2/17/2010 2_4 The Smith Chart 1/ The Smith Chart. 1) A graphical method to solve many transmission line problems.

Transcription

1 2/17/2010 2_4 The Smith Chart 1/3 2.4 The Smith Chart Readg Assignment: pp The Smith Chart An icon of microwave engeerg! The Smith Chart provides: 1) A graphical method to solve many transmission le problems. 2) A visual dication of microwave device performance. The most important fact about the Smith Chart is: It exists on the complex Γ plane. HO: THE COMPEX Γ PANE Q: But how is the complex Γ plane useful? A: We can easily plot and determe values of Γ ( z ) HO: TRANSFORMATIONS ON THE COMPEX Γ PANE Q: But transformations of Γ are relatively easy transformations of le impedance Z is the difficult one.

2 2/17/2010 2_4 The Smith Chart 2/3 A: We can likewise map le impedance onto the complex Γ plane! HO: MAPPING Z TO Γ HO: THE SMITH CHART HO: SMITH CHART GEOGRAPHY HO: THE OUTER SCAE The Smith Chart allows us to solve many important transmission le problems! HO: Z IN CACUATIONS USING THE SMITH CHART EXAMPE: THE INPUT IMPEDANCE OF A SHORTED TRANSMISSION INE EXAMPE: DETERMINING THE OAD IMPEDANCE OF A TRANSMISSION INE EXAMPE: DETERMINING THE ENGTH OF A TRANSMISSION INE An alternative to impedance Z, is its verse admittance Y. HO: IMPEDANCE AND ADMITTANCE

3 2/17/2010 2_4 The Smith Chart 3/3 Expressg a load or le impedance terms of its admittance is sometimes helpful. Additionally, we can easily map admittance onto the Smith Chart. HO: ADMITTANCE AND THE SMITH CHART EXAMPE: ADMITTANCE CACUATIONS WITH THE SMITH CHART

4 2/4/2010 The Complex Gamma Plane present.doc 1/7 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 =5 Ω R =20 Ω R =50 Ω ikewise, sce impedance Z is a complex value, we can dicate specific impedance values as pot on a two dimensional complex impedance plane : R Im { Z } Note each dimension is defed by a sgle real le: Z =30 +j 40 Ω Re { Z } Z =60 -j 30 Ω * The horizontal le (axis) dicatg the real component of Z (i.e., Re{ Z }). * 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.

5 2/4/2010 The Complex Gamma Plane present.doc 2/7 es and Curves on the Complex Z Plane * 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. Im { Z } R =75 Re { Z } X =-30

6 2/4/2010 The Complex Gamma Plane present.doc 3/7 The Validity Region of the Complex Z Plane 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!!

7 2/4/2010 The Complex Gamma Plane present.doc 4/7 The Complex Γ Plane 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

8 2/4/2010 The Complex Gamma Plane present.doc 5/7 es and Curves on the Complex Γ Plane We can plot pots and les on this complex Γ plane exactly as before: Im { Γ} 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 ): Re {Γ}=0.5 Re { Γ} j e θ Γ Γ= Γ Im {Γ} =-0.3 In other words, we express Γ usg polar coordates. 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. Im { Γ } θ Im { Γ } Γ= 06. e j 3π 4 Γ = 135 Γ Γ Γ = 07. θ Γ Re { Γ } Re { Γ } Γ= 07. e j 300

9 2/4/2010 The Complex Gamma Plane present.doc 6/7 The Validity Region of the Complex Γ Plane 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 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 Im { Γ } Therefore, the validity region for the complex Γ plane consists of all pots side the circle Γ= 1--a fite and bounded area! Invalid Region ( Γ > 1) Valid Region ( Γ < 1) Re { Γ } Γ = 1

10 2/4/2010 The Complex Gamma Plane present.doc 7/7 Note that we can plot all valid impedances (i.e., R >0) with this fite validity 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)

11 2/4/2010 Transformations on the Complex G plane present.doc 1/8 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 ): ( ) ( 2 ) Γ z =Γ e = Γ e θ β j2 β z j Γ + z And thus, as we would expect: ( 0) and ( ) -j 2β Γ z = = Γ Γ z = = Γ e = Γ

12 2/4/2010 Transformations on the Complex G plane present.doc 2/8 Transformg Γ to 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. Q: Magnitude Γ and phase θ Γ --aren t those the values used when plottg on the complex Γ plane? Im { Γ } Γ ( z ) 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 = ): Γ = 1 ( z 0) Γ = =Γ Γ ( z ) Γ = = Γ Re { Γ } θ = θ 2β

13 2/4/2010 Transformations on the Complex G plane present.doc 3/8 Graphically Transformg Γ to Γ 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 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

