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Smith Chart Problems Smith Chart Problems 1. The 0:1 length line shown has acharacteristic imp edance of 50 and is terminated with a load imp edance of Z =5+j 25 . L Z L =0:1+j 0:5on the Smith chart. (a) Lo cate z = L Z 0 See the point plotted on the Smith chart. (b) What is the imp edance at ` =0:1? Since we want to move away from the load (i.e., toward the generator), read read 0:074 on the wavelengths toward generator scale and add ` =0:1 to obtain 0:174 on the wavelengths toward generator scale. A radial line from the center of the chart intersects the constant re ec- tion co ecient magnitude circle at z = 0:38 + j 1:88: Hence Z = zZ =50(0:38 + j 1:88) = 19 + 94 . 0 (c) What is the VSWR on the line? Find VSWR = z =13on the horizontal line to the right max of the chart's center. Or use the SWR scale on the chart. (d) What is ? L From the reflection coefficient scale below the chart, nd j j = 0:855. From the angle of reflection coef- L ficient scale on the p erimeter of the chart, nd the angle of j 126:5 = 126:5 . Hence =0:855e . L L (e) What is at ` =0:1 from the load? Note that jj = j j =0:855. Read the angle of the re ection L co ecient from the angle of reflection coefficient j 55:0 scale as 55:0 . Hence =0:855e . Problem 1 l = 0.1λ 01 Z50=Ω =+ Ω 0 ZL Z525[]L j l = 0.1λ 01 0.174λ .TG . 0.074λ .TG . AngΓ= 55.00 Γ= 0 AngL 126.5 =+ z0.10.5L j z=+ 0.38j 1.88 Γ=0.855 55.00 Γ= 0 L 0.855126.5 == VSWRzmax 13 VSWR= 13 Γ=Γ= L 0.855 2. A transmission line has Z =1:0, Z = z =0:2 j 0:2 : 0 L L (a) What is z at ` = =0:25? 4 From the chart, read 0:467 from the wavelengths to- ward generator scale. Add 0:25 to obtain 0:717 on the wavelengths toward generator scale. This is not on the chart, but since it rep eats every half wavelength, it is the same as 0:717 0:500 = 0:217. Drawing a ra- dial line from the center of the chart, we nd an intersec- tion with the constant re ection co ecient magnitude circle at z = Z =2:5+j 2:5. (b) What is the VSWR on the line? From the intersection of the constant re ection co ecientcir- cle with the right hand side of the horizontal axis, read VSWR = z =5:3: max (c) How far from the load is the rst voltage minimum? The rst current minimum? The voltage minimum o ccurs at z which is at a distance of min 0:500 0:467 =0:033 from the load. Or read this distance directly on the wavelengths toward load scale. The current minimum o ccurs at z which is a quarter of a max wavelength farther down the line or at 0:033 +0:25 =0:283 from the load. Problem 2 l = 0.25λ Z=Ω 1.0 == − ΩΩ 0 ZL ZLL z 0.2j 0.2 [ / ] l = 0.25λ 0.217λ .TG . zZ2.52.5== +j zmin λ == 0.033 VSWRzmax 5.3 =− z0.20.2L j 0.467λ .TG . VSWR= 5.3 3. The air- lled two-wire line has a characteristic imp edance of 50 and is op erated at f =3 GHz. The load is Z = 100 + j 40 . L (a) For the line ab ove, nd z on the chart. L 100 + j 40 Z L = The normalized load is z = = 2:0+ j 0:8. L Z 50 0 See the Smith chart for lo cation of p oint. (b) What is the line imp edance 2:5cm from the load? 10 3 10 cm/sec c = = 10 cm: Since we are Note that = 9 f 3 10 Hz going to move toward the generator (away from the load), at the normalized load p osition, rst read 0:217 on the wave- lengths toward generator scale. Then add 2.5 cm/10 cm = 0:25 to this value to obtain 0:467 on the wave- lengths toward generator scale. A radial line from the center at this p oint intersects the constant re ection co- ecient magnitude circle at z =0:43 j 0:17, so Z = zZ = 0 50(0:43 + j 0:17) = 21:5 j 8:5 . (c) What is the VSWR on the line? From the intersection of the re ection co ecient circle and the horizontal axis on the right hand side of the chart, read VSWR = 2.4. Or use the SWR scale belowthe chart. Problem 3 l = 2.5 [cm] Z=Z z → Z=Ω= 50 ,f 3GHz =+ Ω 0 0 ZL ZL 100j 40 [ ] 0.217λ .TG . =+ z2.00.8L j == VSWRzmax 2.4 z=− 0.43j 0.17 0.467λ .TG . l = 0.25λ 4. The line shown is 3=8 long and its normalized input imp edance is z = j 2:5: (a) What is the normalized receiving or load end imp edance, z ? L Since wemust movetoward the load to nd the load imp edance, lo cate z = j 2:5 on the chart and nd the intersection on the wavelengths toward load scale at 0:189. Add the dis- tance 3=8 = 0:375 to get 0:064 on the wavelengths toward load scale. At the intersection of the line from the center of the chart and the constant re ection co ecient circle, read z = j 0:425. L (b) What is the distance from the load to the rst voltage minimum? The voltage minimum o ccurs at the imp edance minimum on the horizontal line to the left of the chart's center. This dis- tance can be read directly o the wavelengths toward load scale (why?) as 0:064. Note that the line imp edance at the voltage minimum would b e that of a short circuit. (c) What length of op en-circuited line could b e used to replace z ? L From the p osition of z ,wewanttomovetoward the new load L (i.e., toward the end of the op en-circuited line) until wereach the op en circuit condition z = 1. This o ccurs at 0:25 on the wavelengths toward load scale and the total distance moved is hence 0:25 0:064 =0:186, which is the length of op en-circuited line needed to replace z . L Problem 4 λ l = 3 8 z2.5=−j → zL """ """! open circuited line section replacing original load zL λ l = 3 8 Vmin z =∞ 0.064 λ z2.5=−j =− zL j 0.425 0.064λ .T.L. 0.189λ .T.L. 0.186 λ 5. An air- lled 50 coaxial line has a loaded VSWR of 3.3 at a frequency of 3GHz. Replacing the load with a short causes the voltage minimum to move 1.0 cm towards the generator. What is the normalized load imp edance? For an air- lled line, the wavelength is 10 c 3 10 cm/sec = = = 10 cm: 9 f 3 10 Hz With the VSWR lo cated at z =3:3, draw the circle of con- max stant re ection co ecient magnitude. With the load in place, our only information is the lo cation of a voltage minimum (equivalent to an imp edance minimum). But thinking of the Smith chart as a crank diagram, we know that the voltage minimum must be at the intersection of the constant re ection co ecient circle and the horizontal axis on the left hand side of the chart between z =0 and z =1:0, (i.e., at z =0:3). With this information, min we have thus established a corresp ondence between a lo cation on the line and a point on the chart. Now, move from this 1:0cm minimum voltage point on the chart =0:1 toward the 10:0cm generator to the lo cation where the voltage minimum was mea- sured in the shorted line measurement. Since it was obtained from a shorted line measurement, the lo cation of this p ointmust be an integral number of half-wavelengths from the load (and hence must have the same line imp edance as at the load). The normalized imp edance here is therefore z =0:44 + j 0:63. For a L line with a 50 characteristic imp edance, the load imp edance is thus Z = 50(0:44 + j 0:63) = 22 + j 31:5: L Problem 5 0.10 λ =+ zL 0.44j 0.63 Vmin VSWR= 3.3 6. A slotted line measurement on an air- lled TEM line yields a VSWR of 1.6 at a frequency of 1 GHz. When the load is replaced by a short, the voltage minimum moves 3 cm towards the load. Find the normalized load imp edance. Lo cate the re ection co ecient circle by its intersection with z = VSWR = 1:6. The wavelength for an air- lled TEM max line is 10 3 10 cm/s = c=f = =30cm: 9 10 Hz Hence the shift in the voltage minimum is 3cm =0:1: 30 cm= Lo cate the voltage minimum at the imp edance minimum, z = min 0:625 and move 0:1 toward the load to a p osition an integral numb er of half-wavelengths from the load.
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