TUTORIAL #7 Using the Smith Chart

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1 TUTORIAL #7 Using the Smith Chart. [.9 P expanded] Use the Smith chart to find the following quantities for the transmission-line circuit in the figure below. L Z0 Z j L Z in (a) The SWR on the line. (b) The reflection coefficient at the load. (c) The load admittance. (d) The input impedance of the line. (e) The distance from the load to the first voltage minimum. (f) The distance from the load to the first voltage maximum. (a) ZL j zl. j ; plot SWR circle on Smith chart read out SWR from Z0 the resistance value on the right, SWR =.6. (b) Read out Γ either from the nomograph or as the ratio ρ/r. / 79. Read out using the degree scale on the outer rim of the chart. If a degree scale is not available, you can use a protractor or the scale of distance-in-wavelengths (toward load, counter-clockwise), which is always available on the outer rim of the Smith chart. ( ) rad (c) Find diametrically opposite point of z L on Smith chart and read out values. yl yl 9 j YL 9.8 j8. ms (d) Follow SWR circle in direction toward generator (clockwise) moving a distance of λ. As the load z L corresponds to relative-to-short position of 76λ, the relative-to-short position of z in would be at Since the chart goes only to λ, you will have to go beyond λ and continue to 6λ. This corresponds to input impedance of

2 z 9 j Z z. j.. in in in (e) The location of a voltage minimum (which is also the location of a current maximum) corresponds to a minimum resistance value. This location is found on the Smith chart as the intersection point of the SWR circle and the abscissa on the left where the scale for the resistance circles resides. Conveniently, this location coincides with the Chart s zeroreference location. Move in the direction toward load (counter-clockwise) until you reach the location of the load; read out this location from the toward load scale: lmin. (f) The location of a voltage maximum (which is also the location of a current minimum) corresponds to a maximum resistance value. This location is found on the Smith chart as the intersection point of the SWR circle and the abscissa on the right where the scale for the resistance circles resides. It corresponds to a quarter-wavelength-reference location on the Chart. Move from this location toward the load: lmax ( ).

3 7 The Complete Smith Chart Black Magic Design ± - 7 > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) ANGLE OF TRANSMISSION COEFFICIENT IN DEGREES 8-9 ANGLE OF REFLECTION COEFFICIENT IN DEGREES SWR dbs RTN. LOSS [db] RFL. COEFF, P RFL. COEFF, E or I RADIALLY SCALED PARAMETERS TOWARD LOAD >... 7 < TOWARD GENERATOR CENTER ATTEN. [db] S.W. LOSS COEFF RFL. LOSS [db] S.W. PEAK (CONST. P) TRANSM. COEFF, P TRANSM. COEFF, E or I ORIGIN

6 . [Impedance match with a shunt stub] A transmission line of characteristic impedance Z0 Ω is terminated with a load impedance of ZL j as shown in the figure. (a) What is the admittance Y 0 of the transmission line? (b) Find the length L (in terms of λ) of the piece of this transmission line such that at L L the input admittance Y in satisfies Y Y Z. What is ImY in at this position? Re in 0 0 (c) Use a short-circuited or an open-circuited piece of a -Ω transmission line (whichever is shorter) to cancel ImY in by connecting it in parallel at the position L as shown in the figure. This piece of transmission line is called a shunt stub. What is the length of the stub L stub? (d) What is now the input impedance (e) What is the input impedance at L L? Z in at L L? L? Zin? stub Z 0 Z0 ZL j? Lstub? SOLUTION: (a) Y0 / Z0 / S (b) ZL j ZL zl. j (same as in Problem ) Z 0 Find y L from impedance Smith chart. Same as in Problem : yl 9 j ( Y 9.86 j8.97 ms) L We now start using the Smith chart as an admittance chart. Go toward the generator on the Smith chart until intersecting the unity admittance circle. L ( 6) 6 L y j in

7 ImY Y 6 S in 0 (c) We now need to connect to Y in a purely reactive load of susceptance ImY ImY Y 6 S (inductive susceptance) stub in 0 A short-circuited stub would achieve the above susceptance with a shorter line than the opencircuited stub. Lstub (8 ) (d) At L L, Yin Yin Ystub ReYin S Z / Y Ω in in (e) At L L, the input impedance is still Ω because the line is matched at the position L L.

8 7 The Complete Smith Chart Black Magic Design ± - 7 > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) ANGLE OF TRANSMISSION COEFFICIENT IN DEGREES 8-9 ANGLE OF REFLECTION COEFFICIENT IN DEGREES SWR dbs RTN. LOSS [db] RFL. COEFF, P RFL. COEFF, E or I RADIALLY SCALED PARAMETERS TOWARD LOAD >... 7 < TOWARD GENERATOR CENTER ATTEN. [db] S.W. LOSS COEFF RFL. LOSS [db] S.W. PEAK (CONST. P) TRANSM. COEFF, P TRANSM. COEFF, E or I ORIGIN

9 . [Impedance match with a series stub] A transmission line of characteristic impedance Z0 Ω is terminated with a load impedance of ZL j as shown in the figure.? Lstub? stub Z0 L? Zin? Z0 ZL j (a) Find the length L (in terms of λ) of the piece of this transmission line such that at L L the real part of the input impedance Z in satisfies Re Z in Z0. What is Im Z in at this position? (b) Use a short-circuited or an open-circuited piece of a -Ω transmission line (whichever is shorter) to cancel Im Z in by connecting it in series at the position L as shown in the figure. This piece of transmission line is called a series stub. What is the length of the stub L stub? (c) What is now the input impedance SOLUTION: Z in at L L? ZL (a) zl. j (same as in Problems and ) Z0 Trace the SWR circle toward generator (clockwise) until it crosses the unity-resistance circle. The normalized impedance at this position is z( L) j. The length L is L ( 76) 6 Im Z in 6. (b) The stub must have purely reactive impedance of value zstub j Im z( L) ( Zstub j6. ) - inductive reactance A short-circuit stub would be shorter than an open-circuit one. The length of the short-circuit stub must be Lstub. (c) Z Z0 in

10 7 The Complete Smith Chart Black Magic Design ± - 7 > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) ANGLE OF TRANSMISSION COEFFICIENT IN DEGREES 8-9 ANGLE OF REFLECTION COEFFICIENT IN DEGREES SWR dbs RTN. LOSS [db] RFL. COEFF, P RFL. COEFF, E or I RADIALLY SCALED PARAMETERS TOWARD LOAD >... 7 < TOWARD GENERATOR CENTER ATTEN. [db] S.W. LOSS COEFF RFL. LOSS [db] S.W. PEAK (CONST. P) TRANSM. COEFF, P TRANSM. COEFF, E or I ORIGIN

Impedance Calculations

Impedance Calculations Revisiting a T-ine With Any Termination In the general case, where a transmission line is terminated in Z, the impedance along the line is given by: Z Z j z j z e e e Z Z Z( z) Z Z j z j z e e Z Z e Z