Lecture Outline 9/27/2017. EE 4347 Applied Electromagnetics. Topic 4a

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1 9/7/17 Course Instructor Dr. Raymond C. Rumpf Office: A 337 Phone: (915) E Mail: rcrumpf@utep.edu EE 4347 Applied Electromagnetics Topic 4a Transmission Lines Transmission These Lines notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited Slide 1 Lecture Outline Introduction Transmission Line Equations Transmission Line Wave Equations Transmission Line Parameters and Characteristic Impedance, Z Special Cases of Transmission Lines General transmission lines Lossless lines Weakly absorbing lines Distortionless lines Examples RG 59 coaxial cable Microstrip design Transmission Lines Slide 1

2 9/7/17 Introduction Transmission Lines Slide 3 Map of Waveguides (LI Media) Single Ended Differential Transmission Lines Contains two or more conductors. No low frequency cutoff. Thought of more as a circuit clement Homogeneous Has TEM mode. Has TE and TM modes. coaxial stripline buried parallel plate Inhomogeneous Supports only quasi (TEM, TE, & TM) modes. microstrip coplanar coplanar strips Waveguides Confines and transports waves. Supports higher order modes. Metal Shell Pipes Homogeneous Inhomogeneous Pipes Enclosed by metal. Does not support TEM mode. Has a low frequency cutoff. Supports TE and TM modes rectangular circular dual ridge Supports TE and TM modes only if one axis is uniform. Otherwise supports quasi TM and quasi TE modes. Has one or less conductors. Usually what is implied by the label waveguide. Dielectric Pipes Composed of a core and a cladding. Symmetric waveguides have no low frequency cutoff. Channel Waveguides Confinement along two axes. TE & TM modes only supported in circularly symmetric guides. optical Fiber photonic crystal Slab Waveguides rib Confinement only along one axis. Supports TE and TM modes. Interfaces can support surface waves. shielded pair Transmission Lines slotline no uniform axis (no TE or TM) uniform axis (has TE and TM) dielectric Slab large area parallel plate interface Slide 4

3 9/7/17 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 wavelength 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

4 9/7/17 L Type Equivalent Circuit Model Distributed Circuit Parameters R (/m) Resistance per unit length. Arises due to resistivity in the conductors. L (H/m) Inductance per unit length. Arises due to stored magnetic energy around the line. G (1/m) Conductance per unit length. Arises due to conductivity in the dielectric separating the conductors. 1 G R C (F/m) Capacitance per unit length. Arises due to stored electric energy between the conductors. z There are many possible circuit models for transmission lines, but most produce the same equations after analysis. R z L z Gz Cz z z Transmission Lines Slide 7 Relation to Electromagnetic Parameters,, Every transmission line with a homogeneous fill has: LC G C Transmission Lines Slide 8 4

5 9/7/17 Fundamental Vs. Intuitive Parameters Fundamental Parameters Intuitive Parameters Electromagnetics Electromagnetics,, n,,,, tan Transmission Lines R, L, G, C Transmission Lines Z,,, VSWR The fundamental parameters are the most basic parameters needed to solve a transmission line problem. However, it is difficult to be intuitive about how they affect signals on the line. An electromagnetic analysis is needed to determine R, L, G, and C from the geometry of the transmission line. The intuitive parameters provide intuitive insight about how signals behave on a transmission line. They isolate specific information to a single parameter. The intuitive parameters are calculated from R, L, G, and C. Transmission Lines Slide 9 Example RLGC Parameters RG 59 Coax CAT5 Twisted Pair Microstrip R 36 mω m L 43 nh m G 1 m C 69 pf m Z 75 R 176 mω m L 49 nh m G m C 49 pf m Z 1 R 15 mω m L 364 nh m G 3 m C 17 pf m Z 5 Surprisingly, almost all transmission lines have parameters very close to these same values. Transmission Lines Slide 1 5

6 9/7/17 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. V Ed I H d 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! L Transmission Lines Slide 1 6

7 9/7/17 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 H E t E H t Transmission Line Equations V I RI L z t I V GV C z t 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 ) + V z, t 1 R z L z 3 I zt, Gz Cz 4 + V zz, t z z z Apply Kirchoff s voltage law (KVL) to the outer loop of the equivalent circuit: Iz, t V z, t I z, t RzLz 1 V zz, t t 3 Transmission Lines Slide

