Solutions to Problems in Chapter 5

Transcription

1 Appendix E Solutions to Problems in Chapter 5 E. Problem 5. Properties of the two-port network The network is mismatched at both ports (s 0 and s 0). The network is reciprocal (s s ). The network is symmetrical (s s and s s ) The network is lossless, since its matrix is a unitary matrix. (S T S I). The unitary matrix condition yields S T S ( 5 j j (( 5 Reflection loss and insertion loss 5 ) + ( 0 ) ( 5 j j 5 ) 0 ( 5 ) ) + ( ) ) (E.) ( ) 0 I (E.) 0 The reflection losses at port and port are RL 0 lg 8.3 db and RL 0 lg 8.3 db (E.3) s s The insertion loss is given by IL 0 lg db and IL 0 lg db (E.4) s s Renormalization of scattering matrix from 50 Ω to 00 Ω The set of equations required for the renormalization is given in Section (S S 50 Ω and S new S 00 Ω ).

2 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 S new ( ( (s r) ( rs ) + rs s s r ) ) ( det S s r ) (s r) ( rs ) + rs s (E.5) where det S ( rs ) ( rs ) r s s (E.6) and r,new,new + (E.7) Using the given values we get r 3 Hence, the renormalized matrix is S new ( ) 7 j4 5 j4 7 and det S 00 7 ( 0.8 ) j0.96 j (E.8) (E.9) Validation by circuit simulation (ADS) We will validate our result by using a circuit simulator (ADS from Agilent, Inc.). First, we write our initial scattering parameter in a snp-file (see Section 5.7). snp-files (n number of ports) contain frequency dependent scattering parameters from simulation or measurement in ASCII format. Here, we use the file to define scattering parameters. Figure E.: sp-file for s-parameter definition Figure E. shows the sp-file. Lines with an exclamation mark (!) are comment line. The line with the sharp symbol (#) defines the format of the subsequent data lines: frequency in GHz (GHz), s-parameters (S) are given with magnitude and phase (MA). The port reference impedance (R) is 50 Ω (50). In our problem the s-parameters do not vary with frequency. In order to calculate and plot the parameters with ADS an arbitrary frequency range from GHz to GHz is chosen. The s-parameter data is give in decimal nomination. We use the previously defined sp-file in a circuit simulation. Figure E. shows the schematic. The two-port network is connected to 50 Ω as well as 00 Ω terminals. Figure E.3 shows the simulation results. The upper two plots show the s-parameters normalized to 50 Ω. The absolute values (magnitude) correspond to the given values of 5/ 0.385

3 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 3 Figure E.: Schematic for renormalization of scattering parameters and / The phases are zero degree (positive real value) for the reflection coefficients and 90 degree (positive imaginary value) for the transmission coefficients. The lower two plots show the s-parameters for a reference value of 00 Ω. The absolute values are 0.8 and The faces are 80 degree (negative real values) for the reflection coefficients and 90 degree (positive imaginary values) for the transmission coefficients. The results agree with our calculated values in Equation E.9.

4 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 4 Figure E.3: Simulation results: s-parameters for 50 Ω (upper plots) and s-parameters for,new 00 Ω (lower plots)

5 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 5 E. Problem 5. Antenna input impedance The reflection coefficient r A of an antenna is determined by its input impedance Z A and the port reference impedance. Solving Equation E.0 for Z A yields r A Z A Z A + (E.0) We rewrite the given reflection coefficient r A Z A + r A r A (E.) r A 0.4e j0 0.4 (cos(0 ) j sin(0 )) Re {r A } + j Im {r A } j0.7 Substituting our result in Equation E. yields + r A + Re {r A } + j Im {r A } Z A r A Re {r A } j Im {r A } (E.) (E.) ( + Re {r A } + j Im {r A }) ( Re {r A } + j Im {r A }) ( Re {r A }) + (Im {r A }) (E.4) (Re {r A }) (Im {r A }) + j Im {r A } ( Re {r A }) + (Im {r A }) (0.9 j33.5) Ω (E.5) Reflection coefficient for port reference impedance of,new 75 Ω The reflection coefficient with respect to a port reference impedance of,new 75 Ω is r A,new Z A,new Z A +,new Re {Z A} + j Im {Z A },new Re {Z A } + j Im {Z A } +,new (Re {Z A},new + j Im {Z A }) (Re {Z A } +,new j Im {Z A }) (Re {Z A } +,new ) + (Im {Z A }) (E.6) (E.7) j0.534 (E.8) Hence, magnitude and phase of the reflection coefficient are r A,new (Re {r A,new }) + (Im {r A,new }) 0.4 (E.9) ( ) Im {ra,new } r A,new arctan (E.0) Re {r A,new } Reflected and accepted power The reflected power is The accepted power is P acc P b s P a W (E.) ( s ) P a W (E.)

