2.4 The Smith Chart. Reading Assignment: pp The Smith Chart. The Smith Chart provides: The most important fact about the Smith Chart is:

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1 2/7/2005 2_4 The Smith Chart 1/2 2.4 The Smith Chart Readg Assignment: pp The Smith Chart The Smith Chart provides: 1) 2) The most important fact about the Smith Chart is: HO: The Complex Γ plane Q: But how is the complex Γ plane useful? A: HO: Transformations on the Complex Γ Plane Q: But transformations of Γ are relatively easy transformations of le impedance Z is the difficult one.

2 2/7/2005 2_4 The Smith Chart 2/2 A: HO: Mappg Z to Γ HO: The Smith Chart HO: Z Calculations usg the Smith Chart Example: The Input Impedance of a Shorted Transmission e Example: Determg the oad Impedance of a Transmission e Example: Determg the ength of a Transmission e Expressg a load or le impedance terms of its admittance is sometimes helpful. Additionally, we can easily map admittance onto the Smith Chart HO: Impedance and Admittance Example: Admittance Calculations with the Smith Chart

3 2/7/2005 The Complex Gamma Plane 1/6 The Complex Γ Plane Resistance R is a real value, thus we can dicate specific resistor values as pots on the real le: R =0 R =20 Ω R =50 Ω R R =5 Ω ikewise, sce impedance Z is a complex value, we can dicate specific impedance values as pot on a two dimensional complex plane: Im { Z } Z =30 +j 40 Ω Re { Z } Z =60 -j 30 Ω Note each dimension is defed by a sgle real le: the horizontal le (axis) dicatg the real component of Z (i.e., Re{ Z }), and the vertical le (axis) dicatg the imagary component of impedance Z (i.e., Im { Z }). The tersection of these two les is the pot denotg the impedance Z = 0. * Note then that a vertical le is formed by the locus of all pots (impedances) whose resistive (i.e., real) component is equal to, say, 75. * ikewise, a horizontal le is formed by the locus of all pots (impedances) whose reactive (i.e., imagary) component is equal to -30.

4 2/7/2005 The Complex Gamma Plane 2/6 Im { Z } R =75 Re { Z } X =-30 If we assume that the real component of every impedance is positive, then we fd that only the right side of the plane will be useful for plottg impedance Z pots on the left side dicate impedances with negative resistances! Invalid Region (R<0) Im { Z } Valid Region (R>0) Re { Z } Moreover, we fd that common impedances such as Z = (an open circuit!) cannot be plotted, as their pots appear an fite distance from the orig. Im { Z } Z =0 (short) Z =Z 0 (matched) Re { Z } Z = (open) Somewhere way the heck over there!!

5 2/7/2005 The Complex Gamma Plane 3/6 Q: Yikes! The complex Z plane does not appear to be a very helpful. Is there some graphical tool that is more useful? A: Yes! Recall that impedance Z and reflection coefficient Γ are equivalent complex values if you know one, you know the other. We can therefore defe a complex Γ plane the same manner that we defed a complex impedance plane. We will fd that there are many advantages to plottg on the complex Γ plane, as opposed to the complex Z plane! Γ =-0.5 +j 0.1 Im { Γ } Γ =0.3 +j 0.4 Re { Γ } Γ =0.6 -j 0.3 Note that the horizontal axis dicates the real component of Γ (Re{ Γ }), while the vertical axis dicates the imagary component of Γ (Im { Γ }). We could plot pots and les on this plane exactly as before: Im { Γ } Re {Γ}=0.5 Re { Γ } Im {Γ} =-0.3

6 2/7/2005 The Complex Gamma Plane 4/6 However, we will fd that the utility of the complex Γ pane as a graphical tool becomes apparent only when we represent a complex reflection coefficient terms of its magnitude ( Γ ) and phase (θ Γ ): j e θ Γ Γ= Γ In other words, we express Γ usg polar coordates: Im { Γ } Γ= 06. e j 3π 4 Γ Γ θ Γ Re { Γ } Γ= 07. e j 300 Note then that a circle is formed by the locus of all pots whose magnitude Γ equal to, say, 0.7. ikewise, a radial le is formed by the locus of all pots whose phase θ Γ is equal to 135. θ Γ = 135 Im { Γ } Γ = 07. Re { Γ }