14 2/4/2010 Transformations on the Complex G plane present.doc 4/8 Example: Graphically Transformg Γ to Γ The reflection coefficient of a short circuit is Γ = 1 = 1 j e π 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 ). ( ), and therefore we beg at Γ ( z ) Im { Γ } Γ = 1 e + j π 2 Re { Γ } Γ = 1 e + jπ Γ=1 When we stop, we fd we are at the pot for Γ ; this case o is one, phase is 90 ). 2 Γ 1 j = e π (i.e., magnitude

15 2/4/2010 Transformations on the Complex G plane present.doc 5/8 Example: Now with l = λ/4 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 } 0 1 e + j Γ = Re { Γ } Γ = 1 e + jπ Γ=1 For this case, the put reflection coefficient is of an open circuit! j 0 Γ = 1e = 1 : the reflection coefficient Our short-circuit load has been transformed to an open circuit with a quarterwavelength transmission le!

16 2/4/2010 Transformations on the Complex G plane present.doc 6/8 You re not surprised are you? 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. Im { Γ } 0 1 e + j Γ = Γ = 1 Re { Γ } 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: Γ = 1 e + jπ Γ ( z )

17 2/4/2010 Transformations on the Complex G plane present.doc 7/8 Example: Now with l = λ/2 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! Thus, we fd that Γ = Γ if = λ 2--but you knew this too! Γ ( z ) Im { Γ } Recall that the halfwavelength transmission le is likewise a special case, where we found that Z = Z. This result, of course, likewise means that Γ = Γ. Γ = 1 e + jπ Γ = 1 e + jπ Re { Γ } Γ = 1

18 2/4/2010 Transformations on the Complex G plane present.doc 8/8 Example: Now transform Γ to Γ 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? Γ ( z ) Im { Γ } θ A: In this case, we beg at Γ and rotate COUNTER-COCKWISE along a circular arc (radius 0.5) 2β = π 2 θ = θ + 2β 0.5 radians (i.e., 60 ). Essentially, we are removg the phase shift associated with the transmission le! Γ = 05. e j 150 Γ = 05. e j 60 Re { Γ } The reflection coefficient of the load is therefore: Γ = e 0.5 j 150 Γ = 1

19 2/4/2010 Mappg Z to Gamma present.doc 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: Γ ( z ) ( ) ( ) ( ) ( ) 0 0 = = 0 0 ( z ) ( z ) Z z Z Z z Z 1 z 1 = Z z + Z Z z Z + 1 z + 1

20 2/4/2010 Mappg Z to Gamma present.doc 2/8 Normalized Impedance 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. case Z z Γ Some values we already know specifically 3 Z j Z 0 j j 5 j Z 0 j j

21 2/4/2010 Mappg Z to Gamma present.doc 3/8 Mappg pots on both the Γ and Z planes Therefore, we fd that these five normalized impedances map onto five specific pots on the complex Γ plane Invalid Region z = 0 ( Γ= 1) x z = j ( Γ= j ) Invalid Region z = 0 ( Γ= 1) z = 1 ( Γ= 0) Γ i z = j ( Γ= j ) z = j ( Γ= j ) z = ( Γ= 1) Γ=1 Γ r r z = j ( Γ= j ) z = 1 ( Γ= 0) z = ( Γ= 1) Or, the five complex Γ map onto five pots on the normalized impedance plane.

22 2/4/2010 Mappg Z to Gamma present.doc 4/8 Mappg contours on both the Γ and Z planes 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: where we recall that r = R Z0. ( ) z = r + j0 i.e.,x = 0 Remember, this real-valued impedance results a real-valued reflection coefficient: r 1 Γ= r + 1 r 1 I.E.,: r Re { } i Im { } r 1 Γ Γ = Γ Γ = + 0

23 2/4/2010 Mappg Z to Gamma present.doc 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 = 0 i Invalid Region Γ i Γ = 1 x = 0 ( Γ = 0) i Γ r Invalid Region x r r = x = 0 ( Γ = 0) i

24 2/4/2010 Mappg Z to Gamma present.doc 6/8 Z = jx In other words, the case where impedance is purely imagary, with no resistive component (i.e., R = 0). Meang that normalized impedance is: where we recall that x = X Z0. ( ) z = 0+ jx i.e.,r = 0 Remember, this imagary impedance results a reflection coefficient with unity magnitude: Γ = 1

25 2/4/2010 Mappg Z to Gamma present.doc 7/8 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 Invalid Region Γ i Γ = 1 r = 0 ( Γ = 1) Γ r Invalid Region x x = j r r = 0 ( Γ = 1) x = j

26 2/4/2010 Mappg Z to Gamma present.doc 8/8 What about r=0.5, or x=-1.5?? Q: These two mappgs may very well be fascatg an academic sense, but they are 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!