6 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 6 Validation by circuit simulation We apply a circuit simulator (ADS from Agilent, Inc.) in order to validate our results. First, we define a sp-file for the reflection coefficient with a reference impedance of 50 Ω (see Figure E.4). Once again, we chose an arbitrary frequency range from to GHz for the s- parameter representation. Figure E.4: sp-file with reflection coefficient (port reference impedance 50 Ω) Figure E.5: ADS schematic The schematic of the circuit is shown in Figure E.5. The s-parameter and impedance results are displayed in Figure E.6. There is good agreement between circuit simulation and calculation.

7 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 7 Figure E.6: (a) Magnitude and (b) phase of reflection coefficients for port reference impedances of 50 Ω (red) and 75 Ω (blue). (c) Real and imaginary part of reflection coefficients. (d) Real and imaginary part of input impedance.

8 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 8 E.3 Problem 5.3 Scattering matrix of a two port network with series impedance Figure E.7 shows a circuit with a two-port network consisting of a series impedance Z. Example 5.3 (see book page 7) gives us the procedure to calculate the scattering matrix. First, we need the input impedance. Z in Z + (E.3) The reflection coefficient then becomes s Z in Z in + + Z s (E.4) Since the network is symmetrical we get s s. Z in Two-port network U0 U Z0 Z Figure E.7: Two-port network with a series impedance Z The transmission coefficient s is given as s U U 0 Z0 + Z + + Z s (E.5) The voltage ratio U /U 0 is determined by the voltage divider rule. Due to reciprocity we get s s. Hence, the scattering matrix S is ( ) Z Z0 S (E.6) + Z Z Special cases We evaluate our result by looking at two special cases. First, we consider a short circuit (Z 0), i.e. the two-port network becomes a direct through-connection. The scattering matrix then is ( ) 0 S (E.7) 0

9 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 9 As expected there is no reflection (s ii 0) and full transmission ( s ij ). Next, we consider an open connection (Z ). The scattering matrix now reads ( ) 0 S (E.8) 0 As expected there is full reflection ( s ii ) and no transmission (s ij 0). Scattering matrix of T-network Z in T-network Z U Z 0 U U 0 x Z 3 Z Figure E.8: Two-port circuit with a T-network Figure E.8 shows the circuit. First, we need the input impedance. Z in Z + Z 3 (Z + ) Z + Z 3 (Z + ) Z 3 + Z + After a short calculation the input reflection coefficient is s Z in Z in + Z Z + Z Z 3 + Z Z 3 + Z Z Z0 Z Z + Z Z 3 + Z Z 3 + Z + Z + Z 3 + Z0 By swapping the indices and we get the output reflection coefficient. (E.9) (E.30) (E.3) s Z Z + Z Z 3 + Z Z 3 Z + Z Z 0 Z Z + Z Z 3 + Z Z 3 + Z + Z + Z 3 + Z 0 (E.3) The calculation of the transmission factors involves the voltage ratio U /U 0. With an auxiliary quantity U x we first determine the following voltage ratios. Finally, we get U U x s U U 0 Z + and Z0 U U x Ux U x U 0 Z 3 (Z + ) + Z + Z 3 (Z + ) Z 3 (Z + ) + Z + Z 3 (Z + ) (E.33) U 0 Z + (E.34) Z 3 Z Z + Z Z 3 + Z Z 3 + Z + Z + Z 3 + Z0 s (E.35) Due to reciprocity we find s s.

