1 Module 17 Stub Matching Technique in Transmission Lines 1. Introduction 2. Concept of matching stub 3. Mathematical Basis for Single shunt stub matching 4.Designing of single stub using Smith chart 5. Series single stub 6. Concept of Double stub matching 7. Mathematical Basis for Double Stub Matching 8. Design of Double stub using Smith Chart: 9. Summary Objectives After completing this module, you will be able to 1. Understand the concept of matching stub and different types of it. 2. Understand the mathematical principle behind the stub matching technique. 3. Design single stub and double stub using Smith chart. 4. Know about advantages and limitations of stub matching technique.
2 1. Introduction In the earlier modules, we have discussed about how matching networks and QWT can be used for matching the load impedance with the lossless transmission line. Both these techniques suffer from some drawbacks.especially the losses like, insertion loss, mismatch loss are of much more prominent losses. There is another technique known as Stub Matching Technique to match the load impedance with the Characteristic impedance of the given transmission line so as to have zero reflection coefficient. The technique is marvelous and the designing can be done using Smith chart in a very simple way. Let us discuss it with adequate details rather qualitatively avoiding mathematical rigor in it. 2. Concept of Matching Stub A section of a transmission line having small length is called as a stub. Two types namely single stub and double stub comprising of one or two stubs are in common use. A section of a two wire transmission line of small length l connected in parallel to the line to be matched at a distance d from the load as shown is known as a single stub. The stub may or may not have its characteristic impedance the same as that of the given transmission line. Generally the stub is chosen so as to have the same characteristic impedance as that of the given transmission line.
Similarly the stub may be short circuited or open circuited at its open end. However, short circuited stub is preferred because practically it is easier to achieve a reliable perfect sho rt circuit rather than a perfect open circuit. Thus, a short circuited stub of proper length l connected at proper distance d serves the purpose of a perfect match. Determining precise values of I and d are the main computations involved in stub matching technique. The values of distance d and the length l of the stub depend upon whether the stub is desired to be inductive or capacitive. Different types of arrangements of stubs such as Single shunt stub,single series stub,double stub, Triple stub etc. can be found to be documented in literature. 3. Mathematical Basis for Single shunt stub matching A simple concept of impedances in parallel is used to derive the necessary equation needed to compute the value of normalized stub impedance ys. Once the normalized stub impedance ys is known, the values of d and l can be calculated easily using Smith Chart. So, let us see how to derive the equation needed to compute the value of normalized stub impedance ys. Consider the circuit with a loss less transmission line of characteristic impedance Z0, terminated in a load with impedance ZL as shown. Let Vg and Zg be the voltage and impedance of the generator respectively. Assume that the line is matched with the generator.hence we have, Zg = Z0. A single stub of length l and characteristic impedance Z0s is connected across the points P and P in parallel to the transmission line at a distance d from the load. Let Zd be the impedance of the line at a distance d from the load and Zs - the impedance of the stub. 3 Now for matching of the impedances at the connection of the stub with the line, the impedance between P and P seen from both sides must be equal to each other. It can be easily understood that the impedance between P and P seen from the load towards the generator is equal to the characteristic impedance Z0 of the transmission line. Let Zpp, be the impedance across the points P and P seen from the generator side towards the load. Let us determine Zpp.