27 2/7/2010 The Smith Chart present.doc 1/12 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 r =0 simply maps to the circle Γ = 1 on the complex Γ plane, and x = 0 simply maps to the le Γ i = 0. But, for the examples given above, the mappg is not so straight forward. The contours 2 2 will general be functions of both 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

28 2/7/2010 The Smith Chart present.doc 2/12 Vertical contours on the complex Ζ plane map 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: a = the radius of the circle ( ) P x = x,y = y pot located at the center of the circle c c c Thus, the vertical le r = c r maps to a circle on the complex Γ plane!

29 2/7/2010 The Smith Chart present.doc 3/12 onto circles on the complex G plane By spection, it is apparent that the center of this circle is located at this pot on the complex Γ plane: c r P c Γ r =, Γ i = cr In other words, the center of this circle always lies somewhere along the Γ i = 0 le. Γ ikewise, by spection, we fd the radius of this circle is: r = 03. i Γ = 1 1 a = + c 1 r r = 00. We perform a few of these mappgs and see where these circles lie on the complex Γ plane r = 03. r = 10. r = 30. Γ r r = 50.

30 2/7/2010 The Smith Chart present.doc 4/12 Some important stuff to notice 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 Γ= 1. r = 03. Γ i Γ = 1 2. If c r < 0 then the circle lies entirely outside the circle Γ = 1. r = 00. r = If c r = 0 (i.e., a reactive impedance), the circle lies on circle Γ = 1. Γ r r = 03. r = If c r =, then the radius of the circle is zero, and its center is at the pot 0 Γ r = 1, Γ i = 0 (i.e., Γ= 1e j ). In 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.

31 2/7/2010 The Smith Chart present.doc 5/12 Horizontal contours on the complex Ζ plane map 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 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!

32 2/7/2010 The Smith Chart present.doc 6/12 onto circles on the complex G 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 r = 1 le. Γ Γ r = 1 i ikewise, by spection, the radius of this circle is: x = 05. x = 10. x = 30. x = 20. a = 1 c We perform a few of these mappgs and see where these circles lie on the complex Γ plane i Γ = 1 Γ r x = 30. x = 05. x = 10. x = 20.

33 2/7/2010 The Smith Chart present.doc 7/12 Some more important stuff to notice We see that as the magnitude of constant c i creases, the radius of the circle decreases, and its center moves toward the pot ( Γ r = 1, Γ i = 0). Note: Γ 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. x = 05. i x = 10. x = 30. Γ r = x = 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. Γ = 1 Γ r 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 x = 05. x = 10. x = 30. x = If c i =±, then the radius of the circle is zero, and its center is at the pot 0 Γ r = 1, Γ i = 0 (i.e., Γ= 1e j ). In other 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!

34 2/7/2010 The Smith Chart present.doc 8/12 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. 5. Note also that much of the circle formed by mappg x = c onto the complex Γ plane lies outside the circle Γ= 1. i x = 05. Γ i x = 10. x = 30. Γ r = x = 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). Γ = 1 Thus, we typically can completely ignore the portions of the circles that lie outside the Γ = 1 circle! x = 05. x = 10. x = 30. x = 20. Mappg many les of the form results tool called the Smith Chart r = cr and x ci = onto circles on the complex Γ plane

35 2/7/2010 The Smith Chart present.doc 9/12 Im{ Γ } Re{ Γ }

36 2/7/2010 The Smith Chart present.doc 10/12 Rectilear and Curvilear Grids Note the Smith Chart is simply the vertical les r = cr and horizontal les x ci = of the normalized impedance plane, mapped onto the two types of circles on the complex Γ plane. For the normalized impedance plane, a vertical le r = c r and a horizontal le x = ci are always perpendicular to each other when they tersect. We say these les form a rectilear grid. r = 0 x x = 1 r 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. r = 1 x = 0 x = 1 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!

37 2/7/2010 The Smith Chart present.doc 11/12 The proverbial square peg.. The rectilear grid of the complex impedance plane: x x = 1 r = 0 r x r = 1 x = 0 x = 1 r Distortg this rectilear grid

38 2/7/2010 The Smith Chart present.doc 12/12 And then distortg some more we have the curvilear grid of the Smith Chart! x r

39 2/16/2010 Smith Chart Geograhpy present.doc 1/5 Smith Chart Geography We have located specific pots on the complex impedance plane, such as a short circuit or a matched load. We ve also identified contours, such as r = 1 or x = 2. We can likewise identify whole regions (!) of the complex impedance plane, providg a bit of a geography lesson of the complex impedance plane. r =-1 Im { z } r =+1 For example, we can divide the complex r 1 1 r 0 0 r 1 1 r impedance plane to four regions based on normalized resistance value r: Re { z } r =0

40 2/16/2010 Smith Chart Geograhpy present.doc 2/5 Mappg onto the Γ Plane Just like pots and contours, these regions of the complex impedance plane can be mapped onto the complex gamma plane! Γ i r =0 r = r 0 0 r 1 r 1 Γ r r 1 r = 10.