7 2/7/2005 The Complex Gamma Plane 5/6 Perhaps the most important aspect of the complex Γ plane is its validity region. Recall for the complex Z plane that this validity region was the right-half plane, where Re{ } 0 Z > (i.e., positive resistance). The problem was that this validity region was unbounded and fite extent, such that many important impedances (e.g., open-circuits) could not be plotted. Q: What is the validity region for the complex Γ plane? A: Recall that we found that for Re{ Z } > 0 (i.e., positive resistance), the magnitude of the reflection coefficient was limited: 0< Γ< 1 Therefore, the validity region for the complex Γ plane consists of all pots side the circle 1 Γ = --a fite and bounded area! Im { Γ } Invalid Region ( Γ> 1) Valid Region ( Γ < 1) Re { Γ } Γ = 1

8 2/7/2005 The Complex Gamma Plane 6/6 Note that we can plot all valid impedances (i.e., R >0) with this fite region! Im { Γ } j Γ= e π = 10. (short) Γ = 0 (matched) Re { Γ } j 0 Γ= e = 10. (open) Γ= 1 ( Z = jx purely reactive)

9 2/7/2005 Transformations on the Complex 1/7 Transformations on the Complex Γ Plane The usefulness of the complex Γ plane is apparent when we consider aga the termated, lossless transmission le: z = z = 0 Z, β Γ 0 Z, β 0 Γ Recall that the reflection coefficient function for any location z along the transmission le can be expressed as (sce z = 0 ): ( z ) Γ =Γ =Γ e j 2βz e j ( θ 2βz ) Γ + And thus, as we would expect: -j 2β Γ ( z = 0) = Γ and Γ ( z = ) = Γ e = Γ Recall this result says that addg a transmission le of length to a load results a phase shift θ Γ by 2β radians, while the magnitude Γ remas unchanged.

10 2/7/2005 Transformations on the Complex 2/7 Q: Magnitude Γ and phase θ Γ --aren t those the values used when plottg on the complex Γ plane? θ A: Precisely! In fact, plottg the transformation of Γ to Γ along a transmission le length has an terestg graphical terpretation. et s parametrically plot Γ ( z ) from z = z (i.e., z = 0 ) to z = z (i.e., z = ): Im { Γ } Γ ( z ) ( z ) Γ = = Γ 0 Γ Re { Γ } Γ= 1 ( z ) Γ = =Γ θ = θ 2β Sce addg a length of transmission le to a load Γ modifies the phase θ Γ but not the magnitude, we trace a circular arc as we parametrically plot Γ ( z )! This arc has a radius Γ and an arc angle 2β radians. Γ

11 2/7/2005 Transformations on the Complex 3/7 With this knowledge, we can easily solve many terestg transmission le problems graphically usg the complex Γ plane! For example, say we wish to determe Γ for a transmission le length = λ 8 and termated with a short circuit. z = z = 0 Γ Z β Z β Γ = 1 0, 0, = λ 8 j The reflection coefficient of a short circuit is Γ = 1 = 1e π, and therefore we beg at that pot on the complex Γ plane. We then move along a circular arc 2β = 2( π 4) = π 2 radians (i.e., rotate clockwise 90 ). Im { Γ } Γ ( z ) Γ = 1 e + j π 2 Re { Γ } Γ = 1 e + jπ Γ=1

12 2/7/2005 Transformations on the Complex 4/7 When we stop, we fd we are at the pot for Γ ; this case 2 Γ = 1e j π o (i.e., magnitude is one, phase is 90 ). Now, let s repeat this same problem, only with a new transmission le length of = λ 4. Now we rotate clockwise 2 β = π radians (180 ). Γ ( z ) Im { Γ } Γ = 1 e + j π 2 Re { Γ } Γ = 1 e + jπ Γ=1 For this case, the put reflection coefficient is the reflection coefficient of an open circuit! j 0 Γ = 1e = 1 : Our short-circuit load has been transformed to an open circuit with a quarter-wavelength transmission le! But, you knew this would happen right?