At the point P there is a division of signal. A part of the signal enters the stub and the remaining signal passes towards the load. Thus the two branches namely the stub and the line of length d between the stub and load are in parallel. It is easy to understand that, when seen from generator side, Zpp equals the impedance of the parallel combination of Zd and Zs as shown in diagram. Thus we have, Zpp = Zd // Zs = (Zd x Zs) / (Zd + Zs) Hence, 1/ Zpp = Ypp = (Zd + Zs)/ (Zd x Zs) = (1 / Zs) + (1 / Zd) = Ys +Yd Where Ys,Yd and Ypp represent the corresponding admittances. If we divide by Y0, we obtain the relation in normalized form as follows. Ypp / Y0 = (Ys / Y0) + (Yd/ Y0) ypp = (yos ys)/y0 + yd Note the notations used in the derivation. i) Y0 represents the characteristic admittance of the transmission line. ii) Ys = yos ys is the admittance, Y0s is the characteristic admittance and ys is the normalized admittance of the single stub. iii) Yd = yd y0 is the admittance of the line and yd is normalized admittance of the line at a distance d from the load. Thus we have two results for admittance: i) Admittance across P and P = Ypp = Ys +Yd as seen from generator towards load ii) Admittance across P and P = Y0 = 1/ Z0 as seen from the load towards generator. For satisfying the condition of matching, the two impedances must be equal. Thus we have, Ypp = Ys +Yd = Y0 ------- (1) Dividing both sides of this equation by Y0, we obtain the equation in normalized form as shown. -------------- (2) 4
5 + = 1 ------ (3) Equation (2) can be rearranged to obtain expression for as follows. = (1- ) ---------- (4) Equation (4) plays an important role in determining stub position and its distance from the load using Smith chart. If the characteristic impedance of the stub is equal to the characteristic impedance of the line, that is if Yo = Yos, then equation (4) reduces to = (1- ) ----------- (4 ) This much background is sufficient for designing a single stub. 4. Designing of single stub using Smith chart Let us try to understand how the two quantities namely stub length and its distance d from the load can be determined using Smith chart. Instead of discussing theoretically, let us discuss with the help of a numerical example. The statement of the problem is as shown. Problem: - A lossless transmission line of characteristic impedance Z0 =50 Ω is terminated in a load of impedance ZL=100 + j50 Ω. Determine the smallest length and the distance d from the load of a single capacitive stub having characteristic impedance Zos = 75 Ω shorted at its ends, which when connected, will serve the purpose of a perfect match of load with the transmission line. The step wise procedure for the solution of this problem is as follows. Step (i) : Compute and plot the normalized load admittance = 1/ on smith chart.it be remembered that while using Smith chart for calculations involving impedances, constant r circles become constant g circles while, constant x circles become constant s circles.let point P represent. Here, = ZL/Z0 = 2+j, =1/ = 0.4-0.2 j. Thus point P has coordinates g= 0.4, s = -0.2j. We may describe the same fact as P (0.4,-0.2j).
6 Step (ii): Draw a circle of radius OP with O as its center. As we know the circle drawn in this way is known as the VSWR circle. Note the points Q and Q in which the circle drawn intersects the r=0 circle now called as g= 0 of the Smith chart. Note the coordinates ( g, s ) of point Q and Q. In this problem we have Q ( 1, - j ) and Q (1, j ).The point Q and Q represent the value of yd in equation (4). If we want to have inductive stub, the point Q is to be used for yd while for capacitive stub, yd is to be taken from point Q. As per the requirement in the problem, we need a capacitive stub. Hence we will select point Q. Accordingly, we have yd = 1 - j. Step (iii) : Calculate the value of normalized admittance using equation (4) and plot on the smith chart. Let S represent = (1- ) ---------- (4). Note that point S will always lie on the r= 0 i.e. g = 0 circle. = (1- (1-j)) = 1.5 j Step (iv): Join the center O with points P, Q and S.Extend OP, OQ and OS to meet the radial distance scale marked with Towards Generator to meet in points K,M and N respectively. and note the readings k, m and n respectively. In the problem under consideration, we have k= 0.464, m = 0.338, and n = 0.157λ. Step (v): For determination of distance of the stub from the load proceed as follows.