41 2/16/2010 Smith Chart Geograhpy present.doc 3/5 Reactive Boundaries and Borders Instead of dividg the complex impedance plane to regions based on normalized resistance r, we could divide it based on normalized reactance x: Im { z } x =1 x 1 x 0 1 Re { z } x =0 1 x 0 x =-1 x 1

42 2/16/2010 Smith Chart Geograhpy present.doc 4/5 Mappg onto the Γ Plane These four regions can likewise be mapped onto the complex gamma plane: x 0 1 Γ i x = 10. x x = x =1 x = 30. x =0 Γ r x =-1 x 1 1 x 0

43 2/16/2010 Smith Chart Geograhpy present.doc 5/5 Smith Chart Geography Note the four resistance regions and the four reactance regions combe to from 16 separate regions on the complex impedance and complex gamma planes! Eight of these sixteen regions lie the valid region (i.e., r > 0), while the other eight lie entirely the valid region. Make sure you can locate the eight impedance regions on a Smith Chart this understandg of Smith Chart geography will help you understand your design and analysis results! r < 1 0 < x < 1 r < 1 1 < x < 0 r < 1 x > 1 r > 1 r > 1 x > 1 0< x < 1 < 1 x < 0 r > 1 r > 1 x < 1 r < 1 x < 1

44 2/18/2010 The Outer Scale present.doc 1/25 The Outer Scale Note that around the outside of the Smith Chart there is a scale dicatg the phase angle θ Γ (i.e., j Γ= Γ e θ Γ ), from 180 < < 180 θ Γ.

45 2/18/2010 The Outer Scale present.doc 2/25 e position z and phase angle are related! Recall however, for a termated transmission le, the reflection coefficient function is: j 2 z j 2 z 0 ( z ) e β e β + θ Γ = Γ = Γ 0 0 Thus, the phase of the reflection coefficient function depends on transmission le position z as: ( ) 2 π z θγ z = 2βz + θ0 = 2 z θ0 4π + = + θ 0 λ λ As a result, a change le position z (i.e., Δ z ) results a change reflection coefficient phase θ Γ (i.e., Δ θ Γ ): z θγ 4π Δ Δ = λ For example, a change of position equal to one-quarter wavelength Δ z = λ results a 4 phase change of π radians we rotate half-way around the complex Γ plane (otherwise known as the Smith Chart).

46 2/18/2010 The Outer Scale present.doc 3/25 A second outer scale The Smith Chart thus has a second scale (besides θ Γ ) that surrounds it one that relates transmission le position wavelengths (i.e., Δ z λ ) to the reflection coefficient phase: z 1 θγ = + λ 4 4π z θ = 4π Γ λ 1 4

47 2/18/2010 The Outer Scale present.doc 4/25 This second scale is very useful! Sce the phase scale on the Smith Chart extends from 180 < θ Γ < 180 (i.e., π < θ < π ), this electrical length scale extends from: Γ 0 < z < 0.5 λ Note for this mappg the reflection coefficient phase at location z = 0 is θ = π. Therefore, θ 0 = π, and we fd that: j θ j π 0 0 e 0 e 0 Γ = Γ 0 = Γ = Γ In other words, Γ 0 is a negative real value. Q: But, Γ 0 could be anythg! What is the likelihood of Γ 0 beg a real and negative value? Most of the time this is not the case this second Smith Chart scale seems to be nearly useless!? A: Quite the contrary! This electrical length scale is fact very useful you just need to understand how to utilize it!

48 2/18/2010 The Outer Scale present.doc 5/25 The first of many analogies This electrical length scale is very much like the mile markers you see along an terstate highway; although the specific numbers are quite arbitrary, they are still very useful. Take for example Interstate 70, which stretches across Kansas. The western end of I-70 (at the Colorado border) is denoted as mile 1. At each mile along I-70 a new marker is placed, such that the eastern end of I-70 (at the Missouri border) is labeled mile 423 Interstate 70 runs for 423 miles across Kansas!

