EE468G NOTES (8B) Reading assignment: Chapter 11 Contents: Transmission line: time harmonic analysisTime harmonic wave on transmission line The transmission line equation in frequency domain can be obtained by replacing t with j: ()()VRI j LI R j L IzIGV j CV G j C Iz           In the above, V and I are complex functions of “z”. Lossless case: , (1), (2)Vj L IzIj CVz Take z on both sides of eq. (1),  222Iuse eq.(2) to replace zVIjLzzj L j CV LCV             Let 22LC, LC [1/m] is the wavenumber. The equation for the voltage function V(z): 2220VVz General solution:  j z j zV z V e V e   Using eq.(1), we get the solution for the current I(z):  0011,j z j zj z j zVjI z V e V ej L z j LLV e V e ZZ          Parameters of a lossless transmission line: Characteristic impedance: 0, [ ]L L LZCLC    Wavenumber: 2[rad/m] or [1/m]LC Phase velocity: 1puLC LC   Group velocity: gdud For lossy line: the general solution can be written as:  z j z z j zV z V e e V e e        is the attenuation constant, its unit is [1/m] or [Np/m] (Np = Nepper). Sometimes, the attenuation constant is expressed in the unit of [dB/m]. The relation of the two units are: [dB/m] = 8.686 [Np/m][Np/m] = 0.1151 [dB/m] The attenuation constant is important for long-distance transmission. For example, if a cable has an attenuation constant of 0.3 [dB/km], then the attenuation would be 30 [dB] at 100 km distance away. This means the signal power is reduced by a factor of 1000. Special case for lossless TEM line: LCGroup velocity: The velocity of information/energy propagation. Consider the amplitude modulated signal. Phase velocity is the velocity of the carrier Group velocity is the velocity of the envelope. General definition of group velocity: gdud Compare with phase velocity: pugu pu Envelope CarrierExample: Calculate the phase velocity and group velocity at f=300MHz on a coaxial cable that is filled with dielectric 92(3 10 )r. Assume that the coaxial cable’s inner and outer conductor radius to be 2 [mm] and 6 [mm]. Solution: For coaxial cable,     292012 92 3 10240.13 10 3 10ln lnCb a b a         7704 10ln ln 8/2 2.77 10 [H/m]22L b a    7 12 9 29 9 22.77 10 40.13 10 (3 10 )3.33 10 (3 10 )LC             9 9 99 9 8998g93.33 10 (3 10 ) 103.33 10 3 2 10 2 3 103.33 10 6.7722.5 10 [1/m]1u 0.44 10 [m/s]22.5 10dddd                 In the general solution, the V and V are two constants that will be determined by the load and source conditions (boundary conditions). At z=0, the ratio of reflected voltage and incident voltage is called voltage reflection coefficient (at load location): LVV, This means: LVV, With this definition, the general solution can be written as:      00()1j z j z j z j zLLj z j z j z j zLLV z V e V e V e eVI z V e V e e eZZ                       z LZ 0Z ()gVt -l 0  j z j zV z V e V e   +z propagation (Incident wave) -z propagation (Reflected wave)At z=0:     0(0) 101LLVVVIZ    , Since (0)(0)LVZI, we have   0011(0)(0) 11LLLLLVVZZVIZ   Solve for L from the above, we have: 00LLLZZZZ Example: Given 0100 30 [ ], 50 [ ]LZ j Z    , calculate the voltage reflection coefficient at the load. Solution: 000100 30 50100 30 5050 30150 300.359 0.1280.381 ( 19.66 )LLLZZZZjjjjj  The load reflection coefficient Lis an alternative representation of the load impedance. In other words, when we know the load impedance, we can calculate the reflection coefficient. We can also determine the load impedance if the reflection coefficient is given. In fact, this is how the impedance is measured Example: An unknown load impedance is attached to a transmission line with 075[ ]Z . It is found that 0.25 63oL  . Determine the load impedance. Solution: From: 00LLLZZZZ, we find that    6306311 0.257511 0.251 0.25 0.454 0.891751 0.25 0.454 0.8911.114 0.223750.887 0.22375 1.122 0.533(84.16 40) [ ]oojLLjLeZZejjjjjj      Input impedance at a reference point (z=-l) on a transmission line Consider the general solution:    0()j z j zLj z j zLV z V e eVI z e eZ     Substitute z=-l into the above two equations, and take the ratio of V(z) and I(z):    002002()()11 ( )1 1 ( )j l j lj l j lLLj l j lj l j lLLjlLjlLV e eeeVlZ l ZVI l e eeeZelZZel            In the above, the reflection coefficient at z=-l is defined as 2()jlLle    Using the identity:cos( ) sin( )jle l j l, and the expression for    00L L LZ Z Z Z    we can write the above in another format:    000tantanLLZ jZ lZ l ZZ jZ l z LZ 0Z -l 0 V(z)) ()Zl I(z)Example: Given 0 eff20 [ ], 50[ ], 1.5,LZZ     f=50 MHz, l=1 [m], calculate , , ( ),Ll   , and ()Zl. Solution:     6eff8L2 2 1.28 12 2 50 101.5 1.28[1/m]3 1020 50 300.42920 50 70tan tan 1.28 1 3.37( ) 0.429 0.429 0.358 0.23620 50 3.3750 87.7 50.2 [ ]50 20