Chapter 1 - Transmission Lines PDF

Preview

Summary

First chapter of book of transmission media and antenna systems...

Description

TRANSMISSION LINES LEARNING OBJECTIVES By the end of the chapter, the student is expected to learn:

Image Courtesy of http://support.brighthouse.com/media/video/equipment/wires/Coaxial_Cable.jpg

Coaxial cables are one of the most popular transmission lines we see today, especially in our television sets. Suppose that this coaxial cable has a polyethylene dielectric, determine how long (in nanoseconds) will a signal be transmitted with 500 m of this cable.

T

ransmission lines have always been a fundamental part of electronic communications system. These are everywhere. From telephone lines, Ethernet cable, CATV cables to LAN cables and even fiber optic cables (which will be discussed thoroughly in future lessons). All of these can be classified, in one way or another as transmission lines. And hence, it is very important to understand the basic principles of these lines in the course of the study in Transmission Media and Antenna Systems. This chapter focuses on the different characteristics of transmission lines and the properties of electromagnetic waves as these propagate through them. This chapter also includes exercises in the form of problem solving and conceptual questions which will widen the understanding of the student on the topic.

• The transmission line equivalent circuit and its primary and secondary line constant. • The behavior of an electromagnetic wave as it propagates in a transmission line. • The special cases of short and open transmission lines. • The transmission line impedance matching via quarter-wave transformer and stub – matching. • The transmission line measuring instruments and the Smith chart.

Chapter 1: Transmission Lines

Transmission Media and Antenna Systems

Unit 1

TRANSMISSION LINE EQUIVALENT CIRCUIT AND LINE CONSTANTS In this lesson, we will be focusing on the electrical/electronic equivalent of these transmission lines and their characteristics. We will discover how these lines differ from the ordinary copper wires we use and study in our previous electronics engineering subjects. Transmission lines characteristics are primarily based on both its physical and electrical properties. Physical properties like diameter and conductor spacing and electrical properties such as conductivity and relative permittivity (otherwise known as dielectric constant) on the other hand determine what we call the primary line constants . These line constants in turn are used to represent a transmission line equivalent model. A typical transmission line equivalent circuit is shown in Figure 1.1.

Figure 1.1 Transmission line equivalent circuit with primary line constants.

As we can see in the figure, the four primary line constants are: 1. Series resistance (R) 2. Series inductance (L) 3. Shunt capacitance (C) 4. Shunt conductance (G)1 These line constants are uniformly distributed throughout the line, i.e., their values depend on how long the transmission line is. That is why, sometimes, these units are termed as distributed parameters. And to simplify the analysis, a transmission line is analyzed not as a whole but as several sections which is a unit length of the cable. Hence, their units are expressed in units per length. (For example, the series resistance is commonly expressed in ohms per meter or ohms per foot) These primary line constants can be expressed and/or calculated quantitatively. However, the computations vary from transmission line to transmission line.

1Usually,

PITFALL ALERT! A common mistake of the students studying transmission lines is the failure to differentiate a primary line constant to a load. If the problem refers to the transmission line, it is a line constant and therefore in units per length (ex: ohms/m, F/m). The load, on the other hand, is not distributed, therefore not expressed in units per length (it is expressed usually in ohms)

the shunt conductance value is zero, especially in applications in communications engineering; hence, its calculation will not be discussed in this text. But you may refer to engineering electromagnetics books for discussions of shunt conductance on transmission lines if you wish so

2 TMA

Chapter 1: Transmission Lines

Transmission Media and Antenna Systems

Hence, this text focuses only on the two of the most common types of transmission lines. The parallel – wire line and the coaxial cable.

D

Series Resistance S

The series resistance of a parallel – wire line and the coaxial cable is given by equations 1.1 and 1.2 respectively.

√𝑓 𝑅 = 0.1 𝑟 1 1 𝑅 = 0.1√𝑓 ( + ) 𝑑 𝐷

Figure 1.2 Parallel – wire line

(1.1) d (1.2)

The series resistances in both cases are both expressed in ohms/100 ft. The radius (r) of the parallel wire line and the inner (d) and outer (D) diameters of the coaxial cable are all expressed in in. and the frequency (f) is in MHz. It is notable that frequency is a factor in the computation for the distributed resistance of a transmission line. As we can see in the formula, it is evident that frequency is directly proportional to the square root of the frequency. And this phenomenon is the consequence of what we call as the skin effect.

D

Figure 1.3 Coaxial cable

Skin effect not only affects the resistance of a transmission line. It has also the effect of making the electromagnetic wave travel closer to surface as frequency increases, thus arising to name of the phenomenon skin effect for this makes wave propagation literally in the skin of the transmission line. In our future lessons, we will revisit the skin effect and how it is vital in the operation of antennas.

EXAMPLE 1.1 Resistance of a 10Base – 5 Cable A coaxial cable with an outer diameter of 3/8” and a wall thickness of 0.134 inch is used as an Ethernet cable. (This is the typical 10Base – 5 coaxial cable dimensions) Determine the resistance of 500 meters (typical length of 10Base – 5) of this cable. Assume a frequency of 1 GHz.

SOLUTION We determine first the inner diameter. From Figure 1.2, we can easily say the relationship between the wall thickness and the outer diameter.

𝑑 = 𝐷 − 2𝑡 𝑑 = 0.375 − 2(0.134) = 0.107 𝑖𝑛 1 1 + ) 0.107 0.375 𝑅 = 0.38 Ω/100 𝑓𝑡

Using equation 1.2

𝑅 = 0.1√0.1 (

Finally

R 500 m = (0.38 Ω/100 𝑓𝑡)(500 𝑚 ≫ 𝑓𝑡) 𝐑 𝟓𝟎𝟎 𝐦 = 𝟔. 𝟐𝟑 𝛀

3 TMA

Chapter 1: Transmission Lines

Transmission Media and Antenna Systems

CHECK YOUR UNDERSTANDING 1. A parallel – wire line has a diameter of 1/2 inch and a conductor spacing of 2 inches. It has air as its dielectric. Determine the distributed resistance of this transmission line in ohms/km if it operates in a frequency of 800 MHz. 2. What will happen if the operating frequency of a line increases to a tremendously high – level? Is the effect of this favorable for the user of the transmission line?

RECALL… Distributed Series Inductance and Shunt Capacitance For parallel – wire line and coaxial cable transmission lines, the following equations can be used to calculate the distributed series inductance and shunt capacitance. For parallel – wire lines

Series Inductance

Shunt Capacitance

2𝑆 𝜇 ln ( ) 𝑑 𝜋 𝜋𝜖 𝐶= 2𝑆 ln ( 𝑑 )

𝐿=

(1.3)

(1.4)

𝜇 = 𝜇𝑜 𝜇𝑟

Where: μo - absolute permeability (4π x 10-7 H/m) μr - relative permeability (unitless)

𝜖 = 𝜖 𝑜 𝜖𝑟

Where: εo - absolute permittivity (8.854 x 10-12 F/m) εr - relative permittivity (unitless)

For coaxial cables

Series Inductance

Shunt Capacitance

𝐿=

𝜇

2𝜋

𝐶=

𝐷 ln ( ) 𝑑

2𝜋𝜖 𝐷 ln (𝑑 )

(1.5)

(1.6)

The distributed inductances in the equations above are in H/m while the distributed capacitances are in F/m. The units for the dimensions of the transmission lines may be any unit of length as long as they are consistent with each other. Finally, the permeability μ and permittivity ε are also in H/m and F/m respectively. These two parameters are of very much importance in the study of transmission line especially in applications related to communications engineering. As we will see later in this lesson, the distributed inductance and capacitance will be used very often so it pays to know the basic equations needed for calculating the values of these primary line constants.

Figure 1.4 William Thomson (Lord) Kelvin. Together with James Clerk Maxwell and Oliver Heaviside, Lord Kelvin first worked in the mathematical analysis of transmission lines. (Image Courtesy of http://www.bbc.co.uk/history/historic_figures/i mages/kelvin_lord.jpg)

4 TMA

Chapter 1: Transmission Lines

Transmission Media and Antenna Systems

EXAMPLE 1.2 Capacitance of a No.22 AWG Wire No.22 AWG wire is used as a parallel – wire line transmission line. Determine the capacitance of three miles of this transmission line if the conductors used are spaced by 8 inches polyethylene dielectric. (From AWG tables, No.22 wire diameter is 0.0253 inches)

SOLUTION We first determine the distributed capacitance of the transmission line using equation 1.4. Take note that the value of the dielectric constant is not given

readily which is oftentimes the case. The value of the dielectric constant for polyethylene is 2.27 𝐶= Finally

𝜋(8.854𝑥10−12 )(2.27) = 9.79 𝑝𝐹/𝑚 2⋅8 ln (0.0253)

C3 mi = (9.79

𝑝𝐹 𝑚

) (3 𝑚𝑖 ⋅

5280 𝑓𝑡 ≫ 𝑚) 𝑚𝑖

𝐂𝟑 𝐦𝐢 = 𝟎. 𝟎𝟒𝟕 𝝁𝑭

EXAMPLE 1.3 Distributed Inductance of a RG-58U The outer diameter of an RG-58U (a low power RF coaxial cable commonly used as generic carrier of signals in laboratory instruments like oscilloscope) coaxial cable measures 0.195 inch. If it has a polyethylene dielectric, determine its inner diameter so that it will have a distributed inductance of approximately 76.8 nH/ft.

76.8𝑥10−9 ⋅

(4𝜋𝑥10−7 ) 0.195 3.28 𝑓𝑡 ) ln ( = 𝑑 2𝜋 1𝑚

Solving for d, we have: 𝒅 = 𝟎. 𝟎𝟓𝟓 𝒊𝒏

SOLUTION Since there is no relative permeability given in the problem, which is usually the case, we use a relative permeability of 1. Using equation 1.5

CHECK YOUR UNDERSTANDING 1. Determine the distributed capacitance of a Teflon (k = 2.1) dielectric coaxial cable with an inner – to – outer radius ratio of 0.22. 2. A No. 6 AWG (d = 0.162 in) copper wire is used as a parallel – wire transmission line with air dielectric. If the line is spaced 12 inches, determine the distributed inductance L of the transmission line in μH/m. 3. Using your knowledge on AWG tables, what is the relation of the AWG number to the distributed inductance and capacitance when used as parallel – wire air – dielectric transmission lines?

Characteristic Impedance Looking back to the transmission line equivalent circuit, let us examine a single section of the equivalent model as seen in Figure 1.5. Here, we used an unbalanced transmission line to make the analysis much clearer.

5 TMA

Chapter 1: Transmission Lines

Transmission Media and Antenna Systems One of the assumptions that we have to understand in transmission lines is that they are considered infinetly long. Yes! Although transmission lines are not really infinite, they are very long compared to other wires used in electronics engineering (another reason for why we consider this infinite will be explained later). And as the length of a wire increases, as we are about to see here and in the future lessons, the current that travels through it behaves differently.

Δx Figure 1.5 One section of equivalent circuit model of an unbalanced transmission line.

In our analysis, we get a single section of a transmission line with length Δx. From the figure, it is easy to say that Z1 = (R + jωL)Δx and (1/Z2)= (G + jωC)Δx. We multiply the values of R, L, G and C by Δx because as we recall, they are distributed, hence their values are in units/length. It is implied then that to get a certain value of resistance, inductance, conductance and capacitance, we need to multiply these quantities by a certain length. We can use any length in this case, but using a differential length for a single section is more appropriate. Why? This is to show that even in the smallest section of the transmission line; these four primary constants are equal. Yes. Let us take note then that every differential section of this line is equal since again, the four primary constants are distributed. Now consider a transmission line, using the section with a differential length Δx and the assumption that is infinite, we can actually derive the value of the impedance (in ohms) seen looking into the line. (The derivation is left for the student as an exercise; see Short Quiz 1.1 Problem 3) This is given by equation 1.7

𝑍𝑜 = √

𝑅 + 𝑗𝜔𝐿 𝐺 + 𝑗𝜔𝐶

(1.7)

The value of the characteristic impedance is in ohms while ω (2πf) is the angular frequency in rad/s and j being the imaginary operator.

PITFALL ALERT! One common error that may be encountered when finding the value of Zo using equation 1.7 is by forgetting that the values inside the square root are complex numbers. And most calculators could not evaluate the square root of complex numbers. This leaves students often confused why they are getting MATH ERROR whenever they try to calculate the Zo. The best remedy for that is first to evaluate the inner part of the equation first (without the square root), then convert it to polar form. By De Moivre’s theorem, the square root of a complex number in polar form can be calculated by getting the square root of the and then magnitude getting the half of the argument. These will be the magnitude and argument of the answer, and in turn can be easily converted back to rectangular form.

6 TMA

Chapter 1: Transmission Lines

Transmission Media and Antenna Systems This impedance seen looking into the line is what we call the characteristic impedance or sometimes called as the surge impedance. The characteristic impedance, as we see equation 1.7, does not depend on the length of the line but rather on the four primary line constants discussed previously. Thus, the characteristic impedance can be considered as a secondary line constant. (It is one of the two secondary line constants, the other one will be discussed later). Secondary line constants describe the transmission characteristics of a transmission line. It gives us the idea and generalization of the behavior of current as it travels along the line.

EXAMPLE 1.4 Characteristic Impedance I Given the following primary line constants: R = 0.1 ohm/ft L = 76.8 nH/ft G = 0.05 S/ft C = 31.4 pF/ft Determine the characteristic impedance of the line at (a) 60 Hz; (b) 1 MHz; (c) 2.45 GHz

SOLUTION

𝑍𝑜 = √

(b) at 1 MHz 𝑍𝑜 = √

0.1 + 𝑗(2𝜋 ⋅ 1𝑥106 )(76.8𝑥10−9 ) 0.05 + 𝑗(2𝜋 ⋅ 1𝑥106 )(31.4𝑥10−12 ) 𝒁𝒐 = 𝟐. 𝟒 + 𝒋𝟐. 𝟎 𝛀

(c) at 2.45 GHz

Using equation 1.7 to determine the value of the characteristic impedance, we have: (a) at 60 Hz

𝒁𝒐 = 𝟏. 𝟒 + 𝐣𝟐. 𝟏𝐱𝟏𝟎−𝟒 𝛀

𝑍𝑜 = √

0.1 + 𝑗(2𝜋 ⋅ 2.45𝑥109 )(76.8𝑥10−9 ) 0.05 + 𝑗(2𝜋 ⋅ 2.45𝑥109 )(31.4𝑥10−12 )

0.1 + 𝑗(2𝜋 ⋅ 60)( 76.8𝑥10−9 ) 0.05 + 𝑗(2𝜋 ⋅ 60)(31.4𝑥10−12 )

𝒁𝒐 = 𝟒𝟗. 𝟑 + 𝒋𝟐. 𝟓 𝛀

The characteristic impedance is of very much importance in the study and application of transmission lines. Why? Because this determines whether the transmission we intend to do with the line is successful or not. As we can recall from circuit theory, we have this theorem called the maximum power transfer theorem which states that in a circuit, to achieve maximum power transfer from source to a load, the impedance looking in to the source must be equal to the impedance of the load. Using this theorem, if the characteristic impedance of our transmission line is terminated to a load impedance equal to it, maximum power transfer will be achieved. In other words, no power will be loss in the line and all the power sent will be dissipated in the load (which is the true goal in communications engineering). Hence, it is very important for us to know the characteristic impedance of a transmission line to be able to compensate it with the load impedance and consequently achieve a lossless transmission.

ZL = Zo

Figure 1.6 Source, transmission line and load set-up. It shows that an infinitely long transmission line is equal to a transmission line terminated to a load impedance equal to Zo

7 TMA

Chapter 1: Transmission Lines

Transmission Media and Antenna Systems

As mentioned earlier, there is one more reason why a transmission line is considered to be infinite in length. Recall from the last section that if a transmission line is terminated by a load impedance equal to its characteristic impedance, lossless transmission will occur. All the power from the generator will be absorbed by the load, no power will be returned or reflected back to the source. And apparently, that is also the case when the line is infinite. The signal will travel infinitely throughout the line with no reflections. Thus, terminating a transmission line with an impedance equal to Zo makes the line look like infinite. Now, having established the idea about the characteristic impedance, let us look back again to equation 1.7.

𝑍𝑜 = √

𝑅 + 𝑗𝜔𝐿 𝐺 + 𝑗𝜔𝐶

(1.7)

Evidently, the equation is in complex domain due to the presence of reactive elements L and C. With this, we expect that Z o will also be in complex domain. However, there are two operating regions of concern in the actual application of transmission lines that renders the characteristic impedance to be real and purely resistive. Notice that if the frequency wherein the transmission line operates is too low, it is evident that R >> jωL and G >> jωC thus turning the equation 1.7 to equation 1.8 shown below.

𝑍𝑜 = √

𝑅 𝐺

(1.8)

Equation 1.8 shows the characteristic impedance of a transmission line used at low frequencies. This is typically used in electrical engineering where the frequency of the line could be as low as 60 Hz. Now, if the frequency on the other hand is too high, it is then evident that jωL >> R and jωC >> G thus turning the original characteristic impedance equation to equation 1.9, which is real and purely resistive

𝑍𝑜 = √

𝐿 𝐶

(1.9)

This equation for the characteristic impedance is the one most used for communications engineering applications.

8 TMA

Chapter 1: Transmission Lines

Transmission Media and Antenna Systems

EXAMPLE 1.5 Characteristic Impedance II A transmission line has L = 32.5 nH/m and C = 12.2 pF/m. What value of load resistance should this line be terminated to ensure a lossless transmission?

SOLUTION The problem refers to the characteristic impedance. Using equation 1.9, we have

𝑍𝑜 = √

32.5𝑥10−9

12.2𝑥 10−12

Finally, we can use equation 1.9 in conjunction with equations 1.3 to 1.6 and change of base to get equations for the values of characteristic impedance of parallel – wire transmission line and coaxial cable. The derivation is left for the students as an exercise (see Check Your Understanding, problem 1 and Short Quiz 1.1, problem 6).

Table 1.1 Dielectric constants of some materials

For parallel – wire line:

276 2𝑆 𝑍𝑜 = log ( ) 𝐷 √𝜖𝑟

Dielectric

Relative Permittivity

Vacuum