49 2/18/2010 The Outer Scale present.doc 6/25 A Kansas geography lesson The location of various towns and burgs along I-70 can thus be specified terms of these mile markers. For example, along I-70 we fd: Oakley at 76 Hays at 159 Russell at 184 Sala at 251 Junction City 296 Topeka at 361 awrence at 388

50 2/18/2010 The Outer Scale present.doc 7/25 Mile markers: the key to successful navigation So say you are travelg eastbound ( ) along I-70, and you want to know the distance to Topeka. Topeka is at mile marker 361, but this does not of course mean you are 361 miles from Topeka. Instead, you subtract from 361 the value of the mile marker denotg your position along I-70. For example, if you fd yourself the lovely borough of Russell (mile marker 184), you have precisely = 177 miles to go before reachg Topeka! Q: I m confused! Say I m awrence (mile marker 388); usg your logic I am a distance of = -27 miles from Topeka! How can I be a negative distance from somethg?? A: The mile markers across Kansas are arranged such that their value creases as we move from west to east across the state. Take the value of the mile marker denotg to where you are travelg, and subtract from it the value of the mile marker where you are. If this value is positive, then your destation is East of you; if this value is negative, it is West of your current position (hopefully you re the westbound lane!).

51 2/18/2010 The Outer Scale present.doc 8/25 Its not rocket science! For example, say you re travelg to Sala (mile marker 251). If you are Oakley (mile marker 76) then: = 175 Sala is 175 miles East of Oakley If, on the other hand, you beg your journey from Junction City (mile marker 296), we fd: = -45 Sala is 45 miles West of Junction City miles miles

52 2/18/2010 The Outer Scale present.doc 9/25 Please tell me this is useful Q: But just what the &()#$% does this discussion have to do with SMITH CHARTS!!?!? A: The electrical length scale (z λ ) around the perimeter of the Smith Chart is precisely analogous to mile markers along an terstate! 3λ 8 Recall that the change phase ( Δ θ Γ ) of the reflection coefficient function is related to the change distance ( Δ z ) along a transmission le as: z θγ 4π Δ Δ = λ λ 2 0 λ 4 The value Δ z λ can be determed from the outer scale of the Smith Chart, simply by takg the difference of the two mile markers values. λ 8

53 2/18/2010 The Outer Scale present.doc 10/25 For example For example, say you re at some location z = z1 along a transmission le. The value of the reflection coefficient function at that pot happens to be: 65 Γ z = z = e j ( ) 1 Fdg the phase angle of θ Γ = 65 on the outer scale of the Smith Chart, we note that the correspondg electrical length value is: 0.160λ Note this tells us nothg about the location z = z1. This does not mean that z 1 = 0.160λ, for example!

54 2/18/2010 The Outer Scale present.doc 11/25 Contued Now, say we move a short distance Δ z (i.e., a distance less than λ 2) along the transmission le, to a new location denoted as z = z2. We fd that this new location that the reflection coefficient function has a value of: ( ) Γ z = z = e + j Now fdg the phase angle of θ Γ =+ 74 on the outer scale of the Smith Chart, we note that the correspondg electrical length value is: λ Note this tells us nothg about the location z = z2. This does not mean that z 1 = 0.353λ, for example!

55 2/18/2010 The Outer Scale present.doc 12/25 See the analogy? Q: So what do the values 0.160λ and 0.353λ tell us? A: They allow us to determe the distance between pots z 2 and z 1 on the transmission le: Δ z z2 z = 1!!! λ λ λ Thus, for this example, the distance between locations z 2 and z 1 is: Δ z = λ λ = λ The transmission le location z 2 is a distance of 0.193λ from location z 1!

56 2/18/2010 The Outer Scale present.doc 13/25 The power of negative thkg Q: But, say the reflection coefficient at some pot z 3 has a phase value of θ Γ = 112. This maps to a value of: 0.094λ on the outer scale of the Smith Chart. The distance between z 3 and z 1 would then turn out to be: Δ z = = λ What does the negative value mean?? A: Just like our I-70 mile marker analogy, the sign (plus or mus) dicates the direction of movement from one pot to another.

57 2/18/2010 The Outer Scale present.doc 14/25 This isn t rocket science either In the first example, we fd that Δ z > 0, meang z 2 > z 1 : z = z λ. 2 1 Clearly, the location z 2 is further down the transmission le (i.e., closer to the load) than is location z 1. For the second example, we fd that Δ z < 0, meang z 3 < z 1 : z = z λ 3 1 Conversely, this second example, the location z 3 is closer to the begng of the transmission le (i.e., farther from the load) than is location z 1.

58 2/18/2010 The Outer Scale present.doc 15/25 You shouldn t have be surprised This is completely consistent with what we already know to be true! In the first case, the positive value Δ z = λ maps to a phase Δ = = 139 change of θ Γ ( ). In other words, as we move toward the load from location z 1 to location z 2, we rotate counter-clockwise around the Smith Chart. ikewise, the negative value Δ z = λ maps to a phase change of Δ = = 47 θ Γ ( ). In other words, as we move away from the load (toward the source) from a location z 1 to location z 3, we rotate clockwise around the Smith Chart.

