Transmission Lines (Bounce Diagram) 1

ECE 3317 Applied Electromagnetic Waves Notes 8 Transmission Lines (Bounce Diagram) 1 Step Response The concept of the bounce diagram is useful to find a step response on a terminated lossless line: t=0 Lossless line Rg + vg ( t ) - V0 [ V ] Z0 z=L z=0 V0 vg ( t ) u(t) = unit step function vg ( t ) = V0 u ( t ) t Generator voltage RL Note: The bounce diagram is useful if the source is a step function or a rectangular pulse (discussed later in these notes). If the source is something else, it is better to use the general theory presented in Notes 7. 2 Step Response (cont.) The voltage wave is shown approaching the load. t=0 V+ t = t1 t = t2 t=0 Rg V0 = 1 [ V ] + - vg ( t ) cd RL Z0 z=L z=0  Z0  V = V  R + Z  0 0   g + (from voltage divider) 3 Bounce Diagram t=0 Rg + V0 = 1 [ V ]  RL − Z 0  ΓL =   R Z + 0   L  Rg − Z 0  Γg =   + R Z 0   g z=L Γg L T= cd  Z0  V = V  R + Z  0 0   g Z0 z=0 Basic constants + vg ( t ) - RL ΓL t =0 T V 0 + V+ ΓL V + t 2T ↓ 3T ΓgΓL V + (1 + Γ L ) V + 4T (1 + Γ L + Γ g Γ L ) V + Γ g Γ 2L V + (1 + Γ L + Γ g Γ L + Γ g Γ 2L ) V + 5T (1 +  + Γ 2g Γ 2L ) V + 6T z The purple values give the total voltage in each triangular region. Γ 2g Γ 2L V + Γ 2g Γ3L V + (1 +  + Γ 2g Γ3L ) V + 4 Steady-State Solution Adding all infinite number of bounces (t = ∞), 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 After some math, we have (please see the Appendix):  RL V ( z , ∞) =   RL + Rg   V0  Note: The steady-state solution does not depend on the transmission line length or the characteristic impedance! This is the DC circuit-theory voltage divider equation! 5 Example = Rg 225 [ Ω ] t=0 V0 = 4 [ V ] + - vg ( t ) Z 0 75 [ Ω ] = = RL 25 [ Ω ] T = 1 [ ns] z=L z=0 0 V +  Z0  = V0 1 [V] R Z + 0   g  RL − Z 0  1 Γ L = = −  2 R Z + 0   L Γg  Rg − Z 0  1  =   Rg + Z 0  2 1 t [ns] ↓ 1 Γg = 2 Γ L =− 1 [V] 1 [V] 2 3 4 5 6 0.25 [V] − 1 [V] 2 1 [V] 4 1 + [V] 8 0 1 2 z [m] 0.5 [V] − 1 + [V] 16 0.4375 [V] 1 − [V] 32 1 − [V] 64 0.390625 [V] 0.375 [V] 0.40625 [V] 6 Example (cont.) The bounce diagram can be used to get an “oscilloscope trace” of the voltage at any point on the line. 0 1 t [ns] ↓ z= 1 Γg = 2 1 [V] 2 3 4 5 6 0.25 [V] − 3 L 4 Γ L =− 1 [V] 0 0.75 [ns] 1 [V] 2 1.25 [ns] 1 − [V] 4 1 + [V] 8 1 + [V] 16 0.4375 [V] 1 − [V] 32 1 − [V] 64 0.390625 [V] z= 1 2 z [m] v( 34 L , t ) 0.5 [V] 2.75 [ns] 3 L 4 [V] (oscilloscope trace) 1 [V] 3.25 [ns] 0.375 [V] 0.5 [V] 0.375 [V] 4.75 [ns] 0.4375 [V] 0.25 [V] 0.40625 [V] 1 2 3 4 5 t [ns]  RL  Steady state voltage: = v( z , ∞)  = V 0.400 [V]  R + R  0 g   L 7 Example (cont.) Final “oscilloscope trace” of the voltage at z= 3 L 4 t=0 + V0 = 4 [ V ] - vg ( t ) Z0 RL z = 0.75 L z=0 z=L 1 [V] 0.5 [V] 0.375 [V] 0.4375 [V] 0.25 [V] 1 2 3 t [ns] 4 5 8 Example (cont.) The bounce diagram can also be used to get a “snapshot” of the line voltage at any point in time. t = 3.75 [ ns] v ( z , 3.75 [ns]) (snapshot ) [V] Wavefront is moving to the left L/4 t = 3.75 [ns] 0.375 [V] 0.25 [V] L 4 L 2 z [m] 3L 4 L 9 Example (cont.) This set of snapshots shows the waves bouncing back and forth. Initial wave Γ L =− 1.0 [ V ] First reflected wave 1 2 1.0 [ V ] 1.0 [ V ] 0.5 [ V ] 1.0 [ V ] 1.0 [ V ] 1.0 [ V ] 1.0 [ V ] 0.5 [ V ] 0.5 [ V ] 0.5 [ V ] 10 Bounce Diagram for Current We just change the signs of the reflection coefficients, as shown. Γ Ig =− 0 1 t [ns] ↓ 1 1 Γ IL = 2 Normalized Current 2 1 1 1 2 2 3 1 4 1 − 8 1.5 − 1.25 4 5 0 1.1875 6 1.203125 1 16 1 32 1 − 64 1.125 z [m] Γ I = −Γ i − ( L, t ) Γ =+ i ( L, t ) I L v − ( L, t )( −1 / Z 0 ) = + v ( L, t )(1 / Z 0 ) v − ( L, t ) = − + v ( L, t ) = −Γ L 1.21875 Note: The normalized current is defined as Z0 i (z,t). 11 Example for Current Oscilloscope Trace for Current z= Normalized Current Γ Ig =− 0 1 t [ns] ↓ 1 2 1 1.25 4 5 3 L 4 1 2 3 z= 1.1875 6 1.203125 0.75 [ns] 1 2 1.25 [ns] 1 Γ IL = 2 0 z [m] 3 L 4 Z 0i ( 43 L , t ) (oscilloscope trace of current) 1.5 1.5 1 − 4 2.75 [ns] 1 − 83.25 [ns] 1.125 1 16 4.75 [ns] 1 32 1.21875 1 − 64  V0  ( z, ∞)  = 0.016 [A] Steady state current: i=  R + R  g   L 1.25 1 1.125 (units are volts) 1 2 3 4 1.1875 5 t [ns ] Z= 0i ( z , ∞) = ( 0.016 )( 75) 1.20 [ V ] 12 Example for Current (cont.) t = 3.75 [ ns] Snapshot for Current Γ Ig =− 0 1 t [ns] ↓ 1 2 Z 0i ( z , 3.75 [ns]) 1 I Γ = Normalized Current L 2 1 1 1 2 2 3 1.25 Wavefront is moving to the left z [m] 1.25 1.125 1.5 − L/4 4 t = 3.75 [ns] 5 1 4 1 − 8 0 (snapshot of current) 1.1875 6 1.203125 1 16 1 32 1 − 64 (units are volts) 1.125 1.21875 L 4 L 2 3L 4 L z [m] 13