Chp1 Transmission line theory with examples-part2
- 1. 1 EKT 441 MICROWAVE COMMUNICATIONS CHAPTER 1: TRANSMISSION LINE THEORY (PART II)
- 2. 2 •The Smith Chart – Intro •Using the Smith Chart •Reflection Coefficient Mag & Angle •VSWR •Impedance Matching •Quarter-Wave Transformer •Single/double stub Tuner •Lumped element tuner •Multi-section transformer TRANSMISSION LINE THEORY PART II
- 3. 3 THE SMITH CHART - INTRO Figure 8: The Smith Chart The Smith Chart parameters: • The reflection coefficient: Γ = |Γ|ejθ (|Γ| ≤ 1), (-180°≤ θ ≤180°) • The normalized impedance: z = Z / Z0 • The normalized admittance: y = 1 / z • SWR and RL scale
- 4. 4 • Graphical tool for use with transmission line circuits and microwave circuit elements. • Only lossless transmission line will be considered. • Two graphs in one ; Plots normalized impedance at any point. Plots reflection coefficient at any point. THE SMITH CHART - INTRO
- 5. 5 The transmission line calculator, commonly referred as the Smith Chart THE SMITH CHART - INTRO
- 6. 6 USING THE SMITH CHART The Smith Chart is a plot of normalized impedance. For example, if a Z0 = 50 Ω transmission line is terminated in a load ZL = 50 + j100 Ω as below:
- 7. 7 To locate this point on Smith Chart, normalize the load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2 Ω USING THE SMITH CHART
- 8. 8 The normalized load impedance is located at the intersection of the r =1 circle and the x =+2 circle. USING THE SMITH CHART
- 9. 9 The reflection coefficient has a magnitude and an angle : Γ Γ=Γ θj L e LΓ Γ θ Where the magnitude can be measured using a scale for magnitude of reflection coefficient provided below the Smith Chart, and the angle is indicated on the angle of reflection coefficient scale shown outside the circle on chart.1=ΓL USING THE SMITH CHART
- 10. 10 USING THE SMITH CHART Scale for magnitude of reflection coefficient Scale for angle of reflection coefficient
- 11. 11 For this example, 0 45 7.0 j j L e e = Γ=Γ Γθ USING THE SMITH CHART
- 12. 12 After locating the normalized impedance point, draw the constant circle. For example, the line is 0.3λ length: Γ Γ θj L e USING THE SMITH CHART
- 13. 13 • Move along the constant circle is akin to moving along the transmission line. Moving away from the load (towards generator) corresponds to moving in the clockwise direction on the Smith Chart. Moving towards the load corresponds to moving in the anti-clockwise direction on the Smith Chart. Γ Γ θj L e USING THE SMITH CHART
- 14. 14 • To find ZIN, move towards the generator by: Drawing a line from the center of chart to outside Wavelengths Toward Generator (WTG) scale, to get starting point a at 0.188λ Adding 0.3λ moves along the constant circle to 0.488λ on the WTG scale. Read the corresponding normalized input impedance point c, ZNIN = 0.175 - j0.08Ω Γ Γ θj L e USING THE SMITH CHART
- 15. 15 Denormalizing, to find an input impedance, Ω−= = 475.8 0 jZ ZZZ IN NININ USING THE SMITH CHART VSWR is at point b, 9.5=VSWR
- 16. 16 For Z0 = 50Ω , a ZL = 0 (short cct) b ZL = ∞ (open cct) c ZL = 100 + j100 Ω d ZL = 100 - j100 Ω e ZL = 50 Ω USING THE SMITH CHART
- 17. 17 EXAMPLE 1.5 ZL= 50 - j25 and Z0=50 Ohm. Find Zin, VSWR and ΓL using the Smith Chart. LΓLΓ
- 18. 18 SOLUTION TO EXAMPLE 1.5 (i) Locate the normalized load, and label it as point a, where it corresponds to (ii) Draw constant circle. (iii)It can be seen that 5.01 jZNL −= Γ Γ θj L e 0 76 245.0 − =Γ j L e and 66.1=VSWR
- 19. 19 (iv) Move from point a (at 0.356λ) on the WTG scale, clockwise toward generator a distance λ/8 or 0.125λ to point b, which is at 0.481λ. We could find that at this point, it corresponds to Denormalizing it, 07.062.0 jZNIN −= Ω−= 5.331 jZIN SOLUTION TO EXAMPLE 1.5 (Cont’d)
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- 21. 21 EXAMPLE 1.6 The input impedance for a 100 Ω lossless transmission line of length 1.162 λ is measured as 12 + j42Ω. Determine the load impedance.
