Intrductin t Smith Charts Dr. Russell P. Jedlicka Klipsch Schl f Electrical and Cmputer Engineering New Mexic State University as Cruces, NM 88003 September 2002
EE521 ecture 3 08/22/02 Smith Chart Summary A Smith Chart is a cnfrmal mapping between the nrmalized cmplex impedance plane ( z = r + j x ) and the cmplex reflectin cefficient plane. First, the nrmalized impedance is given as z = = R + jx Cnsider the right-hand prtin f the nrmalized cmplex impedance plane. All values f impedance such that R $ 0 are represented by pints in the plane. The impedance f all passive devices will be represented by pints in the right-half plane. The cmplex reflectin cefficient may be written as a magnitude and a phase r as real and imaginary parts. Γ Γ = Γ e = Γ + jγ r i Remember that 0 1and the p ' is measured with respect t the psitive real ' r axis. Γ The reflectin cefficient in terms f the lad,, terminating a line, is defined as Rearranging the abve equatin we get Γ = + Page 1 f 22
EE521 ecture 3 08/22/02 = 1 + 1 Γ Γ S we get the cnfrmal mapping by dividing thrugh by (remember ' is cmplex). z = r + jx = 1 + Γ 1 Γ Substituting in the cmplex expressin fr ' and equating real and imaginary parts we find the tw equatins which represent circles in the cmplex reflectin cefficient plane. Γ r r 1 + r 2 ( Γ 0) i 2 + = 1 1 + r 2 The first circle is centered at 2 1 + Γ r i x = 1 x ( Γ ) 1 r 1 + r 1 1 + r, 0 2 2 whse lcatin is always inside the unit circle in the cmplex reflectin cefficient plane. The crrespnding radius is S it is bserved that this circle will always be fully cntained within the unit circle. The radius can never by greater than unity. The secnd circle is centered at 1 1, x whse lcatin is always utside the unit circle in the cmplex reflectin cefficient plane. Nte that the center f these circles will always be t the right f the unit circle. The crrespnding radius is 1 x. The radius can vary between 0 and infinity. Page 2 f 22
EE521 ecture 3 08/22/02 The first circles, thse centered n the real axis, represent lines f cnstant real part f the lad impedance (r is cnstant, x varies). The circles whse centers reside utside the unit circle represent lines f cnstant imaginary part f the lad impedance (x is cnstant, r varies). Circles centered at the match pint ( =, r ' = 0 ) are equidistance frm the rigin ( ' = cnstant) and are called cnstant VSWR circles. This related quantity is defined as VSWR V max = = V min 1 + 1 Γ Γ Nte that the VSWR can vary between 1 VSWR. The phase f the reflectin cefficient is given by the angle frm the right-hand hrizntal axis. We have -180 # p' # +180. Abve the hrizntal axis is psitive, belw negative. Page 3 f 22
EE521 ecture 3 08/22/02 When we place these ther circles n the REFECTION COEFFICIENT PANE we call it a SMITH CHART. It can als be referred t as an: impedance chart admittance chart immittance chart (impedance and admittance) Page 4 f 22
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EE521 ecture 3 08/22/02 Nrmalized Impedance The transmissin line f characteristic impedance,, is terminated in a lad impedance,. If the characteristic impedance is = 50 + j 0 S and the lad impedance is = 100 + j 100 S, the nrmalized lad impedance is j z = 100 + 100 50 = 2 + j2 ( dimensinless) Nw enter this n the Smith Chart as shwn in belw. Page 6 f 22
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EE521 ecture 3 08/22/02 VSWR Circles T cnstruct the VSWR circle fr a lssless line, place the cmpass center at the center f the Smith Chart ( =, r ' = 0 ) and cnstruct a circle thrugh z. T read ff the VSWR n the line, nte the intercept f the VSWR circle with the right-half zer reactance line (it is the hrizntal line that bisects the tp and bttm f the chart. Fr the previus example, this is shwn in Figure 2 where we read ff the VSWR as Γ = VSWR = 43. et s check this against the calculated VSWR. First, cmpute the reflectin cefficient. + = ( j ) 100 + 100 50 ( 100 + j100) + 50 Taking the magnitude we find Γ = 062.. Then the VSWR is VSWR = 1 + 1 Γ Γ = 4. 265 This tells us that the graphically btained value (VSWR = 4.3) is accurate t abut ne significant figure. Page 8 f 22
