36 Smith Chart and VSWR Generator Transmission line I d ( ) Wire 2 + Load Z + g V d Zo • Consider the general phasor expressions - ( ) ZL F = Vg - Wire 1 l d V +ejβd(1 − Γ e−j2βd) 0 + jβd −j2βd L |V (d)|max V (d)=V e (1 + ΓLe ) and I(d)= |V (d)| Zo describing the voltage and current variations on TL’s in sinusoidal |V (d)|min dmax steady-state. dmin – Complex addition displayed Unless ΓL =0,thesephasorscontainreflectedcomponents,which graphically superposed on a means that voltage and current variations on the line “contain” Smith Chart standing waves. SmithChart 1 .5 2 In that case the phasors go through cycles of magnitude variations as a function of d,andinthevoltagemagnitudeinparticular(seemargin) .2 z(dmaxx!5) ( d) Γ(d) varying as 1+Γ =VSWR .2 .5 1 2 r!5 + −j2βd + 0 VSWR |V (d)| = |V ||1+ΓLe | = |V ||1+Γ(d)| 1 Γ(dmax)=|ΓL| takes maximum and minimum values of ".2 x!"5 + + |V (d)|max = |V |(1 + |ΓL|) and |V (d)|min = |V |(1 −|ΓL|) ".5 "2 "1 at locations d = dmax and dmin such that |1+Γ(d)| maximizes for d = dmax −j2βdmax −j2βdmin Γ(dmax)=ΓLe = |ΓL| and Γ(dmin)=ΓLe = −|ΓL|, |1+Γ(d)| minimizes for d = dmin and such that Γ(dmin)=−Γ(dmax) λ d − d is an odd multiple of . max min 4 1 Generator Transmission line I d – These results can be most easily understood and verified graphi- ( ) Wire 2 + Load cally on a SC as shown in the margin. + Zg Z - V (d) o ZL F = Vg - Wire 1 l d • We define a parameter known as voltage standing wave ratio,or 0 |V (d)|max VSWR for short, by |V (d)| |V (d)|min dmax |V (dmax)| 1+|ΓL| VSWR − 1 VSWR ≡ = ⇔|ΓL| = . dmin |V (dmin)| 1 −|ΓL| VSWR +1 Complex addition displayed Notice that the VSWR and |ΓL| form a bilinear transform pair just graphically superposed on a like Smith Chart 1+Γ z − 1 SmithChart z = ⇔ Γ= . 1 1 − Γ z +1 .5 2 Since 1+Γ(dmax) Γ(dmax)=|ΓL|⇒VSWR = , .2 z(dmaxx!5) ( d) Γ(d) 1 − Γ(dmax) 1+Γ =VSWR ! this analogy between the transform pairs also implies that 0 .2 .5 1 2 VSWR r 5 1 Γ(dmax)=|ΓL| z(dmax)=VSWR, ".2 x!"5 as explicitly marked on the the SC shown in the margin . Consequently, ".5 "2 "1 – the VSWR of any TL can be directly read offfrom its SC plot |1+Γ(d)| maximizes for d = dmax as the normalized impedance value z(dmax) on constant-|ΓL| circle |1+Γ(d)| minimizes for d = dmin crossing the positive real axis of the complex plane. such that Γ(dmin)=−Γ(dmax) 2 • The extreme values the VSWR can take are: 1. VSWR=1 if |ΓL| =0and the TL carries no reflected wave. 2. VSWR=∞ if |ΓL| =1corresponding to having a short, open, or apurelyreactiveloadthatcausesatotalreflection. SmithChart 1 .5 2 d d d ! |1+Γ( )| maximizes for = max .2 z(dmaxx 5) ( d) Γ(d) 1+Γ =VSWR ! 0 .2 .5 1 2 VSWR r 5 1 Γ(dmax)=|ΓL| " x!"5 .2 |1+Γ(d)| minimizes for d = dmin such that Γ(dmin)=−Γ(dmax) ".5 "2 "1 3 Generator Transmission line I d • In the lab it is easy and useful to determine the VSWR and dmax or ( ) Wire 2 + Load dmin of a TL circuit with an unknown load, since + Zg Z - V (d) o ZL F = Vg - Wire 1 1. given the VSWR, l d 0 |V (d)|max VSWR − 1 V d |Γ | = | ( )| L VSWR +1 |V (d)|min d is easily determined, and max dmin 2. given dmax or dmin the complex ΓL or its transform zL can be easily Complex addition displayed obtained. graphically superposed on a Smith Chart SmithChart 1 Say dmax is known: then, .5 2 • since (as we have seen above) .2 z d x!5 d) Γ(d) ( max) Γ( =VSWR −j2βdmax 1+ Γ(dmax)=ΓLe = |ΓL| ! 0 .2 .5 1 2 VSWR r 5 1 Γ(dmax)=|ΓL| it follows that ".2 x!"5 j2βdmax 1+ΓL ΓL = |ΓL|e ⇒ zL = . 