EKT 441 MICROWAVE COMMUNICATIONS CHAPTER 1 TRANSMISSION LINE

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EKT 441 MICROWAVE COMMUNICATIONS CHAPTER 1: TRANSMISSION LINE THEORY (PART II) 1

EKT 441 MICROWAVE COMMUNICATIONS CHAPTER 1: TRANSMISSION LINE THEORY (PART II) 1

TRANSMISSION LINE THEORY PART II • The Smith Chart – Intro • Using the

TRANSMISSION LINE THEORY PART II • 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 2

THE SMITH CHART - INTRO The Smith Chart parameters: • The reflection coefficient: Γ

THE SMITH CHART - INTRO The Smith Chart parameters: • The reflection coefficient: Γ = |Γ|ejθ (|Γ| ≤ 1), (-180°≤ θ ≤ 180°) • The normalized impedance: z = Z / Z 0 • The normalized admittance: y=1/z • SWR and RL scale Figure 8: The Smith Chart 3

THE SMITH CHART - INTRO • Graphical tool for use with transmission line circuits

THE SMITH CHART - INTRO • Graphical tool for use with transmission line circuits and microwave circuit elements. • Only lossless transmission line will be considered. • Two graphs in one ; Ø Plots normalized impedance at any point. Ø Plots reflection coefficient at any point. 4

THE SMITH CHART - INTRO The transmission line calculator, commonly referred as the Smith

THE SMITH CHART - INTRO The transmission line calculator, commonly referred as the Smith Chart 5

USING THE SMITH CHART The Smith Chart is a plot of normalized impedance. For

USING THE SMITH CHART The Smith Chart is a plot of normalized impedance. For example, if a Z 0 = 50 Ω transmission line is terminated in a load ZL = 50 + j 100 Ω as below: 6

USING THE SMITH CHART To locate this point on Smith Chart, normalize the load

USING THE SMITH CHART To locate this point on Smith Chart, normalize the load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j 2 Ω 7

USING THE SMITH CHART The normalized load impedance is located at the intersection of

USING THE SMITH CHART The normalized load impedance is located at the intersection of the r =1 circle and the x =+2 circle. 8

USING THE SMITH CHART The reflection coefficient has a magnitude angle and an :

USING THE SMITH CHART The reflection coefficient has a magnitude angle and an : 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. 9

USING THE SMITH CHART Scale for magnitude of reflection coefficient Scale for angle of

USING THE SMITH CHART Scale for magnitude of reflection coefficient Scale for angle of reflection coefficient 10

USING THE SMITH CHART For this example, 11

USING THE SMITH CHART For this example, 11

USING THE SMITH CHART After locating the normalized impedance point, draw the constant circle.

USING THE SMITH CHART After locating the normalized impedance point, draw the constant circle. For example, the line is 0. 3λ length: 12

USING THE SMITH CHART • Move along the constant circle is akin to moving

USING THE SMITH CHART • Move along the constant circle is akin to moving along the transmission line. Ø Moving away from the load (towards generator) corresponds to moving in the clockwise direction on the Smith Chart. Ø Moving towards the load corresponds to moving in the anti-clockwise direction on the Smith Chart. 13

USING THE SMITH CHART • To find ZIN, move towards the generator by: ØDrawing

USING THE SMITH CHART • To find ZIN, move towards the generator by: ØDrawing a line from the center of chart to outside Wavelengths Toward Generator (WTG) scale, to get starting point a at 0. 188λ ØAdding 0. 3λ moves along the constant circle to 0. 488λ on the WTG scale. ØRead the corresponding normalized input impedance point c, ZNIN = 0. 175 - j 0. 08Ω 14

USING THE SMITH CHART Denormalizing, to find an input impedance, VSWR is at point

USING THE SMITH CHART Denormalizing, to find an input impedance, VSWR is at point b, 15

