The Smith Chart: The Smith chart is a graphical aide used to simplify the solution of Tx-line problems More importantly, the Smith chart allows us to visualize the periodic nature of the line impedance or the line voltage along the line length (useful for specialized design). The Smith Chart is nothing more than a mapping between the line impedance and the generalized reflection coef. Z Zo z jφ Z Γ= = =Γ e,where z = L Z + Z z + Z L L L L o L Smith Chart Review o. Similarly: z L = + Γ e Γ e Note, that Re z 0. Therefore, all values of z map into the unit circle of L L the complex Γ plane Given z = r + jx, the mapping consists of the locust of points L ( ) r ( ) Re Γ + Im ( Γ ) =, ( ( ) ) ( ) ( ) ( ) Re Γ + Im ( Γ) = + r + r x x The Smith chart consists of circles centered on the real axis representing the locust of points of constant r values, and dissected circles centered off the real axis representing constant x values (positive above, and negative below). Γ for ( ) jφ jφ
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 r,0 + r + r Center = Radius = As the normalized resistance r varies from 0 to, we obtain a family of circles completely contained inside the domain of the reflection coefficient Γ. r = 0 Im(Γ ) r = Re(Γ ) r = 5 r = r
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, x x Center = Radius = As the normalized reactance x varies from - to, we obtain a family of arcs contained inside the domain of the reflection coefficient Γ. x = Im(Γ ) x = x ± x = 0 x = - Re(Γ ) x = -
Basic Smith Chart techniques for loss-less transmission lines Given Z(d) Find Γ(d) Given Γ(d) Find Z(d) Given Γ R and Z R Find Γ(d) and Z(d) Given Γ(d) and Z(d) Find Γ R and Z R Find d max and d min (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)
. Normalize the impedance Given Z(d) Find Γ(d) ( d) Z R X zn ( d) = = + j = r+ j x Z Z Z 0 0 0. Find the circle of constant normalized resistance r. Find the arc of constant normalized reactance x 4. 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 ( ) 0 Z d = 5 + j00 Ω with Z = Ω
. Normalization z n (d) = (5 + j 00)/ = + j.0. Find normalized reactance arc x =.0. Find normalized resistance circle r = 05.906 0 5 4. This vector represents the reflection coefficient - Γ (d) = + j4 Γ (d) = 46 Γ (d) = 885 rad -0 5 =.906 - - -. 46
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 and of the normalized reactance x that correspond to the reflection coefficient point.. The normalized impedance is n ( d) z = r+ jx and the actual impedance is ( d) ( ) Z(d) = Z z = Z r+ j x = Z r+ j Z x 0 n 0 0 0
d Given Γ R and Z R Find Γ(d) and Z(d) Z s Ix ( ) + Vx ( ) Z o, Z R x =-d x = 0 NOTE: the magnitude of the reflection coefficient is constant along a loss-less transmission line terminated by a specified load, since ( d) exp( j d) Γ = Γ β = Γ R Therefore, on the complex plane, a circle with center at the origin and radius Γ R represents all possible reflection coefficients found along the transmission line. When the circle of constant magnitude of the reflection coefficient is drawn on the Smith chart, one can determine the values of the line impedance at any location. The graphical step-by-step procedure is:. Identify the load reflection coefficient Γ R and the normalized load impedance Z R on the Smith chart. R