- 22. 22 SOLUTION TO EXAMPLE 1.6 (i) Normalize the input impedance: (ii) Locate the normalized input impedance and label it as point a 42.012.0 100 4212 0 j j Z Z z in in += + ==
- 23. 23 (iii) Take note the value of wavelength for point a at WTL scale. At point a, WTL = 0.436λ (iv) Move a distance 1.162λ towards the load to point b WTL = 0.436λ + 1.162λ = 1.598λ But, to plot point b, 1.598λ – 1.500λ = 0.098λ Note: One complete rotation of WTL/WTG = 0.5λ SOLUTION TO EXAMPLE 1.6 (Cont’d)
- 24. 24 (v) Read the point b: 7.015.0 jZNL −= Denormalized it: Ω−= = 7015 0 j ZZZ NLL SOLUTION TO EXAMPLE 1.6 (Cont’d)
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- 26. 26 EXAMPLE 1.7 On a 50 Ω lossless transmission line, the VSWR is measured as 3.4. A voltage maximum is located 0.079λ away from the load (towards generator). Determine the load.
- 27. 27 (i) Use the given VSWR to draw a constant circle. (ii) Then move from maximum voltage at WTG = 0.250λ (towards the load) to point a at WTG = 0.250λ - 0.079λ = 0.171λ. (iii)At this point we have ZNL = 1 + j1.3 Ω, or ZL = 50 + j65 Ω. SOLUTION TO EXAMPLE 1.7 Γ Γ θj L e
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- 29. 29 THE IMPEDANCE MATCHING The important of impedance matching or tuning: Maximum power is delivered when the load is matched to the line. The power loss in the feed line is minimized. (increase power handling capability by optimizing VSWR) Improved the signal-to-noise ratio (SNR). (e.g with controlled mismatch, an amplifier can operate with minimum noise generation) Reduced the amplitude and phase errors.
- 30. 30 THE IMPEDANCE MATCHING Factors in the matching network selection: Complexity Bandwidth Implementation Adjustability Figure 9: A lossless network matching an arbitrary load impedance to a transmission line.
- 31. 31 IMPEDANCE MATCHING • The transmission line is said to be matched when Z0 = ZL which no reflection occurs. • The purpose of matching network is to transform the load impedance ZL such that the input impedance Zin looking into the network is equal to Z0 of the transmission line.
- 32. 32 Adding an impedance matching networks ensures that all power make it or delivered to the load. IMPEDANCE MATCHING (Cont’d)
- 33. 33 Techniques of impedance matching : Quarter-wave transformer Single / double stub tuner Lumped element tuner Multi-section transformer IMPEDANCE MATCHING (Cont’d)
- 34. 34 • It much more convenient to add shunt elements rather than series elements Easier to work in terms of admittances. • Admittance: Z Y 1 = IMPEDANCE MATCHING (Cont’d)
- 35. 35 Adding shunt elements using admittances: With Smith chart, it is easy to find normalized admittance – move to a point on the opposite side of the constant circle. Γ Γ θj L e IMPEDANCE MATCHING (Cont’d)
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- 37. 37 QUARTER WAVE TRANSFORMER The quarter wave transformer matching network only can be constructed if the load impedance is all real (no reactive component)
- 38. 38 To find the impedance looking into the quarter wave long section of lossless ZS impedance line terminated in a resistive load RL: ljRZ ljZR ZZ LS SL Sin β β tan tan + + = , 24 2 πλ λ π β ==lBut, for quarter wavelength, ∞=∴ lβtan QUARTER WAVE TRANSF. (Cont’d)
- 39. 39 So, 0 2 Z R Z Z L S in == Rearrange to get impedance matched line, LS RZZ 0= QUARTER WAVE TRANSF. (Cont’d)
- 40. 40 EXAMPLE 1.8.5 Calculate the position and characteristic impedance of a quarter wave transformer that will match a load impedance, RL = 15Ω; to a 50 Ω line.
