INF5481 RF Circuit, Theory and Design Assignment #3

Transcription

1 INF548 RF Circuit, Theory and Design ssignment #3 Problem We have the two port network in figure, which is a cascade of two identical networks consisting of a single shunt and series impedance. Find the Z and Y representation of the entire circuit. BCD Z B BCD Z B Z Z BCD series BCD shunt Figure : Diagram for problem Theory We now define the BCD, Z and Y matrix representations for a two port network. BCD ( V I ) B V2 C D I 2 V I2 C I I2 B V I 2 D I I 2 V2 V2 Two useful BCD matrices to have in out toolbox is the series and shunt impedance. ()

2 INF548: ssignment #3 Series impedance We first set I 2 and determine and C. s there is no current through the impedance V and, since I I 2, C. Setting we use ohms law to express the current I I 2 V Z. Inserting this into the expressions for B and D we obtain B Z and D. Z BCD series (2) Shunt impedance We again set I 2, noting that V, has to be. From ohms law we express I V Z Z, which substituted into the expression for C becomes Z. Setting we have B since V and D since I I 2. BCD shunt Z (3) Z ( V ) Z Z 2 I Z 2 Z 22 I 2 (4) To go from BCD to Z we can use table 4-2 in [] Z D BC C D (5) Y The Y matrix can be found as the inverse of (4) I Z Z 2 V I 2 Z 2 Z 22 I Y Y 2 V I 2 Y 2 Y 22 To go from BCD to Y we again use table 4-2 in [] Y D (D BC) B (6) (7) (8) Page 2 of 5

3 INF548: ssignment #3 Solution We first find BCD as a cascading of a shunt (3) and series impedance (2) BCD BCD shunt BCD series ZB Z ZB Z Z B Z Before we proceed we find that the determinant of this matrix is det(bcd ) Z B Z Z B Z Z BZ Z BZ. Matrices with a determinant of ± is particularly kind beasts and are called unimodular [2], if you take the product of two unimodular matrices you get another unimodular matrix and the inverse of a unimodular matrix is also unimodular! We now find the full BCD representation as a cascade of two BCD networks BCD BCD BCD ZB ZB Z Z B Z Z Z B Z ( Z B Z Z B Z B (Z B Z ) ) Z Z (Z BZ ) Z BZ (Z BZ )2 Z ZB (2 Z) Z (2 Z) Z (Z )2 where Z Z B Z We first find the determinant before using (5) and (8) to find the Z and Y representation respectively. Since it is unimodular we know that det(bcd) Z D BC C D Z Z (2 Z) Z (Z ) 2 Y D (D BC) B Z B (2 Z) ( Z (Z ) 2 ) Z Page 3 of 5

4 INF548: ssignment #3 Problem 2 We have the circuit in figure 2, where we want to find the voltage over the load impedance. Using an BCD representation for the generator load, transformer and the transmission line; find V L. Z G 5 Ω V G 3 V : 2 Z 5 Ω βl 9 V L Z L 25 Ω Figure 2: Diagram for problem 2 Theory We have one new circuit element, namely the transformer. BCD representation of an ideal transformer ssuming an ideal transformer we have V I 2 I N N 2. where the minus comes from the way we define the current (you missed this in the problem solving class). From the above it follows that V N 2 N, V I I 2. Setting it is clear from the above that V must be so B, finding an expression for the currents we get which means that D N 2 N. V I I 2 I I 2 V I 2 N 2 N Page 4 of 5

5 INF548: ssignment #3 Setting I 2, I so C and V I2 N N 2. Defining N N N 2 we now have the BCD representation N BCD transformer N (9) Solution The BCD representation can be found as a cascade of a series impedance Z G, a transformer and a transmission line. We developed the representation for series impedance in section, the transformer is expressed in (??) and we find the transmission line from []. BCD BCD ZG BCD transformer BCD Z ZG N cos(βl) jz sin(βl) N jz sin(βl) cos(βl) N ZG N cos(π/2) jz sin(π/2) N jz sin(π/2) cos(π/2) N ZG N jz N jz ( ZG N jz NjZ ZG ) N jz j NZ NZ NZ We now have VG j I G ( ZG ) NZ NZ VL NZ I L Z G V G jv L j V L NZ NZ Z L jv L ( ZG NZ NZ Z L ) V L jv G ( ZG NZ NZ j 3 V Z L ) ( 5 Ω /2 5 Ω /2 5 Ω 25 Ω j3 V (2 ) j V ) Page 5 of 5

6 INF548: ssignment #3 we also have a second solution V G Z G j V L NZ V L j V GNZ V L j Z G 3 V /2 5 Ω 5 Ω j3/2 V Correction: Sage gets to two solutions (see src/oppg2.sage) { j3v V L jv and DS obtains V 9 Problem 3 Express the S parameters for a series resistance R Ω, assume Z 5 Ω. The setup for setting a 2 is shown in figure 3. Z G Z Z R Ω Z V G Z L Z S Figure 3: Diagram for problem 3. Setup for finding the S-parameters for a Ω resistor. Page 6 of 5

7 INF548: ssignment #3 Z G Z Z R Ω Z V G Z L Z Z G Z Z S Z in R Ω V G Z in S Figure 4: Replacing transmision line with equivalent input impedance, where Z in Z Theory We start by defining the S matrix representation (assuming line impedance is the same on both sides) b S S 2 a () b 2 S 2 S 22 a 2 S b S 2 b a a2 S 2 b 2 a a2 a 2 a S 22 b 2 a 2 a () where a i and b i are normalized power waves and can be expressed in a variety of ways a i 2 Z (V i Z I i ) V i Z I i Z b i 2 Z (V i Z I i ) V i Z I i Z Page 7 of 5