USING THE SMITH CHART For Z 0 = 50Ω , a ZL = 0

USING THE SMITH CHART For Z 0 = 50Ω , a ZL = 0 (short cct) b ZL = ∞ (open cct) c ZL = 100 + j 100 Ω d ZL = 100 - j 100 Ω e ZL = 50 Ω 16

EXAMPLE 1. 5 ZL= 50 - j 25 and Z 0=50 Ohm. Find Zin,

EXAMPLE 1. 5 ZL= 50 - j 25 and Z 0=50 Ohm. Find Zin, VSWR and ΓL using the Smith Chart. 17

SOLUTION TO EXAMPLE 1. 5 (i) Locate the normalized load, and label it as

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 and 18

SOLUTION TO EXAMPLE 1. 5 (Cont’d) (iv) Move from point a (at 0. 356λ)

SOLUTION TO EXAMPLE 1. 5 (Cont’d) (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. (v) We could find that at this point, it corresponds to Denormalizing it, 19

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EXAMPLE 1. 6 The input impedance for a 100 Ω lossless transmission line of

EXAMPLE 1. 6 The input impedance for a 100 Ω lossless transmission line of length 1. 162 λ is measured as 12 + j 42Ω. Determine the load impedance. 21

SOLUTION TO EXAMPLE 1. 6 (i) Normalize the input impedance: (ii) Locate the normalized

SOLUTION TO EXAMPLE 1. 6 (i) Normalize the input impedance: (ii) Locate the normalized input impedance and (iii) label it as point a 22

SOLUTION TO EXAMPLE 1. 6 (Cont’d) (iii) Take note the value of wavelength for

SOLUTION TO EXAMPLE 1. 6 (Cont’d) (iii) Take note the value of wavelength for point a at (iv) 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 23

SOLUTION TO EXAMPLE 1. 6 (Cont’d) (v) Read the point b: Denormalized it: 24

SOLUTION TO EXAMPLE 1. 6 (Cont’d) (v) Read the point b: Denormalized it: 24

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EXAMPLE 1. 7 On a 50 lossless transmission line, the VSWR is measured as

EXAMPLE 1. 7 On a 50 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. 26

SOLUTION TO EXAMPLE 1. 7 (i) Use the given VSWR to draw a constant

SOLUTION TO EXAMPLE 1. 7 (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 + j 1. 3 Ω, or ZL = 50 + j 65 Ω. 27

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THE IMPEDANCE MATCHING The important of impedance matching or tuning: Maximum power is delivered

THE IMPEDANCE MATCHING The important of impedance matching or tuning: Maximum power is delivered when the load is matched to the line. l The power loss in the feed line is minimized. (increase power handling capability by optimizing VSWR) l Improved the signal-to-noise ratio (SNR). (e. g with controlled mismatch, an amplifier can operate with minimum noise generation) l Reduced the amplitude and phase errors. l 29

THE IMPEDANCE MATCHING Figure 9: A lossless network matching an arbitrary load impedance to

THE IMPEDANCE MATCHING Figure 9: A lossless network matching an arbitrary load impedance to a transmission line. Factors in the matching network selection: l Complexity l Bandwidth l Implementation l Adjustability 30

IMPEDANCE MATCHING • The transmission line is said to be matched when Z 0

IMPEDANCE MATCHING • The transmission line is said to be matched when Z 0 = 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 Z 0 of the transmission line. 31

IMPEDANCE MATCHING (Cont’d) Adding an impedance matching networks ensures that all power make it

IMPEDANCE MATCHING (Cont’d) Adding an impedance matching networks ensures that all power make it or delivered to the load. 32

IMPEDANCE MATCHING (Cont’d) l Techniques of impedance matching : Ø Quarter-wave transformer Ø Single

IMPEDANCE MATCHING (Cont’d) l Techniques of impedance matching : Ø Quarter-wave transformer Ø Single / double stub tuner Ø Lumped element tuner Ø Multi-section transformer 33

IMPEDANCE MATCHING (Cont’d) • It much more convenient to add shunt elements rather than

IMPEDANCE MATCHING (Cont’d) • It much more convenient to add shunt elements rather than series elements Easier to work in terms of admittances. • Admittance: 34

IMPEDANCE MATCHING (Cont’d) Adding shunt elements using admittances: With Smith chart, it is easy

IMPEDANCE MATCHING (Cont’d) 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. 35

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QUARTER WAVE TRANSFORMER The quarter wave transformer matching network only can be constructed if

QUARTER WAVE TRANSFORMER The quarter wave transformer matching network only can be constructed if the load impedance is all real (no reactive component) 37

QUARTER WAVE TRANSF. (Cont’d) To find the impedance looking into the quarter wave long

QUARTER WAVE TRANSF. (Cont’d) To find the impedance looking into the quarter wave long section of lossless ZS impedance line terminated in a resistive load RL: But, for quarter wavelength, 38

QUARTER WAVE TRANSF. (Cont’d) So, Rearrange to get impedance matched line, 39

QUARTER WAVE TRANSF. (Cont’d) So, Rearrange to get impedance matched line, 39

EXAMPLE 1. 8. 5 Calculate the position and characteristic impedance of a quarter wave

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. 40

SOLUTION TO EXAMPLE 1. 8. 5 To get the transformer’s impedance, use To find

SOLUTION TO EXAMPLE 1. 8. 5 To get the transformer’s impedance, use To find the position of quarter-wave transformer from the load: d = 0. 25 λ ZT = 27. 39 Ω 15 Ω Qwave transformer 41

EXAMPLE 1. 8. 6 A transistor has an input impedance of ZL = 25

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. 42

EXAMPLE 1. 8. 6 We can directly apply that ℓ=λ/4 Z 0=50Ω Zin w

EXAMPLE 1. 8. 6 We can directly apply that ℓ=λ/4 Z 0=50Ω Zin w Zline To transistor Zin = 25Ω ZL The characteristic of the line is 43

EXAMPLE 1. 8. 6 (cont) Para meter 2 -wire line Coaxial Line Parallel Plate

EXAMPLE 1. 8. 6 (cont) Para meter 2 -wire line Coaxial Line Parallel Plate Line Unit R Ω/m L H/m G S/m C F/m 44

EXAMPLE 1. 8. 6 (Cont) Thus, the width of the line is From previous

EXAMPLE 1. 8. 6 (Cont) Thus, the width of the line is From previous table, we find the values for capacitance and inductance as; 45

EXAMPLE 1. 8. 6 (Cont) The line length l follows from the condition The

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: Where d = l = λ/4, and the reflection coefficient is given by 46

EXAMPLE 1. 8. 6 (Cont) Quarter Wave Impedance Matching • Designed to achieve matching

EXAMPLE 1. 8. 6 (Cont) Quarter Wave Impedance Matching • Designed to achieve matching at a single frequency/narrow bandwidth • Easy to build and use 47

STUB MATCHING Stub Matching • Single stub or Double Stub • Parallel Stub or

STUB MATCHING Stub Matching • Single stub or Double Stub • Parallel Stub or Series Stub • Open stub or Shorted Stub 48

SINGLE STUB TUNING Figure 11: Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.

SINGLE STUB TUNING Figure 11: Single-stub tuning circuits. (a) Shunt stub. (b) Series stub. 49

SHUNT STUB MATCHING NETWORK The matching network has to transform the real part of

SHUNT STUB MATCHING NETWORK The matching network has to transform the real part of load impedance, RL to Z 0 and reactive part, XL to zero Use two adjustable parameters – e. g. shunt-stub. 50

SHUNT STUB MATCHING NET. (Cont’d) Thus, the main idea of shunt stub matching network

SHUNT STUB MATCHING NET. (Cont’d) 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. 51