8 9/7/17 Derivation of First TL Equation ( of ) 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. I z, t V z, tiz, trzlz V zz, t t Vzz, tv z, t Iz, t RI z, tl z t In the limit as z, the expression on the left hand side becomes a derivative with respect to z. V z, t I z, t RI z, tl z t Transmission Lines Slide 15 Derivation of Second TL Equation (1 of ) + V z, t z R z L z I zt, Gz Cz I zz, t z + V zz, t z Apply Kirchoff s current law (KCL) to the main node the equivalent circuit: V zz, t I z, t Izz, tgzv zz, t Cz t 1 3 Transmission Lines Slide

9 9/7/17 Derivation of Second TL Equation ( of ) 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. V zz, t Iz, tizz, tgzv zz, tcz t Izz, tiz, t V zz, t GV z z, tc z t In the limit as z, the expression on the left hand side becomes a derivative with respect to z. I z, t V z, t GV z, tc z t Transmission Lines Slide 17 Transmission Line Wave Equations Transmission Lines Slide 18 9

10 9/7/17 Starting Point Telegrapher Equations We start with the transmission line equations derived in the previous section. V z, t I z, t RI z, tl z t I z, t V z, t GV z, tc z t time domain For time harmonic (i.e. frequency domain) analysis, we Fourier transform the equations above. dv z R jl I z di z G jcv z frequency domain Note: Our derivative d/ 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 R jl I z To derive a wave equation in terms of V(z), we first differentiate Eq. (1) with respect to z. dv z di z R jl Eq. (3) Transmission Lines Slide di z Eq. (1) G jcv z Eq. () Second, we substitute Eq. () into the right hand side of Eq. (3) to eliminate I(z) from the equation. dv z R jlg jcv z Last, we rearrange the terms to arrive at the final form of the wave equation. dvz R jlg jcv z 1

11 9/7/17 Wave Equation in Terms of I(z) dv z R jl I z To derive a wave equation in terms of just I(z), we first differentiate Eq. () with respect to z. d I z dv z G jc Eq. (3) Transmission Lines Slide 1 di z Eq. (1) G jcv z Eq. () Second, we substitute Eq. (1) into the right hand side of Eq. (3) to eliminate V(z) from the equation. d I z G jcr jliz Last, we rearrange the terms to arrive at the final form of the wave equation. d Iz G jcr jliz Propagation Constant, Define the propagation constant to be j G jc R jl G jcr jl In our wave equations, we have a common term. Given this definition, the transmission line equations are written as dv z V z d I z I z Transmission Lines Slide 11

12 9/7/17 Solution to the Wave Equations If we hand the wave equations off to a mathematician, they will return with the following solutions. dv z V z d I z I z z V z V e V e z I z I e I e z z 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 3 Transmission Line Parameters: Attenuation Coefficient, Phase Constant, Transmission Lines Slide 4 1

13 9/7/17 Derivation and (1 of 7) Step 1 Start with our expression for. j G jc R jl Square this expression to get rid of square root on right hand side. j G jcr jl Expand this expression. j RG jrc jlg LC Collect real and imaginary parts on the left hand and right hand sides. j RG LC j RC LG Transmission Lines Slide 5 Derivation and ( of 7) Step Generate two equations by equating real and imaginary parts. j RG LC j RC LG RCLG RG LC We now have two equations and two unknowns. Transmission Lines Slide 6 13

14 9/7/17 Derivation and (3 of 7) Step 3 Derive a quadratic equation for. RC LG Eq. (1a) RG LC Eq. (1b) Solve Eq. (1a) for. RC LG Eq. () Substitute Eq. () into Eq. (1b) and simplify. RC LG RG LC RC LG RG 4 Transmission Lines Slide RCLG 4 RG4 LC 4 LC RG RC LG LC Derivation and (4 of 7) Step 4 Solve for using the quadratic formula. b b 4ac Recall the quadratic formula: ax bx c x a Our equation for is in the form of a quadratic equation where a 1 LC RG RC LG 4 x The solution is 4 LC RG LC RG RC LG RG LC R L G C b LCRG c RCLG Transmission Lines Slide 8 14