- 41. 41 To get the transformer’s impedance, use SOLUTION TO EXAMPLE 1.8.5 39.27)15)(50(0 === LS RZZ To find the position of quarter-wave transformer from the load: d = 0.25 λ ZT = 27.39 Ω ZT = 27.39 Ω 15 Ω Qwave transformer
- 42. 42 EXAMPLE 1.8.6 A transistor has an input impedance of ZL = 25 Ω, at an operating frequency of 500 MHz. Find: a)The length, l b)Width, w c)characteristic impedance of the quarter-wave parallel plate line transformer for which matching is achieved. Assume thickness of the dielectric is d = 1 mm, and the relative dielectric constant is εr = 4. Assume that surface resistance, R and shunt conductance, G, can be neglected.
- 43. 43 EXAMPLE 1.8.6 We can directly apply that 355.35)25)(50(0 === LS RZZ To transistorw Zline ZL Zin Z0=50Ω ℓ=λ/4 The characteristic of the line is ε µ w d C L Z p Line ==
- 44. 44 EXAMPLE 1.8.6 (cont) Para mete r 2-wire line Coaxial Line Parallel Plate Line Unit R Ω/m L H/m G S/m C F/m δσπ conda 1 a D a 2 cosh π µ ( )a Da diel 2 cosh πσ ( )a Da 2 cosh πε + bacond 11 2 1 δπσ a b ln 2π µ ( )a b diel ln 2πσ ( )a bln 2πε δσcondw 2 w d µ d w dielσ d w ε
- 45. 45 EXAMPLE 1.8.6 (Cont) Thus, the width of the line is From previous table, we find the values for capacitance and inductance as; mm Z d w rline p 329.5 0 0 == εε µ mnH w d L p /8.235== µ mpF d w C p /6.188== ε
- 46. 46 EXAMPLE 1.8.6 (Cont) The line length l follows from the condition The input impedance of the combined transmission line and the load is: mm LCf l 967.74 4 1 4 === λ ( ) ( ) )(1 )(1 tan tan d d Z djZZ djZZ ZZ line Lline lineL linein Γ− Γ+ = + + = β β Where d = l = λ/4, and the reflection coefficient is given by − + − =Γ=Γ − d v f j ZZ ZZ ed plineL lineLdj πβ 2 2exp)( 2 0
- 47. 47 EXAMPLE 1.8.6 (Cont) Quarter Wave Impedance Matching • Designed to achieve matching at a single frequency/narrow bandwidth • Easy to build and use
- 48. 48 STUB MATCHING Stub Matching • Single stub or Double Stub • Parallel Stub or Series Stub • Open stub or Shorted Stub
- 49. 49 SINGLE STUB TUNING Figure 11: Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.
- 50. 50 SHUNT STUB MATCHING NETWORK The matching network has to transform the real part of load impedance, RL to Z0 and reactive part, XL to zero Use two adjustable parameters – e.g. shunt- stub.