10 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 0 Special case Z 3 0 In order to validate our formula we consider the case Z 3 0. In this case we expect zero transmission s ij 0 and indeed Equation E.35 is zero. Equation E.3 becomes s Z in Z Z + Z Z Z0 Z in + Z Z + Z + Z + Z0 Z (Z ) + (Z ) Z (Z + ) + (Z + ) (Z + ) (Z ) (Z + ) (Z + ) Z Z + (E.36) (E.37) Since the input impedance is Z in Z our result is correct. Scattering matrix of a π-network Z in -network Z 3 U Z 0 Z U Z U 0 x Figure E.9: Two-port circuit with a π-network Figure E.9 shows the π-network. First, we calculate the input impedance. Z in Z (Z 3 + Z ) (E.38) The input reflection coefficient becomes s Z in Z in + Z Z Z 3 + Z Z 3 Z Z 3 Z Z0 Z Z0 Z 3Z0 Z Z Z 3 + Z Z + Z Z 3 + Z Z 3 + Z Z0 + Z Z0 + Z 3Z0 (E.39) (E.40) In order to calculate the output reflection coefficient we swap indices and. s Z Z Z 3 Z Z 3 + Z Z 3 Z Z 0 Z Z 0 Z 3Z 0 Z Z Z 3 + Z Z + Z Z 3 + Z Z 3 + Z Z 0 + Z Z 0 + Z 3Z 0 (E.4) The calculation of the transmission factors involves the voltage ratio U /U 0. With an auxiliary quantity U x we first determine the following voltage ratios. U U x Z Z 3 + Z and U x U 0 Z (Z 3 + Z ) + Z (Z 3 + Z ) (E.4)

11 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 Finally, we get Z0 s U U Ux U 0 U x U 0 Z Z (Z 3 + Z ) Z 3 + Z + Z (Z 3 + Z ) Z Z Z Z Z 3 + Z Z + Z Z 3 + Z Z 3 + Z Z0 + Z Z0 + Z 3Z0 (E.43) (E.44) s (E.45) Due to reciprocity we find s s. Special case Z 0 Let us consider the case Z 0 (short circuit). Now, the input reflection coefficient in Equation E.40 becomes s Z Z 3 Z Z 0 Z 3Z 0 Z Z 3 + Z Z 0 + Z 3Z 0 (E.46) This represents the short circuit (r ) at port. The output reflection coefficient in Equation E.4 is now s Z Z 3 Z Z 3 Z Z 3 (Z + Z 3 ) Z Z 3 + Z + Z 3 Z Z 3 + (Z + Z 3 ) Z Z 3 Z + Z 3 Z Z 3 Z Z 3 Z + Z Z Z + Z 3 (E.47) (E.48) This result is correct since the output impedance is Z Z 3. Finally, Equation E.45 gives us zero transmission which is correct. E.4 Problem 5.4 Figure E.0 shows us the signal flow graph of the circuit in Figure 5.5 (book page 85). a e j s x, e j b e j s x, s x, e j b s x, a Figure E.0: Signal flow graph

12 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 The signal flow graph gives us the following scattering parameters. s b a s x, e jβ(l +l ) s x, e jβl (E.49) a 0 s b a s x, e jβ(l +l ) s x, e jβl (E.50) a 0 s b s x, e jβ(l +l ) (E.5) s b a a 0 a a 0 s x, e jβ(l +l ) (E.5) The transmission lines only modify the phase of the scattering parameters. If the lengths of the lines are known the effect can be compensated. E.5 Problem 5.5 We will derive the equation for the wave source in Figure 5.4 (book page 77). We repeat the representation in Figure E. for convenience. Voltage source Signal flow graph Equations U 0 b a b0 b a r b b r a 0 U Z b0 Z Z Z Z r Z Z Figure E.: Signal flow graph of a voltage source Reflection coefficient r The reflection coefficient is determined by the input impedance of the circuit. The ideal voltage source represents a short circuit. Therefore, the input impedance is and the reflection coefficient is r + (E.53) Wave source The circuit in Figure E. includes an ideal voltage source and is therefore an active circuit. Consequently, there is an outgoing wave b even though there is no incoming wave a. This is expressed by the term b 0 in the following equation. b b 0 + r a (E.54)