Find the radial distance in terms of wavelength between point K and point M measured along the scale Towards the Generator. This distance gives the value of d. In this problem we have d = KBM =0.5-(k m) = 0.5-(0.464-0.338) λ = 0.374 λ. 7 Step (vi): For determination of length of the stub proceed as follows. Starting from point B on the Smith chart, move along the scale Towards Generator to reach the point N. Calculate the radial distance travelled in terms of wavelength from B to S. This distance gives the value of stub length l in terms of wave length. In the problem under consideration we have, Thus finally, we have the required values as shown. i) Distance of the stub from the load (d) = 0.374 λ ii) Length of the short circuited capacitive stub ( ) = 0.407 λ One should note the following points. 1) If we need to have capacitive stub, the point Q is to be used to determine yd. 2) For having inductive stub, the point Q is to be used to determine yd. 3) In the case of short circuited stub, length is to be measured from point B up to S along the scale Towards the generator. 4) In the case of open circuited stub, length is to be measured from point A up to S along the scale Towards the generator. verify the following results. there might be a little deviation from the expected values. Sr. no. Stub Type Z0 Ω ZL Ω d 1 Short circuited 80 120+80 j 0.368λ 0.399λ Open circuited 80 120+80 j 0.168λ 0.399λ 2 Short circuited 50 75+100 j 0.415λ 0.371 λ Open circuited 50 75+100 j 0.165λ 0.371 λ Single stub matching technique, gives two solutions for the stub length. Out of these two, the solution giving smaller stub length is preferred over the other. It is because it results in smaller
quality factor Q which in turn provides larger matching bandwidth. It means that the same stub arrangement can be used over a larger frequency range. Due to smaller stub length, the inevitable loss along the line is reduced. In addition, it is possible to have a smaller and economically feasible matching network. Though simple and useful single stub approach suffers from few drawbacks. For example, with single stub, two parameters namely the distance d from the load and the stub length are to be cared for simultaneously. In addition change of signal frequency can affect one of the two solutions notably. In such a case it needs to redesign the stub. Due to this and many other reasons, a system of stubs is used depending upon the need and type of application. 5. Series single stub In our discussion made so far, a single stub was connected in parallel to the main transmission line. Instead, we can have the stub connected in series with the main transmission line as shown. Generally, the characteristic impedance of the stub is taken to have the same value as that of the main transmission line. 8 DIAGRAM 1:- Single series open stub The stub can be open circuited or short circuited. Accordingly, the design parameters change As an example, see the problem as stated here. Statement of the problem: - Design an open circuit series single stub to match a load with impedance ZL=90 +60 j so as to provide a match with transmission line having characteristic impedance Z0= 75.Assume the characteristic impedance of the stub to have the same value as that of the main transmission line.
Analytical solution can be obtained using the formulae documented in literature. However the calculations are much involved and tedious. Smith chart gives quick and sufficiently accurate solutions. Using Smith chart, we obtain the two solutions for open circuit series single stub under consideration as shown in the following table. Table: - Open circuit Series single stub Given: Z L =90 + 60 j, Z 0 = 75. 9 Solution No Stub length and distance d from load 1 d = 2 d = Designers and fabricators take care about slights refinement, if necessary, of the values obtained from Smith chart. In series single stub tuning the given transmission line is to be cut at some point which can be an undesirable point. The design may not work satisfactorily for Variable load. Due to these and some other reasons, this type of stub though simple to design, is rarely used in practice. 6. Concept of Double stub matching Now let us discuss the Double stub matching which is one of the most commonly used techniques. Generally, in double stub matching, characteristic impedance of the two stubs is taken the same as that of the main lossless transmission line. The two short circuited stubs separated by a suitably chosen fixed distance d 2 are connected in parallel to the main line. The values of distance d 2 are generally chosen as 1 /8, 3 /8, 5 /8 etc as per the convenience.the distance of the first stub d 1 from the load is also held fixed and is kept adjustable if needed to have proper tuning.thus, here knowing the distances d1, d2, the load impedance Z L and characteristic impedance Z 0 ; it is expected to determine the lengths and of the two stubs.
