Lecture Notes - EEEC6430310 Electromagnetic Fields and Waves - Smith Chart
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The Smith chart is a graphical tool used to analyze high frequency circuits. It represents all possible complex impedances in terms of the reflection coefficient. Circles of constant resistance and arcs of constant reactance intersect on the chart to indicate impedance values. The chart can be used to determine impedances, reflection coefficients, standing wave ratios, and more from various circuit parameters. It provides a clever way to visualize complex impedance functions that continues to be popular for high frequency applications.
Lecture Notes - EEEC6430310 Electromagnetic Fields and Waves - Smith Chart
- 1. EEEC6430310 ELECTROMAGNETIC FIELDS AND WAVES Smith Chart FACULTY OF ENGINEERING AND COMPUTER TECHNOLOGY BENG (HONS) IN ELECTRICALAND ELECTRONIC ENGINEERING Ravandran Muttiah BEng (Hons) MSc MIET
- 2. Introduction to Smith Chart The smith chart is one of the most useful graphical tools for high frequency circuit applications. The chart provides a clever way to visualize complex functions and it continuous to endure popularity decades after its original conception. From a mathematical point of view, the Smith chart is simply a representation of all possible complex impedances with respect to coordinates defined by the reflection coefficient. The domain of definition of the reflection coefficient is a circle of radius 1 in the complex plane as shown in figure 1. This is also the domain of the Smith chart. 1 Figure 1 Re Γ Im Γ 1 1
- 3. 2 The goal of the Smith chart is to identify all possible impedances on the domain of existence of the reflection coefficient. To do so, we start from the general definition of line impedance (which is equally applicable to the load impedance). 𝑍 𝑑 = 𝑉 𝑑 𝐼 𝑑 = 𝑍o 1 + Γ 𝑑 1 − Γ 𝑑 This provides the complex function 𝑍 𝑑 = 𝑓 Re Γ , Im Γ that we want to graph. It is obvious that the result would be applicable only to lines with exactly characteristic impedance 𝑍o. In order to obtain universal curves, we introduce the concept of normalised impedance, 𝑧 𝑑 = 𝑍 𝑑 𝑍o = 1 + Γ 𝑑 1 − Γ 𝑑
- 4. 3 The normalised impedance is represented on the Smith chart by using families of curves that identify the normalised resistance, 𝑟 (real part) and the normalised reactance, 𝑥 (imaginary part), 𝑧 𝑑 = Re 𝑧 + jIm 𝑧 = 𝑟 + j𝑥 Let’s represent the reflection coefficient in terms of its coordinates, Γ 𝑑 = Re Γ + jIm Γ Now we can write, 𝑟 + j𝑥 = 1 + Re Γ + jIm Γ 1 − Re Γ − jIm Γ = 1 − Re2 Γ − Im2 Γ + j2Im Γ 1 − Re Γ 2 + jIm2 Γ
- 5. 4 The real part gives, 𝑟 = 1 − Re2 Γ − Im2 Γ 1 − Re Γ 2 + Im2 Γ 𝑟 Re Γ − 1 2 + Re2 Γ − 1 + 𝑟Im2 Γ + Im2 Γ + 1 1 + 𝑟 − 1 1 + 𝑟 = 0 𝑟 Re Γ − 1 2 + Re2 Γ − 1 + 1 1 + 𝑟 + 1 + 𝑟 Im2 Γ = 1 1 + 𝑟 1 + 𝑟 Re2 Γ − 2Re Γ 𝑟 1 + 𝑟 + 𝑟2 1 + 𝑟 2 + 1 + 𝑟 Im2 Γ = 1 1 + 𝑟 ⇒ Re Γ − 𝑟 1+𝑟 2 + Im2 Γ = 1 1+𝑟 2 Equation of a circle Add a quantity equal to zero
- 6. 5 The imaginary part gives, 𝑥 = 2Im Γ 1 − Re Γ 2 + Im2 Γ 𝑥2 1 − Re Γ 2 + Im2 Γ − 2𝑥Im Γ + 1 − 1 = 0 1 − Re Γ 2 + Im2 Γ − 2 𝑥 Im Γ + 1 𝑥2 = 1 𝑥2 1 − Re Γ 2 + Im2 Γ − 2 𝑥 Im Γ + 1 𝑥2 = 1 𝑥2 ⇒ Re Γ − 1 2 + Im Γ − 1 𝑥 2 = 1 𝑥2 Equation of a circle Multiply by 𝑥 and add a quantity equal to zero
- 7. 6 The result for the real part indicates that on the complex plane with coordinate Re Γ , Im Γ all the possible impedances with a given normalised resistance, 𝑟 are found on a circle with, Center = 𝑟 1 + 𝑟 , 0 Radius = 1 1 + 𝑟 As the normalised resistance, 𝑟 varies from 0 to ∞, we obtain a family of circles completely contained inside the domain of the reflection coefficient Γ ≤ 1 as shown in figure 2.
- 8. 7 Re Γ Im Γ 𝑟 = 1 𝑟 = 5 𝑟 = 0 𝑟 = 0.5 𝑟 → ∞ Figure 2
- 9. 8 The result for the imaginary part indicates that on the complex plane with coordinate Re Γ , Im Γ all the possible impedances with a given normalised reactance, 𝑥 are found on a circle with, Center = 1, 1 𝑥 Radius = 1 𝑥 As the normalised reactance,𝑥 varies from −∞ to ∞, we obtain a family of arcs contained inside the domain of the reflection coefficient Γ ≤ 1 as shown in figure 3.