59 2/18/2010 The Outer Scale present.doc 16/25 A graphical summary of what I just said Δ z = λ 74 ( z z ) e + j Γ = = ( z z ) e j Γ = = 3 65 ( z z ) e j Γ = = 1 Δ z = λ

60 2/18/2010 The Outer Scale present.doc 17/25 Yet another outer scale Q: I notice that there is a second electrical length scale on the Smith Chart. Its values crease as we move clockwise from an itial value of zero to a maximum value of 05λ.. What s up with that? A: This scale uses an alternative mappg between θ Γ and z λ : z 1 θ 1 z Γ = θ = 4π Γ λ 4 4π 4 λ This scale is analogous to a situation where a second set of mile markers were placed along I-70. These mile markers beg at the east side of Kansas (at the Missouri border), and end at the west side of Kansas (at the Colorado border)

61 2/18/2010 The Outer Scale present.doc 18/25 What s the pot? Q: What good would this second set of markers do? Does it serve any purpose? A: Not much really. After all, this second set is redundant it does not provide any new formation that the origal set already provides. Yet, if we were to place this new set along I-70, we almost certaly would place the origal mile markers along the eastbound lanes, and this new set along the westbound lanes. In this manner, all I-70 motorists (eastbound or westbound) would see an crease the mile markers as they traverse the Sunflower State. As a result, a positive distance to their destation dicates to all drivers that their destation is front of them ( the direction they are drivg), while a negative distance dicates to all drivers that their destation is behd the (they better turn around!).

62 2/18/2010 The Outer Scale present.doc 19/25 The power of positive thkg Thus, it could be argued that each set of mile markers is optimized for a specific direction of travel the origal set if you are travelg east, and this second set if you are travelg west Similarly, the two electrical length scales on the Smith Chart are meant for two different directions of travel. If we move down the transmission le toward the load, the value Δ z will be positive. Conversely, if we move up the transmission le and away from the load (i.e., toward the generator ), this second electrical length scale will also provide a positive value of Δ z. Aga, these two electrical length scales are redundant you will get the correct answer regardless of the scale you use, but be careful to terpret negative signs properly

63 2/18/2010 The Outer Scale present.doc 20/25 Oh, so you noticed Q: Wait! I just used a Smith Chart to analyze a transmission le problem the manner you have just explaed. At one pot on my transmission le the phase of the reflection coefficient is θ Γ =+ 170, which is denoted as 0486λ. on the wavelengths toward load scale. I then moved a short distance along the le toward the load, and found that the reflection coefficient phase was θ Γ = 144, which is denoted as λ on the wavelengths toward load scale. Accordg to your struction, the distance between these two pots is: Δ z = λ λ = λ A large negative value! This says that I moved nearly a half wavelength away from the load, but I know that I moved just a short distance toward the load! What happened?

64 2/18/2010 The Outer Scale present.doc 21/25 Here s the problem A: Note the electrical length scales on the Smith Chart beg and end where θ π Γ = ± (by the short circuit!). ( z ) Γ = z 1 ( z ) Γ = z 2 In your example, when rotatg counter-clockwise around the chart (i.e., movg toward the load) you passed by this transition. This makes the calculation of Δ z a bit more problematic.

65 2/18/2010 The Outer Scale present.doc 22/25 Yet another enlighteng analogy To see why, let s aga consider our I-70 analogy. Say we are awrence, and wish to drive eastbound on Interstate 70 until we reach Columbia, Missouri. The mile marker for awrence is of course 388, and Columbia Missouri is located at mile marker 126. We might conclude that the distance from awrence to Columbia is: = 262 miles Q: Yikes! Accordg to this, Columbia is 262 miles west of awrence should we turn the car around? A: Columbia, Missouri is most decidedly east of awrence, Kansas. The problem is that mile markers reset to zero once we reach a state border, and then aga crease as we travel eastward Kansas Missouri 1 2 6

66 2/18/2010 The Outer Scale present.doc 23/25 The pafully obvious* Thus, to accurately determe the distance between awrence and Columbia, we need to break the problem to two steps: Step 1: Determe the distance between awrence (mile marker 388), and the last mile marker before the state le (mile marker 423): = 35 miles Step 2: Determe the distance between the first mile marker after the state le (mile marker 0) and Columbia (mile marker 126): = 126 miles Thus, the distance between awrence and Columbia is the distance between awrence and the state le (35 miles), plus the distance from the state le to Columbia (126 miles): = 161 miles Columbia, Missouri is 161 miles east of awrence, Kansas! * Don t compla; it s far superior to the obviously paful.

