9/27/2017

Course Instructor Dr. Raymond C. Rumpf Office: A‐337 Phone: (915) 747‐6958 E‐Mail: [email protected]

EE 4347 Applied Electromagnetics

Topic 4a Transmission Lines

Transmission Lines These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited Slide 1

Lecture Outline • Introduction • Equations • Transmission Line Wave Equations • Transmission Line Parameters –  and 

– Characteristic Impedance, Z0 • Special Cases of Transmission Lines – General transmission lines – Lossless lines – Weakly absorbing lines – Distortionless lines • Examples – RG‐59 – Microstrip design

Transmission Lines Slide 2

1 9/27/2017

Introduction

Transmission Lines Slide 3

Map of (LI Media)

• Confines and transports waves. Waveguides • Supports higher‐order modes. • Has one or less conductors. • Usually what is implied by Transmission Lines “Pipes” the label “.” • Contains two or more conductors. • No low‐frequency cutoff. • Thought of more as a circuit clement Metal Shell Pipes Dielectric Pipes Homogeneous Inhomogeneous • Enclosed by metal. • Composed of a core and a cladding. • Has TEM mode. • Supports only • Does not support TEM mode. • Symmetric waveguides have no • Has TE and TM quasi‐(TEM, TE, • Has a low frequency cutoff. low‐frequency cutoff. modes. & TM) modes. Homogeneous Channel Waveguides • Supports TE and TM modes • Confinement along two axes. coaxial microstrip • TE & TM modes only supported in circularly symmetric guides.

rectangular circular dual‐ridge Single‐Ended stripline coplanar optical Fiberphotonic crystal rib Inhomogeneous • Supports TE and TM modes Slab Waveguides buried parallel coplanar strips only if one axis is uniform. plate • Otherwise supports quasi‐TM • Confinement only along one axis. and quasi‐TE modes. • Supports TE and TM modes. • Interfaces can support surface waves. Differential

shielded pair slotline no uniform axis uniform axis dielectric Slab large‐area interface (no TE or TM) (has TE and TM) parallel plate Transmission Lines Slide 4

2 9/27/2017

Transmission Line Parameters RLGC

We can think transmission lines as being composed of millions of tiny little circuit elements that are distributed along the length of the line.

In fact, these circuit element are not discrete, but continuous along the length of the transmission line.

Transmission Lines Slide 5

RLGC Circuit Model

It is not technically correct to represent a transmission line with discrete circuit elements like this. However, if the size of the circuit z is very small compared to the of the signal on the transmission line, it becomes an accurate and effective way to model the transmission line.

z

Transmission Lines Slide 6

3 9/27/2017

L‐Type Equivalent Circuit Model

Distributed Circuit Parameters There are many possible circuit models for R (/m) Resistance per unit length. transmission lines, but most produce the same Arises due to resistivity in the equations after analysis. conductors. L (H/m) Inductance per unit length. Arises due to stored magnetic R z L z energy around the line. G (1/m) Conductance per unit length. Arises due to conductivity in the dielectric separating the conductors. Gz Cz 1 G  R C (F/m) Capacitance per unit length. Arises due to stored electric energy between the conductors. z zz

Transmission Lines Slide 7

Relation to Electromagnetic Parameters

Every transmission line with a homogeneous fill has: ,  , LC   G   C 

Transmission Lines Slide 8

4 9/27/2017

Fundamental Vs. Intuitive Parameters

Fundamental Parameters Intuitive Parameters Electromagnetics Electromagnetics ,  , n,  ,  , , tan 

Transmission Lines Transmission Lines

RLGC, , , Z0 ,  ,  , VSWR

The fundamental parameters are the The intuitive parameters provide intuitive most basic parameters needed to solve a insight about how signals behave on a transmission line problem. transmission line. However, it is difficult to be intuitive They isolate specific information to a about how they affect signals on the line. single parameter. An electromagnetic analysis is needed to determine R, L, G, and C from the The intuitive parameters are calculated geometry of the transmission line. from R, L, G, and C .

Transmission Lines Slide 9

Example RLGC Parameters

RG‐59 Coax CAT5 Twisted Pair Microstrip

R  36 mΩ m R 176 mΩ m R 150 mΩ m L  430 nH m L  490 nH m L  364 nH m G 10  m G  2  m G  3  m C  69 pF m C  49 pF m C 107 pF m

Z0 75 Z0 100 Z0 50

Surprisingly, almost all transmission lines have parameters very close to these same values.

