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

2. Exercise 11.3
A 40-m long TL has Vg=15 Vrms, Zo=30+j60 W, and VL=5e-j48o Vrms. If the line is matched to the load and the generator, find: the input impedance Zin, the sending-end current Iin and Voltage Vin, the propagation constant g.
Answers: ZL Zg Vg +
Vin -
Iin VL
Zo=30+j60
g=a +j b
40 m

3. Transmission Lines TL parameters TL Equations
Input Impedance, SWR, power
Smith Chart
Applications
Quarter-wave transformer
Slotted line
Single stub
Microstrips

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

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

6. Other TL (higher order) [Chapter 11]

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

8. 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

9. 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.
Parameter
Two-Wire Line
Coaxial Line
Parallel-Plate Line
Unit R L G C

10. TL representation

11. Distributed line parameters
Using KVL:

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

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

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

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

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

17. Example:
An air filled planar line with w=30cm, d=1.2cm, t=3mm, sc=7x107 S/m.
Find R, L, C, G for 500MHz
Answer d w

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

19. Different cases of TL Lossless Distortionless Lossy Transmission line

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

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

22. Summary
g = a + jb Zo
=General
Lossless
Distortionless
RC = GL

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

24. Define reflection coefficient at the load, GL

25. Terminated, Lossless TL
Then,
Similarly,
The impedance anywhere along the line is given by
The impedance at the load end, ZL, is given by

26. 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

27. Example
A generator with 10Vrms and Rg=50, is connected to a 75W load thru a 0.8l, 50W-lossless line.
Find VL

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

29. Impedance (Lossy line)
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.

30. Impedance (Lossless line)
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.

31. Exercise : using formulas
A 2cm lossless TL has Vg=10 Vrms, Zg=60 W, ZL=100+j80 W and Zo=40W, l=10cm . find: the input impedance Zin, the sending-end Voltage Vin, ZL Zg Vg +
Vin -
Iin VL Zo
g=j b
2 cm
Voltage Divider:

32. 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

33. 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 G=0

34. Three common Cases of line-load combinations:
Shorted Line (ZL=0)
Open-circuited Line (ZL=∞)
Matched Line (ZL = Zo)

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

36. Standing Waves -Open Open Line (ZL=∞) ,we had So substituting in V(z)
Voltage minima
|V(z)| -z
-l/4
-l/2 -l

37. Standing Waves -Matched
Matched Line (ZL = Zo), we had
So substituting in V(z)
|V(z)| -z
-l/4
-l/2 -l

38. Java applets http://www.amanogawa.com/transmission.html

39. The Smith Chart

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

41. What can be seen on the screen?
Network Analyzer
What can be seen on the screen?

42. Smith Chart
Suppose you use as coordinates the reflection coefficient real and imaginary parts.
and define the normalized ZL: Gi
|G| Gr

43. 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

44. Examples of circles of r and x
Circles of x

45. Examples of circles of r and x
Circles of x

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

47. Fun facts about the Smith Chart
A lossless TL is represented as a circle of constant radius, |G|, 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

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

49. Fun facts : Admittance in the SC
The admittance, y=YL/Yo where Yo=1/Zo, can be found by moving ½ turn (l/4) on the TL circle

50. Fun facts about the Smith Chart
The Gr +axis, where r > 0 corresponds to Vmax
The Gr -axis, where r < 0 corresponds to Vmin
Vmin
Vmax (Maximum impedance)

51. Exercise: using S.C.
A 2cm lossless TL has Vg=10 Vrms, Zg=60 W, ZL=100+j80 W and Zo=40W, l=10cm . find: the input impedance Zin, the sending-end Voltage Vin,
Load is at S.C.
Move .2l and arrive to .4179l
Read ZL Zg Vg +
Vin -
Iin VL Zo
g=j b
2 cm
Voltage Divider:

52. zL
zin
0.2l

53. Exercise: cont….using S.C.
A 2cm lossless TL has Vg=10 Vrms, Zg=60 W, ZL=100+j80 W and Zo=40W, l=10cm . find: the input impedance Zin, the sending-end Voltage Vin,
Distance from the load (.2179l) to the nearest minimum & max
Move to horizontal axis toward the generator and arrive to .5l (Vmax) and to .25l for the Vmin.
Distance to min= =.282l
Distance to 2st voltage maximum is .282l +.25l=.482 See drawing ZL Zg Vg +
Vin -
Iin VL Zo
g=j b
2 cm

54. Exercise : using formulas
A 2cm lossless TL has Vg=10 Vrms, Zg=60 W, ZL=100+j80 W and Zo=40W, l=10cm . find: the input impedance Zin, the sending-end Voltage Vin, ZL Zg Vg +
Vin -
Iin VL Zo
g=j b
2 cm
Voltage Divider:

55. Distance to first Vmax:
Another example:
A 26cm lossless TL is connected to load ZL=36-j44 W and Zo=100W, l=10cm . find: the input impedance Zin
Load is at S.C.
Move .1l and arrive to .527l (=.027l)
Read ZL Zg Vg +
Vin -
Iin VL Zo
g=j b
26cm
Distance to first Vmax:

56. Exercise 11.4
A 70 W lossless line has s =1.6 and qG =300o. If the line is 0.6l long, obtain G, ZL, Zin and the distance of the first minimum voltage from the load.
Answer
The load is located at:
Move to l and draw line from
center to this place, then read where it crosses you TL circle.
Distance to Vmin in this case, lmin =.5l-.3338l=

58. Java Applet : Smith Chart

59. Applications Slotted line as a frequency meter Impedance Matching
If ZL is Real: Quarter-wave Transformer (l/4 Xmer)
If ZL is complex: Single-stub tuning (use admittance Y)
Microstrip lines

60. HP Network Analyzer in Standing Wave Display
Slotted Line
Used to measure frequency and load impedance
HP Network Analyzer in Standing Wave Display

61. Slotted line example
Given s, the distance between adjacent minima, and lmin for an “air” 100W transmission line, Find f and ZL
s=2.4, lmin=1.5 cm, lmin-min=1.75 cm
Solution: =8.6GHz
Draw a circle on r=2.4, that’s your T.L.
move from Vmin to zL

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

63. 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
Zin=Zo therefore z =1 or y=1 (this is our goal!)

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

65. 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 Yo or yline = 1+jb
Then the length of the shorted stub is adjusted so that it's susceptance cancels that of the line, or ystub= -jb

66. Example: Single Stub
A 75W lossless line is to be matched to a 100-j80 W load with a shorted stub. Calculate the distance from the load, the stub length, and the necessary stub admittance.
Answer:
Change to:
=0.094l (1+j.96) or next intersection:0.272l,
Short stub: =0.126l
With ystub= -j.96/75 =-j.0128 mhos

68. Example: Single Stub
A 50W lossless T.L. is 20 m long and terminated into a 120+j220W load.   To perfectly match , what should be the length and location of a short-circuited stub line. Assume an operating frequency of 10MHz.
Answer:  l=30m, zL= 2.4+j4.4 (circulito rojo). 
Trazo raya por el centro y leo yL al otro lado (circulito azul).
La yL está en la posición 0.472l.
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 0.214l.
Por tanto la distancia desde la carga al segmento(stub) = (debido a cambio de escala) +.214= .242 l.
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 .2958l. El segmento empieza en carga de corto circuito (ver palabra roja que dice short) está en .25l. Por lo tanto el largo del segmento es
= .0485l

69. Microstrips

70. Microstrips analysis equations & Pattern of EM fields

71. Microstrip Design Equations
Falta un radical en eeff

72. Microstrip Design Curves

73. Example
A microstrip with fused quartz (er=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: eeff=2.75, Zo=86.03 W, l=18.09 mm

74. Diseño de microcinta:
Dado (er=4) para el substrato, y h=1mm halla w para Zo=50 W y cuánto es eeff?
Solución: Suponga que como Z es pequena w/h>2