Exercises for the Antenna Matching Course

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1 Exercises for the Antenna Matching Course Lee Vishloff, PEng, IEEE WCP C RELEASE 1

2 Notifications 2016 Services, Inc. All rights reserved. The and Services Inc. stylized text belongs to tech-knows Services Inc. All other marks belong to their respective owners. This document previous versions and is believed to be correct at time of publication. Services Inc. and its affiliates assume no responsibility for errors or omissions. No warranties of any nature are created, modified, or extended by the contents contained herein. Document Part Number: C Document Version Number: Release 1 Document Release Date: March 29,

3 1 Introduction & Basics 1. Which of the following source and loads are conjugate matched? a. b. c. d. e. 3

4 f. 2. What is the reflection coefficient for the following combination of sources and loads? a. b. c. d. e. f. 4

5 3. What is the impedance seen at the cable input for the following conditions: Length = 150 mm Velocity of propagation is m/sec Frequency of operation is 1 GHz Characteristic Impedance of cable is 50 Ω. 4. An antenna is specified as having a VSWR of 3. What is it s return loss? 5

6 2 Smith Chart 1. What are the complex impedance values for points A through E on the following chart assuming a 50 Ohm Z0? 2. What is the reflection coefficient for points A through E in terms of magnitude and angle? 6

7 3. Can this stub move the impedance to the Z0 origin? 4. With a load impedance of 25 + j25 what shunt lumped element (i.e. inductor or capacitor) will bring the impedance seen by the source closest to the Z0 origin at 850 MHz? (Use whole numbers of nh or pf as appropriate). 3 Matching Networks 1. Use the following network to match 200 Ohms to a 100 Ohms source at 100 MHz 2. Use the same network to match 100 Ohms to a 50 Ohms source at 100 MHz 3. Use the following network to match 100 Ohms to 50 Ohms using the following network at 800 MHz. 4. Which network structure would you use if you were concerned about harmonic output from your product? The one shown in question 1 or question 3? 5. What inductance would you use to move an impedance of 92 + j140 to 50 Ohm line? Start by considering whether you need a shunt or series element. 7

8 4 Dual Band Antenna 1. In this exercise will use an antenna with the following reflection coefficient data. (Also available from a text file labelled antenna_a.sp1) which is in magnitude angle format. # MHZ S MA R For this exercise we will match a band from 850 to 950 for the low band and 2400 to 2500 for the high band. The raw antenna looks like this on the Smith Chart. 8

9 Remembering that series inductors and shunt capacitors move the high band more than the low band we will start by using a shunt capacitor to move the high band to the 50 Ohm circle rather than a shunt inductor. Perform the final match with a resonant circuit so that the lower band does not move, or only moves slightly, as it will be close to the center of the chart already. 9

10 5 Answer Key Introduction & Basics 1. a. Yes b. No c. Yes d. Yes e. No f. No a. 1/3 b. 0 c. 1/3 d. 1/2 e Arg 45 f db 10

Smith Chart 1. Complex impedances a. 25 + j25 b. 25 + j0 c. 100 + j0 d. 20 j100 e. 15 j25 2. Gamma - Γ Arg Γ a. 0.

11 Smith Chart 1. Complex impedances a j25 b j0 c j0 d. 20 j100 e. 15 j25 2. Gamma - Γ Arg Γ a Arg 116 b Arg 180 c Arg 0 d Arg -52 e Arg -123 See Smith Chart below for solutions. 11

12 3. No The stub does not move the impedance to the Z0 origin. A 0 length open stub leaves the Load impedance unperturbed at 25Ω and the minimum impedance is when the stub presents a short circuit resulting in a 0Ω impedance being presented to the source. The path of a stub matching element is a circle touching the load impedance and the 0. A stub can bring an impedance to the Z0 circle if it falls outside a circle that touches the real axis at 0 and Z0. (i.e. outside the shaded area to the left). 4. A shunt capacitor of value 4 pf. (25 + j25) -1 = 20 j20 ms. A positive susceptance is capacitive. 2 π 850 MHz * 4 pf = 21.4 ms recall the unit S = Ω -1 Matching Networks 1. Matching 200 Ω to 100 Ω a. First off we set Z0 as 100Ω instead of the usual 50Ω. This results in a relative impedance of 2.0 on the Smith Chart b. Using a shunt inductor the impedance will head northward in a counter-clockwise direction hitting the Z0 circle at an impedance of 1+j1 and admittance of 0.5-j.5 c. The inductor needs to have an susceptance of j0.5 - j0 (end point mines start point) at 1 GHz. Y0 (1/Z0) 1/100 = 10 ms so the inductor has a susceptance of 5 ms. L = 318 nh at 1 GHz. Using standard values we would use 300 or 330 nh. d. The impedance is now at 1 +j1 on the Smith Chart, or 100 +j100 Ω. A reactance of - j100 is now required, which requires capacitor value of 15.9 pf. Using a standard value we would use 15 pf. 12

13 2. Use the same network to match 100 Ohms to a 50 Ohms source at 100 MHz. a. The approach will be exactly the same but the shunt susceptance required will be -j10 ms instead of -j5 ms making the inductor half as large 159 nh (use 150 nh) b. Similarly the reactance for the series capacitor will now be j50 instead of j100 making the capacitor twice as large at 31.8 pf (use 33 pf). 3. Use the same approach as above, but the shunt capacitor will intersect the Z0 circle 1 j1 (0.5 +j0.5 on Admittance chart) instead of 1+ j1. (Remember we are now operating at 800 MHz). C = 2 pf L = 10 nh 4. The network in question 3 as it is a low-pass filter. 5. Moving 92 + j140 to 50 Ohm line requires following a constant conductance so the inductor is a shunt inductor. The Admittance for this point is (92 + j140) -1 = j0.005 S or j0.25 when normalized to Y0. The 0.16 conductance circle intercepts the Z0 circle at approximately 0.35 (or 7 ms). We need to move about 2 ms (7 ms 5 ms) which requires a GHz. 13

14 Dual-Band Antenna 1. The raw antenna looks like this on the Smith Chart. We may wish to move the high band northward (with a shunt inductor) or move the band southward with a shunt capacitor. We will use the capacitor since this has less impact on the low band, which is already near the center of the Smith Chart. I picked a value of capacitor that centers 2.4 and 2.45 around the 50 Ohm circle. Using ideal components and ignoring preferred values this is about 8.5 pf. ( j139 ms at GHz) We observe the following: a) A series inductor is needed to move the high band to the origin. b) A small bit of capacitance would move the low band down a touch. c) A series resonant circuit is inductive above resonance. 14

15 Thus a network like this should work for this match. Interpolating the data, a point on the 50 Ohm circle in the high band would be approximately 50 j110 at 2.42 GHz => we want a net series reactance of +j110 to reach the origin in the high band and a resonant circuit between 900 and 950 MHz (making it slightly capacitive at 850 and 900 MHz). Dual-Band Matching Network and its Performance. 15

16 Using normal value components we have several many choices. Here is an example. Dual-Band Example with Standard Value Components 16

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