Transmission Lines Slide 10

5 9/27/2017

Transmission Line Equations

Transmission Lines Slide 11

E & H  V and I

Fundamentally, all circuit problems are electromagnetic problems and can be solved as such.

All two‐conductor transmission lines either support a TEM wave or a wave very closely approximated as TEM.

An important property of TEM waves is that E is uniquely related to V and H and uniquely related to E.     VEd   I  Hd L L

This let’s us analyze transmission lines in terms of just V and I. This makes analysis much simpler because these are scalar quantities!

Transmission Lines Slide 12

6 9/27/2017

Transmission Line Equations

The transmission line equations do for transmission lines the same thing as Maxwell’s curl equations do for unguided waves.

Maxwell’s Equations Transmission Line Equations   H VI E  RI L t zt   E I V H  GV  C t zt

Like Maxwell’s equations, the transmission line equations are rarely directly useful. Instead, we will derive all of the useful equations from them.

Transmission Lines Slide 13

Derivation of First TL Equation (1 of 2) R z L z

+ 2 3 + I zt, Vzt ,  1 Gz Cz 4 Vz  zt, 

‐ ‐

z zz

Apply Kirchoff’s law (KVL) to the outer loop of the equivalent circuit: Izt, Vzt ,  IztRz ,  Lz Vz zt,   0   t  1 2 4 3

Transmission Lines Slide 14

7 9/27/2017

Derivation of First TL Equation (2 of 2)

We rearrange the equation by bringing all of the voltage terms to the left‐hand side of the equation, bringing all of the current terms to the right‐hand side of the equation, and then dividing both sides by z. Izt, Vzt,, IztRz   Lz Vz zt ,0 t  Vz zt,,  Vzt  Izt , RI z, t L zt

In the limit as z  0, the expression on the left‐hand side becomes a derivative with respect to z.

Vzt,, Izt RI z, t L zt

Transmission Lines Slide 15

Derivation of Second TL Equation (1 of 2)

R z L z 1 2 + I  zt,  I  zzt ,  + 3 4 Vzt ,  Gz Cz Vz  zt, 

‐ ‐

z zz

Apply Kirchoff’s current law (KCL) to the main node the equivalent circuit: Vz zt, Izt ,  Iz zt, GzVz  zt,  Cz 0     t   1 2 3 4

Transmission Lines Slide 16

8 9/27/2017

Derivation of Second TL Equation (2 of 2)

We rearrange the equation by bringing all of the current terms to the left‐hand side of the equation, bringing all of the voltage terms to the right‐hand side of the equation, and then dividing both sides by z. Vz zt, IztIzztGzVzztCz,,  ,  0 t  Iz zt,,  Izt  Vz  zt , GV z z, t C zt

In the limit as z  0, the expression on the left‐hand side becomes a derivative with respect to z.

Izt,, Vzt GV z, t  C zt

Transmission Lines Slide 17

Transmission Line Wave Equations

Transmission Lines Slide 18

9 9/27/2017

Starting Point – Telegrapher Equations

We start with the transmission line equations derived in the previous section.

Vzt ,, Izt  Izt ,, Vzt  RI z, t L GV z, t  C time‐domain ztzt

For time‐harmonic (i.e. frequency‐domain) analysis, we Fourier transform the equations above.

dV z dI z R jLIz GjCVz frequency‐domain dz dz

Note: Our derivative d/dz became an ordinary derivative because z is the only independent variable left. These last equations are commonly referred to as the telegrapher equations.

Transmission Lines Slide 19

Wave Equation in Terms of V(z)

dV z dI z R jLIz Eq. (1)GjCVz Eq. (2) dz dz To derive a wave equation in terms of V(z), we first differentiate Eq. (1) with respect to z. dV2  z dIz  RjL Eq. (3) dz2 dz

Second, we substitute Eq. (2) into the right‐hand side of Eq. (3) to eliminate I(z) from the equation. dV2  z R jL G jCVz dz 2 Last, we rearrange the terms to arrive at the final form of the wave equation. dV2  z RjLGjCVz  0 dz 2

Transmission Lines Slide 20

10 9/27/2017

Wave Equation in Terms of I(z)

dV z dI z R jLIz Eq. (1)GjCVz Eq. (2) dz dz To derive a wave equation in terms of just I(z), we first differentiate Eq. (2) with respect to z. dI2  z dVz  GjC Eq. (3) dz2 dz

Second, we substitute Eq. (1) into the right‐hand side of Eq. (3) to eliminate V(z) from the equation. dI2  z GjCRjLIz dz 2 Last, we rearrange the terms to arrive at the final form of the wave equation. dI2  z GjCRjLIz  0 dz 2

Transmission Lines Slide 21

Propagation Constant, 

In our wave equations, we have a common term GjCRjL . Define the  to be j  GjCRjL    

Given this definition, the transmission line equations are written as dV2  z  2Vz 0 dz2 dI2  z  2 Iz 0 dz2

Transmission Lines Slide 22

11 9/27/2017

Solution to the Wave Equations

If we hand the wave equations off to a mathematician, they will return with the following solutions.

2 dV z zz   2Vz 0  Vz  Ve Ve dz2 00

2 dI z zz   2 Iz 0  I  zIeIe dz2 00

Forward wave Backward wave

Both V(z) and I(z) have the same differential equation so it makes sense they have the same solution.

Transmission Lines Slide 23

Transmission Line Parameters:

Attenuation Coefficient,  Constant, 

Transmission Lines Slide 24

12 9/27/2017

Derivation  and  (1 of 7)

Step 1 – Start with our expression for .

j  GjCRjL    

Square this expression to get rid of square‐root on right‐hand side.

 j2 GjCRjL 

Expand this expression.

22j2 RG j  RC j  LG  2 LC

Collect real and imaginary parts on the left‐hand and right‐hand sides. 22 j2  RG   2 LC  j  RC  LG

Transmission Lines Slide 25

Derivation  and  (2 of 7)

Step 2 – Generate two equations by equating real and imaginary parts.

2  RCLG

22 j2  RG   2 LC  j  RC  LG

22RGLC   2

We now have two equations and two unknowns.

Transmission Lines Slide 26

13 9/27/2017

Derivation  and  (3 of 7)

Step 3 – Derive a quadratic equation for 2.

2  RC LG Eq. (1a) 22RG   2 LC Eq. (1b)

Solve Eq. (1a) for .   RC LG Eq. (2) 2

Substitute Eq. (2) into Eq. (1b) and simplify.

2  22RC LG RG LC 2  2 RC LG 2 22RG LC 4 2 44442RCLG2  2 RG  22 LC 2  422LC RG RC LG 0 2 Transmission Lines Slide 27

Derivation  and  (4 of 7)

Step 4 – Solve for 2 using the quadratic formula.

bb2 4 ac Recall the quadratic formula: ax2  bx c0 x  2a Our equation for  is in the form of a quadratic equation where a 1 2 2 bLCRG 422  2 LC RG RC LG 0  2 cRCLG  2 x   2 The solution is

222  LC RG LC RG4  RC  LG 2 2   2 RG2222222 LC R L G  C  2

Transmission Lines Slide 28

14 9/27/2017

Derivation  and  (5 of 7)

Step 5 – Resolve the sign of the square‐root.

RGLCR2222222 LGC   2  2

In order for this expression to always give a real value for , the sign of the square‐root must be positive.

The final expression is

RGLCR2222222 LGC   2  2

Transmission Lines Slide 29

Derivation  and  (6 of 7)

Step 6 – Solve for 2 using our expression for 2.

Recall Eq. (1b): 22RGLC   2

RG2222222 LC R L G  C   2  2

We obtain an equation for 2 by substituting our expression for 2 into Eq. (1b).

RG2222222 LC R L G  C  22RGLC  2  RG2222222 LC R L G  C  2  2

Transmission Lines Slide 30

15 9/27/2017

Derivation  and  (7 of 7)

Step 7 – We arrive at our final expressions for  and  in terms of the fundamental parameters R, L, G, and C by taking the square‐root of our latest expressions for 2 and 2.

RG2222222 LC R L G  C   2 RGLCR2222222   LGC    2 Both  and must be positive quantities for passive materials. This means we take the positive sign for the square‐root.

