Lecture 13 Date:

Transcription

1 ecture 3 Date: The Signal Flow Graph (Contd.) Impedance Matching and Tuning Tpe Matching Network Example

2 Signal Flow Graph (contd.) Splitting Rule Now consider the three equations SFG a a b 2 3 a b b a b a 2 a 3 These equations can be equivalentl written as a b a a 2 3 a a b b SFG a 2 a3 Rule 4 Splitting Rule If a node has one (and onl one!) incoming branch, and one (or more) exiting branches, the incoming branch can be split, and directl combined with each of the exiting branches.

3 Signal Flow Graph (contd.) Example j a b 0.3 a 2 Splitting Rule Gives Series Rule Gives j a 3 a j0.3 j b a a3 a 2 a j b j0.2 a 3

4 Signal Flow Graph (contd.) The splitting rule is particularl useful when we encounter signal flow graphs of the kind: 0.3 j j Note this node has two incoming branches!! We can split the -0.2 branch, and rewrite the graph as: Note this node has onl one incoming branch!! j 0.2(0.3) j0. Now appl self-loop rule to simplif it further 0.2

5 Signal Flow Graph (contd.) Q: Can we split the other branch of the loop? 0.3 j j0. j (0.3) the equations represented b the two signal flow graphs are not equivalent j A: NO!! Do not make this mistake! We cannot split the 0.3 branch because it terminates in a node with two incoming branches (i.e., -j and 0.3). This is a violation of rule 4.

6 Example- Indraprastha Institute of Consider the basic 2-port network, terminated with load Γ : a S 2 b 2 2 Γ S S 22 b determine the value: a b S 2 a 2 Γ Solution: Isn t this simpl S? Onl if Γ = 0 (and in this situation it is not!)

7 Example- (contd.) let s decompose (simplif) the signal flow graph and find out! Step-: splitting rule on node a 2 Step-2: self-loop rule on node b 2 a S S 2 S22Γ b 2 a S S2 - S Γ 22 b 2 S 2 Γ S 2 Γ b a 2 b a 2 Step-3: series rule gives a S b S S - S Γ Step-4: parallel rule gives a b S S S - S Γ b a

8 Example 2 Below is a single-port device (with input at port x) constructed with two two-port devices (S x and S ), a quarter wavelength transmission line, and a load impedance. j2 l = λ/4 Z 0 Z 0 Γ =0.5 Given port x Z 0 = 50Ω S x port 2x port S port 2 Draw the complete signal flow graph of this circuit, and then reduce the graph to determine: a) The total current through load Γ ; b) The power delivered to (i.e., absorbed b ) port x.

9 Example 2 (contd.) The signal flow graph describing this network is: x a x b 2 a b 2 x S 2 j l e S 2 x S x S 22 S S 22 x b x S 2 x a 2 j l e b S 2 a 2 Γ We know that the value of the wave incident on port of device S x is: a V z z j2 j 2 Z 50 5 x x x xp 0

10 Example 2 (contd.) et us place the given numeric values of branches on this SFG: a x j x b 2 j a 0.8 b j x b x a 2 b a 2

11 Example 2 (contd.) Remove the zero valued branches: a x j x b 2 j a 0.8 b j x b x a 2 b a 2

12 Example 2 (contd.) Now appl splitting rule at node a 2 a x j x b 2 j a 0.8 b * 0.5 = j 0.8 x b x a 2 b a 2

13 Example 2 (contd.) Then appl self-loop rule at node b 2 a x j x b 2 j a 0.8 = b j 0.8 x b x a 2 b a 2

14 Example 2 (contd.) let s use this simplified signal flow graph to find the solutions to our questions! a a) The total current through load Γ V z z V z z I I z2 z2p Z0 a2 b2 b2 a2 I Z P P Using the series rule on the SFG gives x b x j / 5 0.5* -j* = -j * -j* 0.8 = -j0.4 b 2 a Thus, we need to determine the value of nodes a 2 and b 2 Therefore, x j 2 b 2 = -j0.5* a j0.5* a 2= 0.5* b

15 Example 2 (contd.) Thus the total current through Γ is: I b2 a ma b) The power delivered to (i.e., absorbed b ) port x is: 2 2 x x xp x x xp V z z V z z Pabs P P 2Z 2Z 0 0 P abs a 2 2 b x x 2 Requires knowledge of nodes a x and b x