13 2/7/2005 Transformations on the Complex 5/7 z = z = 0 Z β Γ = 1 Z β Γ = 1 0, (open) 0, (short) = λ 4 Recall that a quarter-wave transmission le was one of the special cases we considered earlier. Recall we found that the put impedance was proportional to the verse of the load impedance. Thus, a quarter-wave transmission le transforms a short to an open. Conversely, a quarter-wave transmission can also transform an open to a short: Im { Γ } Γ=1 Γ = 1 e + j π 2 Re { Γ } Γ = 1 e + jπ Γ ( z )

14 2/7/2005 Transformations on the Complex 6/7 Fally, let s aga consider the problem where Γ = 1 (i.e., short), only this time with a transmission le length = λ 2 ( a half wavelength!). We rotate clockwise 2β = 2 π radians (360 ). Hey look! We came clear around to where we started! Γ ( z ) Im { Γ } Γ = 1 e + jπ Re { Γ } Γ = 1 e + jπ Γ=1 Thus, we fd that Γ = Γ if = λ 2--but you knew this too! Recall that the half-wavelength transmission le is likewise a special case, where we found that Z = Z. This result, of course, likewise means that Γ =Γ.

15 2/7/2005 Transformations on the Complex 7/7 Now, let s consider the opposite problem. Say we know that the put impedance at the begng of a transmission le with length = λ 8 is: Γ = 0.5e j 60 Q: What is the reflection coefficient of the load? A: In this case, we beg at Γ and rotate COUNTER- COCKWISE along a circular arc (radius 0.5) 2β = π 2 radians (i.e., 60 ). Essentially, we are removg the phase shift associated with the transmission le! Im { Γ } θ = θ + 2β Γ ( z ) θ 0.5 Γ = 05. e j 60 Re { Γ } Γ = 05. e j 150 Γ= 1 The reflection coefficient of the load is therefore: Γ = 0.5e j 150

16 2/7/2005 Mappg Z to Gamma 1/8 Mappg Z to Γ Recall that le impedance and reflection coefficient are equivalent either one can be expressed terms of the other: ( ) ( ) ( ) ( ) Z z Z 0 1 +Γ z Γ ( z ) = and Z ( z ) = Z0 Z z + Z0 1 Γ z Note this relationship also depends on the characteristic impedance Z 0 of the transmission le. To make this relationship more direct, we first defe a normalized impedance value z (an impedance coefficient!): Z ( z ) R ( z ) X ( z ) z ( z) = = + j = r ( z ) + j x ( z ) Z Z Z Usg this defition, we fd: ( ) ( ) ( ) ( ) ( z ) ( z ) Z z Z0 Γ ( z ) = Z z + Z Z z = Z z z 1 = z Z0 1 Z + 1

17 2/7/2005 Mappg Z to Gamma 2/8 Thus, we can express Γ ( z ) explicitly terms of normalized impedance z --and vice versa! ( ) ( ) ( ) ( ) z z 1 1+Γ z Γ ( z ) = z ( z ) = z z Γ z The equations above describe a mappg between coefficients z and Γ. This means that each and every normalized impedance value likewise corresponds to one specific pot on the complex Γ plane! For example, say we wish to mark or somehow dicate the values of normalized impedance z that correspond to the various pots on the complex Γ plane. Some values we already know specifically: case Z z Γ Z j Z 0 j j 5 j Z 0 j j

18 2/7/2005 Mappg Z to Gamma 3/8 Therefore, we fd that these five normalized impedances map onto five specific pots on the complex Γ plane: Γ i Γ=1 z = 1 ( Γ= 0) z = j ( Γ= j ) z = ( Γ= 1) Γ r z = 0 ( Γ= 1) z = j ( Γ= j ) Or, the five complex Γ map onto five pots on the normalized impedance plane: x z = 0 z = j ( Γ= j ) z = ( Γ= 1) ( Γ= 1) r z = j ( Γ= j ) z = 1 ( Γ= 0)

19 2/7/2005 Mappg Z to Gamma 4/8 Now, the precedg provided examples of the mappg of pots between the complex (normalized) impedance plane, and the complex Γ plane. We can likewise map whole contours (i.e., sets of pots) between these two complex planes. We shall first look at two familiar cases. Z = R In other words, the case where impedance is purely real, with no reactive component (i.e., X = 0 ). Meang that normalized impedance is: ( ) z = r + j0 i.e.,x = 0 where we recall that r = R Z0. Remember, this real-valued impedance results a real-valued reflection coefficient: r 1 Γ= r + 1 I.E.,: r 1 Re { } i Im { } r 1 Γr Γ = Γ Γ = + 0

20 2/7/2005 Mappg Z to Gamma 5/8 Thus, we can determe a mappg between two contours one contour ( x = 0 ) on the normalized impedance plane, the other ( Γ i = 0) on the complex Γ plane: x = 0 Γ i = 0 Γ i Γ=1 x = 0 ( Γ = 0) i Γ r x r x = 0 ( Γ = 0) i

