Lines Here we have a reflection and transmission coefficient at the junction between two lines. t=0 = Rg 225 [ Ω ] + V0 = 4 [ V ] - vg ( t ) T = 1 [ ns] Junction T = 1 [ ns] = Z 0 150 [ Ω ] = Z 0 75 [ Ω ] z=0 Note: When the wave first hits the junction, it sees the characteristic impedance of the line on the other side as a load. 150 − 75 1 = 225 3 4 TJ+ = 1 + Γ +J = 3 = Γ +J KVL: TJ = 1 + ΓJ = RL 50 [ Ω ] z=L TJ+ Γ +J TJ− Γ −J 75 − 150 1 Γ −J = =− 225 3 2 TJ− = 1 + Γ −J = 3 (This follows from the fact that voltage must be continuous across the junction.) 14 Two Transmission Lines (cont.) Bounce Diagram for Two Lines t=0 = Rg 225 [ Ω ] + V0 = 4 [ V ] - T = 1 [ ns] vg ( t ) 1 Γg = 2 1 t [ns] ↓ 1 [V] 4 = RL 50 [ Ω ] z=L 1 Γ +J = 3 2 TJ− = 3 TJ+ = 4 3 Γ −J =− 0[V] 1 3 Γ L =− 0.3333 [V] 1.3333 [V] 1.3333 [V] 1.5000 [V] 1.3333 [V] −0.6667 [V] 0.1667[V] 0.0555[V] -0.4444 [V] -0.3888 [V] 1 2 z [m] 0 [V] 1 [V] 2 3 T = 1 [ ns] = Z 0 150 [ Ω ] Z 0 75 [ Ω ] = z=0 0 Junction 0.2222 [V] 0.2222 [V] 0.4444 [V] 1.1111 [V] 0.6667 [V] 1.1111 [V] 15 Pulse Response Superposition can be used to get the response due to a rectangular pulse. Rg vg ( t ) + +- - vg ( t ) Z0 RL z=0 V0 vg ( t ) z=L vg (= t ) V0 ( u ( t ) − u ( t − W ) ) u(t) = unit step function t 1.0 u (t ) W We thus subtract two bounce diagrams, with the second one being a shifted version of the first one. t 16 Example: Pulse Find an “oscilloscope trace” of the voltage at z = 0.75 L = Rg 225 [ Ω ] Oscilloscope trace = Z 0 75 [ Ω ] vg ( t ) +- z=L z=0 z = 0.75L Given pulse: V0 vg ( t ) V0 = 4 [ V ] V + = 1 [V] W = 0.25 [ ns] W = RL 25 [ Ω ] T = 1 [ ns] t  Z0  V 1 [V] = Recall : V + =  R + Z  0 0 g   17 Example (cont.) (Oscilloscope trace for original step function) [V] 1 [V] 0.5 [V] 0.375 [V] 0.4375 [V] 0.25 [V] 1 2 3 4 5 t [ns] Subtract [V] 1 [V] Shifted trace 0.5 [V] 0.375 [V] 0.4375 [V] 0.25 [V] 1 2 3 4 5 (shifted by 0.25 [ns]) t [ns] 18 Example: Pulse (cont.) V0 vg ( t ) V0 = 4 [ V ] W = 0.25 [ ns] W t = Rg 225 [ Ω ] Oscilloscope trace = Z 0 75 [ Ω ] vg ( t ) +- = RL 25 [ Ω ] T = 1 [ ns] z=L z=0 z = 0.75L v ( 43 L , t ) 1 [V] Oscilloscope trace of voltage 0.125 [V] 1 −0.5 [V] 2 3 −0.25 [V] 0.0625 [V] 4 5 t [ns] −0.03125 [V] 19 Example (cont.) Final “oscilloscope trace” of the voltage at z= vg ( t ) + - 3 L 4 RL Z0 z=0 z = 0.75 L 1 [V] z=L v ( 43 L , t ) 0.125 [V] 1 −0.5 [V] 2 3 −0.25 [V] 0.0625 [V] 4 5 t [ns] −0.03125 [V] 20 Example: Pulse (cont.) Find a “snapshot” of the voltage at t = 1.5 [ns]. = Rg 225 [ Ω ] = Z 0 75 [ Ω ] vg ( t ) +- z=0 V0 vg ( t ) = RL 25 [ Ω ] T = 1 [ ns] t = 1.5 [ ns] V0 = 4 [ V ] z=L V + = 1 [V] W = 0.25 [ ns] W t 21 Example: Pulse (cont.) Shifted numbers 0 1 Original step 1 Γg = 2 1 [V] 1 [V] 3 0.25 [V] 4 5 6 1 [V] 2 L/2 2 t [ns] − 1 − [V] 4 1 + [V] 8 1 [V] 16 1 − [V] 32 Γ L =− 0 1 − [V] 64 0.390625 [V] 0.25 0.5 [V] 1 [V] 2.25 3.25 0.375 [V] 1 Γg = 2 Delayed step Γ L =− 0.25 [ nS] 0 1 [V] 1.25 + 0.4375 [V] 1 2 0.25 [V] 4.25 − 1 [V] 2 1 [V] 4 1 + [V] 8 3L / 4 1 2 z [m] 0.5 [V] − 1 [V] 16 1 − [V] 32 0.375 [V] + 5.25 0.40625 [V] 6.25 0.4375 [V] 1 − [V] 64 0.390625 [V] 0.40625 [V] 22 Example: Pulse (cont.) Snapshot from original step function [ V] 1.0 0.5 0.25L [ V] 0.5L 0.75L L z [m] Snapshot from shifted step function Subtract 1.0 0.5 0.25L 0.5L 0.75L L z [m] 23 Example: Pulse (cont.) V0 vg ( t ) V0 = 4 [ V ] Snapshot W = 0.25 [ ns] t W = Rg 225 [ Ω ] Z 0 75 [ Ω ] = vg ( t ) +- = RL 25 [ Ω ] T = 1 [ ns] z=L z=0 Snapshot of voltage v ( z , 1.5 [ns]) 0.25L 0.5L 0.75L L z [m] Pulse is moving to the left −0.5 [V] 24 Capacitive Load Rg = Z 0 t=0 V0 [ V ] + - CL vg ( t ) Z0 z=0 z=L Note: The generator is assumed to be matched to the transmission line for convenience (we wish to focus on the effects of the capacitive load). 0 Hence Γ g = The reflection coefficient is now a function of time. 25 Capacitive Load (cont.) ( L − z ) / cd ( delay = td ) t= d td t=0 V0 [ V ] Rg = Z 0 CL + v (t ) g - Z0 z=0 z=L z (time = t)  Z0  V = V  R + Z  0 0   g = V0 / 2 t =0 t V + = V0 / 2 ΓL (t ) Γ g =0 + T 2T 3T V+ 0 V+ z Γ L ( t − td ) V + (1 + Γ ( t − t ) ) V L d + 26