SHUNT STUB USING SMITH CHART • Locate the normalized load impedance ZNL. • Draw

SHUNT STUB USING SMITH CHART • Locate the normalized load impedance ZNL. • Draw constant SWR circle and locate YNL. • Move clockwise (WTG) along j. B value of yd. circle to intersect with 1 ± • The length moved from YNL towards yd is the through line length, d. • Locate yl at the point j. B. • 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. 52

SHORTED SHUNT STUB MATCHING Generic layout of the shorted shunt stub matching network: 53

SHORTED SHUNT STUB MATCHING Generic layout of the shorted shunt stub matching network: 53

EXAMPLE 1. 9 Construct the shorted shunt stub matching network for a 50Ω line

EXAMPLE 1. 9 Construct the shorted shunt stub matching network for a 50Ω line terminated in a load ZL = 20 – j 55Ω 54

SOLUTION TO EXAMPLE 1. 9 1. 2. Locate the normalized load impedance, ZNL =

SOLUTION TO EXAMPLE 1. 9 1. 2. Locate the normalized load impedance, ZNL = ZL/Z 0 = 0. 4 – j 1. 1Ω 2. Draw constant 3. Locate YNL. (0. 112λ at WTG) 4. Moving to the first intersection with the circle. 1 ± j. B circle, which is at 1 + j 2. 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 55

SOLUTION TO EXAMPLE 1. 9 (Cont’d) 6. Locate the location of short on the

SOLUTION TO EXAMPLE 1. 9 (Cont’d) 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 8. j. B, which is at 0 - j 2. 0, located at WTG= 0. 324λ yl 8. Determine the stub length, l 0. 324λ – 0. 25λ = 0. 074 λ 56

SOLUTION TO EXAMPLE 1. 9 (Cont’d) Thus, the values are: d = 0. 075

SOLUTION TO EXAMPLE 1. 9 (Cont’d) Thus, the values are: d = 0. 075 λ l = 0. 074 λ yd = 1 + j 2. 0 Ω yl = -j 2. 0 Ω Where YTOT = yd + yl = (1 + j 2. 0) + (-j 2. 0) = 1 57

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OPEN END SHUNT STUB MATCHING Generic layout of the open ended shunt stub matching

OPEN END SHUNT STUB MATCHING Generic layout of the open ended shunt stub matching network: 59

EXAMPLE 1. 10 Construct an open ended shunt stub matching network for a 50Ω

EXAMPLE 1. 10 Construct an open ended shunt stub matching network for a 50Ω line terminated in a load ZL = 150 + j 100 Ω 60

SOLUTION TO EXAMPLE 1. 10 1. 2. Locate the normalized load impedance, ZNL =

SOLUTION TO EXAMPLE 1. 10 1. 2. Locate the normalized load impedance, ZNL = ZL/Z 0 = 3. 0 + j 2. 0Ω 2. Draw constant 3. Locate YNL. (0. 474λ at WTG) 4. Moving to the first intersection with the circle. 1 ± j. B circle, which is at 1 + j 1. 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 61

SOLUTION TO EXAMPLE 1. 10 (Cont’d) 6. Locate the location of open end on

SOLUTION TO EXAMPLE 1. 10 (Cont’d) 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 8. j. B, which is at 0 – j 1. 6, located at WTG= 0. 339λ yl 8. Determine the stub length, l 0. 339λ – 0. 00λ = 0. 339 λ 62

SOLUTION TO EXAMPLE 1. 10 (Cont’d) Thus, the values are: d = 0. 204

SOLUTION TO EXAMPLE 1. 10 (Cont’d) Thus, the values are: d = 0. 204 λ l = 0. 339 λ yd = 1 + j 1. 6 Ω yl = -j 1. 6 Ω Where YTOT = yd + yl = (1 + j 1. 6) + (-j 1. 6) = 1 63

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IMPORTANT!! In both previous example, we chose the first intersection with the 1 ±

IMPORTANT!! In both previous example, we chose the first intersection with the 1 ± j. B 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. 65

EXERCISE (TRY THIS!) Determine through line length and stub length for both example above

EXERCISE (TRY THIS!) Determine 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 66