- 51. 51 Thus, the main idea of shunt stub matching network is to: (i) Find length d and l in order to get yd and yl . (ii) Ensure total admittance ytot = yd + yl = 1 for complete matching network. SHUNT STUB MATCHING NET. (Cont’d)
- 52. 52 • Locate the normalized load impedance ZNL. • Draw constant SWR circle and locate YNL. • Move clockwise (WTG) along circle to intersect with 1 ± jB value of yd. • The length moved from YNL towards yd is the through line length, d. • Locate yl at the point jB . • Depends on the shorted/open stub, move along the periphery of the chart towards yl (WTG). • The distance traveled is the length of stub, l . SHUNT STUB USING SMITH CHART Γ Γ θj L e
- 53. 53 SHORTED SHUNT STUB MATCHING Generic layout of the shorted shunt stub matching network:
- 54. 54 EXAMPLE 1.9 Construct the shorted shunt stub matching network for a 50Ω line terminated in a load ZL = 20 – j55Ω
- 55. 55 1. Locate the normalized load impedance, ZNL = ZL/Z0 = 0.4 – j1.1Ω 2. Draw constant circle. 3. Locate YNL. (0.112λ at WTG) 4. Moving to the first intersection with the 1 ± jB circle, which is at 1 + j2.0 yd 5. Get the value of through line length, d from 0.112λ to 0.187λ on the WTG scale, so d = 0.075λ SOLUTION TO EXAMPLE 1.9 Γ Γ θj L e
- 56. 56 6. Locate the location of short on the Smith Chart (note: when short circuit, ZL = 0, hence YL = ∞) on the right side of the chart with WTG=0.25λ 7. Move clockwise (WTG) until point jB, which is at 0 - j2.0, located at WTG= 0.324λ yl 8. Determine the stub length, l 0.324λ – 0.25λ = 0.074 λ SOLUTION TO EXAMPLE 1.9 (Cont’d)
- 57. 57 Thus, the values are: d = 0.075 λ l = 0.074 λ yd = 1 + j2.0 Ω yl = -j2.0 Ω Where YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1 SOLUTION TO EXAMPLE 1.9 (Cont’d)
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- 59. 59 OPEN END SHUNT STUB MATCHING Generic layout of the open ended shunt stub matching network:
- 60. 60 EXAMPLE 1.10 Construct an open ended shunt stub matching network for a 50Ω line terminated in a load ZL = 150 + j100 Ω
- 61. 61 SOLUTION TO EXAMPLE 1.10 1. Locate the normalized load impedance, ZNL = ZL/Z0 = 3.0 + j2.0Ω 2. Draw constant circle. 3. Locate YNL. (0.474λ at WTG) 4. Moving to the first intersection with the 1 ± jB circle, which is at 1 + j1.6 yd 5. Get the value of through line length, d from 0.474λ to 0.178λ on the WTG scale, so d = 0.204λ Γ Γ θj L e
- 62. 62 6. Locate the location of open end on the Smith Chart (note: when open circuit, ZL = ∞, hence YL = 0) on the left side with WTG = 0.00λ 7. Move clockwise (WTG) until point jB, which is at 0 – j1.6, located at WTG= 0.339λ yl 8. Determine the stub length, l 0.339λ – 0.00λ = 0.339 λ SOLUTION TO EXAMPLE 1.10 (Cont’d)
- 63. 63 Thus, the values are: d = 0.204 λ l = 0.339 λ yd = 1 + j1.6 Ω yl = -j1.6 Ω Where YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1 SOLUTION TO EXAMPLE 1.10 (Cont’d)
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- 65. 65 In both previous example, we chose the first intersection with the1 ± jB circle in designing our matching network. We could also have continued on to the second intersection. Thus, try both intersection to determine which solution produces max/min length of through line, d or length of stub, l. IMPORTANT!!
- 66. 66 Determine the through line length and stub length for both example above by using second intersection. For shorted shunt stub (example 1.9): d = 0.2 λ and l = 0.426 λ For open ended shunt stub (example 1.10): d = 0.348 λ and l = 0.161 λ EXERCISE (TRY THIS!)