Bài giảng: Kỹ thuật siêu cao tần (CT389)Chương II: Đồ thị Smith(Smith Chart)

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3/21/2012 Bài giảng: Kỹ thuật siêu cao tần (CT389) Chương II : Đồ thị Smith (Smith Chart) Giảng viên: GVC.TS Lương Vinh Quốc Danh Bộ môn Điện tử Viễn thông, Khoa Công Nghệ E-mail: lvqdanh@ctu.edu.vn Luong Vinh Quoc Danh Introduction:  The Smith chart is one of the most useful graphical tools for high frequency circuit applications  From a mathematical point of view, the Smith chart is a representation of all possible complex impedances with respect to coordinates defined by the complex reflection coefficient  The domain of definition of the reflection coefficient for a lossless line is a circle of unitary radius (vòng tròn đơn vị) in the complex plane (mặt phẳng phức) This is also the domain of the Smith chart The goal of the Smith chart is to identify all possible impedances on the domain of existence of the reflection coefficient Luong Vinh Quoc Danh 3/21/2012 Introduction: (Cont’d) We start from the general definition of line impedance:  ( x )  ( x ) Z ( x)  Z (2.1) or normalized line impedance: Reflection coefficient: z( x)   ( x )  ( x ) ( x )  Z ( x)  Z Z ( x)  Z (2.3) z( x)  z( x)  (2.4) ( x )  (2.2) Luong Vinh Quoc Danh Introduction: (Cont’d) Let’s represent the reflection coefficient in terms of its coordinates (x) = r(x) + j i(x) or  = r + ji Where (2.6) r = Re() and i = Im() Normalized line impedance: z = r + jx (2.8) r = R/Z0: normalized resistance x = X/Z0: normalized reactance Luong Vinh Quoc Danh 3/21/2012 Introduction: (Cont’d) Now we can rewrite (2.2) as follows: r  jx  The real part r gives: r  r  ji  r  ji (2.10)  r2  i2 (1  r )  i2 r      r    i    1 r   1 r  (2.11) (2.13) Equation of a circle Luong Vinh Quoc Danh Introduction: (Cont’d) The result for the real part indicates that on the complex plane with coordinates (Re(Γ), Im(Γ)) all the possible impedances with a given normalized resistance r are found on a circle with As the normalized resistance r varies from to ∞ , we obtain a family of circles completely contained inside the domain of the reflection coefficient | Γ | ≤ Luong Vinh Quoc Danh 3/21/2012 Introduction: (Cont’d) The imaginary part x gives: x 2i (1  r )  i2 r  12   i  (2.12) 1 1     x  x (2.15) Equation of a circle The result for the imaginary part indicates that on the complex plane with coordinates (Re(Γ), Im(Γ)) all the possible impedances with a given normalized reactance x are found on a circle with Luong Vinh Quoc Danh Introduction: (Cont’d) As the normalized reactance x varies from -∞ to ∞ , we obtain a family of arcs contained inside the domain of the reflection coefficient | Γ | ≤ Luong Vinh Quoc Danh 3/21/2012 Introduction (Cont’d) Basic Smith Chart techniques for loss-less transmission lines  Given Z(d)  Find Γ(d) Given Γ(d)  Find 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) Luong Vinh Quoc Danh Smith Chart Phillip Hagar Smith (1905–1987): graduated from Tufts College in 1928, invented the Smith Chart in 1939 while he was working for the Bell Telephone Laboratories Luong Vinh Quoc Danh 10 3/21/2012 Smith Chart (Cont’d) Toward Generator Constant Reflection Coefficient Circle Away From Generator Scale in Wavelengths Full Circle is One Half Wavelength Since Everything Repeats K A Connor Luong VinhDepartment Quoc Danh RPI ECSE 11 Smith Chart (Cont’d) (2.20) where d=l–x  = (lossless) (2.21) Luong Vinh Quoc Danh 12 3/21/2012 Smith Chart (Cont’d) SWR: Standing Wave Ratio RFL COEFF P : Reflection Coefficient Power ||2 RFL COEFF E or I : Reflection Coefficient Voltage || RTN LOSS [dB] : Return Loss S11 Luong Vinh Quoc Danh 13 Smith Chart : Admittances j Luong Vinh Quoc Danh 14 3/21/2012 Smith Chart : Admittances (Cont’d) 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 Analytically, the normalized impedance and admittance are related as 15 Luong