67 2/18/2010 The Outer Scale present.doc 24/25 Back to the real world Now back to the Smith Chart problem; as we rotate counter-clockwise around the Smith Chart, the wavelengths toward load scale creases value, until it reaches a maximum value of 05λ. (at θ =± Γ π ). At that pot, the scale resets to its mimum value of zero. We have metaphorically crossed the state le of this scale. Thus, to accurately determe the electrical length moved along a transmission le, we must divide the problem to two steps: Step 1: Determe the electrical length from the itial pot to the end of the scale at 05λ.. Step 2: Determe the electrical distance from the begng of the scale (i.e., 0) and the second location on the transmission le. Add the results of steps 1 and 2, and you have your answer!

68 2/18/2010 The Outer Scale present.doc 25/25 Your problem is solved For example, let s look at the case that origally gave us the erroneous result. The distance from the itial location to the end of the scale is: 0.014λ ( z ) Γ = z 1 And the distance from the begng of the scale to the second pot is: λ λ = λ Thus the distance between the two pots is: ( z ) Γ = z λ λ = λ The second pot is just a little closer to the load than the first! 0.050λ

69 2/18/2010 Z Calculations usg the Smith Chart present.doc 1/8 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: Q: Evaluatg this unattractive expression looks not the least bit pleasant. Isn t there a less disagreeable method to determe z? z = Z Z 0 1 Z + j Z0 tan β = Z0 Z0 Z0 + j Z tan β Z Z0 + j tan β = 1 + jz Z tanβ z + j tan β = 1 + jz tan β 0

70 2/18/2010 Z Calculations usg the Smith Chart present.doc 2/8 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 Z Z Z 1 z z Γ = = = Z Z0 Z Z0 2. 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?

71 2/18/2010 Z Calculations usg the Smith Chart present.doc 3/8 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 Γ plane! 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 65

72 2/18/2010 Z Calculations usg the Smith Chart present.doc 4/8 2. 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! 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: Γ= 0.685e + j 74 This is the value of Γ!

73 2/18/2010 Z Calculations usg the Smith Chart present.doc 5/8 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 + words: 74 is located at the tersection of circles r = 0.5 and x =1.2. In other z = j1.2

74 2/18/2010 Z Calculations usg the Smith Chart present.doc 6/8 Step 1 Γ = Γ = e j 65 θ Γ = 65

75 2/18/2010 Z Calculations usg the Smith Chart present.doc 7/8 Step 2 2 = λ = = λ λ = λ Γ = e j 74 2β = 221 Γ = Γ = e j 65 1 = 016. λ

76 2/18/2010 Z Calculations usg the Smith Chart present.doc 8/8 Step 3 z = j 12.

77 2/16/2010 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

78 2/16/2010 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 ( z ) Γ = 1 = e j 180

79 2/16/2010 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 ( z ) Γ = 1 = e j 180 z = j

80 2/16/2010 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

81 2/16/2010 Example The oad Impedance.doc 2/2 = λ 2β = 965. Γ ( z ) z = j 024. z = 1+ j 14.

82 2/16/2010 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.

83 2/16/2010 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 ) z = j 0 x = 0

84 2/16/2010 Example Determg the tl length.doc 3/7 Solution 2: z = 2+ j 2 z = j 0 x = 0 Γ ( z ) 2β = 210 = λ

85 2/16/2010 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!

86 2/16/2010 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

87 2/16/2010 Example Determg the tl length.doc 6/7 Solution 2: z = j 16. Γ ( z ) z = 2+ j 2 r = 1 2β = 339 = λ

88 2/16/2010 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.

89 2/13/2009 Admittance present 1/4 Impedance & 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 1 Y = Z

90 2/13/2009 Admittance present 2/4 Normalized Admittance 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: 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

91 2/13/2009 Admittance present 3/4 Susceptance and Conductance Now sce admittance is a complex value, it has both a real and imagary component: where: Y = G + j B { Y} G = { } Re Conductance Im Z B = Susceptance 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!

92 2/13/2009 Admittance present 4/4 Be Careful! A: NOOOO! We fd that G 1 R and B 1 X (generally). Do not make this mistake! In fact, we fd that: 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 I wish I had a nickel for every time my software has crashed oh wait, I do! And that IF R = 0 (i.e., Z = R ), we get, as expected: G = 0 and B = 1 X

93 2/17/2010 Admittance and the Smith Chart present 1/7 Admittance and the Smith Chart Just like the complex impedance plane, we can plot pots and contours on the complex admittance plane: Im {Y} = B G =75 Re {Y} = G B =-30 Y = 120 j 60 Q: Can we also map these pots and contours onto the complex Γ plane? A: You bet! et s first rewrite the refection coefficient function terms of le admittance Y ( z ): Y0 Y ( z ) Γ ( z ) = Y + Y z 0 ( )