15 9/7/17 Derivation and (5 of 7) Step 5 Resolve the sign of the square root. RG LC R L G C The final expression is In order for this expression to always give a real value for, the sign of the square root must be positive. RG LC R L G C Transmission Lines Slide 9 Derivation and (6 of 7) Step 6 Solve for using our expression for. Recall Eq. (1b): RG LC We obtain an equation for by substituting our expression for into Eq. (1b). RG LC R L G C RG LC RG LC R L G C RG LC R L G C Transmission Lines Slide 3 15

16 9/7/17 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 and. RG LC R L G C RG LC R L G C Both and must be positive quantities for passive materials. This means we take the positive sign for the square root. RG LC R L G C RG LC R L G C Transmission Lines Slide 31 Transmission Line Parameters: Characteristic Impedance, Z Transmission Lines Slide 3 16

17 9/7/17 Characteristic Impedance, Z () The characteristic impedance Z of a transmission line is defined as the ratio of the voltage to the current at any point of a forward travelling wave. V V Z I I Definition for a forward travelling wave. Definition for a backward travelling wave. Notice the negative sign! Most characteristic impedance values fall in the 5 to 1 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 Z (1 of 5) Step 1 Substitute our solution into the transmission line equations. V z V e V e z z I z I e I e z z dv z R jl I z d V e V e z z z z R jlie Ie di z d I e z z G jcv z I e z z G jcv e V e Transmission Lines Slide 34 17

18 9/7/17 Derivation of Z ( of 5) Step Expand the equations and calculate the derivatives. d V e V e z z z z R jlie Ie d I e I e z z z z G jcv e V e V e V e z z z z z R jli e R jli e I e I e z z G j C V e G j C V e z Transmission Lines Slide 35 Derivation of Z (3 of 5) Step 3 Equate the expressions multiplying the common exponential terms. V R jl I V e V e R j L I e R jl I e z z z z V R jl I I G jc V I e I e G j C V e G jc V e z z z z I G jc V Transmission Lines Slide 36 18

19 9/7/17 Derivation of Z (4 of 5) Step 4 Solve each of our four equations for V /I to derive expressions for Z. V R jl I V R jl I I G jc V I G jc V V R jl Z I V R jl Z I V Z I G jc V Z I G jc Transmission Lines Slide 37 Derivation of Z (5 of 5) Step 5 Put Z in terms of just R, L, G, and C. Recall our expression for : j G jcr jl We can substitute this into either of our expressions for Z. R jl Z G jc Proceed with the first expression. G jcr jl R jl R jl R jl Z G jc R jl G jc Transmission Lines Slide 38 19

20 9/7/17 Final Expression for Z () We have derived a general expression for the characteristic impedance Z of a transmission line in terms of the fundamental parameters R, L, G, and C. Definition: Z V I V I Expression: Z R jl R jl G jc G jc Transmission Lines Slide 39 Dissecting the Characteristic Impedance, Z The characteristic impedance describes the amplitude and phase relation between voltage and current along a transmission line. With this picture in mind, the characteristic impedance can be written as Z Z Z z V z V e V I z I e e Z V e e z z z jz Z The characteristic impedance can also be written in terms of its real and imaginary parts. Z R jx Reactive part of Z. This is not equal to jl or 1/jC. Resistive part of Z. This is not equal to R or G. Transmission Lines Slide 4

21 9/7/17 Special Cases of Transmission Lines: General Transmission Line Transmission Lines Slide 41 Parameters for General TLs Propagation Constant, j G jc R jl Attenuation Coefficient, RG LC R L G C Phase Constant, RG LC R L G C Characteristic Impedance, Z Z R jx R jl G jc Transmission Lines Slide 4 1

22 9/7/17 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 R G Transmission Lines Slide 44

23 9/7/17 Parameters for Lossless TLs Propagation Constant, j j LC Attenuation Coefficient, Phase Constant, LC Characteristic Impedance, Z Z R jx R L C L C X Transmission Lines Slide 45 Special Cases of Transmission Lines: Weakly Absorbing Line Transmission Lines Slide 46 3

24 9/7/17 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 R L and G C Ensures low ohmic loss for signals propagating through the line. Ensures very little conduction between the lines through the dielectric. Transmission Lines Slide 47 Parameters for Weakly Absorbing TLs Attenuation Coefficient, 1 R GZ Z 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 4