Vinh Quoc Danh Smith Chart : Nodes and Anti-nodes (anti-node) r max = S Load impedance ZL can be deduced graphically if we know: + SWR + Distance from node (anti-node) to load (node) r = 1/S Luong Vinh Quoc Danh 16 3/21/2012 Basic Applications  Given Z(d) ⇒ Find Γ(d) Normalize the impedance z Find the circle of constant normalized resistance r Find the arc of constant normalized reactance x 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: Find Γ(d), given Z(d)=25+ j100 [Ω] with Z0 = 50Ω Luong Vinh Quoc Danh 17 Basic Applications (Cont’d) Luong Vinh Quoc Danh 18 3/21/2012 Basic Applications (Cont’d) • Given Γ(d) ⇒ Find Z(d) Determine the complex point representing the given reflection coefficient Γ(d) on the chart Read the values of the normalized resistance r coefficient point and of the normalized reactance x that correspond to the reflection coefficient point The normalized impedance is and the actual impedance is Luong Vinh Quoc Danh 19 Basic Applications (Cont’d)  Given ΓR and ZR ⇐⇒ Find Γ(d) and Z(d) Identify the load reflection coefficient ΓR and the normalized load impedance ZR on the Smith chart Draw the circle of constant reflection coefficient amplitude |Γ(d)| = |ΓR| Starting from the point representing the load, travel on the circle in the clockwise direction, by an angle 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 Luong Vinh Quoc Danh 20 10 3/21/2012 Basic Applications (Cont’d) Example: Find dmax and dmin for ZR =25+j100 (Ω); and ZR = 25−j100 (Ω) with Z0 = 50 (Ω) Luong Vinh Quoc Danh 23 Basic Applications (Cont’d) Luong Vinh Quoc Danh 24 12 3/21/2012 Basic Applications (Cont’d)  Given ΓR and ZR ⇒ Find the Voltage Standing Wave Ratio (VSWR) Identify the load reflection coefficient ΓR and the normalized load impedance ZR on the Smith chart Draw the circle of constant reflection coefficient amplitude |Γ(d)| = |ΓR| Find the intersection of this circle with the real positive axis for the reflection coefficient (corresponding to the transmission line location dmax) A circle of constant normalized resistance will also intersect this point Read or interpolate the value of the normalized resistance to determine the VSWR Luong Vinh Quoc Danh 25 Basic Applications (Cont’d) Example: Find the VSWR for ZR1= 25 + j100 (Ω) ; ZR2= 25 − j100 (Ω) with Z0 = 50 Ω Luong Vinh Quoc Danh 26 13 3/21/2012 Basic Applications (Cont’d)  Given Z(d) ⇐⇒ Find Y(d) Identify the load reflection coefficient ΓR and the normalized load impedance ZR on the Smith chart Draw the circle of constant reflection coefficient amplitude |Γ(d)| =|ΓR| The normalized admittance is located at a point on the circle of constant |Γ| which is diametrically opposite to the normalized impedance Luong Vinh Quoc Danh 27 Basic Applications (Cont’d) Example: Given ZR = 25+ j 100 (Ω) with Z0 = 50 Ω Find YR Luong Vinh Quoc Danh 28 14 3/21/2012 Basic Applications (Cont’d)  Find circuit impedance Given R = 50 ; C1 = 10 pF; C2 = 12 pF; L = 22.5 nH;  = 109 rad/s Find Z Z Choose Z0 = 50  (point A) (point B) (point C) (point D) (point E) Luong Vinh Quoc Danh 29 Impedance matching Power transferred to a voltage anti-node on the lossless TX line: (2.52) (2.55) where : Power generated by signal source : Reflected power When || = 0, there is no reflected waves, therefore, no reflected power Power generated by signal source is totally received by load Luong Vinh Quoc Danh 30 15 3/21/2012 Impedance matching (Cont’d)  For lossy transmission line: P L : incident power at load P  : incident power at source S P L : reflected power at load PS : reflected power at source PL  PS e   l (2.57) PL  |  |2 PL  |  |2 PS e   l (2.58) PS  PL e   l  |  |2 P  e   l (2.59) S Power generated by source: PS  PS  PS  PS (  |  |2 e   l ) (2.60) Power consumption at load: PL  PL  PL  PS e   l (   ) (2.61) 31 Luong Vinh Quoc Danh Impedance matching (Cont’d) Power efficiency:  PL PS  e   l (  |  |2 )  |  |2 e   l When      e   l Luong Vinh Quoc Danh (2.62) (2.63) 32 16 3/21/2012 Impedance matching using lumped elements Matching circuit Z0 ZL -circuit -circuit X1 Z0 X2 X1 ZL Z0 X2 ZL X1, X2 : inductance, or capacitance 33 Luong Vinh Quoc Danh Impedance matching using lumped elements (Cont’d) Chip capacitors Chip resistors (size = mm x mm) Luong Vinh Quoc Danh 34 17 3/21/2012 Impedance matching: -circuit Example: ZL = 10 - j40 (); Z0 = 50 ;  = 109 rad/s Luong Vinh Quoc Danh 35 Impedance matching: -circuit (cont’d) Zt’ = ZL + jx1 Zt = ZL + jx1 ZL = 0.2 – j0.8 Luong Vinh Quoc Danh 36 18 3/21/2012 Impedance matching: -circuit (cont’d) Zt’ = ZL + jx1 Zt = ZL + jx1 yt = + j2 yt‘= - j2 y = yt + jb2 jx1 = zt – zL = 0.2 –j0.4 – (0.2 – j0.8) jx1 = zt‘ – zL = 0.2 + j0.4 – (0.2 – j0.8) = j1.2 = j0.4  L1 =0.4R0 / = 20 nH  L1 = 1.2R0/ = 60 nH jb2 = 1- yt = - j2  L2 = R0/2 = 25 nH  jb2 = 1- yt‘ = j2  C2 = 2/R0 = 40 pF R0 = 50  ZL = 10 – j40 [] Luong Vinh Quoc Danh 37 Impedance matching: -circuit  Example: ZL = 10 - j40 (); Z0 = 50 ;  = 109 rad/s Luong Vinh Quoc Danh 38 19 3/21/2012 Impedance matching: -circuit (cont’d) yt = yL + jb2 zL = 0.2 – j0.8 39 Luong Vinh Quoc Danh Impedance matching: -circuit (cont’d) yt’ = 0.3 – j0.46 yt = 0.3 + j0.46 zt = – j1.55 zt‘= + j1.55 z = zt + jx1 jb2 = yt‘ – yL = 0.3 - j0.46 – (0.3 + j1.18) = -j1.64 jb2 = yt – yL = 0.3 + j0.46 – (0.3 + j1.18)  L2 = R0/1.64 = 30.5 nH = - j0.72  L2 = R0 /0.72  = 70 nH  jx1 = 1- zt‘ = - j1.55  C1 = 1/1.5R0 = 13 pF jx1 = 1- zt = j1.55  L1 = 1.55R0/ = 77.5 nH R0 = 50  R0 = 50  ZL = 10 – j40 [] Luong Vinh Quoc Danh ZL = 10 – j40 [] 40 20 3/21/2012 Impedance matching: forbidden regions Forbidden region of -circuit Forbidden region of  -circuit Luong Vinh Quoc Danh 41 Impedance matching: Single stub method  Example: Z0 = 50 , ZL = 50/[2+j(2+3)] () Short-circuited stub: length = l, Rs0 = 100 () Find l, d Luong Vinh Quoc Danh 42 21 3/21/2012 Impedance matching: Single stub (cont’d) yL = Z0/ZL = 2+ j3.732 yL  moving toward Gen  yd = 1- j2.6 Luong Vinh Quoc Danh 43 Impedance matching: Single stub (cont’d)  yd = – j2.6 Yd = yd G0 = (1- j2.6)/50 = 0.02 – j0.052 [S] Therefore Bd = - 0.052 Bs = - Bd = 0.052 Normalized susceptance bs = Bs/Gs0 = Bs Rs0 = 0.052 x 100 = 5.2 Short-circuited stub: l = (0.5 – 0.031)  = 0.469 y = j5.2 d = (0.302 – 0.215)  = 0.087  y=∞ Luong Vinh Quoc Danh 44 22 3/21/2012 Impedance matching: Single stub (cont’d) An RF Filter (1.1 GHz) Luong Vinh Quoc Danh 45 Impedance matching: Double stub method  Example: Z0 = 50 ; ZL = (100 + j100) [] Stub #1: l1, shortcircuited line, d = 0.4 Stub #2: l2; d12 = 3/8 Find l1, l2 Luong Vinh Quoc Danh 46 23 3/21/2012 Impedance matching: Double stub (cont’d) zL = 2+j2 yd1= 0.55 – j1.08 yt1 = yd1 + ys1 yt1’ = yd1’ + ys1 yL rotated CW 0.4 yd1 47 Luong Vinh Quoc Danh Impedance matching: Double stub (cont’d) ys1 = yt1 – yd1 = j0.97 ys1’ = yt1’ – yd1’ = -j0.8 ys1  l1 = 0.373 ys1’  l2 = 0.143 yt1 = 0.55 – j0.11 yt1’ = 0.55 – j1.88 Luong Vinh Quoc Danh 48 24 3/21/2012 Impedance matching: Double stub (cont’d) ys2 = yt2 – yd2 = j0.61 yd2’ = 1+ j2.6 ys2  l2 = 0.337 yt1’ rotated CW 3/8  yd2’ ys2’ = yt2’ – yd2’ = -j2.6 ys2’  l2’ = 0.058 yt1 rotated CW 3/8  yd2 yd2 = 1- j0.61 Luong Vinh Quoc Danh 49 Impedance matching: Double stub (cont’d)    The advantage of this technique is the position of stubs (d12 and x) are fixed The matching are done by changing the length of stubs d12 = /8, /4, or 3/8 The disadvantage of this technique is not all impedances can be matched (forbidden regions depend on values of d12 and x) Luong Vinh Quoc Danh 50 25 3/21/2012 Impedance matching: Double stub (cont’d) 3.1GHz Microstrip Solid-State Amplifier The amplifier uses two double-stub matching networks to achieve conjugate matching between the 50Ω source/load and the transistor Luong Vinh Quoc Danh 51 26
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