- 10. 9 Re Γ Im Γ 𝑥 = 1 𝑥 → ±∞ 𝑥 = 0.5 𝑥 = −0.5 𝑥 = −1 Figure 3 𝑥 = 0
- 11. 10 Basic Smith Chart techniques for loss-less transmission line. Given 𝑍 𝑑 ⇒ Find Γ 𝑑 Given Γ 𝑑 ⇒ Find 𝑍 𝑑 Given ΓR and 𝑍R ⇒ Find Γ 𝑑 and 𝑍 𝑑 Given Γ 𝑑 and 𝑍 𝑑 ⇒ Find Γ 𝑑 and 𝑍 𝑑 Find 𝑑max and 𝑑min (maximum and minimum locations for the voltage standing wave pattern Find the Voltage Standing Wave Ratio (VSWR) Given 𝑍 𝑑 ⇒ Find 𝑌 𝑑 Given 𝑌 𝑑 ⇒ Find 𝑍 𝑑
- 12. 11 Given 𝑍 𝑑 ⇒ Find Γ 𝑑 (1) Normalise the impedance, 𝑍 𝑑 = 𝑍 𝑑 𝑍o = 𝑅 𝑍o + j 𝑋 𝑍o = 𝑟 + j𝑥 (2) Find the circle of constant normalised resistance 𝑟 (3) Find the arc of constant normalised reactance 𝑥 (4) The intersection of the two curves indicates the reflection coefficient in the complex plane. The chart provides directly the magnitude and the phase angle of Γ 𝑑 . Example 1: Find Γ 𝑑 , given 𝑍 𝑑 = 25 + j100 Ω with 𝑍o = 50 Ω
- 13. 12 50.906° (1) Normalisation 𝑧 𝑑 = 25+j100 50 = 0.5 + j2.0 (2) Find normalized resistance circle 𝑟 = 0.5 (3) Find normalized reactance arc 𝑥 = 2.0 (4) This vector represents the reflection coefficient Γ 𝑑 = 0.52 + j0.64 Γ 𝑑 = 0.8246 Γ 𝑑 = 0.8885 rad = 50.906° 0.8246 Figure 4
- 14. 13 Given Γ 𝑑 ⇒ Find 𝑍 𝑑 (1) Determine the complex point representing the given reflection coefficient Γ 𝑑 on the chart. (2) Read the values of the normalised resistance 𝑟 and of the normalized reactance 𝑥 that correspond to the reflection coefficient point. (3) The normalized impedance is, 𝑧 𝑑 = 𝑟 + j𝑥 and the actual impedance is, 𝑍 𝑑 = 𝑍o𝑧 𝑑 = 𝑍o 𝑟 + j𝑥 = 𝑍o𝑟 + j𝑍o𝑥
- 15. 14 Given ΓR and 𝑍R ⇔ Find Γ 𝑑 and 𝑍 𝑑 Note: The magnitude of the reflection coefficient is constant along a loss- less transmission line terminated by a specified load, since, Γ 𝑑 = ΓRexp −j2𝛽𝑑 = ΓR Therefore, on the complex plane, a circle with center at the origin and radius ΓR represents all possible reflection coefficients found along the transmission line. When the circle of constant magnitude of the reflection coefficient is drawn on the Smith chart, one can determine the values of the line impedance at any location. The graphical step-by step procedure is, (1) Identify the load reflection coefficient ΓR and the normalized load impedance 𝑍R on the Smith chart.
- 16. 15 (2) Draw the circle of constant reflection coefficient amplitude Γ 𝑑 = ΓR . (3) Starting from the point representing the load, travel on the circle in the clockwise direction, by an angle, 𝜃 = 2𝛽𝑑 = 2 2π 𝜆 𝑑 (4) The new location on the chart corresponds to location 𝑑 on the transmission line. Here, the values of Γ 𝑑 and 𝑍 𝑑 can be read from the chart as before. Example 2: Given, 𝑍R = 25 + j100 Ω with 𝑍o = 50 Ω Find 𝑍 𝑑 and Γ 𝑑 for 𝑑 = 0.18 𝜆
- 17. 16 Circle with constant Γ Γ 𝑑 = 0.8246 ∠ − 78.7° = 0.161 − j0.809 Γ 𝑑 ΓR ∠ΓR 𝑧R 𝑧 𝑑 𝑧 𝑑 = 0.236 − j1.192 𝑍 𝑑 = 𝑧 𝑑 × 𝑍o = 11.79 − j59.6 Ω 𝜃 = 2𝛽𝑑 = 2 2π 𝜆 0.18 𝜆 = 2.262 rad = 129.6° Figure 5
- 18. 17 Given, ΓR and 𝑍R ⇒ Find 𝑑max and 𝑑min (1) Identify on the Smith chart the load reflection coefficient ΓR or the normalized load impedance 𝑍R. (2) Draw the circle of constant reflection coefficient amplitude Γ 𝑑 = ΓR . The circle intersects the real axis of the reflection coefficient at two points which identify 𝑑max (when Γ 𝑑 = Real positive) and 𝑑min (when Γ 𝑑 = Real negative). (3) A commercial Smith chart provides an outer graduation where the distances normalised to the wavelength can be read directly. The angles, between the vector ΓR and the real axis, also provide a way to compute 𝑑max and 𝑑min. Example 3: Find 𝑑max and 𝑑min for 𝑍R = 25 + j100 Ω; 𝑍R = 25 − j100 Ω 𝑍o = 50 Ω
- 19. 18 Im 𝑍R > 0 ΓR ∠ΓR 𝑧R 𝑧 𝑑 Figure 6 2𝛽𝑑max = 50.9° 𝑑max = 0.0707𝜆 2𝛽𝑑min = 230.9° 𝑑min = 0.3207𝜆 𝑍R = 25 + j100 Ω 𝑍o = 50 Ω
- 20. 19 Im 𝑍R < 0 ΓR ∠ΓR 𝑧R Figure 7 2𝛽𝑑max = 309.1° 𝑑max = 0.4293𝜆 2𝛽𝑑min = 129.1° 𝑑min = 0.1793𝜆 𝑍R = 25 − j100 Ω 𝑍o = 50 Ω
- 21. 20 Given, ΓR and 𝑍R ⇒ Find the VSWR The VSWR is defined as, 𝑉𝑆𝑊𝑅 = 𝑉 max 𝑉min = 1 + ΓR 1 − ΓR The normalized impedance at a maximum location of the standing wave pattern is given by, 𝑧 𝑑max = 1 + Γ 𝑑max 1 − Γ 𝑑max = 1 + ΓR 1 − ΓR = 𝑉𝑆𝑊𝑅 This quantity is always real and ≥ 1. The VSWR is simply obtained on the Smith chart, by reading the value of the (real) normalized impedance, at the location 𝑑max where Γ is real and positive.
