EE468G NOTES (8) Contents: Transmission line: Time domain analysis Transmission line parameters (0, Z) Transmission line equations Time-domain solution (general) Phase velocity, group velocity Step response Pulse responseFunction of transmission lines: Convey electromagnetic signal and power Transmission line examples Parallel Line Coaxial Cable Stripline Wavewgude Microstrip LineCircuit model and parameter: 0,Z Distributed parameters: R, G, L, C Circuit parameters: Voltage: V(z,t) – Time varying case, V(z): Time harmonic Current: I(z) – Time varying case, I(z): Time harmonic. Significance of line length: on microwave transmission line, the voltage and current are different from location to location along the length direction. This is different from traditional circuit network. In the following sketch, 1 2 3 4 5 6 7 8,,V V V V V V V V     z LZ 0,Z 0 Source Transmission line Load l 1 2 3 4 5 6 7 8 gRTransmission line equation (derived from Maxwell equation): General case: VIRI LztIVGV Czt     Lossless case: G=0, R=0. , (1), (2)VILztIVCzt Wave equation (lossless case): 2222From eq.(1), take on both sizeUse eq.(1) for IzV I IL L zz z t t zVLCttVLCt                            General solution:    , ( )V z t V t z u V t z u    2 2 2 22 2 2 2 211, or ,V I V ILC uz t z u tLC        Verify that  V t z u satisfies the wave equation. Let t z u Left hand side:       22222 2 21( ) 1 ( )1 1 1VVzuVVz u zVVu u u          Right hand side:    222222 2 2 2( ) ( )1 ( ) 1 ( )VVtVVtVVu t u Similar verification can be performed for the second component in the general solution.Let a special solution be given by cos , 1( , )2 3 30, OtherwisezzttV z t   Sketch this solution as function of z for t=0, 4, 8, 12. It can be seen that at different time instance, we saw the waveform in different locations. This indicates that the waveform is moving (traveling) in z-direction. The speed of traveling is 12 24 363 [m/s]4 8 12u     In the same way, we can find that the second component in the solution, ()V t z u travels in the –z direction at the same speed, u. z -3 0 3 z z z V(t-z/4) V(t-z/4) V(t-z/4) 9 12 15 21 24 27 33 36 39 (t=0) (t=4) (t=8) (t=12) V(t-z/4)Characteristic impedance: 00[ ], or [ ]VVZZII     For lossless transmission line: 0LZC For lossless TEM line: 11uLC Using the characteristic impedance, we can determine current given voltage function.  00( , ) ( ) ( )11, ( ) ( )zzV z t V t V tuuzzI z t V t V tuuZZ       Example: Calculate the characteristic impedance of the coaxial cable RG-58/U for which a=0.406 [mm], b=1.553 [mm], 2.26r. Solution:    002 2 2.26From eq. (6.12): 93.73 [pF/m]ln ln 1.553 0.406rCba         00From eq.(9.46): ln ln 1.553 0.406 0.268 [ H/m]22L b a   Hence 6090.268 1053.47 [ ]93.73 10LZC   Launching wave on transmission line: source condition At z=0: the input impedance looking into the line is : in 0RZ because the line on the right side extends to infinite and there is no reflected wave (no left-going wave). Using voltage division principle on the two resistors, inin g(0, ) ( )gRV t V tRR    in 00, 0,(0, )V t V tItRZ The solution at any location is:  , ( / )1( , ) ( / )ingin ggin gRV z t V t z uRRI z t V t z uRR From this solution, we can see that the voltage waveform on the line is the same as that of the source voltage. gR inR gV z 0 0Z gR inR Extend to Reflection from a resistive load The general voltage and current solution on the transmission line is  00, ( ) ( )11( , ) ( ) ( )zzV z t V t V tuuzzV z t V t V tuuZZ       At z=l,  00, ( ) ( )11( , ) ( ) ( )LzlLzlllV z t V t V tuullI z t V t V tuuZZ       Ohm’s law on the load resistor: /L L LV I R, hence 00( ) ( )11( ) ( )LLLllV t V tuuV V VR Z ZI V VllV t V tuu       Let ()()LlVtulVtu, we have 011LLLRZ or 00LLLRZRZ z 0 LR 0Z lExample: Calculate the reflection coefficient at z=l if 40[ ]LR  and 050[ ]Z . Solution: 40 50 100.1140 50 90L    Step response: Analysis procedure: Start from t=0 (the time to switch on) Calculate the voltage at z=0: V(0,0) Propagate this wave into z=l Find reflected wave at z=l Propagate the reflected wave to z=0 Calculate the reflected wave at z=0 Propagate the reflected wave to z=l ………. We will explain this procedure via an example: 08g100 , 10[ ], 50[ ]3 10 [m/s], 3[m]V 12[ ]LgR R ZulV        z 0 LR 0Z l gV 0t  gRAt t=0, z=0 (switch turned on):  010500,0 ( ) 12 10 [ ]50 10ggZV V V t VZR     The line voltage is then expressed as    , 10 [ ]V z t U t z u V The wave-front reaches z=l at the time of 89/ 3/3 10 10 10 [ ] 10 [ns]t l u s      Between t=0 and t=10 [ns], the only voltage wave on the line is this incident wave (+z propagating wave), as shown below. The current on the line is determined by the traveling wave equation: 0( , )/ ( , )Z V z t I z t Let I(z,t) be the current flowing in +z direction. When 0 10[ ]t ns: 0( , )( , ) ( , )10 ( / )0.2 ( / ) [A], 0 10[ ]50V z tI z t I z tZU t z uU t z u t ns     0 3 0 10 [ns]tAt t=10 [ns], the wave-front reaches the load at z=l: the incident wave is 10 [V]. Because the