RG2222222 LC R L G  C   2

RG2222222 LC  R  L G  C   2

Transmission Lines Slide 31

Transmission Line Parameters:

Characteristic Impedance, Z0

Transmission Lines Slide 32

16 9/27/2017

Characteristic Impedance, Z0 ()

The characteristic impedance Z0 of a transmission line is defined as the ratio of the voltage to the current at any point of a forward travelling wave.  VV00 Z0  I00I

Definition for a forward Definition for a backward travelling travelling wave. wave. Notice the negative sign!

Most characteristic impedance values fall in the 50  to 100  range. The specific value of impedance is not usually of importance. What is important is when the impedance changes because this causes reflections, standing waves, and more.

Transmission Lines Slide 33

Derivation of Z0 (1 of 5)

Step 1 – Substitute our solution into the transmission line equations.

 z  z Vz   Ve00 Ve z  z Iz Ie00 Ie

dV z dI z R jLIz GjCVz dz dz

d zz  d zz   Ve00 Ve  Ie00 Ie dz dz z  z z  z RjL Ie00 Ie GjC Ve00 Ve

Transmission Lines Slide 34

17 9/27/2017

Derivation of Z0 (2 of 5)

Step 2 – Expand the equations and calculate the derivatives.

d zz  d zz  Ve00 Ve Ie00 Ie dz dz z  z z  z R jL  Ie00  Ie GjCVe 00  Ve

zz  zz  Ve00 Ve Ie00 Ie  z  z  z  z R jLIe00  R jLIe GjCVe00  GjCVe

Transmission Lines Slide 35

Derivation of Z0 (3 of 5)

Step 3 – Equate the expressions multiplying the common exponential terms.  VRjLI00 

    Ve z  Ve zzz R jLIe    R jLIe   00   0 0

 VRjLI00 

 IGjCV00 

    Ie z  Ie zzz G jCVe    G jCVe   00   0  0

 IGjCV00 

Transmission Lines Slide 36

18 9/27/2017

Derivation of Z0 (4 of 5)

Step 4 – Solve each of our four equations for V0/I0 to derive expressions for Z0.

 V RjL  VRjLI 0 00    Z0 I0 

 V0 RjL  VRjLI00   Z0 I0 

V    0 Z IGjCV00 0 IGjC0  

V    0  Z IGjCV00  0 IGjC0  

Transmission Lines Slide 37

Derivation of Z0 (5 of 5)

Step 5 – Put Z0 in terms of just R, L, G, and C.

Recall our expression for : j  GjCRjL    

We can substitute this into either of our expressions for Z0. RjL   Z  0  GjC 

Proceed with the first expression.

2 R jL RjL  R jL Z0  GjCRjLGjCRjL  GjC 

Transmission Lines Slide 38

19 9/27/2017

Final Expression for Z0 ()

We have derived a general expression for the characteristic impedance

Z0 of a transmission line in terms of the fundamental parameters R, L, G, and C. VV Definition: 00 Z0  I00I R jL  R jL Expression: Z   0  GjC GjC

Transmission Lines Slide 39

Dissecting the Characteristic Impedance, Z0

The characteristic impedance describes the and phase relation between voltage and current along a transmission line. With this picture in mind, the characteristic impedance can be written as

Vz Ve z ZZ  0 00Z0  j zzV0    zZ0 Iz Ie000 e  ZVe e Z0 The characteristic impedance can also be written in terms of its real and imaginary parts.

Z00RjX 0

Reactive part of Z0. This is not equal to jL or 1/jC.

Resistive part of Z0. This is not equal to R or G.

Transmission Lines Slide 40

20 9/27/2017

Special Cases of Transmission Lines:

General Transmission Line

Transmission Lines Slide 41

Parameters for General TLs

Propagation Constant,  jGjCRjL      

Attenuation Coefficient,  RG2222222 LC R L G  C   2 Phase Constant,  RG2222222 LC  R  L G  C   2

Characteristic Impedance, Z0 RjL  ZRjX  00 0 GjC 

Transmission Lines Slide 42

21 9/27/2017

Special Cases of Transmission Lines:

Lossless Lines

Transmission Lines Slide 43

Definition of Lossless TL

When we think about transmission lines, we tend to think of the special case of the lossless line because the equations simplify considerably.

For a transmission line to be lossless, it must have RG0

Transmission Lines Slide 44

22 9/27/2017

Parameters for Lossless TLs

Propagation Constant,  jjLC   

Attenuation Coefficient, 

  0

Phase Constant, 

 LC

Characteristic Impedance, Z0 L ZRjX  00 0 C L RX 0 00C

Transmission Lines Slide 45

Special Cases of Transmission Lines:

Weakly Absorbing Line

Transmission Lines Slide 46

23 9/27/2017

Definition of Weakly Absorbing TL

Most practical transmission lines have loss, but very low loss making them weakly absorbing.

We will define a weakly absorbing line as

RL  and G  C

Ensures very little conduction between the lines through the dielectric.

Ensures low ohmic loss for signals propagating through the line.

Transmission Lines Slide 47

Parameters for Weakly Absorbing TLs

Attenuation Coefficient,  1 R  GZ0 2 Z0 Conductance through the dielectric dominates attenuation in high‐impedance transmission lines.

Resistivity in the conductors dominates attenuation in low‐impedance transmission lines.

In weakly absorbing transmission lines, there usually exists a “sweep spot” for the impedance where attenuation is minimized.

Transmission Lines Slide 48

24 9/27/2017

Special Cases of Transmission Lines:

Distortionless Lines

Transmission Lines Slide 49

Definition of Distortionless TL

In a real transmission line, different frequencies will be attenuated differently because  is a function of . This causes distortion in the signals carried by the line.

RG2222222 LC R L G  C    2 To be distortionless, there must be a choice of R, L, G, and C that eliminates  from the expression of , effectively making  independent of frequency .

The necessary condition to be distortionless is

RG  L C

Transmission Lines Slide 50

25 9/27/2017

Parameters for Distortionless TLs

Propagation Constant, 

jRGjLC    

Attenuation Coefficient, 

  RG

Phase Constant,  To be distortionless, we must have   .  is a measure of how quickly a signal accumulates phase. Different frequencies have  LC different and therefore must accumulate different phase through the same length of line.

Characteristic Impedance, Z0 RL ZRjX   00 0 GC RL RX  0 00GC

Transmission Lines Slide 51

Example:

Properties of RG-59 Coax

Transmission Lines Slide 52

26 9/27/2017

The Lossless Circular Coax

Fundamental Parameters (derived in EE 3321) 2 C  Fm ln ba  1 b L ln  H m 24 a a Attenuation Coefficient,    0 b Phase Constant,      and 

Characteristic Impedance, Z0  b Z00RjX 0  ln ab 2 a  b RX00ln 0 2 a

Transmission Lines Slide 53

Typical RLGC for RG‐59 Coax at 2 GHz

The typical RG‐59 coaxial cable operating at 2.0 GHz has the following RLGC parameters:

R  36 mΩ m L  430 nH m G 10  m C  69 pF m

Calculate the transmission line parameters , , , and Z0.

Classify the line as lossless, weakly absorbing, distortionless, etc.

Transmission Lines Slide 54

27 9/27/2017

Solution (1 of 3)

Our equations mostly utilize the angular frequency  instead of the ordinary frequency f. 2f 2  2.0  1091 s   12.5664  10 9 rad s

The characteristic impedance Z0 is RjL  Z  0 GjC  36 mΩ mj  12.5664 109 rad s 430 nH m  10  mj  12.5664 109 rad s 69 pF m 78.94j 1.92 104

Note the imaginary part of Z0 is very small indicating that our line is very low loss.

Transmission Lines Slide 55

Solution (2 of 3)

The complex propagation constant  is RjLGjC  9 36 mΩ mj  12.5664 10 rad s 430 nH m  9 10  mj  12.5664 10 rad s 69 pF m 6.23 1041j 68.45 m

From this result, we read off  and . jj  6.23  1041  68.45 m

 6.23 104 Np m Np is

  68.45 rad m rad is

Transmission Lines Slide 56

28 9/27/2017

Solution (3 of 3)

Is the line lossless?  NO

No because R ≠ 0 and G ≠ 0. Also, we can determine this because  ≠ 0 .