25 9/7/17 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 LC R L G C 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 R G L C Transmission Lines Slide 5 5

26 9/7/17 Parameters for Distortionless TLs Propagation Constant, j RG j LC Attenuation Coefficient, RG Phase Constant, LC To be distortionless, we must have. is a measure of how quickly a signal accumulates phase. Different frequencies have different wavelengths and therefore must accumulate different phase through the same length of line. Characteristic Impedance, Z R L Z R jx G C R L R X G C Transmission Lines Slide 51 Example: Properties of RG-59 Coax Transmission Lines Slide 5 6

27 9/7/17 The Lossless Circular Coax Fundamental Parameters (derived in EE 331) C ln 1 b L ln H m 4 a a Attenuation Coefficient, Phase Constant, F m ba Characteristic Impedance, Z b Z R jx ln a b a b R ln X a and b Transmission Lines Slide 53 Typical RLGC for RG 59 Coax at GHz The typical RG 59 coaxial cable operating at. GHz has the following RLGC parameters: R 36 mω m L 43 nh m G 1 m C 69 pf m Calculate the transmission line parameters,,, and Z. Classify the line as lossless, weakly absorbing, distortionless, etc. Transmission Lines Slide 54 7

28 9/7/17 Solution (1 of 3) Our equations mostly utilize the angular frequency instead of the ordinary frequency f f. 1 s rad s The characteristic impedance Z is Z R jl G jc 9 36 mω m j rad s43 nh m 9 1 m j rad s69 pf m j Note the imaginary part of Z is very small indicating that our line is very low loss. Transmission Lines Slide 55 Solution ( of 3) The complex propagation constant is R j LG j C 9 36 mω m j rad s43 nh m 9 1 m j rad s69 pf m 6.31 j68.45 m 4 1 From this result, we read off and. j 6.31 j68.45 m Np m rad m Np is Nepers rad is radians Transmission Lines Slide 56 8

29 9/7/17 Solution (3 of 3) Is the line lossless? NO No because R and G. Also, we can determine this because. Is the line weakly absorbing? YES? RL? 9 36 mω m rad s43 nh m? Yes Is the line distortionless? NO, but close? RC LG? 36 mω m69 pf m43 nh m1 m? No, but close G C? 9 1 m rad s69 pf m Transmission Lines Slide 57?? Yes 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 =.), the minimum loss occurs at 51. (b/a = 3.6). fifty ohms Transmission Lines Slide 58 9

30 9/7/17 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 5 which we interpret as the point of best voltage handling capability. We observe the lowest peak current at around 3 which we interpret as the point of best power handling capability. fifty ohms Transmission Lines Slide 59 Why 5 Impedance is Best? Two researchers, Lloyd Espenscheid and Herman Affel, working at Bell Labs produced this graph in 199. 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: Z = 6 Best for High Power: Z = 3 Best for Attenuation: Z = 75 5 seems like the best compromise. Transmission Lines Slide 6 3

31 9/7/17 Why 75 Impedance Standard for Coax? Nobody really knows!! The ideal impedance is closer to 5, 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 6 31

32 9/7/17 The Lossless Microstrip Attenuation Coefficient, Phase Constant, r,eff k r 1 r 1 11hw r,eff h w r Characteristic Impedance, Z 6 8h w ln wh1 thin lines 4 eff w h Z R jx 1 1 wh1 wide lines wh ln eff wh1.444 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 5 microstrip transmission line in FR 4, which as a dielectric constant of 4.5, to operate at.4 GHz. w h? Transmission Lines Slide 64 3

33 9/7/17 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 r 1 r 1.11 A.3 6 r 1 r 6 B Z r A 8e A w e h r 1.61 B1ln B1 ln B1.39 r r wh thin lines wh wide lines Transmission Lines Slide 65 Design Solution (1 of ) Applying our design equations, we get A B w wh thin lines h wh 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

34 9/7/17 Design Solution ( of ) We learn from our manufacturing engineer that a convenient choice for substrate thickness h is.5 mm. From this, to get 5 the width w of the microstrip should be w1.88h mm.94 mm The phase constant for this line will be eff f.41 s k 5.3 m c c m s m m 1 Transmission Lines Slide 67 34