Smith Chart The quarter-wave transformer

Transcription

1 Smith Chart The quarter-wave transformer We will cover these topics The Smith Chart The Quarter-Wave Transformer Watcharapan Suwansan8suk #3 EIE/ENE 450 Applied Communica8ons and Transmission Lines King Mongkut s University of Technology Thonburi 2 AOer this lecture, you will be able to Find the reflec8on coefficient, standing wave ra8o, and the input impedance using Smith chart Design a quarter-wave transformer 2.4 THE SMITH CHART 3 4

2 Smith chart is a graphical aid for solving T-line problems Developed by P. Smith (939, Thai calendar 2482) Usage: Past (WW II era) for microwave-system design Present: a part of computer-aid design (CAD) sooware Smith chart combines 2 graphs together: normalized impedance and reflec8on coefficient (in polar coordinate) We will consider the simplest case: Smith chart for lossless transmission lines Smith Chart source: hvp://sss-mag.com/smith.html 5 6 Many T-line equa8ons are of the form, or equivalently, = z z+ z = + for complex numbers and such that and z z 6= 6= means that we don t divide by zero in the equa8on form Ex : An equa8on for the reflec8on coefficient is of this form SMITH CHART DERIVATION Ex 2: An equa8on for the SWR is of this form SWR = + normalized load impedance: 7 8

3 Z in Ex 3: The T-line impedance equa8on is of this form z in = ` + e 2j ` e 2j ` Smith chart helps us solve graphically this form of solu8on Z L normalized input impedance: z in = Z in /Z 0 Variables z and are related by circle-to-circle intersec8on y 0 Recall that in the xy-plan, the equa8on is the circle centered at (x 2) 2 +(y + ) 2 = (2, ) (x a) 2 +(y b) 2 = r 2 (a, b) and of radius x r 9 0 Given z, this is the corresponding z i i axis Intersec8on of two circles, shown for z r > 0,z i > 0 A Smith chart superimposes these circles An arc of the circle for the The loca8on of z =0.2+j.0 imaginary part z +j.0 i =.0 on the Smith chart +j0.5 +j2.0 0 z r +z r r axis An arc of the circle for the real part z r =0.2 +j j We can rewrite = z as 2 circles in the r, i plane: r real part of z r +z r z i 2 = ( r ) 2 + +z r and imaginary part of real part of z i 2 = z i imaginary part of z z i 2 j0.2 j5.0 j0.5 j2.0 j.0 In Matalb: command smithchart draws a Smith chart 2

4 Example : find characteris8c impedance Z 0 = 00 at the load, given Z L Z L load impedance = 40 + j70 USING THE SMITH CHART Solu8on (Arithme8c): From the defini8on, = Z L Z 0 Z L + Z 0 = (40 + j70) 00 (40 + j70) + 00 = j0.57 =0.59 e j j0.4 +j0.7 +j j The actual Smith chart contains marks for angles and a scale for length Read off the degree from the ring angle of reflec8on coefficient in degrees j0.4 Use the REL COEFF or ORIGIN scales to convert the measured length to the magnitude j0.7 j.0 Solu8on 2 (Smith chart): The normalized load impedance is REL COEFF (length from previous slide) 0.59 The intersec8on of 2 circles ( 0.4 and j0.7 ) gives =0.59 e j04 ORIGIN

5 Example (cont d): Find the SWR Solu8on (Arithme8c): +j.0 +j0.7 +j0.4 Subs8tute + SWR = = 0.59 ej04 +j = 3.87 j3.9 j0.4 j0.7 j.0 Solu8on 2 (Smith Chart): Rotate the point = 0.59 ej04 onto the + r axis, and read off the value on the axis The value is the SWR, so SWR = This rota8on yields the SWR because... 8 Example (cont d): Find Zin for a line of length ` = 0.3 +j.0. +j Z0 = 00 ZL = 40 + j70 ` = 0.3 Zin j3.9 Solu8on (Arithme8c): j.0 This point means that So + z= for z = 0.39 and Zin = = = = SWR 3.9 = = 0.59 Defini8on + e e Subs8tute 2 ` = 2 2j ` 2j ` 2 Z0 = e e j26 j26 Subs8tute 00 = 37 j6 = 0.59 ej04 =.2 radian =

6 +j0.6 +j.0 This rota8on yields Z in because... +j0.4 +j.0 +j j j0.4 j0.6 j.0 Solu8on 2 (Smith Chart): Rotate clockwise the point for 2 ` = 26 =0.59 e j04 The end point is the normalized input impedance: z in =0.37 j0.6 So the input impedance is Z in = z in Z 0 = 37 j6 2 j0.4 j0.6 This point means: for and So 0.37 j0.6 = z = + j.0 z =0.37 j0.6 = e j26 j26 + e = + e 2j ` = z e j26 e 2j ` in Defini8on 22 The actual Smith chart contains WTG scale for angle rota8on WTG direc8on Z L This a way to use the WTG scale =0.59 e j The angle 04 maps to the WTG scale of 0.06λ Z in 3. Draw a radical line Wavelengths toward generator (WTG) = moving away from the load As we rotate clockwise on the WTG scale, we get the input impedance at the posi8on further away from the load 4. Read the Smith chart: this point is z in =0.37 j0.6, which is the normalized input impedance Moving down the line 0.3λ brings us to 0.06λ + 0.3λ = 0.406λ on the WTG scale 23 24