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EE521 ecture 3 08/22/02 ad Admittance The lad admittance is the reciprcal f the lad impedance. Fr the previus example, we find Y = 1 Y = 1 100 + j100 = 0005. j0005. Siemens Nrmalized ad Admittance The nrmalized lad admittance is the reciprcal f the nrmalized lad impedance which can be expressed as y Y = = Y 1 100 + j100 1 500 = 025. j025. ( dimensinless) We can find this mre easily by nting that y is n the cnstant VSWR circle at a pint diametrically ppsed frm z. This is shwn in Figure 3. Page 10 f 22
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EE521 ecture 3 08/22/02 Input Impedance The input impedance t the line f characteristic impedance,, a length R frm the lad,, is fund via the equatin. in = + + j j tan tan 2π l λ 2π l λ Where 8 is the wavelength n the transmissin line. Cnsider the fllwing example with = 100 + j 100 S. in = π ( 100 + j100) + j100tan 100 4 = π 100 + j( 100 + j100) tan 4 + j 100 1 2 j1 = 200 j100 Ω Nw cnsider this prcess n the Smith Chart. 1. Enter the nrmalized lad impedance, z = 1 + j 1 2. Cnstruct the cnstant VSWR circle. 3. Travel clckwise (transfrm twards the generatr) n the circle by a distance equivalent t the line length, starting at z. Remember the distance n the Smith Chart is in terms f wavelength. Specifically, wavelength n the transmissin line, which is nt necessarily the free space wavelength. One full revlutin is 8/2. The uter scale is calibrated in wavelengths. 4. Read ff z in. In this case, z in = 2 - j 1. 5. Cmpute in = z in * = ( 2 - j 1 ) * 100 = 200 - j 100 S. Page 12 f 22
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EE521 ecture 3 08/22/02 Input Admittance Use the Smith Chart t find the input admittance. Cnsider the example abve. 1. Enter the nrmalized lad admittance n the VSWR circle. This is accmplished by entering the nrmalized lad impedance, drawing the VSWR circle and marking the pint diametrically ppsed. 2. Travel clckwise (twards the generatr) frm y n the VSWR circle by a distance equivalent t the line length. 3. Read ff y in = 0.44 + j 0.2 4. Find Y in = y in * Y = ( 0.44 + j 0.2 ) / (100) = 0.004 + j 0.002 Seimens. et s cmpare the result frm the Smith Chart n the next page t the analytical result. Frm the example abve we fund in = 200 - j 100 S. Y in 1 1 = = 200 j100 in = 0. 004 j0. 002 Siemens Page 14 f 22
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EE521 ecture 3 08/22/02 Drill Prblem # 1 Cnsider a lad impedance, = 100 - j 25 S, cnnected t a transmissin line f characteristic impedance = 50 S. The line is R = 3/8 8 lng. Find the fllwing quantities using the Smith Chart n the next page. a) z = b) VSWR = c) y = d) z in = e) in = f) y in = g) Y in = \ h) Check the answer in part e) using the analytical result. in = + + j j 2π tan l λ 2π tan l λ Page 16 f 22
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EE521 ecture 3 08/22/02 Drill Prblem # 2 Cnsider a lad impedance, = 0 - j 0 S, cnnected t a transmissin line f characteristic impedance = 50 S. The line is R = 1/4 8 lng. Find the fllwing quantities using the Smith Chart n the next page. a) z = b) VSWR = c) y = d) z in = e) in = f) y in = g) Y in = h) Check the answer in part e) using the analytical result. in = + + j j 2π tan l λ 2π tan l λ Page 18 f 22
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EE521 ecture 3 08/22/02 Drill Prblem #3 Use the Smith Chart n the next page t find the input impedance t the parallel cnnected lines 1 = 2 = 70.7 S. Furthermre, 1 = 2 = 50 + j 0 S. The line lengths are R 1 = R 2 = 8/4. The characteristic impedance f the main line is = 50 S t which the tw parallel lines are cnnected, what is the VSWR n this main line? Drill Prblem #4 Use the Smith Chart t find the input impedance t the parallel cnnected lines 1 = 50 S and 2 = 70.7 S. Furthermre, 1 = 50 + j 0 S and 2 = 100 + j 0 S. The line lengths are R 1 = 8/2 and R 2 = 8/4. The characteristic impedance f the main line is = 50 S t which the tw parallel lines are cnnected, what is the VSWR n this main line? Page 20 f 22
INSERT SMITH CHART GRAPHIC HERE
INSERT SMITH CHART GRAPHIC HERE