8 INF548: ssignment #3 We now recognize the diagonal terms S and S 22 as S b a2 a S 22 b 2 a a 2 V Z V Z V 2 Z V 2 Z V V Γ, Γ 2. nd the off diagonal terms S 2 b a a 2 S 2 b 2 a2 a V Z V 2 Z V 2 Z V Z V V,. We now have the tools we need to find S R, to account for the transmission lines we remember that for a lossless transmission line V in (d l) V e jβl V e jβl, where this can be extended to the output port as well We can split this and write V out (d l) V 2 e jβl V 2 e jβl. V in (l) V e jβl V in (l) V e jβl V out(l) e jβl Vout(l) V2 e jβl. Gathering these equations in two matrices we get ( V in (l) ) e jβl V V out(l) e jβl ( V in (l) ) e jβl V Vout(l) e jβl V2 (2) (3) Page 8 of 5

9 INF548: ssignment #3 We now return to our definition of the S parameters (see (??)) and express it in terms of voltages b S S 2 a b 2 S 2 S 22 a 2 ( V ) V Z S S 2 Z V 2 S 2 S V 22 2 ( Z ) ( Z ) V S S V2 2 V S 2 S 22 (4) What we want to express is ( V in (l) ) ( V Vout(l) in S (l) ) shift V out(l) inserting (??) and (??) and finally using (??) ( e jβl V e jβl e jβl V2 S shift e jβl ( e jβl S S 2 V e jβl e jβl S 2 S 22 S shift e jβl e jβl S S 2 e jβl e jβl S 2 S 22 e jβl S shift e jβl S S 2 e jβl e jβl S 2 S 22 e jβl S shift. ) ( V ) ( V V 2 ) ) Page 9 of 5

10 INF548: ssignment #3 Solution We start by finding the S R parameters for the series resistor, we could do this by converting our previous BCD series representation but we will instead use the definitions given above. By proper termination we can set a 2, see dashed wires in figure 3. We then see an input impedance of Z in R Z L Ω 5 Ω 5 Ω we are now ready to find the reflection coefficient S Z in Z Z in Z Ω 2 Ω 2 We find S 2 by inserting our definition of b 2 and a, the rest is ohms law. S 2 b 2 lternativly a a2 V V RI V V R V RZ Z L V R R Z Z L Ω S 2 2 V G Ω 5 Ω 5 Ω 2 Z L Z V G V G Z G R Z L 2Z R Z S 2 2 2Z R Page of 5

11 INF548: ssignment # Our circuit is symmetric so S R ( ) Problem 4 Repeat the previous problem, with the finite transmission lines connected to a series inductor Z L j Ω as shown in figure 5. Z 5 Ω βl 45 Z L j Ω L.59 nh Z 5 Ω βl 45 S Figure 5: Diagram for problem 4 The frequency is Z L jωl f Ω.59 nh 2π GHz Page of 5

12 INF548: ssignment #3 Theory Solution We start with the input impedance and reflection coefficient Z in Z L Z load j Ω 5 Ω S Z in Z Z in Z Z L Z load Z Z L Z load Z j Ω j Ω 2 5 Ω j Ω( Ω 2 5 Ω) Ω Ω j( Ω Ω) Ω Ω j(9 Ω) Ω Ω j. S 2 can be found the same way as in exercise S 2 b 2 a a2 V V Z L I V V V Z L Z L Z Z load V Z L Z L Z Z load j Ω j Ω 5 Ω 5 Ω j j j j 2 Page 2 of 5

13 INF548: ssignment #3 j j j. gain, the circuit is symmetric so when we switch the Z load around the answer will be the same j. j. S j. j. To get the final S parameters we need to consider the two transmission lines, where 2βl 2 45 π/2 gives e j2βl j e jβl j. j. e jβl S e jβl j. j. e jβl (j.)e j2βl ( j.)e j2βl ( j.)e j2βl (.j)e 2jβl. j. j j.. Page 3 of 5

14 INF548: ssignment #3 Problem 5 Measuring on a short GCPW transmission line you obain the following S parameters at 2.5 GHz. The S-parameters are normalized to Z 5 Ω j j S GCPW j.47.82j (a) Using the fact that S Γ, calculate the VSWR and the input impedance of the device. Using S 2, calculate the insertion loss in db. Γ.53, VSWR., Z L 49.5 Ω, IL.34 db (b) Since the line is relativly well matched to 5 Ω, S 2 e γl. Ignoring the low loss, calculate βl as the (negative) phase of S 2. Can the length l be uniquely determined? Ignoring missmatch we have S 2 e γl e (αjβ)l ignoring loss as well (α ), this reduces to e jβl The angle of S 2 can be found by the tangens function, as long as we correct the sign βl tan (.964/.294) 88.3 To find the length l, we would need the phase velocity, which we lack, so: No, need to extract ε eff. (c) You have designed our own nifty gadget and is convinced it will save the world. Before proceeding to fabrication, you would like to asses the performance when using the GCPW transmission lines to connect to the outside world. You have the S parameters of your gadget (S g ), describe how you would find the total S parameters as depicted in figure 6. Grounded CoPlanar Waveguide Page 4 of 5

15 INF548: ssignment #3 Your nifty gadget goes here? S GCPW S g S GCPW Figure 6: Cascading S parameters. Mandatory for INF948 students: Insert your own circuit and carry out the required calculations. Note: it does not have to save the world, nor do you have to do it by hand. References [] R. Ludwig and G. Bogdanov, RF circuit design: theory and applications, 2nd ed. N.J.:Prentice-Hall, 29. [2] E. W. Weisstein. unimodular matrix. from mathworld a wolfram web resource. [Online]. vailable: UnimodularMatrix.html Page 5 of 5