. Draw the circle of constant reflection coefficient amplitude Γ(d) = Γ R.. Starting from the point representing the load, travel on the circle in the clockwise direction, by an angle π θ= β d= d λ 4. 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. Example: Given Z = 5 + j 00 Ω with Z = Ω R find Zd ( ) and Γ ( d) for d= 0.8λ 0
Circle with constant Γ Γ R Γ R z R 0 5 θ = β d = (π/λ) 0.8 λ =.6 rad = 9.6 - Γ(d) = 46-78.7 = 0.6 j 09 - Γ (d) θ - z n (d) = 6 j.9 - Z(d) = z(d) Z 0 =.79 j59.6 Ω - z(d)
Given Γ R and Z R Find d max and d min. Identify on the Smith chart the load reflection coefficient Γ R or the normalized load impedance Z R.. Draw the circle of constant reflection coefficient amplitude Γ(d) = Γ R. The circle intersects the real axis of the reflection coefficient at two points which identify d max (when Γ(d) = Real positive) and d min (when Γ(d) = Real negative). A commercial Smith chart provides an outer graduation where the distances normalized to the wavelength can be read directly. The angles, between the vector Γ R and the real axis, also provide a way to compute d max and d min. Example: Find d max and d min for Z = 5 + j00 Ω ; Z = 5 j00 Ω ( Z = Ω ) R R 0
Im(Z R ) > 0 ZR = 5 + j 00 Ω ( Z0 = Ω) β d max =.9 d max = 0.0707λ Γ R z nr Γ R 0 5 - - - - - β d min = d min = 7λ
Im(Z R ) < 0 ZR = 5 j 00 Ω ( Z0 = Ω) β d max = 09. d max = 9 λ 0 5 - Γ R Γ R - β d min = 9. d min = 9 λ - - - z nr
Given Γ R and Z R Find the Voltage Standing Wave Ratio (VSWR) The Voltage standing Wave Ratio or VSWR is defined as Vmax +Γ VSWR = = R V min Γ R The normalized impedance at a maximum location of the standing wave pattern is given by ( dmax ) ( d ) + Γ +Γ z R n ( dmax ) = = = VSWR!!! Γ Γ max This quantity is always real and. The VSWR is simply obtained on the Smith chart, by reading the value of the (real) normalized impedance, at the location d max where Γ is real and positive. R
The graphical step-by-step procedure is:. Identify the load reflection coefficient Γ R and the normalized load impedance Z R on the Smith chart.. Draw the circle of constant reflection coefficient amplitude Γ(d) = Γ R.. Find the intersection of this circle with the real positive axis for the reflection coefficient (corresponding to the transmission line location d max ). 4. A circle of constant normalized resistance will also intersect this point. Read or interpolate the value of the normalized resistance to determine the VSWR. Example: Find the VSWR for Z = 5 + j00 Ω ; Z = 5 j00 Ω ( Z = Ω ) R R 0
Circle with constant Γ z R Γ R Circle of constant conductance r = 0 5 - Γ R z(d max )= - - - - z R For both loads VSWR =
Remember that Given Z(d) Find Y(d) λ zn d + = yn d 4 ( ) is only valid for normalized impedance and admittance. The actual values are given by λ λ Z d + = Z0 zn d + 4 4 y ( ) ( ) 0 ( ) n d Yd = Y yn d = Z where Y 0 = /Z 0 is the characteristic admittance of the transmission line. The graphical step-by-step procedure is: 0
. Identify the load reflection coefficient Γ R and the normalized load impedance Z R on the Smith chart.. Draw the circle of constant reflection coefficient amplitude Γ(d) = Γ R.. The normalized admittance is located at a point on the circle of constant Γ which is diametrically opposite to the normalized impedance. Example: Given Z = 5 + j 00 Ω with Z = Ω R find Y R. 0
Circle with constant Γ z n (d) = + j.0 Z(d) = 5 + j00 [ Ω ] 0 5 - θ = 80 = β λ/4 - - y n (d) = 65 j 706 Y(d) = 0.005 j 0.0094 [ S ] z n (d+λ/4) = 65 j 706 - Z(d+λ/4) = 5.884 j.594 [ Ω ] -