10 7. Mathematical Basis for Double Stub Matching Since, we are using the two stubs in parallel to the main transmission line; the mathematical analysis is done using admittances rather than using impedances. The following two equations form the basis for the mathematical formulation of double stub matching. The capital letter Y is used to represent actual admittance while small letter y is used to represent normalized admittance. =,. (1) (At junction p1p1 ) = = (2) (At junction p2p2 ) Note that, the equations involve admittances since the stubs are in parallel with the transmission line. One can easily understand the symbols of the admittances in these two equations.
Now, since, the characteristic impedance of the stubs is the same as that of the main transmission line, the two equations in normalized form can be written as shown. =,. (1 ) (at junction p1p1 ) = = 1 (2 ) (at junction p2p2 ) It be noted that the stubs may be open circuited or short circuited. The analytical treatment for determining the values of stub lengths is tedious and somewhat complicated. The computer simulation software developed by some researchers for designing double stub can be found to be documented in literature. 8. Design of Double stub using Smith Chart: Instead of dealing with analytical treatment, let us try to understand the step wise procedure which will enable us to solve a simple problem related with double stub matching using Smith chart. As an illustration consider a problem which can be stated as shown. Problem: - consider a transmission line with characteristic impedance Z 0 = 50 Ω terminated in a load impedance ZL = 120 +105 j Ω.Design a short circuited double stub with following specifications. i) Distance of the nearest from the load = d 1 =u λ. ii) Separation between the two stubs = d 2 = t λ, iii) characteristic impedance of the both the stubs = Z 0 Given: u= 0.380 λ, t = 0,375, Z 0 = 50 In literature, the procedure for solving such problems can be found to be described in a very brief, precise and technical manner. In comparison, the stepwise solution given here may appear somewhat lengthy. However, for a beginner it can serve the purpose as a systematic approach for understanding various steps in obtaining the solution avoiding any serious mistakes. Solution on Smith chart is as shown. 11
12 Step (i): Calculate and plot the normalized load impedance zl = Z L / Z 0 Point P [ 2.4+2.1j ] on the Smith chart. step (ii): Draw a circle with O as its center and OP as radius. As is known to us, this circle is known as VSWR circle. Extend PO towards O to meet the VSWR circle drawn at point Q. The point Q represents the normalized impedance yl = 1 / zl 0.24-0.2j Step (iii): Calculate the angle θ (= 720 x t 0 = 720 0.375=270 0 ) in degree.rotate the radius OB of r = 1 circle about O through an angle θ in anticlockwise direction and obtain the point B1 on r = 0 circle. Obtain the midpoint O of the line OB1. Draw a circle with O as the center and OO as its radius. This circle is known as Spacing circle or auxiliary circle in literature. Step (iv): Extend the line OQ towards Q to meet the radial distance scale at point T on the scale marked with Towards Generator. Note the reading at point T ( =0.468 ). Let us adopt the convention of representing the measurements at any point by the letter representing the point itself. Accordingly, the reading at point T will be T. step (v): From point T move a distance uλ = (0.38 ) along the distance scale marked with Towards Generator in the clockwise direction to reach the point T. Note the reading (T ) at point T. If (T+u) exceeds 0.5, then calculate T by subtracting the integral multiples of 0.5 from
13 the sum (T+u) until T <0.5.That is T = (T + u-m) λ, where m is a positive integer. In problem ( T+u = 0.468+0.380 = 0.848 =0.348) this Step (vi): Join points O and T. Note the point V of intersection of the line OT with the VSWR circle, i.e. the circle with OP as the radius. The point V represents the value of y d1.it be noted that yd1 has only one value while, the other normalized admittances and hence the stub lengths will have two possible values. (V=0.6-1.22j) Step (vii): Note the real part say Vr of V. Find the points F and F of intersection of the r =Vr circle with the spacing circle. Then F=y11 and F = y11. [ Vr = 0.6, F = 0.6-0.1 j, F = 0.6-1.9 j ] Step (viii): calculate and represent the difference (F V) by point H on the r = 0 circle. Similarly, calculate and represent the difference (F V) by point H on the r = 0 circle. The points H and H are purely Imaginary and represent the two values of normalized admittance of the first stub namely ys1 and ys1 respectively. [ H = 1.12 j, H = - 0.68 j ] Step (ix): Extend the lines OH and OH to meet the distance scale marked Towards Load at points K and K respectively.note the readings at points K and K on the scale [ k = 0.134, k = 0.44 ] Step (x) : Since we are dealing with short circuited stub, for determination of stub lengths, the measurements are to be made wrt the point B of the Smith Chart. Move from point K along the distance scale Towards the Load in the anticlockwise direction and measure the distance up to the point B. This distance gives 1 which is the length of the first stub. [ 1 = 0.25 + 0.134 = 0.384 ] Step (xi): Move from point K along the distance scale Towards Load in the anticlockwise direction and measure the distance up to the point B. This distance gives 1 which is the second solution for the length of the first stub. [0.44-0.25]=0.19 Step (xii): Now, draw a circle with O as its center and OF as the radius. Note the point D of intersection of this circle with r = 1 circle.the point D represents the value of the normalized admittance y d2 of the second stub.[d = yd2=1-0.55j). Note that the angle between OF and OD measured from OF in clockwise sense must be θ. [ θ = 270 0 ] Step (xii): Again, draw another circle with O as its center and OF as the radius. Note the point D of intersection of this circle with r=1 circle.the point D represents the second solution for the value of the normalized admittance y d2 of the second stub. [D = yd2 = 1+ 2.4 j ] Note that, here also the angle between OF and OD measured from OF in clockwise sense must be θ. { θ = 270 0 }
Step (xiii): Calculate S = 1- D and S = 1-D. Plot points S and S on the smith chart which will obviously lie on the r = 0 circle. [S = +0.55j, S = -2.4j] Step (xiv): Extend the lines OS and OS to meet the distance scale marked with Towards Load at points E and E respectively. [ E = 0.08λ, E = 0.312 λ ] Step (xv): Move from point E along the distance scale in the direction Towards Load to reach the point B. The distance measured in this way from E to B gives the length 2 of the second stub. [ 2 = 0.25+0.08 = 0.33 ] Step (xv): Move from point E along the distance scale in the direction Towards Load to reach the point B1.The distance measured in this way from E to B1 gives the second solution for the length 2 of the second stub.[ 2 = 0.312-0.25 = 0.062 ] Thus finally we can record the two solutions in a tabular form as shown. 14 Solution No Length of first stub Length of second stub 1 1 = 0.384 2= 0.33 2 1 =0.19 2 = 0.062 We know that due to impedance matching, the reflected voltage (or current) along the given transmission line is eliminated.maximum power transfer from the source to the load is also achieved. The VSWR becomes unity implying the fact that there are no voltage peaks on the transmission line. Due to impedance matching, for any length of transmission line the input impedance equals the characteristic impedance, In other words, impedance matching makes the input impedance independent of the length of transmission line. It be noted that, though after impedance matching is no reflected voltage (or current) along the given transmission line, there are reflections on the stub and also along the small part of the transmission line of length d between the stub and load. Matching devices known as stub tuners which are basically impedance transformers designed to introduce a variable shunt susceptance in to coaxial transformation line, find various applications.power and attenuation measurements, tuned reflectometric measurements, providing a dc return in single ended mixers and detectors etc are some of such applications. Friends, different types of matching techniques have been discussed by researchers and
designers working in this field. The introductory aspects discussed in this module can be useful for understanding the contents therein. 9. Summary 15 Stub matching technique is practically easy to design and implement. The mismatch loss arising in matching networks is reduced in stub matching technique. Double stub matching has advantage of dealing with stub lengths only. Single stub matching involves with stub length and location of stub as the parameters With smaller stub length, the inevitable loss along the line is reduced.