16 Example 2 (contd.) We have: appl series rule ax j 2 / 5 -j0.5 b 2 ax j 2 / 5 b x j0.4 a b x -j0.5* 0.5* -j0.4 = -0. ax j 2 / 5 appl parallel rule = 0.25 b x Therefore: x b 0.25a 0.25* j 2 / 5 j x

17 Example 2 (contd.) Therefore, the power delivered to (i.e., absorbed b ) port x is: 2 2 j 2 / 5 j Pabs 37.5 mw 2 2

18 Introduction Impedance Transformation One of the most important and fundamental two-port networks that microwave engineers design is a lossless matching network (otherwise known as an impedance transformer). Q: In high frequenc circuits, a source and load are connected b a T. Can we implement matching networks in transmission line circuits? Furthermore, these matching networks seem too good to be true can we reall design and construct them to provide a perfect match? A: We can easil provide a near perfect match at precisel one frequenc But, since lossless matching and transmission lines are made of entirel reactive elements (not to mention the reactive components of source and load impedance), we find that changing the frequenc will tpicall unmatch our circuit!

19 Introduction (contd.) Therefore, a difficult challenge for an RF/microwave design engineer is to design a wideband matching network a matching network that provides an adequate match over a wide range of frequencies. Generall speaking, matching network design requires a tradeoff between following parameters for desirable attributes:. Bandwidth 2. Complexit 3. Implementation 4. Adjustabilit

20 Matching with umped Components let s begin to examine how matching networks are built! we begin with the simplest solution: An -network, consisting of a single capacitor and a single inductor. Q: Just two elements! That seems simple enough. Do we alwas use these -networks when constructing lossless matching networks? A: Nope. -networks have two major drawbacks:. The are narrow-band. 2. Capacitors and inductors are difficult to make at microwave frequencies! Soon we will see how these -networks actuall work

21 Matching Network Analsis Consider following circuit where a passive load is attached to an active source: V g Z g Z R jx It will absorb power power that is delivered to it b the source given b expression P 2 V I * Z V Re g P Vg 2 Z g Z Z g Z * Re Z R Vg 2 Z Z 2 Z Z 2 2 V 2 g 2 g g Recall that the power delivered to the load will be maximized (for a given V g and Z g ) if the load impedance is equal to the complex conjugate of the source impedance (Z =Z g )

22 Matching Network Analsis (contd.) We call the maximized power, the available power P avl from the source it is, after all, the largest amount of power that the source can ever deliver! R R Vg P P V V 2 2 8R max 2 g 2 g avl g * 2 g 2 Zg Zg 2Rg Note the available power of the source is dependent on source parameters onl (i.e., V g and R g ). This makes sense! Do ou see wh? Thus, we can sa that to take full advantage of all the available power of the source, we must make the load impedance the complex conjugate of the source impedance. Otherwise, the power delivered to the load will be less than power made available b the source! In other words : 2 g P P avl

23 Matching Networks and Transmission ines Note: we can construct a network to transform the input impedance of the transmission line into the complex conjugate of the source impedance: V g Q: But, do we have to place the matching network between the source and the transmission line? A: Nope! We could also place a (different) matching network between the transmission line and the load. Z g * Z g l Z 0 Z Z g l V g * Z g Z 0 Z

24 Matching Networks and Transmission ines (contd.) Q: So which method should we choose? Do engineers tpicall place the matching network between the source and the transmission line, or place it between the transmission line and the load? A: Actuall, the tpical solution is to do both! We find that often there is a matching network between the a source and the transmission line, and between the line and the load. Z g l V g Z 0 Z

25 Matching Networks and Transmission ines (contd.) The first network matches the source to the transmission line in other words, it transforms the output impedance of the equivalent source to a value numericall equal to characteristic impedance Z 0 : Z out 0 g V g Z Z Z 0 The second network matches the load to the transmission line in other words it transforms the load impedance to a value numericall equal to characteristic impedance Z 0 : Z 0 Zin Z 0 Z Q: Yikes! Wh would we want to build two separate matching networks, instead of just one?