- 22. 21 The graphical step-by-step procedure is: (1) Identify the load reflection coefficient, ΓR and the normalised load impedance, 𝑍R on the Smith chart. (2) Draw the circle of constant reflection coefficient amplitude, Γ 𝑑 = ΓR . (3) Find the intersection of this circle with the real positive axis for the reflection coefficient (corresponding to the transmission line location 𝑑max). (4) A circle of constant normalized resistance will also intersect this point. Read or interpolate the value of the normalized resistance to determine the VSWR. Example 4: Find the VSWR for; 𝑍R1 = 25 + j100 Ω; 𝑍R2 = 25 − j100 Ω 𝑍o = 50 Ω
- 23. 22 Figure 8 Circle with constant Γ 𝑍R1 𝑍R2 ΓR2 ΓR1 Circle of constant Conductance 𝑟 = 10.4 𝑧 𝑑max = 10.4 For both loads 𝑉𝑆𝑊𝑅 = 10.4
- 24. 23 Given 𝑍 𝑑 ⇔ Find 𝑌 𝑑 Note: The normalized impedance and admittance are defined as, 𝑧 𝑑 = 1 + Γ 𝑑 1 − Γ 𝑑 𝑦 𝑑 = 1 − Γ 𝑑 1 + Γ 𝑑 Since, Γ 𝑑 + 𝜆 4 = −Γ 𝑑 𝑧 𝑑 + 𝜆 4 = 1 + Γ 𝑑 + 𝜆 4 1 − Γ 𝑑 + 𝜆 4 = 1 − Γ 𝑑 1 + Γ 𝑑 = 𝑦 𝑑
- 25. 24 We know that the equality, 𝑧 𝑑 + 𝜆 4 = 𝑦 𝑑 is only valid for normalized impedance and admittance. The actual values are given by, 𝑍 𝑑 + 𝜆 4 = 𝑍o ∙ 𝑧 𝑑 + 𝜆 4 𝑌 𝑑 = 𝑌o ∙ 𝑦 𝑑 = 𝑦 𝑑 𝑍o where, 𝑌o = 1 𝑍o is the characteristic admittance of the transmission line.
- 26. 25 The graphical step-by-step procedure is: (1) Identify the load reflection coefficient, ΓR and the normalised load impedance, 𝑍R on the Smith chart. (2) Draw the circle of constant reflection coefficient amplitude, Γ 𝑑 = ΓR . (3) The normalized admittance is located at a point on the circle of constant Γ which is diametrically opposite to the normalised impedance. Example 5: Given, 𝑍R = 25 + j100 Ω with 𝑍o = 50 Ω Find 𝑌R.
- 27. 26 Circle with constant Γ 𝑧 𝑑 = 0.5 + j2.0 𝑍 𝑑 = 25 + j100 Ω 𝜃 = 180° = 2𝛽 ∙ 𝜆 4 𝑦 𝑑 = 0.11765 − j0.4706 𝑌 𝑑 = 0.002353 − j0.009421 S 𝑧 𝑑 + 𝜆 4 = 0.11765 − j0.4706 𝑍 𝑑 + 𝜆 4 = 5.8824 − j23.5294 Ω Figure 9
- 28. 27 The Smith chart can be used for line admittance, by shifting the space reference to the admittance location. After that, one can move on the chart just reading the numerical values as representing admittances. Let’s review the impedance-admittance terminology: Impedance = Resistance +j Reactance 𝑍 = 𝑅 + j𝑋 Admittance = Conductance +j Susceptance 𝑌 = 𝐺 + j𝐵 On the impedance chart, the correct reflection coefficient is always represented by the vector corresponding to the normalized impedance. Charts specifically prepared for admittances are modified to give the correct reflection coefficient in correspondence of admittance. Smith Chart For Admittance
- 29. 28 Figure 10 𝑧 𝑑 = 0.5 + j2.0 𝑦 𝑑 = 0.11765 − j0.4706 Γ Negative Inductive Susceptance Positive Capacitive Susceptance
- 30. 29 Since related impedance and admittance are on opposite sides of the same Smith chart, the imaginary parts always have different sign. Therefore, a positive (inductive) reactance corresponds to a negative (inductive) susceptance, while a negative (capacitive) reactance corresponds to a positive (capacitive) susceptance. Numerically, we have, 𝑧 = 𝑟 + j𝑥 𝑦 = 𝑔 + j𝑏 = 1 𝑟 + j𝑥 𝑦 = 𝑟 − j𝑥 𝑟 + j𝑥 𝑟 − j𝑥 = 𝑟 − j𝑥 𝑟2 + 𝑥2 𝑔 = 𝑟 𝑟2 + 𝑥2 𝑏 = − 𝑥 𝑟2 + 𝑥2
- 31. 30 Impedance Matching A number of techniques can be used to eliminate reflections when line characteristic impedance and load impedance are mismatched. Impedance matching techniques can be designed to be effective for a specific frequency of operation (narrow band techniques) or for a given frequency spectrum (broadband techniques). One method of impedance matching involves the insertion of an impedance transformer between line and load. Impedance Transformer 𝑍R 𝑍o In the following, we neglect effects of loss in the lines. Figure 11