21 2/7/2005 Mappg Z to Gamma 6/8 Z = jx In other words, the case where impedance is purely imagary, with no resistive component (i.e., 0 R = ). Meang that normalized impedance is: ( ) z = 0+ jx i.e.,r = 0 where we recall that x = X Z0. Remember, this imagary impedance results a reflection coefficient with unity magnitude: Γ = 1 Thus, we can determe a mappg between two contours one contour ( r = 0) on the normalized impedance plane, the other ( Γ= 1) on the complex Γ plane: r = 0 Γ = 1

22 2/7/2005 Mappg Z to Gamma 7/8 Γ i Γ = 1 r = 0 ( Γ = 1) Γ r x r = 0 ( Γ = 1) r

23 2/7/2005 Mappg Z to Gamma 8/8 Q: These two mappgs may very well be fascatg an academic sense, but they seem not particularly relevant, sce actual values of impedance generally have both a real and imagary component. Sure, mappgs of more general impedance contours (e.g., r = 05. or x = 15. ) onto the complex Γ would be useful but it seems clear that those mappgs are impossible to achieve!?! A: Actually, not only are mappgs of more general impedance contours (such as r = 05. and x = 15. ) onto the complex Γ plane possible, these mappgs have already been achieved thanks to Dr. Smith and his famous chart!

24 2/7/2005 The Smith Chart 1/11 The Smith Chart Say we wish to map a le on the normalized complex impedance plane onto the complex Γ plane. For example, we could map the vertical le r =2 (Re{ z } = 2) or the horizontal le x =-1 (Im{ z } = 1). Im { z } r =2 Re { z } x =-1 Recall we know how to map the vertical le r =0; it simply maps to the circle Γ= 1 on the complex Γ plane. ikewise, we know how to map the horizontal le x = 0; it simply maps to the le Γ i = 0 on the complex Γ plane. But for the examples given above, the mappg is not so straight forward. The contours will general be functions of both 2 2 Γr and Γ i (e.g., Γ r +Γ i = 05. ), and thus the mappg cannot be stated with simple functions such as Γ = 1 or Γ = 0. i Jim Stiles The Univ. of Kansas Dept. Of EECS

25 2/7/2005 The Smith Chart 2/11 As a matter of fact, a vertical le on the normalized impedance plane of the form: r = c r, where c r is some constant (e.g. r = 2 or r = 05. ), is mapped onto the complex Γ plane as: i cr c r Γr +Γ = 1+ cr 1+ Note this equation is of the same form as that of a circle: ( ) ( ) c c x x + y y = a where: ( ) c c c a = the radius of the circle P x = x,y = y pot located at the center of the circle Thus, the vertical le r = c r maps to a circle on the complex Γ plane! By spection, it is apparent that the center of this circle is located at this pot on the complex Γ plane: cr P c Γ r =, Γ i = 1 + cr 0 Jim Stiles The Univ. of Kansas Dept. Of EECS

26 2/7/2005 The Smith Chart 3/11 In other words, the center of this circle always lies somewhere along the Γ i = 0 le. ikewise, by spection, we fd the radius of this circle is: 1 a = + c 1 r We perform a few of these mappgs and see where these circles lie on the complex Γ plane: r = 03. Γ i Γ = 1 r = 00. r = 10. Γ r r = 03. r = 30. Jim Stiles The Univ. of Kansas Dept. Of EECS

27 2/7/2005 The Smith Chart 4/11 We see that as the constant c r creases, the radius of the circle decreases, and its center moves to the right. Note: 1. If c r > 0 then the circle lies entirely with the circle 1 Γ=. 2. If c r < 0 then the circle lies entirely outside the circle Γ= If c r = 0 (i.e., a reactive impedance), the circle lies on circle 1 Γ=. 4. If c r =, then the radius of the circle is zero, and its 0 center is at the pot Γ = 1, Γ = 0 (i.e., Γ= 1e j ). In r other words, the entire vertical le r = on the normalized impedance plane is mapped onto just a sgle pot on the complex Γ plane! But of course, this makes sense! If r =, the impedance is fite (an open circuit), regardless of what the value of the reactive component x is. i Now, let s turn our attention to the mappg of horizontal les the normalized impedance plane, i.e., les of the form: x = c i Jim Stiles The Univ. of Kansas Dept. Of EECS

28 2/7/2005 The Smith Chart 5/11 where c i is some constant (e.g. x = 2 or x = 05. ). We can show that this horizontal le the normalized impedance plane is mapped onto the complex Γ plane as: Γr + Γi = ci c ( 1) 2 2 i Note this equation is also that of a circle! Thus, the horizontal le x = c i maps to a circle on the complex Γ plane! By spection, we fd that the center of this circle lies at the pot: 1 P c Γ r = 1, Γ i = ci other words, the center of this circle always lies somewhere along the vertical Γ = 1 le. r ikewise, by spection, the radius of this circle is: a = 1 c i Jim Stiles The Univ. of Kansas Dept. Of EECS

29 2/7/2005 The Smith Chart 6/11 We perform a few of these mappgs and see where these circles lie on the complex Γ plane: Γ i Γ r = 1 x = 05. x = 10. x = 20. x = 30. Γ=1 Γ r x = 30. x = 05. x = 10. x = 20. We see that as the magnitude of constant c i creases, the radius of the circle decreases, and its center moves toward the Γ = 1, Γ = 0. pot ( ) r i Note: 1. If c i > 0 (i.e., reactance is ductive) then the circle lies entirely the upper half of the complex Γ plane (i.e., where Γ i > 0) the upper half-plane is known as the ductive region. Jim Stiles The Univ. of Kansas Dept. Of EECS

30 2/7/2005 The Smith Chart 7/11 2. If c i < 0 (i.e., reactance is capacitive) then the circle lies entirely the lower half of the complex Γ plane (i.e., where Γ i < 0) the lower half-plane is known as the capacitive region. 3. If c i = 0 (i.e., a purely resistive impedance), the circle has an fite radius, such that it lies entirely on the le Γ = 0. i 4. If c i =±, then the radius of the circle is zero, and its 0 center is at the pot Γ = 1, Γ = 0 (i.e., Γ= 1e j ). In other r words, the entire vertical le x = or x = on the normalized impedance plane is mapped onto just a sgle pot on the complex Γ plane! But of course, this makes sense! If x =, the impedance is fite (an open circuit), regardless of what the value of the resistive component r is. i 5. Note also that much of the circle formed by mappg x = c i onto the complex Γ plane lies outside the circle Γ= 1. This makes sense! The portions of the circles layg outside Γ= 1 circle correspond to impedances where the real (resistive) part is negative (i.e., r < 0). Thus, we typically can completely ignore the portions of the circles that lie outside the Γ = 1 circle! Jim Stiles The Univ. of Kansas Dept. Of EECS

31 2/7/2005 The Smith Chart 8/11 Mappg many les of the form r c = r and i x = c onto circles on the complex Γ plane results tool called the Smith Chart. Im{ Γ } Re{ Γ } Jim Stiles The Univ. of Kansas Dept. Of EECS

32 2/7/2005 The Smith Chart 9/11 Note that around the outside of the Smith Chart there is a scale dicatg the phase angle θ Γ, from 180 < < 180. θ Γ However, there is another scale that also directly dicates the equivalent transmission le distance z associated with phase shift θγ = 2β z, terms of λ (i.e., the electrical distance). The two scales are related by the equation: θγ θγ z = = λ 2β 4π Note the Smith Chart is simply the vertical les horizontal les i r = c and x = c of the normalized impedance plane, mapped onto the two types of circles on the complex Γ plane. Note for the normalized impedance plane, a vertical le r = cr and a horizontal le x = ci are always perpendicular to each other when they tersect. We say these les form a rectilear grid. However, a similar thg is true for the Smith Chart! When a mapped circle r = cr tersects a mapped circle x = ci, the two circles are perpendicular at that tersection pot. We say these circles form a curvilear grid. In fact, the Smith Chart is formed by distortg the rectilear grid of the normalized impedance plane to the curvilear grid of the Smith Chart! r Jim Stiles The Univ. of Kansas Dept. Of EECS

33 2/7/2005 The Smith Chart 10/11 I.E.,: x x = 1 r = 0 r x = 1 x = 0 x = 1 Distortg this rectilear grid: x r Jim Stiles The Univ. of Kansas Dept. Of EECS

34 2/7/2005 The Smith Chart 11/11 And then distortg some more we have the curvilear grid of the Smith Chart! x r Jim Stiles The Univ. of Kansas Dept. Of EECS

35 2/8/2005 Z Calculations usg the Smith Chart.doc 1/7 Z Calculations usg the Smith Chart z z 0 = 1 z z = z = 0 The normalized put impedancez of a transmission le length, when termated normalized load z, can be determed as: z = Z Z 0 1 Z + j Z tan β Z Z Z0 + j tan β = 1 + jz Z tanβ 0 = 0 Z0 Z0 + j Z tan β z + j tan β = 1 + jz tan β 0 Q: Evaluatg this unattractive expression looks not the least bit pleasant. Isn t there a less disagreeable method to determe z?

36 2/8/2005 Z Calculations usg the Smith Chart.doc 2/7 A: Yes there is! Instead, we could determe this normalized put impedance by followg these three steps: 1. Convert z to Γ, usg the equation: Z Z0 Γ = Z + Z Z = Z z 1 = z Z0 1 Z Convert Γ to Γ, usg the equation: Γ = Γ j 2β e 3. Convert Γ to z, usg the equation: Z 1 z = + Γ Z = 1 Γ 0 Q: But performg these three calculations would be even more difficult than the sgle step you described earlier. What short of dimwit would ever use (or recommend) this approach?

37 2/8/2005 Z Calculations usg the Smith Chart.doc 3/7 A: The benefit this last approach is that each of the three steps can be executed usg a Smith Chart no complex calculations are required! 1. Convert z to Γ Fd the pot z from the impedance mappgs on your Smith Chart. Place you pencil at that pot you have now located the correct Γ on your complex Γplane! For example, say z = 0.6 j1.4. We fd on the Smith Chart the circle for r =0.6 and the circle for x =-1.4. The tersection of these two circles is the pot on the complex Γ plane correspondg to normalized impedance z = 0.6 j1.4. This pot is a distance of units from the orig, and is located at angle of 65 degrees. Thus the value of Γ is: Γ = 0.685e j Convert Γ to Γ Sce we have correctly located the pot Γ on the complex Γ plane, we merely need to rotate that pot clockwise around a circle ( Γ = 0.685) by an angle 2β. When we stop, we are located at the pot on the complex Γ plane where Γ=Γ!

38 2/8/2005 Z Calculations usg the Smith Chart.doc 4/7 For example, if the length of the transmission le termated z = 0.6 j1.4 is = 0.307λ, we should rotate around the Smith Chart a total of 2β = 1.228π radians, or 221. We are now at the pot on the complex Γ plane: e + Γ= j 74 This is the value of Γ! 3. Convert Γ to z When you get fished rotatg, and your pencil is located at the pot Γ=Γ, simply lift your pencil and determe the values r and x to which the pot corresponds! For example, we can determe directly from the Smith Chart that the pot Γ = j e + 74 is located at the tersection of circles r =0.5 and x =1.2. In other words: z = j1.2

39 2/8/2005 Z Calculations usg the Smith Chart.doc 5/7 Step 1 Γ = Γ = e j 65 θ Γ = 65

40 2/8/2005 Z Calculations usg the Smith Chart.doc 6/7 Step 2 2 = λ Γ = e j 74 Γ = Γ = e j 65 1 = 016. λ = = λ λ = λ 2β = 221

41 2/8/2005 Z Calculations usg the Smith Chart.doc 7/7 Step 3 z = j 12.

42 2/8/2005 Example Shorted Transmission e.doc 1/3 Example: The Input Impedance of a Shorted Transmission e et s determe the put impedance of a transmission le that is termated a short circuit, and whose length is: a) = λ 8 = 0.125λ 2β = 90 b) = 3λ 8 = 0.375λ 2β = 270 z z 0 = 1 z = 0 z = z = 0

43 2/8/2005 Example Shorted Transmission e.doc 2/3 a) = λ 8 = 0.125λ 2β = 90 Rotate clockwise 90 from Γ = 10. = e j 180 and fd z = j. z = j Γ = 1 = e j 180

44 2/8/2005 Example Shorted Transmission e.doc 3/3 b) = 3λ 8 = 0.375λ 2β = 270 Rotate clockwise 270 from Γ = 10. = e j 180 and fd z = j. Γ = 1 = e j 180 z = j

45 2/8/2005 Example The oad Impedance.doc 1/2 Example: Determg the oad Impedance of a Transmission e Say that we know that the put impedance of a transmission le length = 0.134λ is: z = j1.4 et s determe the impedance of the load that is termatg this le. z = 1+ j 1 4. z 0 = 1 = 0134λ. z =?? ocate z on the Smith Chart, and then rotate counterclockwise (yes, I said counter-clockwise) 2β = Essentially, you are removg the phase shift associated with the transmission le. When you stop, lift your pencil and fd z! z = z = 0

46 2/8/2005 Example The oad Impedance.doc 2/2 = λ 2β = 965. z = j 024. z = 1+ j 1 4.

47 2/8/2005 Example Determg the tl length.doc 1/7 Example: Determg Transmission e ength A load termatg at transmission le has a normalized impedance z = j2.0. What should the length of transmission le be order for its put impedance to be: a) purely real (i.e., x = 0)? b) have a real (resistive) part equal to one (i.e., r = 1.0 )? Solution: a) Fd z = j2.0 on your Smith Chart, and then rotate clockwise until you bump to the contour x = 0 (recall this is contour lies on the Γ r axis!). When you reach the x = 0 contour stop! ift your pencil and note that the impedance value of this location is purely real (after all, x = 0!). Now, measure the rotation angle that was required to move clockwise from z = j2.0 to an impedance on the x = 0 contour this angle is equal to 2β! You can now solve for, or alternatively use the electrical length scale surroundg the Smith Chart.

48 2/8/2005 Example Determg the tl length.doc 2/7 One more important pot there are two possible solutions! Solution 1: 2β = 30 = λ z = 2+ j 2 Γ ( z ) x = 0 z = j 0

49 2/8/2005 Example Determg the tl length.doc 3/7 Solution 2: z = 2+ j 2 z = j 0 x = 0 Γ ( z ) 2β = 210 = λ

50 2/8/2005 Example Determg the tl length.doc 4/7 b) Fd z = j2.0 on your Smith Chart, and then rotate clockwise until you bump to the circle r = 1 (recall this circle tersects the center pot or the Smith Chart!). When you reach the r = 1 circle stop! ift your pencil and note that the impedance value of this location has a real value equal to one (after all, r = 1!). Now, measure the rotation angle that was required to move clockwise from z = j2.0 to an impedance on the r = 1 circle this angle is equal to 2β! You can now solve for, or alternatively use the electrical length scale surroundg the Smith Chart. Aga, we fd that there are two solutions!

51 2/8/2005 Example Determg the tl length.doc 5/7 Solution 1: z = 2+ j 2 Γ ( z ) r = 1 z = 10. j 16. 2β = 82 = λ

52 2/8/2005 Example Determg the tl length.doc 6/7 Solution 2: z = j 16. Γ ( z ) z = 2+ j 2 r = 1 2β = 339 = λ

53 2/8/2005 Example Determg the tl length.doc 7/7 Q: Hey! For part b), the solutions resulted z = 1 j 1.6 and z = 1+ j 1.6--the imagary parts are equal but opposite! Is this just a cocidence? A: Hardly! Remember, the two impedance solutions must result the same magnitude for Γ--for this example we fd Γ ( z ) = Thus, for impedances where r =1 (i.e., z = 1 + j x ): and therefore: ( jx ) ( ) z j x Γ= = = z jx j x 2 jx Γ = = 2 + jx 2 2 x 4 + x 2 2 Meang: 2 x 4 Γ = 1 Γ 2 2 of which there are two equal by opposite solutions! x = ± 2 Γ 1 Γ 2 Which for this example gives us our solutions x =± 1.6.

54 2/10/2005 Admittance.doc 1/4 Admittance As an alternative to impedance Z, we can defe a complex parameter called admittance Y: I Y = V where V and I are complex voltage and current, respectively. Clearly, admittance and impedance are not dependent parameters, and are fact simply geometric verses of each other: 1 1 Y = Z = Z Y Thus, all the impedance parameters that we have studied can be likewise expressed terms of admittance, e.g.: Y ( z ) = 1 Z ( z ) Y = 1 Z Y = 1 Z Moreover, we can defe the characteristic admittance Y 0 of a transmission le as: + I ( z ) Y0 = + V z ( ) And thus it is similarly evident that characteristic impedance and characteristic admittance are geometric verses:

55 2/10/2005 Admittance.doc 2/4 1 1 Y = Z = Y 0 0 Z0 0 As a result, we can defe a normalized admittance value y : Y y = Y 0 An therefore (not surprisgly) we fd: y Y Z 1 0 = = = Y0 Z z Note that we can express normalized impedance and admittance more compactly as: y = YZ 0 and z = Z Y0 Now sce admittance is a complex value, it has both a real and imagary component: Y = G + jb where: { } Re Y G = Conductance { } Im Z B = Susceptance

56 2/10/2005 Admittance.doc 3/4 Now, sce Z = R + jx, we can state that: G jb 1 + = R + jx Q: Yes yes, I see, and from this we can conclude: 1 G = and R B = 1 X and so forth. Please speed this up and quit wastg my valuable time makg such obvious statements! A: NOOOO! We fd that G 1 R and B 1 (generally). Do not make this mistake! X In fact, we fd that 1 R jx G + jb = R + jx R jx R jx = 2 2 R + X R X = j R + X R + X

57 2/10/2005 Admittance.doc 4/4 Thus, equatg the real and imagary parts we fd: R G = R + X 2 2 X and B = R 2 X 2 + Note then that IF X = 0 (i.e., Z = R ), we get, as expected: 1 G = and B = 0 R And that IF R = 0 (i.e., Z = R ), we get, as expected: G = 0 and B = 1 X I wish I had a nickel for every time my software has crashed oh wait, I do!

58 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 1/9 Example: Admittance Calculations with the Smith Chart Say we wish to determe the normalized admittance y 1 of the network below: = z2 z y 0 = j 17. = 037λ. z = j 26. z = z = 0 First, we need to determe the normalized put admittance of the transmission le: y z 0 = 1 = 037λ. z = j 26. z = z = 0

59 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 2/9 There are two ways to determe this value! Method 1 First, we express the load z = j 2 6. terms of its admittance y = 1 z. We can calculate this complex value or we can use a Smith Chart! z = j 26. y = 017. j 028.

60 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 3/9 The Smith Chart above shows both the impedance mappg (red) and admittance mappg (blue). Thus, we can locate the impedance z = j 2 6. on the impedance (red) mappg, and then determe the value of that same Γ pot usg the admittance (blue) mappg. From the chart above, we fd this admittance value is approximately y = 017. j Now, you may have noticed that the Smith Chart above, with both impedance and admittance mappgs, is very busy and complicated. Unless the two mappgs are prted different colors, this Smith Chart can be very confusg to use! But remember, the two mappgs are precisely identical they re just rotated 180 with respect to each other. Thus, we can alternatively determe y by aga first locatg z = j 2 6. on the impedance mappg : z = j 26.

61 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 4/9 Then, we can rotate the entire Smith Chart while keepg the pot Γ location on the complex Γ plane fixed. y = 017. j 028. Thus, use the admittance mappg at that pot to determe the admittance value of Γ. Note that rotatg the entire Smith Chart, while keepg the pot Γ fixed on the complex Γ plane, is a difficult maneuver to successfully as well as accurately execute. But, realize that rotatg the entire Smith Chart 180 with respect to pot Γ is equivalent to rotatg 180 the pot Γ with respect to the entire Smith Chart! This maneuver (rotatg the pot Γ ) is much simpler, and the typical method for determg admittance.

62 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 5/9 z = j 26. y = 017. j 028. Now, we can determe the value of y by simply rotatg clockwise 2β from y, where = 037. λ :

63 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 6/9 2β y = 017. j 028. y = 07. j 17. Transformg the load admittance to the begng of the transmission le, we have determed that y = 0 7. j Method 2 Alternatively, we could have first transformed impedance z to the end of the le (fdg z ), and then determed the value of y from the admittance mappg (i.e., rotate 180 around the Smith Chart).

64 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 7/9 z = j 05. z = j 26. 2β The put impedance is determed after rotatg clockwise 2β, and is z = j 05.. Now, we can rotate this pot 180 to determe the put admittance value y :

65 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 8/9 2β z = j 05. y = 07. j 17. The result is the same as with the earlier method-- y = 07. j 17.. Hopefully it is evident that the two methods are equivalent. In method 1 we first rotate 180, and then rotate 2β. In the second method we first rotate 2β, and then rotate the result is thus the same! Now, the remag equivalent circuit is:

66 2/17/2005 Example Admittance Calculations with the Smith Chart.doc 9/9 y 1 z = j 17. y = 07. j 17. Determg y 1 is just basic circuit theory. We first express z 2 terms of its admittance y 2 = 1 z 2. Note that we could do this usg a calculator, but could likewise use a Smith Chart (locate z 2 and then rotate 180 ) to accomplish this calculation! Either way, we fd that y 2 = j 03.. y 1 y = j 03. y = 07. j 17. Thus, y 1 is simply: = + y1 y2 y ( 03. j 03. ) ( 07. j 17. ) = + + = 10. j 14.