7 A sloved line is a T-line configura8on (usually a waveguide or a coaxial cable) THE SLOTTED LINE a sloved line Useful for determining the load impedance a VNA is superseded by the modern equipment, a vector network analyzer (VNA) SWR SWR = + Recall these rela8onships = SWR SWR + = angle of in radius = Z L Z 0 Z L + Z 0 ZL = Z 0 + reflec8on coefficient at the load distance from the load to the voltage minimum = +2 `min n `min = for any integer n that makes the right-side 0 Z in (`) = `min 2j ` + e e 2j ` Z0 Typical steps to determine the load impedance Z L are as follows. Find the magnitude and angle of, so 2. Determined from Z L = e j determined from the SWR determined from `min Z L Z L = Z in (` = 0) Z L + jz 0 tan Z in (`) =Z 0 Z 0 + jz L tan ` ` Z in (`) 27 28

8 Example 2: find λ using a sloved line 2cm Answer: the voltage minima repeats every λ/2, so λ = 4 cm Given a 50 Ω coaxial sloved line A short circuit is placed at the load On the arbitrarily posi8oned scale on the sloved line, the voltage minima occur at z = 0.2 cm, 2.2 cm, 4.2 cm Ques8on: What is the wavelength λ on the line? Reason for λ/2: A short circuit implies that =. Hence, V (z) = V e j z ) = V 0 + e j z =2 V 0 + sin z Since period of sin z = half the period of sin the voltage minima repeats every λ/2 z =, 2 2 = Example 2 : find Z L Answer: j9.7 The short circuit is removed & replaced by an unknown load Given the SWR =.5 The voltage minima are recorded at z = 0.72 cm, 2.72 cm, 4.72 cm Ques8on: What is the load impedance Z L? For a load Z L and the corresponding reflec8on coefficient, the voltage is repeated in magnitude very /2: V z + = V (z) = V (z) 2 AOer some algebra So we can effec8vely consider the load terminals to be at any of voltage minima in the st step (short circuit) 3 32

9 2.72cm 4.2cm From the SWR =.5, we get the magnitude So =.5.5+ =0.2 Assume without loss of generality that the load is at 4.2 cm Then, `min = =.48 cm, giving us And the load impedance is = +2 `min = +2 2 `min = We will consider the following setup Z 0 Z R L /4 2.5 THE QUARTER-WAVE TRANSFORMER A feed line of characteris8c impedance Z 0 is connected to a line sec8on of length λ/4 The impedances Z 0 and R L are known real numbers Ques8on: what is the characteris8c impedance Z that makes? =0 means the line sec8on is matched to the feeding line (so there is no standing wave on the feeding line) Answer: We will show that Z = p Z 0 R L using 2 methods: () Impedance viewpoint and (2) mul8-reflec8on viewpoint 35 36

10 Here are some remarks about the answer Z 0 Z = p Z 0 R L /4 R L means is a geometric mean of and Z = p Z 0 R L Z Z 0 R L There is no standing wave on the feed line although there is a standing wave on the matching sec8on The value Z = p Z 0 R L is valid for sec8on length /4 and for odd n =, 3, 5,... n /4 The matching sec8on is designed for a wavelength At other wavelengths (or other frequencies), the impedance mismatched will occur IMPEDANCE VIEWPOINT Here is a reason for. Z = p Z 0 R L Example 3: an effect of the frequency Z 0 Z R L Z 0 = 50 Z R L = 00 Z in The input impedance is /4 ` = 2 4 = 2 Take the limit `! 2 (tan = + or ) To get =0, we must have Z in = Z 0, which yields the stated value of Z A quarter-wavelength transformer is designed for a specific frequency Let so the length of the matching sec8on is for = the corresponding wavelength f = the frequency on the line Ques8on: Plot 0 f 0 0/4 as a func8on of f/f 0 0/4 = phase velocity v p f

11 Example 3 : Answer Example 3 : Deriva8on The necessarily characteris8c impedance is The reflec8on coefficient has a magnitude of = Z in Z 0 Z in + Z 0 Solu8on: We will show that the plot is given above. No8ce that =0 for the frequencies f = f 0, 3f 0, 5f 0,... R L + jz tan ` Z in = Z 50 Ω Z + jr L tan ` 70.7 Ω 00 Ω Subs8tu8on will give us the plot of 4 42 Infinite set of waves travel forward and backward on the matching sec8on Z 0 Z R L MULTIPLE-REFLECTION VIEWPOINT par8al reflec8on coefficient par8al transmission coefficient T 3 T 3 T

12 The coefficients equal these expressions The reflec8on coeff. is an infinite sum /4 Z 0 Z R L = T = Z Z 0 Z + Z 0 (Think of the incident wave seeing only an impedance Z at the junc8on) T T =+ 3 = R L Z R L + Z T 3 T 2 The round-trip path up and down the λ/4 sec8on results in a 80 phase shio (sign change) +T T 2 T 2 T T 2 2 T 2 =+ 2 2 = Z 0 Z Z 0 + Z We get the same conclusion: if then. Z = p Z 0 R L =0 The reflec8on coefficient is Z 0 Z R L Geometric series The Smith Chart Smith chart deriva8on Using Smith chart SloVed line Summary The Quarter-Wave Transformer Impedance viewpoint Mul8ple-reflec8on viewpoint Subs8tute the expressions = 2(Z 2 Z 0 R L ) (Z + Z 0 )(R L + Z )( ) Equals zero if we choose Z = p Z 0 R L 47 48