94 2/17/2010 Admittance and the Smith Chart present 2/7 Rotation around the Smith Chart Thus, Y0 Y Γ = Y + Y 0 Y0 Y and Γ = Y + Y 0 We can therefore likewise express Γ terms of normalized admittance: Y0 Y 1 Y Y0 1 y Γ= = = Y + Y 1 + Y Y 1 + y 0 0 Note this can likewise be expressed as: 1 y y 1 jπ y 1 Γ= = = e 1+ y y + 1 y + 1 Contrast this to the mappg between normalized impedance and Γ: Γ= z z + j The difference between the two is simply the factor e π a rotation of 180 around the Smith Chart!. 1 1

95 2/17/2010 Admittance and the Smith Chart present 3/7 An example For example, let s pick some load at random; z = 1 + j, for stance. We know where this pot is mapped onto the complex Γ plane; we can locate it on our Smith Chart. Now let s consider a different load, and express it terms of its normalized admittance an admittance that has the same numerical value as the impedance of the first load (i.e., y = 1 + j ). Im{ Γ} z = 1 + j Q: Where would this admittance value map onto the complex Γ plane? Re{ Γ} A: Start at the location z = 1 + j on the Smith Chart, and then rotate around the center 180. You are now at the proper location on the complex Γ plane for the admittance y = 1 + j! y = 1 + j 180

96 2/17/2010 Admittance and the Smith Chart present 4/7 We of course could just directly calculate Γ from the equation above, and then plot that pot on the Γ plane. Note the reflection coefficient for z = 1 + j is: z 1 1+ j 1 j Γ= = = z j j while the reflection coefficient for y = 1 + j is: 1 y 1 (1 + j ) j Γ= = = 1+ y 1 + (1 + j) 2+ j Note the two results have equal magnitude, but are separated phase by 180 j ( 1 = e π ). This means that the two loads occupy pots on the complex Γ plane that are a 180 rotation from each other! Moreover, this is a true statement not just for the pot we randomly picked, but is true for any and all values of z and y mapped onto the complex Γ plane, provided that z = y.

97 2/17/2010 Admittance and the Smith Chart present 5/7 Another example Im{ Γ } For example, the g =2 circle mapped on the complex plane can be determed by rotatg the r =2 circle 180 around the complex Γ plane, and the b =-1 contour can be found by rotatg the x =-1 contour 180 around the complex Γ plane. g = 2 x = 1 z = 1 + j Re{ Γ } r = 2 y = 1 + j b = 1

98 2/17/2010 Admittance and the Smith Chart present 6/7 Thus, rotatg all the resistance circles and reactance contours of the Smith Chart 180 around the complex Γ plane provides us a mappg of complex admittance onto the complex Γ plane: The Admittance Smith Chart Im{ Γ} Note that circles and contours have been rotated with respect to the complex Γ plane the complex Γ plane remas unchanged! Re{ Γ }

99 2/17/2010 Admittance and the Smith Chart present 7/7 We re not surprised! This result should not surprise us. Recall the case where a transmission le of length = λ 4 is termated with a load of impedance z (or equivalently, an admittance y ). The put impedance (admittance) for this case is: Z Z Z 1 Z z y = = Z Z0 Z = = z In other words, when = λ 4, the put impedance is numerically equal to the load admittance and vice versa! But note that if = λ 4, then 2β = π --a rotation around the Smith Chart of 180!

100 2/17/2010 Example Admittance Calculations with the Smith Chart 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

101 2/17/2010 Example Admittance Calculations with the Smith Chart 2/9 There are two ways to determe this value! Method 1 First, we express the load z = j 26. 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.

102 2/17/2010 Example Admittance Calculations with the Smith Chart 3/9 The Smith Chart above shows both the impedance mappg (red) and admittance mappg (blue). Thus, we can locate the impedance z = j 26. 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 26. on the impedance mappg : z = j 26.

103 2/17/2010 Example Admittance Calculations with the Smith Chart 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.

104 2/17/2010 Example Admittance Calculations with the Smith Chart 5/9 z = j 26. y = 017. j Now, we can determe the value of y by simply rotatg clockwise 2β from y, where = 037. λ :

105 2/17/2010 Example Admittance Calculations with the Smith Chart 6/ λ Γ ( z ) y = 017. j 028. y = 07. j 17. Transformg the load admittance to the begng of the transmission le, we have determed that y = 07. j 17.. 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).

106 2/17/2010 Example Admittance Calculations with the Smith Chart 7/9 z = j 05. z = j 26. Γ ( z ) 037. λ 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 :

107 2/17/2010 Example Admittance Calculations with the Smith Chart 8/9 180 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: