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Transmission Lines Dr. Sandra Cruz-Pol ECE Dept. UPRM

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Exercise 11.3 A 40-m long TL has V g =15 V rms, Z o =30+j60, and V L =5e -j48 o V rms. If the line is matched to the load and the generator, find: the input impedance Z in, the sending-end current I in and Voltage V in, the propagation constant. Answers: ZLZL ZgZg VgVg + V in - I in +VL-+VL- Z o =30+j60 +j 40 m

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Transmission Lines I.TL parameters II.TL Equations III.Input Impedance, SWR, power IV.Smith Chart V.Applications Quarter-wave transformer Slotted line Single stub VI.Microstrips

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Transmission Lines (TL) TL have two conductors in parallel with a dielectric separating them They transmit TEM waves inside the lines

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Common Transmission Lines Two-wire (ribbon) Coaxial Microstrip Stripline (Triplate)

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Other TL (higher order) [Chapter 11]

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Fields inside the TL V proportional to E, I proportional to H

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Distributed parameters The parameters that characterize the TL are given in terms of per length. R = ohms/meter L = Henries/ m C = Farads/m G = mhos/m

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Common Transmission Lines R, L, G, and C depend on the particular transmission line structure and the material properties. R, L, G, and C can be calculated using fundamental EMAG techniques. ParameterTwo-Wire LineCoaxial LineParallel-Plate Line Unit R L G C

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TL representation

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Distributed line parameters Using KVL:

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Distributed parameters Taking the limit as z tends to 0 leads to Similarly, applying KCL to the main node gives

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Wave equation Using phasors The two expressions reduce to Wave Equation for voltage

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TL Equations Note that these are the wave eq. for voltage and current inside the lines. The propagation constant is and the wavelength and velocity are

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Waves moves through line The general solution is In time domain is Similarly for current, I z

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Characteristic Impedance of a Line, Z o Is the ratio of positively traveling voltage wave to current wave at any point on the line z

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Example: An air filled planar line with w =30cm, d =1.2cm, t=3mm, c =7x10 7 S/m. Find R, L, C, G for 500MHz Answer w d

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Exercise 11.1 A transmission line operating at 500MHz has Z o =80, =0.04Np/m, =1.5rad/m. Find the line parameters R,L,G, and C. Answer: 3.2 /m, 38.2nH/m, S/m, 5.97 pF/m

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Different cases of TL Lossless Distortionless Lossy Transmission line

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Lossless Lines (R=0=G) Has perfect conductors and perfect dielectric medium between them. Propagation: Velocity: Impedance

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Distortionless line (R/L = G/C) Is one in which the attenuation is independent on frequency. Propagation: Velocity: Impedance

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Summary j ZoZo =General Lossless Distortionless RC = GL

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Excersice 11.2 A telephone line has R=30 /km, L=100 mH/km, G=0, and C= 20 F/km. At 1kHz, obtain: the characteristic impedance of the line, the propagation constant, the phase velocity. Solution:

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Define reflection coefficient at the load, L

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Terminated, Lossless TL Then, Similarly, The impedance anywhere along the line is given by The impedance at the load end, Z L, is given by

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Terminated, Lossless TL Then, Conclusion: The reflection coefficient is a function of the load impedance and the characteristic impedance. Recall for the lossless case, Then

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Example A generator with 10V rms and R g =50, is connected to a 75 load thru a lossless line. Find V L

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Terminated, Lossless TL It is customary to change to a new coordinate system, z = - l, at this point. Rewriting the expressions for voltage and current, we have Rearranging, -z-z z = - l

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The impedance anywhere along the line is given by The reflection coefficient can be modified as follows Then, the impedance can be written as After some algebra, an alternative expression for the impedance is given by Conclusion: The load impedance is transformed as we move away from the load. Impedance (Lossy line)

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The impedance anywhere along the line is given by The reflection coefficient can be modified as follows Then, the impedance can be written as After some algebra, an alternative expression for the impedance is given by Conclusion: The load impedance is transformed as we move away from the load. Impedance (Lossless line)

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Exercise : using formulas A 2cm lossless TL has V g =10 V rms, Z g =60, Z L =100+j80 and Z o =40, =10cm. find: the input impedance Z in, the sending-end Voltage V in, ZLZL ZgZg VgVg + V in - I in +VL-+VL- Z o j 2 cm Voltage Divider:

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SWR or VSWR or s Whenever there is a reflected wave, a standing wave will form out of the combination of incident and reflected waves. The (Voltage) Standing Wave Ratio - SWR (or VSWR) is defined as

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Power The average input power at a distance l from the load is given by which can be reduced to The first term is the incident power and the second is the reflected power. Maximum power is delivered to load if =0

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Three common Cases of line-load combinations: Shorted Line ( Z L =0 ) Open-circuited Line ( Z L = ) Matched Line ( Z L = Z o )

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Standing Waves -Short Shorted Line ( Z L =0 ), we had So substituting in V(z) -z /4 /2 |V(z)| Voltage maxima *Voltage minima occurs at same place that impedance has a minimum on the line

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Standing Waves -Open Open Line ( Z L = ),we had So substituting in V(z) |V(z)| -z /4 /2 Voltage minima

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Standing Waves -Matched Matched Line ( Z L = Z o ), we had So substituting in V(z) |V(z)| -z /4 /2

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Java applets html html ?CONTENT_ID=2483 ?CONTENT_ID=2483

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The Smith Chart

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Smith Chart Commonly used as graphical representation of a TL. Used in hi-tech equipment for design and testing of microwave circuits One turn (360 o ) around the SC = to /2

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What can be seen on the screen? Network Analyzer

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Smith Chart Suppose you use as coordinates the reflection coefficient real and imaginary parts. and define the normalized Z L : i r

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Now relating to z=r+jx After some algebra, we obtain two eqs. Similar to general equation of a circle of radius a, center at (h,k) Circles of r Circles of x

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Examples of circles of r and x Circles of r Circles of x

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Examples of circles of r and x Circles of r Circles of x

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The joy of the SC Numerically s = r on the +axis of r in the SC Proof:

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Fun facts about the Smith Chart A lossless TL is represented as a circle of constant radius, | |, or constant s Moving along the line from the load toward the generator, the phase decrease, therefore, in the SC equals to moves clockwisely. To generator

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Fun facts about the Smith Chart One turn (360 o ) around the SC = to /2 because in the formula below, if you substitute length for half- wavelength, the phase changes by 2, which is one turn. Find the point in the SC where =+1,-1, j, -j, 0, 0.5 What is r and x for each case?

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Fun facts : Admittance in the SC The admittance, y=Y L /Y o where Y o =1/Z o, can be found by moving ½ turn ( /4) on the TL circle

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Fun facts about the Smith Chart The r +axis, where r > 0 corresponds to V max The r -axis, where r < 0 corresponds to V min V max (Maximum impedance) V min

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Exercise: using S.C. A 2cm lossless TL has V g =10 V rms, Z g =60, Z L =100+j80 and Z o =40, =10cm. find: the input impedance Z in, the sending-end Voltage V in, Load is S.C. Move.2 and arrive to.4179 Read ZLZL ZgZg VgVg + V in - I in +VL-+VL- Z o j 2 cm Voltage Divider:

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zLzL z in 0.2

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Exercise: cont….using S.C. A 2cm lossless TL has V g =10 V rms, Z g =60, Z L =100+j80 and Z o =40, =10cm. find: the input impedance Z in, the sending-end Voltage V in, Distance from the load (.2179 to the nearest minimum & max Move to horizontal axis toward the generator and arrive to.5 V max and to.25 for the V min. Distance to min= =.282 Distance to 2 st voltage maximum is.282 See drawing ZLZL ZgZg VgVg + V in - I in +VL-+VL- Z o j 2 cm

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Exercise : using formulas A 2cm lossless TL has V g =10 V rms, Z g =60, Z L =100+j80 and Z o =40, =10cm. find: the input impedance Z in, the sending-end Voltage V in, ZLZL ZgZg VgVg + V in - I in +VL-+VL- Z o j 2 cm Voltage Divider:

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Another example: A 26cm lossless TL is connected to load Z L =36-j44 and Z o =100, =10cm. find: the input impedance Z in Load is S.C. Move.1 and arrive to.527 Read ZLZL ZgZg VgVg + V in - I in +VL-+VL- Z o j 26cm Distance to first V max :

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Exercise 11.4 A 70 lossless line has s =1.6 and =300 o. If the line is 0.6 long, obtain, Z L, Z in and the distance of the first minimum voltage from the load. Answer The load is located at: Move to.4338 and draw line from center to this place, then read where it crosses you TL circle. Distance to V min in this case, l min = =

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Java Applet : Smith Chart

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Applications Slotted line as a frequency meter Impedance Matching If Z L is Real: Quarter-wave Transformer ( /4 X mer ) If Z L is complex: Single-stub tuning (use admittance Y) Microstrip lines

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Slotted Line Used to measure frequency and load impedance HP Network Analyzer in Standing Wave Display are/naswave/Stdwave.pdf

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Slotted line example Given s, the distance between adjacent minima, and l min for an air 100 transmission line, Find f and Z L s=2.4, l min =1.5 cm, l min-min =1.75 cm Solution: =8.6GHz Draw a circle on r=2.4, thats your T.L. move from V min to z L

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Quarter-wave transformer …for impedance matching ZLZL Z in Z o, l= /4 Conclusion: **A piece of line of /4 can be used to change the impedance to a desired value (e.g. for impedance matching)

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Single Stub Tuning …for impedance matching A stub is connected in parallel to sum the admittances Use a reactance from a short-circuited stub or open-circuited stub to cancel reactive part Z in =Z o therefore z =1 or y=1 (this is our goal!)

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Single Stub Basics We work with Y, because in parallel connections they add. Y L (=1/Z L ) is to be matched to a TL having characteristic admittance Y o by means of a "stub" consisting of a shorted (or open) section of line having the same characteristic admittance Y o

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Single Stub Steps First, the length l is adjusted so that the real part of the admittance at the position where the stub is attached is equal to Y o or y line = 1+ jb Then the length of the shorted stub is adjusted so that it's susceptance cancels that of the line, or y stub = - jb

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Example: Single Stub A 75 lossless line is to be matched to a 100-j80 load with a shorted stub. Calculate the distance from the load, the stub length, and the necessary stub admittance. Answer: Change to : =0.094 j or next intersection:0.272, Short stub: = With y stub = -j.96/75 j mhos

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Example: Single Stub A 50 lossless T.L. is 20 m long and terminated into a 120+j220 load. To perfectly match, what should be the length and location of a short-circuited stub line. Assume an operating frequency of 10MHz. Answer: =30m, z L = 2.4+j4.4 (circulito rojo). Trazo raya por el centro y leo y L al otro lado (circulito azul). La y L está en la posición Trazo círculo amarillo, esa es mi T.L. donde está la carga. Busco donde interseca el circulo de r =1(circulo turquesa). Lo interseca en 1+j3.2 Ese es mi 1+jb. Y está en la posicion Por tanto la distancia desde la carga al segmento(stub) = (debido a cambio de escala) +.214=.242. Ahora miro el círculo del segmento= circulo grande en el SC (donde Gamma =1).y busco donde jb = -j3.2 (abajo ver flecha roja). Para buscar su posición, trazo línea desde el centro hasta afuera y leo posición está en El segmento empieza en carga de corto circuito (ver palabra roja que dice short) está en.25. Por lo tanto el largo del segmento es =.0485

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Microstrips

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Microstrips analysis equations & Pattern of EM fields

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Microstrip Design Equations Falta un radical en eff

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Microstrip Design Curves

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Example A microstrip with fused quartz ( r =3.8) as a substrate, and ratio of line width to substrate thickness is w/h =0.8, find: Effective relative permittivity of substrate Characteristic impedance of line Wavelength of the line at 10GHz Answer: eff =2.75, Z o =86.03, =18.09 mm

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Diseño de microcinta: Dado ( r =4) para el substrato, y h =1mm halla w para Z o =50 y cuánto es eff ? Solución: Suponga que como Z es pequena w/h>2

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