- 32. 31 A simple narrow band impedance transformer consists of a transmission line section of length 𝜆 4 . The impedance transformer is positioned so that it is connected to a real impedance 𝑍A. This is always positive if a location of maximum or minimum voltage standing wave pattern is selected. 𝑍R Figure 12 𝑍B 𝑍A 𝑍o1 𝑍o2 𝑍o1 𝜆 4 𝑑max or 𝑑min
- 33. 32 Consider a general load impedance with its corresponding load reflection coefficient, 𝑍R = 𝑅R + j𝑋R ΓR = 𝑍R − 𝑍o1 𝑍R + 𝑍o1 = ΓR ej∅ If the transformer is inserted at a location of voltage maximum 𝑑max 𝑍A = 𝑍o1 1 + Γ d 1 − Γ d = 𝑍o1 1 + ΓR 1 − ΓR If it is inserted instead at a location of voltage minimum 𝑑min 𝑍A = 𝑍o1 1 + Γ d 1 − Γ d = 𝑍o1 1 − ΓR 1 + ΓR
- 34. 33 Figure 13 𝑍in 𝑍o 𝐿 = 𝜆 4 𝑍A Consider now the input impedance of a line of length 𝜆 4 . Since: 𝑍A = 𝑍o1 1 + Γ d 1 − Γ d = 𝑍o1 1 − ΓR 1 + ΓR We have, 𝑍in = lim tan 𝛽𝐿 → ∞ 𝑍o 𝑍A + j𝑍otan 𝛽𝐿 j𝑍Atan 𝛽𝐿 + 𝑍o → 𝑍o 2 𝑍A
- 35. 34 Note that if the load is real, the voltage standing wave pattern at the load is maximum when 𝑍R > 𝑍o1 or minimum when 𝑍R < 𝑍o1. The transformer can be connected directly at the load location or at a distance from the load corresponding to a multiple of 𝜆 4 . Figure 14 𝑍B 𝑍A = Real 𝑍o1 𝑍o2 𝑍o1 𝜆 4 𝑍R = Real 𝑑1 = 𝑛 𝜆 4 ; 𝑛 = 0,1,2
- 36. 35 If the load impedance is real and the transformer is inserted at a distance from the load equal to an even multiple of 𝜆 4 then, 𝑍A = 𝑍R 𝑑1 = 2𝑛 𝜆 4 = 𝑛 𝜆 2 but if the distance from the load is an odd multiple of 𝜆 4 , 𝑍A = 𝑍o1 2 𝑍R 𝑑1 = 2𝑛 + 1 𝜆 4 = 𝑛 𝜆 2 + 𝜆 4
- 37. 36 The input impedance of the impedance transformer after inclusion in the circuit is given by, 𝑍B = 𝑍o2 2 𝑍A For impedance matching we need, 𝑍o1 = 𝑍o2 2 𝑍A 𝑍o2 = 𝑍o1𝑍A The characteristic impedance of the transformer is simply the geometric average between the characteristic impedance of the original line and the load seen by the transformer. Let’s now review some simple examples.
- 38. 37 Real Load Impedance, Figure 15 𝑍B 𝑍A 𝑍o1 = 50 Ω 𝑍o2 = ? 𝜆 4 𝑅R = 100 Ω 𝑍B = 𝑍o2 2 𝑅R = 𝑍o1 𝑍o2 = 𝑍o1𝑅R = 50 × 100 ≈ 70.71 Ω
- 39. 38 Note that an identical result is obtained by switching 𝑍01 and 𝑅R, Figure 16 𝑍B 𝑍A 𝑍o1 = 100 Ω 𝑍o2 = ? 𝜆 4 𝑅R = 50 Ω 𝑍B = 𝑍o2 2 𝑅R = 𝑍o1 𝑍o2 = 𝑍o1𝑅R = 100 × 50 ≈ 70.71 Ω
- 40. 39 Another real load case, Figure 17 𝑍B 𝑍A 𝑍o1 = 75 Ω 𝑍o2 = ? 𝜆 4 𝑅R = 300 Ω 𝑍B = 𝑍o2 2 𝑅R = 𝑍o1 𝑍o2 = 𝑍o1𝑅R = 75 × 300 ≈ 150 Ω
- 41. 40 Same impedance as before, but now the transformer is inserted at a distance 𝜆 4 from the load (voltage minimum in this case). Figure 18 𝑍B 𝑍A 𝑍o2 𝑍o1 𝜆 4 𝑅R = 300 Ω 𝜆 4 𝑍o1 = 75 Ω 𝑍A = 𝑍o1 2 𝑅R = 752 300 = 18.75 Ω 𝑍B = 𝑍o2 2 𝑍A = 𝑍o1 ⇒ 𝑍o2 = 𝑍o1𝑍A = 75 × 18.75 ≈ 37.5 Ω
- 42. 41 Complex Load Impedance - Transformer at voltage maximum Figure 19 𝑍B 𝑍A 𝑍o2 𝑍o1 𝜆 4 𝑍R = 100 + j100 Ω 𝑑max 𝑍o1 = 50 Ω ΓR = 100 + j100 − 50 100 + j100 + 50 ≈ 0.62 𝑍A = 𝑍o 1 + ΓR 1 − ΓR ≈ 213.28 Ω 𝑍o2 = 𝑍o1𝑍A = 50 × 213.28 = 103.27 Ω
- 43. 42 Complex Load Impedance - Transformer at voltage minimum Figure 20 𝑍B 𝑍A 𝑍o2 𝑍o1 𝜆 4 𝑍R = 100 + j100 Ω 𝑑min 𝑍o1 = 50 Ω ΓR = 100 + j100 − 50 100 + j100 + 50 ≈ 0.62 𝑍A = 𝑍o 1 − ΓR 1 + ΓR ≈ 11.72 Ω 𝑍o2 = 𝑍o1𝑍A = 50 × 11.72 = 24.21 Ω
- 44. 43 If it is not important to realize the impedance transformer with a quarter wavelength line, we can try to select a transmission line with appropriate length and characteristic impedance, such that the input impedance is the required real value. Figure 21 𝑍A 𝑍o2 𝐿 𝑍R = 𝑅R + j𝑋R 𝑍o1 𝑍o1 = 𝑍A = 𝑍o2 𝑅R + j𝑋R + j𝑍02tan 𝛽𝐿 𝑍o2 + j 𝑅R + 𝑗𝑋R tan 𝛽𝐿
- 45. 44 After separation of real and imaginary parts we obtain the equations, 𝑍o2 𝑍o1 − 𝑅R = 𝑍o1𝑋Rtan 𝛽𝐿 tan 𝛽𝐿 = 𝑍o2𝑋R 𝑍o1𝑅R − 𝑍o2 2 with final solution, 𝑍o2 = 𝑍o1𝑅R − 𝑅R 2 − 𝑋R 2 1 − 𝑅R 𝑍o1 tan 𝛽𝐿 = 1 − 𝑅R 𝑍o1 𝑍o1𝑅R − 𝑅R 2 − 𝑋R 2 𝑋R The transformer can be realized as long as the result for 𝑍o2 is real. Note that this is also a narrow band approach.
- 46. 45 Example 6: A signal generator has an internal impedance of 50 Ω. It needs to feed equal power through a lossless 50 Ω transmission line to two separate resistive loads of 64 Ω and 25 Ω at a frequency of 10 MHz. Design a Quarter-wave transformer to match the loads to the 50 Ω line. Assume the phase velocities of the wave travelling on them is 0.5𝑐. Solution: As the two quarter-wave transformers are connected in parallel to the 50 Ω line, if equal powers are required to the two loads, the input impedances of the two branches looking at the junction from 50 Ω line must be equal to 100 Ω so that when they add together in parallel, the total impedances is 50 Ω.
- 47. 46 𝑅o1 𝑅o2 𝑅in1 = 100 Ω 𝑅in2 = 100 Ω 𝑅L1 = 64 Ω 𝑅L2 = 2𝟓 Ω 𝜆 4 𝜆 4 𝑅o = 50 Ω 100 Ω 100 Ω 50 Ω Figure 22: Quarter-wave transformer.
- 48. 47 Therefore, 𝑍in1 = 𝑅in1 = 100 Ω 𝑍in2 = 𝑅in2 = 100 Ω The characteristic impedances 𝑅o1 ′ = 𝑅in1 × 𝑅L1 = 100 × 64 = 80 Ω 𝑅o2 ′ = 𝑅in2 × 𝑅L2 = 100 × 25 = 50 Ω 𝑉 p = 0.5𝑐 = 𝑐 𝜀r = 3 × 108 𝜀r ∴ 𝜀r= 4 where c is the speed of light. Wavelength along the transformers, 𝜆 = 𝜆o 𝜀r = 30 m 4 = 15 m. Physical length of the transformers, 𝑙 = 𝜆 4 = 15 4 = 3.75 m.
- 49. 48 Example 7: Design a Quarter-wave transformer is to match a 10 Ω resistor to a 50 Ω line at 2 GHz. The transmission line used is coaxial cable whose dielectric has a dielectric constant 𝜀r = 2.25. Sketch the final design, specifying the dimensions and impedance of the transformer. Solution: Transformer impedance: Given 𝑅L = 10, want 𝑅o ′ = 50. So, 𝑅𝜆 4 = 𝑅L𝑅o ′ = 10 × 50 = 22.36. Transformer’s physical length: 𝜆 = 𝑉 p 𝑓 = 𝑐 𝜀𝑟 𝑓 = 3 × 108 2.25 2 × 109 = 0.1 m 𝑙 = 𝜆 4 = 0.1 4 = 0. 025m
- 50. 49 Ratio between Outer and Inner Conductors: For a lossless coaxial cable, 𝑍o = 1 2π 𝜇 𝜀 ln 𝐷 𝑑 = 𝜂o 2π 𝜀r ln 𝐷 𝑑 where 𝜂o = 𝜇o 𝜀o ≅ 377 Ω is the characteristic impedance of free space, while D and d are diameters of the inner and outer conductors respectively. In our case, we have 𝑍o = 22.36, so, 22.36 = 377 2π 2.25 ln 𝐷 𝑑 ln 𝐷 𝑑 = 0.5590 𝐷 𝑑 = 1.7489
- 51. 50 Figure 23 shows the Quarter-wave transformer design and cross-sectional view of the coaxial cable. D 𝑅𝜆 4 = 22.36 𝑍L 𝜀r = 2.25 𝑅o ′ = 50 𝑙 = 𝜆 4 𝐷 = 1.7489𝑑 Figure 23: Quarter-wave transformer d
- 52. 51 Single Stub Impedance Matching Impedance matching can be achieved by inserting another transmission line (stub) as shown in the diagram below. Figure 24 𝑍A = 𝑍o 𝑍R 𝑑stub 𝑍o 𝐿stub 𝑍oS
- 53. 52 There are two design parameters for single stub matching: (1) The location of the stub with reference to the load 𝑑stub. (2) The length of the stub line 𝐿stub. Any load impedance can be matched to the line by using single stub technique. The drawback of this approach is that if the load is changed, the location of insertion may have to be moved. The transmission line realizing the stub is normally terminated by a short or by an open circuit. In many cases it is also convenient to select the same characteristic impedance used for the main line, although this is not necessary. The choice of open or shorted stub may depend in practice on a number of factors. A short circuited stub is less prone to leakage of electromagnetic radiation and is somewhat easier to realize. On the other hand, an open circuited stub may be more practical for certain types of transmission lines, for example microstrips where one would have to drill the insulating substrate to short circuit the two conductors of the line.
- 54. 53 Since the circuit is based on insertion of a parallel stub, it is more convenient to work with admittances, rather than impedances. Figure 25 𝑌A = 𝑌o 𝑌R = 1 𝑍R 𝑑stub 𝑌o = 1 𝑍o 𝐿stub 𝑌oS
- 55. 54 For proper impedance match: 𝑌A = 𝑌stub + 𝑌 𝑑stub = 𝑌o = 1 𝑍o Input admittance of the stub line Line admittance at location 𝑑stub before the stub is applied 𝐿stub 𝑌oS 𝑌R = 1 𝑍R 𝑑stub 𝑌 𝑑stub 𝑌stub Figure 26
- 56. 55 In order to complete the design, we have to find an appropriate location for the stub. Note that the input admittance of a stub is always imaginary (inductance if negative, or capacitance if positive). 𝑌stub = j𝐵stub A stub should be placed at a location where the line admittance has real part equal to 𝑌o, 𝑌 𝑑stub = 𝑌o + j𝐵 𝑑stub For matching, we need to have, 𝐵stub = −𝐵 𝑑stub Depending on the length of the transmission line, there may be a number of possible locations where a stub can be inserted for impedance matching. It is very convenient to analyze the possible solution on a Smith chart.
- 57. 56 𝑦R 𝜃2 𝑧R Figure 27 Load location 𝜃1 First location suitable for stub insertion Unitary conductance circle Second Location suitable for stub insertion Constant Γ 𝑑 circle 𝑦 𝑑stub2 𝑦 𝑑stub1
- 58. 57 The red arrow on the example indicates the load admittance. This provides on the “admittance chart” the physical reference for the load location on the transmission line. Notice that in this case the load admittance falls outside the unitary conductance circle. If one moves from load to generator on the line, the corresponding chart location moves from the reference point, in clockwise motion, according to an angle 𝜃 (indicated by the light green arc). 𝜃 = 2𝛽𝑑 = 4π 𝜆 d The value of the admittance rides on the red circle which corresponds to constant magnitude of the line reflection coefficient. Γ 𝑑 = Γ𝑅 , imposed by the load. Every circle of constant Γ 𝑑 intersects the circle Re 𝑦 = 1 (unitary normalized conductance), in correspondence of two points. Within the first revolution, the two intersections provide the locations closest to the load for possible stub insertion.
- 59. 58 The first solution corresponds to an admittance value with positive imaginary part, in the upper portion of the chart. Line Admittance - Actual: 𝑌 𝑑stub1 = 𝑌o + j𝐵 𝑑stub1 Normalized: y 𝑑stub1 = 1 + j𝑏 𝑑stub1 Stub Location: 𝑑stub1 = 𝜃1 4π 𝜆 Stub Admittance - Actual: −j𝐵 𝑑stub1 Normalized: −j𝑏 𝑑stub1 Stub Length: 𝐿stub = 𝜆 2π tan−1 1 𝑍oS𝐵 𝑑stub1 ⇒ Short 𝐿stub = 𝜆 2π tan−1 𝑍oS𝐵 𝑑stub1 ⇒ Open
- 60. 59 The second solution corresponds to an admittance value with negative imaginary part, in the lower portion of the chart. Line Admittance - Actual: 𝑌 𝑑stub2 = 𝑌o + j𝐵 𝑑stub2 Normalized: y 𝑑stub2 = 1 + j𝑏 𝑑stub2 Stub Location: 𝑑stub2 = 𝜃2 4π 𝜆 Stub Admittance - Actual: j𝐵 𝑑stub2 Normalized: j𝑏 𝑑stub2 Stub Length: 𝐿stub = 𝜆 2π tan−1 − 1 𝑍oS𝐵 𝑑stub2 ⇒ Short 𝐿stub = 𝜆 2π tan−1 −𝑍oS𝐵 𝑑stub2 ⇒ Open
- 61. 60 If the normalized load admittance falls inside the unitary conductance circle (see next figure), the first possible stub location corresponds to a line admittance with negative imaginary part. The second possible location has line admittance with positive imaginary part. In this case, the formulae given above for first and second solution exchange place. If one moves further away from the load, other suitable locations for stub insertion are found by moving toward the generator, at distances multiple of half a wavelength from the original solutions. These locations correspond to the same points on the Smith chart. First set of locations = 𝑑stub1 + 𝑛 𝜆 2 Second set of locations = 𝑑stub2 + 𝑛 𝜆 2
- 62. 61 𝑧R 𝜃2 𝑦R Figure 28 𝜃1 First location suitable for stub insertion Unitary conductance circle Second location suitable for stub insertion Constant Γ 𝑑 circle 𝑦 𝑑stub1 𝑦 𝑑stub2
- 63. 62 Single stub matching problems can be solved on the Smith chart graphically, using a compass and a ruler. This is a step-by-step summary of the procedure: (a) Find the normalized load impedance and determine the corresponding location on the chart. (b) Draw the circle of constant magnitude of the reflection coefficient Γ for the given load. (c) Determine the normalized load admittance on the chart. This is obtained by rotating 180° on the constant Γ circle, from the load impedance point. From now on, all values read on the chart are normalized admittances.
- 64. 63 𝑦R 𝑧R (a) Obtain the normalized load impedance 𝑧R = 𝑍R 𝑍o and find its location on the Smith chart (b) Draw the constant Γ 𝑑 circle (c) Find the normalized load admittance knowing that 𝑦R = 𝑧 𝑑 = 𝜆 4 . From now on the chart represents admittances. 180° = 𝜆 4 Figure 29
- 65. 64 (d) Move from load admittance toward generator by riding on the constant circle, until the intersections with the unitary normalized conductance circle are found. These intersections corresponds to possible locations for stub insertion. Commercial Smith charts provide graduations to determine the angles of rotation as well as the distances from the load in units of wavelength. (e) Read the line normalized admittance in correspondence of the stub insertion locations determined in (d). These values will always be of the form. 𝑦 𝑑stub = 1 + j𝑏 top half of chart 𝑦 𝑑stub = 1 − j𝑏 bottom half of chart
- 66. 65 𝑦R 𝜃1 𝑧R Figure 30 Load location (d) Move from load toward generator and stop at a location where the real part of the normalized line admittance is 1 Unitary conductance circle First location suitable for stub insertion, 𝑑stub1 = 𝜃1 4π 𝜆 (e) Read here the value of the normalized line admittance, 𝑦 𝑑stub1 = 1 + j𝑏
- 67. 66 𝑦R 𝜃2 𝑧R Figure 31 Load location (d) Move from load toward generator and stop at a location where the real part of the normalized line admittance is 1 Unitary conductance circle Second location suitable for stub insertion, 𝑑stub2 = 𝜃2 4π 𝜆 (e) Read here the value of the normalized line admittance, 𝑦 𝑑stub2 = 1 − j𝑏
- 68. 67 (f) Select the input normalized admittance of the stubs, by taking the opposite of the corresponding imaginary part of the line admittance, Line: 𝑦 𝑑stub = 1 + j𝑏 → stub: 𝑦 𝑑stub = −j𝑏 Line: 𝑦 𝑑stub = 1 − j𝑏 → stub: 𝑦 𝑑stub = +jb (g) Use the chart to determine the length of the stub. The imaginary normalized admittance values are found on the circle of zero conductance on the chart. On a commercial Smith chart one can use a printed scale to read the stub length in terms of wavelength. We assume here that the stub line has characteristic impedance 𝑍o as the main line. If the stub has characteristic impedance 𝑍oS ≠ 𝑍o the values on the Smith chart must be re-normalized as, ±j𝑏′ = ±j𝑏 𝑌o 𝑌oS = ±j𝑏 𝑍oS 𝑍o
- 69. 68 Figure 32 𝑦 = ∞ Short circuit (f) Normalized input admittance of stub, 𝑦stub = 0 − j𝑏 (g) Arc to determine the length of a short circuited stub with normalized input admittance, −j𝑏
- 70. 69 Figure 33 𝑦 = 0 Open circuit (f) Normalized input admittance of stub, 𝑦stub = 0 − j𝑏 (g) Arc to determine the length of a short circuited stub with normalized input admittance, −j𝑏
- 71. 70 Figure 34 (f) Normalized input admittance of stub, 𝑦stub = 0 + j𝑏 (g) Arc to determine the length of a short circuited stub with normalized input admittance +j𝑏 𝑦 = ∞ Short circuit
- 72. 71 Figure 35 (f) Normalized input admittance of stub, 𝑦stub = 0 + j𝑏 (g) Arc to determine the length of a short circuited stub with normalized input admittance +j𝑏 𝑦 = 0 Open circuit
- 73. 72 𝑦R Figure 36 After the stub is inserted, the admittance at the stub location is moved to the center of the Smith chart, which corresponds to normalized admittance 1 and reflection coefficient 0 (exact matching condition). Matching condition If you imagine to add gradually the negative imaginary admittance of the inserted stub, the total admittance would follow the yellow arrow, reaching the match point when the complete stub admittance is added. First Solution
- 74. 73 𝑦R Figure 37 If the stub does not have the proper normalized input admittance, the matching condition is not reached First Solution Effect of a stub with positive susceptance Effect of a stub with negative susceptance of insufficient magnitude Effect of a stub with negative susceptance of excessive magnitude
- 75. 74 Double Stub Impedance Matching Impedance matching can be achieved by inserting two stubs at specified locations along transmission line as shown in the diagram below. Figure 38 𝑑stub1 𝐿stub1 𝑌oS1 𝑌A = 𝑌o1 𝑌o = 1 𝑍o1 𝐿stub2 𝑌oS2 𝑌R = 1 𝑍R 𝑑stub2
- 76. 75 There are two design parameters for double stub matching: (1) The length of the first stub line 𝐿stub1. (2) The length of the second stub line 𝐿stub2. In the double stub configuration, the stubs are inserted at predetermine locations. In this way, if the load impedance is changed, one simply has to replace the stubs with another set of different length. The drawback of double stub is that a certain range of load admittances cannot be matched once the stub locations are fixed. Three stubs are necessary to guarantee that match is always possible.
- 77. 76 The length of the first stub is selected so that the admittance at the location of the second stub (before the second stub is inserted) has real part equal to the characteristic admittance of the line. Figure 39 𝑑stub1 𝐿stub1 𝑌oS1 𝑌′A = 𝑌o1 + j𝐵 𝑌o = 1 𝑍o1 𝑌R = 1 𝑍R 𝑑stub2
- 78. 77 Figure 40 𝑑stub1 𝐿stub1 𝑌oS1 𝑌A = 𝑌o1 + j𝐵 − j𝐵 = 𝑌o1 𝑌o = 1 𝑍o1 𝑌R = 1 𝑍R 𝑑stub2 𝐿stub2 𝑌oS2 𝑌stub = −j𝐵 The length of the second stub is selected to eliminate the imaginary part of the admittance at the location of insertion.
- 79. 78 Figure 41 At the location where the second stub is inserted, the possible normalized admittances that can give matching are found on the circle of unitary conductance on the Smith chart The normalized admittance that we want at location 𝑑stub2 is on this circle
- 80. 79 Figure 42 𝐿stub1 𝑌oS1 𝑌A = 𝑌o1 𝑌o1 = 1 𝑍o1 𝑌R 𝑑stub2 𝐿stub2 𝑌oS2 𝑌R 𝑑stub Single stub Double stub The two approaches solve the same problem
- 81. 80 If one moves from the location of the second stub back to the load, the circle of the allowed normalized admittances is mapped into another circle, obtained by pivoting the original circle about the center of the chart. At the location of the first stub, the allowed normalized admittances are found on an auxiliary circle which is obtained by rotating the unitary conductance circle counterclockwise, by an angle. 𝜃aux = 4π 𝜆 𝑑stub2 − 𝑑stub1 = 4π 𝜆 𝑑21
- 82. 81 Figure 43 This is the auxiliary circle for distance between the stubs 𝑑21 = 𝜆 8 + 𝑛 𝜆 2 𝜃aux
- 83. 82 Figure 44 This angle of rotation corresponds to a distance 𝑑12 = 𝑑stub2 −𝑑stub1 The normalized admittance that we want at location 𝑑stub1 is on this auxiliary circle Pivot here 𝜃aux
- 84. 83 Figure 45 This is the auxiliary circle for distance between the stubs 𝑑21 = 𝜆 4 + 𝑛 𝜆 2 𝜃aux
- 85. 84 Figure 46 This is the auxiliary circle for distance between the stubs 𝑑21 = 3 𝜆 8 + 𝑛 𝜆 2 𝜃aux
- 86. 85 Figure 47 This is the auxiliary circle for distance between the stubs 𝑑21 = 𝑛 𝜆 2 This is not a good choice for double stub design. 𝜃aux
- 87. 86 Given the load impedance, we need to follow these steps to complete the double stub design: (a) Find the normalized load impedance and determine the corresponding location on the chart. (b) Draw the circle of constant magnitude of the reflection coefficient Γ for the given load. (c) Determine the normalized load admittance on the chart. This is obtained by rotating −180° on the constant Γ circle, from the load impedance point. From now on, all values read on the chart are normalized admittances. (d) Find the normalized admittance at location 𝑑stub1 by moving clockwise on the constant Γ circle.
- 88. 87 (e) Draw the auxiliary circle. (f) Add the first stub admittance so that the normalized admittance point on the Smith chart reaches the auxiliary circle (two possible solutions). The admittance point will move on the corresponding conductance circle, since the stub does not alter the real part of the admittance. (g) Map the normalized admittance obtained on the auxiliary circle to the location of the second stub 𝑑stub2. The point must be on the unitary conductance circle. (h) Add the second stub admittance so that the total parallel admittance equals the characteristic admittance of the line to achieve exact matching condition.
- 89. 88 𝑦R 𝑧R (a) Obtain the normalized load impedance 𝑧R = 𝑍R 𝑍o and find its location on the Smith chart 180° = 𝜆 4 Figure 48 (b) Draw the constant Γ 𝑑 circle (c) Find the normalized load admittance knowing that 𝑦R = 𝑧 𝑑 = 𝜆 4 . From now on the chart represents admittances. (d) Move to the first stub location
- 90. 89 𝑦R 𝑧R (f) First solution: Add admittance of first stub to reach auxiliary circle Figure 49 (f) Second solution: Add admittance of first stub to reach auxiliary circle (e) Draw the auxiliary circle
- 91. 90 (g) First solution: Map the normalized admittance from the auxiliary circle to the location of the second stub 𝑑stub2 Figure 50 First solution: Admittance at location 𝑑stub2 before insertion of second stub (h) Add the second stub admittance
- 92. 91 (g) Second solution: Map the normalized admittance from the auxiliary circle to the location of the second stub 𝑑stub2 Second solution: Admittance at location 𝑑stub2 before insertion of second stub (h) Add the second stub admittance Figure 51
- 93. 92 Example 8: A 50 Ω lossless transmission line is connected to a load impedance 𝑍L = 35 − j47.5 Ω. Design a short-circuit stub required to match the load at a frequency of 200 MHz. Assume that the transmission line is a coaxial line filled with a dielectric material for which 𝜀r = 9. Solution: Given 𝑍o = 𝑅o = 50 Ω and 𝑍L = 35 − j47.5 Ω Normalized 𝑧L = 𝑍L 𝑍o = 0.7 − j0.95 (i) Enter 𝑧L at point 𝑃1. (ii) Draw a Γ -circle centered at O with radius O𝑃1. (iii) Draw straight line from 𝑃1 through O to point 𝑃2 ′ on the perimeter, intersecting the Γ -circle at 𝑃2, which represents 𝑦L. Note 0.109 at 𝑃2 ′ on the wavelengths toward generator scale.
- 94. 93 (iv) Note the two points of intersection of the Γ -circle with the 𝑔 = 1 circle: At 𝑃3: 𝑦B1 = 1 + j1.2 = 1 + j𝑏B1 At 𝑃4: 𝑦B2 = 1 − j1.2 = 1 + j𝑏B2 (v) Solutions for the position of the stub: For 𝑃3 (from 𝑃2 ′ to 𝑃3 ′ ) 𝑑1 = 0.168 − 0.109 𝜆 = 0.059𝜆 For 𝑃3 (from 𝑃2 ′ to 𝑃4 ′ ) 𝑑2 = 0.332 − 0.109 𝜆 = 0.223𝜆 (vi) Solutions for the length of the short-circuited stub to provide 𝑦𝑠 = −j𝑏𝐵: For 𝑃3 𝑦s = −j𝑏B1 = −j1.2 (from 𝑃sc to 𝑃3 ′′ ) 𝑙1 = 0.361 − 0.250 𝜆 = 0.111𝜆 For 𝑃4 𝑦s = −j𝑏B2 = j1.2 (from 𝑃sc to 𝑃4 ′′ ) 𝑙2 = 0.139 − 0.250 𝜆 = 0.389𝜆
- 95. 94 (vii) To compute the physical lengths of the transmission line sections, we need to calculate the wavelength on the transmission line. Therefore, 𝜆 = 𝑐 𝜀r 𝑓 = 3 × 108 4 200 × 106 ≅ 0.5 m Thus: 𝑑1 = 0.059𝜆 = 29.5 mm 𝑙1 = 0.111𝜆 = 55.5 mm 𝑑2 = 0.223𝜆 = 111.5 mm 𝑙1 = 0.389𝜆 = 194.5 mm
- 96. 95 𝑙1 Figure 52 𝑙2 𝑑1 𝑑2 𝑃1 𝑃4 𝑃2 𝑃3 𝑃2 ′ 𝑃4 ′′ 𝑃3 ′ 𝑃3 ′′ 𝑃4 ′ −j1.2 j1.2 1 + j1.2 1 − j1.2 𝑧L 𝑦L 𝑔 = 1 O
- 97. 96 (1) Amanogawa, Transmission Lines: Smith Chart, Digital Maestro Series, 2000. References