26 Matching Networks and Transmission ines (contd.) A: B using two separate matching networks, we can decouple the design problem. Recall again that the design of a single matching network solution would depend on four separate parameters: Alternativel, the design of the network matching the source and transmission line depends on onl: In addition, the design of the network matching the load and transmission line depends on onl:. the source impedance Z g 2. load impedance Z 3. the T characteristic impedance Z 0 4. the T length l. the source impedance Z g 2. the transmission line characteristic impedance Z 0. the source impedance Z 2. the transmission line characteristic impedance Z 0 Note that neither design depends on the transmission line ength l! Q: How is that possible? A: Remember the case where Z g = Z 0 = Z. For that special case, we found that a conjugate match was the result regardless of the transmission line length.

27 Matching Networks and Transmission ines (contd.) Thus, b matching the source to line impedance Z 0 and likewise matching the load to the line impedance, a conjugate match is assured but the length of the transmission line does not matter! In fact, the tpical problem for microwave engineers is to match a load (e.g., device input impedance) to a standard transmission line impedance (tpicall Z 0 = 50Ω); or to independentl match a source (e.g., device output impedance) to a standard line impedance. A conjugate match is thus obtained b connecting the two with a transmission line of an length! Z g V s l Z 0 Z

28 -Network Analsis et us consider the first matching -network, and denote it as matching network (A): Z jx Y jb Z z 0 Note that this matching network consists of just two lumped elements, which must be purel reactive in other words, a capacitor and an inductor! To make Γ in = 0, the input impedance of the network must be: Zin Z 0

29 -Network Analsis (contd.) Using basic circuit analsis we can find that: Z( z 0) Z in / jb Z Z jx jx / jb Z jbz Note that a matched network, with Z in =Z 0, means that: Re Zin Z0 Zin Im 0 Gives two equations and aids in the determination of two unknowns (X, and B) Essentiall, the -network matching network can be viewed as consisting of two distinct parts, each attempting to satisf a specific requirement.

30 -Network Analsis (contd.) Part : Selecting Y = jb Since the shunt element Y and Z are in parallel, we can combine them into one element Y : Y Y jby Z The impedance of this element is therefore: Z Z Y jb Y Z jbz Z jx Z Z Z jbz z 0

31 -Network Analsis (contd.) To achieve a perfect match, we must set the value of susceptance B such that: Re Z ReZ in Z Z0 Re Z jbz Thus if B is properl selected: Z jx reactive part of matching circuit Z Z jx 0 z 0

32 -Network Analsis (contd.) Part 2: Selecting Z =jx Note that the impedance Z = Z (/jb) has the ideal real value of Z 0. However, it also posses an annoing imaginar part of: X Z Im I Z jbz Z m However, this imaginar component can be easil removed b setting the series element Z =jx to its equal but opposite value! X Z X Im Z jbz It gives Z jx Z Z jx 0 Z Z Z in Z jx Z jx Z in 0 0 z 0 Perfect Match!!!

33 -Network Analsis (contd.) we can solve the preceding equations for the required values X and B to satisf these two equations to create a matched network! B X R / Z R X Z R R X X X Z0 Z0 B R BR Where, Z R jx Note: ) Because of the ±, there are two solutions for B (and thus X) 2) For jb to be purel imaginar (i.e., reactive), B must be real R must be greater than Z 0 (R > Z 0 ) to insure that B and thus X is real. In other words, this matching network (tpe-a) can onl be used when R > Z 0. Notice that this condition means that the normalized load z lies inside the r = circle on the Smith Chart!

34 -Network Analsis (contd.) Now let s consider the second of the two -networks, which we call network (B). Note it is also formed with just two lumped elements. Z jx Y jb Z z 0 To make Γ in = 0, the input admittance of the network must be: Yin Y 0 From circuit theor we can determine that the input admittance for this network is: Yin jb jx Z

35 z 0 Indraprastha Institute of -Network Analsis (contd.) We can describe a matched network, with Y in = Y 0 as: Re Yin Y0 Yin Im 0 Gives two equations and aids in the determination of two unknowns (X, and B) For this design, we set the value of Z =jx such that the admittance Y : Y Z Z jx Z Gives real part as Y 0 Re jx Z Y jb Y Y0 jb

36 -Network Analsis (contd.) It is apparent that a perfect match will occur if the shunt element Y = jb is set to cancel the reactive component of Y : So that: in B ImY Im jx Z Y Y Y jb Y jb Y ( 0 ) 0 Perfect Match!!! Yin Y 0 z 0

37 -Network Analsis (contd.) we can solve the preceding equations for the required values X and B to satisf these two equations to create a matched network! ( Z0 R) / R X R ( Z0 R ) X B Z Where, Z R jx Note: ) Because of the ±, there are two solutions for B (and thus X) 2) For jb and jx to be purel imaginar (i.e., reactive), B and X must be real R must be less than Z 0 (R < Z 0 ) to insure that B and X are real. In other words, this matching network (tpe-b) can onl be used when R < Z 0. Notice that this condition means that the normalized load z lies outside the r = circle on the Smith Chart! 0

38 -Network Analsis (contd.) Smith Chart Regions Depicting the Validit Regions for Networks A and B

39 Tpe Matching Network (contd.) Once the values of X and B are found, we can determine the required values of inductance and/or capacitance C, for the signal frequenc ω 0! and X B If X > 0 0 C If X < 0 0 C If B > 0 0 If B < 0 0 Make sure that ou see and know wh these equations are true As a result, we see that the reactance or susceptance of the elements of our -network will have the proper values for matching at precisel one and onl one frequenc! And this frequenc better be the signal frequenc ω 0!

40 Tpe Matching Network (contd.) If the signal frequenc changes from the design frequenc, the reactance and susceptance of the matching network inductors and capacitors will change As a result the circuit will no longer be matched Therefore the -Tpe matching network has a narrow bandwidth!

41 Tpe Matching Network (contd.) In addition; it becomes ver difficult to build qualit lumped elements with useful values past or 2 GHz. Thus, -network solutions are generall applicable onl in the relativel low RF region (i.e., < 2GHz).

42 Example Indraprastha Institute of The output impedance of a transmitter operating at a frequenc of 2GHz is Z T = (50 + j75)ω. Design an Tpe matching network, as shown below, such that maximum power is delivered to the antenna whose input impedance is Z A = (75 + j5) Ω. = (50 + j75)ω = (75 + j5) Ω. Matching Circuit determine C and

43 Example (contd.) Analtical Approach The condition of the maximum power transfer implies that: Z M * A Z 75 j5 Series connection of and an element (of parallel combination of C and Z T ) ZM Z jb T C jx Z * A BC X C et us write: ZT RT jx T Z A RA jx A RT jx T jb R jx C T T jx R A jx A

44 Example (contd.) Simplification gives: RT RA( BC XT ) ( X A X ) BC RT X R R B ( B X )( X X ) T T A C C T A B C R X R X R T T T T T RA 2 2 RT XT In this example, R T > R A ; therefore the square root is +ve and therefore for positive B C (capacitor) we must choose the plus sign in this expression. Therefore: X RA( BC X T ) B B R C C T X A Insert the given values in the obtained expressions to get: B C B C 9.3mS C 0.73pF X nH X

45 Example (contd.) Smith Chart Based Approach Before we tread this path, let us have a look on Smith chart navigation when series/shunt reactances are added to an impedance (Z) Series Connection of Inductance to a given Impedance Normalized Z

46 Example (contd.) Series Connection of Capacitance to a given Impedance Normalized Z

47 Example (contd.) Shunt Connection of Inductance to a given Impedance Normalized Y

48 Example (contd.) Shunt Connection of Capacitance to a given Impedance Normalized Y

49 Example (contd.) Normalize the transmitter and antenna impedances Since no Z 0 is given, one can choose an. In this example, Z 0 = 75Ω makes simplification easier z Z / Z 2 j ' T T 0 z Z / Z j0.2 ' A A 0 C is in shunt with Z T the movement will be downward on a constant conductance circle starting point will be z T and the end point will be its intersection with z M = j0.2 circle Then the movement will be upward on a constant resistance circle (from z M = (z A )* = j0.2 ) to account for the series inductance

50 Example (contd.) z T = 2 + j Corresponding T = 0.4 j0.2 z M =(z A )* = j0.2 z TC = j.22 Corresponding TC = j0.49

51 Example (contd.) Therefore, the normalized susceptance jb c is: jb 0.4 j0.49 (0.4 j0.2) j0.69 ' ' c TC T Similarl, the normalized reactance jx is: jx z z j0.2 ( j.22) j.02 *' ' A TC Finall, xz nh C b c Z pF