The Smith chart can be used for line admittances, by shifting the space reference to the admittance location. After that, one can move on the chart just reading the numerical values as representing admittances. Let s review the impedance-admittance terminology: Impedance = Resistance + j Reactance Z = R + jx Admittance = Conductance + j Susceptance Y = G + jb
Smith Chart for Admittances y n (d) = 65 j 706 - - - Γ - Negative (inductive) susceptance - 5 0 Positive (capacitive) susceptance y n (d) = +.0
EE5 Set Problem.: For the transmission line network below, you are given Z = 40 + j0 Ω, L Z = Ω, Z = 7Ω, Z = j00 Ω, = 0.5 λ, and = 0.5 λ. Using the o Smith chart, determine: Z a) in b) SWR on line c) SWR on line o sh Zin Zin Zo, β Z sh Zo, β Z L x = x = x = 0 Solution: a) Z = 9. Ω in b) SWR = c) SWR =.4
7 The Complete Smith Chart Black Magic Design 0. 0.0 0.0 0. 9 8 9 8 70 ± 80-70 7 > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < 7 0.04 6 60-60 90-90 6 0.04 85-85 0. 0.05 5-80 -80 5 0.05 40 75-75 0.06-40 0.06 4 70-70 0.07 0.07-0 0 0.08 65-65 0.08-60 -60 0.09 0.09 0-0 0. 0. INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) 4 55-55 0. 9 0. 00-00 - 0. 8 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) 9 8 0..0.0.0 90-90 45-45 7...4.6.8.0.0 4.0 5.0 0. -80 0.4 6 80 40.0.0.0-40 0.4 6.0.4.4 0.5 5 5 0.5 70-5 -70 5.6.6 0.6 4 4 0.6.8.8 60-0 -60 0.0.0 5-5 0.8-0.8.0.0 40 - -40 4.0 4.0 5 5.0 5.0-5 0-0 9 0 0 0-0 ANGLE OF TRANSMISSION COEFFICIENT IN DEGREES 8-9 ANGLE OF REFLECTION COEFFICIENT IN DEGREES 7 8 4 6 5 5 6 4 7 SWR dbs RTN. LOSS [db] RFL. COEFF, P RFL. COEFF, E or I 0040 40 0 0 5 5 4 0.5 8 6.8 5.6 4 RADIALLY SCALED PARAMETERS.4.. 5 0 4 5 6 7 8 9 0 4 0 0 0.0 0. 0. 0.05 0.0 0. TOWARD LOAD > 0 7 5 4 < TOWARD GENERATOR....4.6.8 4 5 0 0..5 4 5 6 0 5 0 0....4.5.6.7.8.9.5 4 5 0 0 9 5 0. 0 CENTER....4.5.6.7.8.9 ATTEN. [db] S.W. LOSS COEFF RFL. LOSS [db] S.W. PEAK (CONST. P) TRANSM. COEFF, P TRANSM. COEFF, E or I ORIGIN
7 NORMALIZED IMPEDANCE AND ADMITTANCE COORDINATES 0 0. 5.0 4.0.0 0 0 0. 9 8 9 8 70 ± 80-70 0.04 6 7 > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < 7 60-60 90-90 6 0.04 85-85 0. 0 0.05 5 0-80 -80 5 0.05 5.0 4.0 40 75-75 0.06-40 0.06 4.0 70-70 0.07 0.07-0 0 5.0 0.08 65-65 0.08.0 4.0.0 -.8.8 60.0-60 0.09 0.09.6.6 0-0 0. 0. INDUCTIVE REACTANCE COMPONENT (+jx/zo),orcapacitivesusceptance(+jb/yo) 4 55.4.0-55.4.8 0. 9.0.0.6.0.0 0. 00. -00.4. - 0. 8 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) EACTANCE COMPONENT (-jx/zo), ORINDUCTIVE SUSCEPTANCE(-jB Yo) VE R CAPA C IT I 9. 8 0..0.0.0 90-90 45.0.0.0-45 7...4.6.8.0.0 4.0 5.0 0. -80 0.4 6 80 40.0.0.0-40 0.4 6.0.4.4 0.5 5 5 0.5 70-5 -70 5.6.6 0.6 4 4 0.6.8.8 60-0 -60 0.0.0 5-5 0.8 0. - 0.8.0.0 40 - -40 4.0 4.0 5 5.0 5.0-5 0-0 9 0. 0 0-0 ANGLE OF TRANSMISS ON COEFFICIENT IN DEGREES - 9 8 ANGLE OF REFLECTION COEFF CIENT IN DEGREES 7 8 4 6 5 5 6 4 7 SWR dbs RTN. LOSS [db] RFL. COEFF, P RFL. COEFF, E or I 0040 40 0 0 5 5 4 0.5 8 6.8 5.6 4 RADIALLY SCALED PARAMETERS.4.. 5 0 4 5 6 7 8 9 0 4 0 0 0 0. 0. 0.05 0.0 0. TOWARD LOAD > 0 7 5 4 < TOWARD GENERATOR....4.6.8 4 5 0 0..5 4 5 6 0 5 0 0....4.5.6.7.8.9.5 4 5 0 0 9 5 0. 0 CENTER....4.5.6.7.8.9 ATTEN. [db] S.W. LOSS COEFF RFL. LOSS [db] S.W. PEAK (CONST. P) TRANSM. COEFF, P TRANSM. COEFF, E or I ORIGIN