1 − ΓL ".5 "2 "1 • alternatively, zL can be obtained directly on the SC by rotating counter- |1+Γ(d)| maximizes for d = dmax clockwise by dmax from the location of |1+Γ(d)| minimizes for d = dmin z(dmax)=VSWR. such that Γ(dmin)=−Γ(dmax) These techniques are illustrated in the next example. 4 SmithChart 1 .5 2 Example 1: An unknown load ZL on a Zo =50ΩTL has .2 x!5 V (dmin)=20V,,dmin =0.125λ and VSWR=4. 0.25 4. ! 0 .2 .5 1 2 r4. 5 Determine (a) the load impedance ZL,and(b)theaveragepowerPL absorbed !0.6, 0.25" by the load. ".2 x!"5 Solution: (a) As shown in the top SC in the margin, VSWR=4 is entered in the SC as ".5 "2 z(dmax)=4+j0,andconstant|ΓL| circle is then drawn (red circle) passing "1 SmithChart through z(dmax)=4. 1 .5 2 Right across z(dmax)=4on the circle is z(dmin)=0.25. 0.470588 $ 0.882353 # Acounter-clockwiserotationfromz(dmin)=0.25 by one fourth of a full circle corre- .2 x!5 sponding to a displacement of dmin =0.125λ (a full circle corresponds to a λ/2 ! displacement) takes us to 0 .2 .5 1 2 r4. 5 zL ≈ 0.4706 − j0.8823 ".2 x!"5 as shown in the second SC. Hence, this gives 0.470588 " 0.882353 # !0.6, "0.125" ZL = ZozL =50(0.4706 − j0.8823) = 23.53 − j44.12 Ω. ".5 "2 "1 (b) We will calculate PL by using V (dmin) and I(dmin).Since 1 z(d )=0.25 it follows that Z(d )= 50 Ω= 12.5Ω. min min 4 5 Therefore the voltage and current phasors at the voltage minimum location are 20 V V (d )=20VandI(d )= . min min 12.5Ω Generator Transmission line Average power transported toward the load at d − dmin is, therefore, I d ( ) Wire 2 + Load 1 1 20 400 Zg ∗ + V (d) Zo Z P (dmin)= Re{V (dmin)I(dmin) } = Re{20 } = W =16W. - L 2 2 12.5 25 F = Vg - Wire 1 l d 0 Since the TL is assumed to be lossless we should have |V (d)|max |V (d)| PL = P (dmin)=16W. |V (d)|min dmax dmin Example 2: If the TL circuit in Example 1 has l =0.625λ,andageneratorwithan internal impedance Zg =50Ω,determinethegeneratorvoltageVg. Solution: Given that l =0.625λ and dmin =0.125λ,wenotethatthereisjustone half-wave transformer between l =0.625λ and dmin =0.125λ.Therefore Vin = −V (dmin)=−20 VandZin = Z(dmin)=12.5Ω. But also Zin Vin = Vg . Zg + Zin Consequently, Zg + Zin 50 + 12.5 62.5 Vg = Vin = −20 = −20 = −100 V. Zin 12.5 12.5 6 SmithChart 1 .5 2 Example 3: Determine V + and V − in the circuit of Examples 1 and 2 above such that .2 x!5 the voltage phasor on the line is given by 0.25 4. + jβd − −jβd ! V (d)=V e + V e . 0 .2 .5 1 2 r4. 5 !0.6, 0.25" Solution: Looking back to Example 1 (also see the SC’s in the margin), we first note ".2 x!"5 that VSWR − 1 4 − 1 |ΓL| = = =0.6=Γ(dmax)=−Γ(dmin). VSWR +1 4+1 ".5 "2 Hence, evaluating V (d) at d = dmin,wehave "1 SmithChart 1 + jβdmin V (dmin)=V e (1 + Γ(dmin)) .5 2 + j 2π λ j π + = V (e λ 8 )(1 + (−0.6)) = 0.4e 4 V =20V, 0.470588 $ 0.882353 # from which .2 x!5 + π V =50e−j 4 V. ! Since 0 .2 .5 1 2 r4. 5 π j2βdmin j 2 ΓL =Γ(0)=Γ(dmin)e = −0.6e , ".2 x!"5 it follows that 0.470588 " 0.882353 # π π π − + j −j j !0.6, "0.125" V =ΓLV = −0.6e 2 × 50e 4 = −30e 4 V. ".5 "2 "1 7 SmithChart 1 .5 2 Example 4: Determine the load voltage and current VL = V (0) and IL = I(0) in the .2 x!5 circuit of Examples 1-3 above. 0.25 4. ! Solution: In general, 0 .2 .5 1 2 r4. 5 !0.6, 0.25" + jβd − −jβd + jβd − −jβd V e − V e V (d)=V e − V e and I(d)= . ".2 x!"5 Zo Therefore, ".5 "2 + − + − V − V "1 VL = V (0) = V + V and IL = I(0) = . SmithChart Zo 1 .5 2 Using Zo =50Ωand 0.470588 $ 0.882353 # + π π V =50e−j 4 VandV − = −30ej 4 V .2 x!5 from Example 3, we find that ! 0 .2 .5 1 2 r4. 5 −j π j π π π 50e 4 +30e 4 π π V =50e−j 4 − 30ej 4 VandI = = e−j 4 +0.6ej 4 A. L L 50 ".2 x!"5 0.470588 " 0.882353 # !0.6, "0.125" ".5 "2 "1 8.