Is the line weakly absorbing?  YES

? ? RL GC ? ? 36 mΩ m 12.5664 109 rad s 430 nH m 10  m 12.5664 109 rad s 69 pF m ? ? 0.036 5403.5 10 106 0.8671 Yes Yes Is the line distortionless?  NO, but close

? RC LG ? 36 mΩ m 69 pF m 430 nH m 10  m ? 2.48 1012 4.30 10 12 No, but close

Transmission Lines Slide 57

Cable Loss Vs. Characteristic Impedance

As we adjust the cable dimensions (i.e. b/a), we change both its impedance and its loss characteristics. This let’s us plot the cable loss vs. characteristic impedance for a coax with different dielectric fills.

For the air‐filled coax, we observe minimum loss at around 77 , where b/a  3.5.

A coaxial cable filled with

polyethelene (r = 2.2), the minimum loss occurs at 51.2  (b/a = 3.6).

https://www.microwaves101.com/encyclopedias/why‐fifty‐ohms Transmission Lines Slide 58

29 9/27/2017

Power Handling Vs. Characteristic Impedance As we adjust the cable dimensions (i.e. b/a), we affect the peak voltage handling capability (breakdown) and its power handling capability (heat). We observe the lowest peak voltage at just over 50  which we interpret as the point of best voltage handling capability.

We observe the lowest peak current at around 30  which we interpret as the point of best power handling capability.

https://www.microwaves101.com/encyclopedias/why‐fifty‐ohms Transmission Lines Slide 59

Why 50  Impedance is Best?

Two researchers, Lloyd Espenscheid and Herman Affel, working at Bell Labs produced this graph in 1929. They needed to send 4 MHz signals hundreds. Transmission lines capable of handling high voltage and high power were needed in order to accomplish this.

Data to the right was generated for an air‐filled coaxial cable.

Best for High Voltage: Z0 = 60  Best for High Power: Z0 = 30  Best for Attenuation: Z0 = 75 

50  seems like the best compromise.

Transmission Lines Slide 60

30 9/27/2017

Why 75  Impedance Standard for Coax?

Nobody really knows!!

The ideal impedance is closer to 50 , however this requires a thicker center conductor. Maybe 75  is a compromise between low loss and mechanical flexibility?

Transmission Lines Slide 61

Example:

Microstrip Design

Transmission Lines Slide 62

31 9/27/2017

The Lossless Microstrip

w Attenuation Coefficient,    0

h  r Phase Constant,   11  rr r,eff 2 2112 hw

 k0,effr

Characteristic Impedance, Z0

 60 8hw  ln wh 1 thin lines   eff wh4 ZRjX00 0  1120  wh 1 wide lines  wh1.393 0.667ln wh  1.444   eff 

Transmission Lines Slide 63

Problem Description

Typically, the manufacturing process fixes the value of dielectric constant r. This means the impedance of microstrips is controlled solely through the ratio w/h.

For this example, design a 50  microstrip transmission line in FR‐4, which as a dielectric constant of 4.5, to operate at 2.4 GHz.

w  ? h

Transmission Lines Slide 64

32 9/27/2017

Design Equations

To solve this problem, we must first derive some design equations. To do this, we solve our microstrip equations for w/h. This gives

Z 1 1 0.11 A 0 rr0.23  60 2rr 1 60 2 B  Z0  r  8e A  2 A wh 2 thin lines w  e  2     h  20.61 r 1 BB1 ln 2  1 ln B  1 0.39   wh 2 wide lines 2  rr

Transmission Lines Slide 65

Design Solution (1 of 2)

Applying our design equations, we get

A 1.5438 B  5.5831 w 1.8799 wh 2 thin lines   h 1.8812 wh 2 wide lines Since the above numbers for w/h are essentially the same, we conclude that w  1.88 h

Transmission Lines Slide 66

33 9/27/2017

Design Solution (2 of 2)

We learn from our manufacturing engineer that a convenient choice for substrate thickness h is 0.5 mm. From this, to get 50  the width w of the microstrip should be

wh1.88 1.88 0.5 mm  0.94 mm The phase constant for this line will be

 eff  3.3941 22.410  91 s 2 f 1 k0   50.3 m cc00299792458 m s  50.3 m11 3.3941 92.67 m

Transmission Lines Slide 67

34