13 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 I U 0 U b a Z 0 Figure E.: Voltage, current and power waves In order to calculate the term b 0, we consider the relation between voltage U and current I and power waves a and b (see Figure E.). The relation between power waves a and b and voltage U and current I is given as b U I and a U + I (E.55) The voltage is determined by the voltage divider rule. The current is given by Ohm s law. U + U 0 (E.56) Therefore, we get and I U 0 + (E.57) a U + I U 0 + ( U 0 ) 0 (E.58) ( + ) b U I U 0 ( U 0 ) ( + ) The term b 0 represents the active part in the signal flow graph. E.6 Problem 5.6 Z0 U0 + b 0 (E.59) Figure E. shows a network with two two-port networks: the first network has a forward gain G f and the feedback network has a gain of G r. From Figure E. we derive the following two relations. This gives us X X + G r Y and Y G f X (E.60) Y G f X (E.6) G cl where G cl G f G r is the closed loop gain. If we apply the result to the signal flow graph in Figure 5. (book page 75), we get s x b a (E.6) s x s y

14 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 4 X X ' G f Y G r Figure E.: Network with feedback E.7 Problem 5.7 We will calculate reflection coefficients and impedances of the circuit in Figure E.4. We assume the following parameters. f GHz The input impedance Z in is (E.63) 50 Ω (Port reference impedance) (E.64) Z A 00 Ω + jωl 00 Ω + j00 Ω where L 5.9 nh (E.65) Z where C 5 pf (E.66) jωc Z in The reflection coefficient is jωc ( jωc ) + (ωc ) (4.4 j.65) Ω (E.67) r Z in Z in + Re {Z in} + Im {Z in } Z 0 + j Im {Z in} (Re {Z in } + ) + Im {Z in } (E.68) e 8.5 (E.69) Zin r Z in r U 0 Z, ZA Figure E.4: Network In order to determine the reflection coefficient r we first consider the reflection coefficient r at the end of the line. r Z A Z A + Re {Z A} + Im {Z A } Z 0 + j Im {Z A} (Re {Z A } + ) + Im {Z A } 0.6 e j9.75 (E.70) (E.7)

15 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 5 At the input of the transmission line we get r r e jβl 0.6 e j9.75 e j e j8.4 r (E.7) If we compare the reflection coefficients r and r we see that r r (complex conjugate matching). Using a Smith-Chart gives us the input impedance Z in (see Figure E.5). We apply a commercial circuit simulator (ADS by Agilent, inc.). The electrical line length is 0.363λ The load impedance is represented by a black diamond the input impedance is depicted by a red square. We read from the diagram Z in (4.8 + j.56) Ω Z in (E.73) Impedances Z in and Z in as well as reflection coefficients r and r show complex conjugate values. Figure E.5: Smith chart and input impedance Z in Circuit simulation We will validate our result by performing a circuit simulation. In our previous calculation we considered a frequeny of f GHz. Now we extend the frequency range from 00 MHz to GHz. Figure E.6 shows the schematic. The results are given in Figure E.7. The upper diagrams in Figure E.7 show reflection coefficients (magnitude and phase) as well as input impedances (real and imaginary part). In order to compare the results with our simulation we have to look at a frequency of f GHz (indicated by markers). The simulation results agree with our previously calculated values.

16 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 6 Figure E.6: Schematic Figure E.7: Simulation results: s-parameters and impedances

17 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 7 E.8 Problem 5.8 A three-port network shall have the following three properties: matching at all ports (s ii 0), reciprocity (s ij s ji ) and no losses (S T S I). From these requirements the matrix equation 0 s s 0 s s 0 0 s 0 s 3 s 0 s s s 3 0 s s gives us the following set of equations (E.74) s + s (E.75) s s 3 0 (E.76) s s 3 0 (E.77) s 3 s 0 (E.78) s + s 3 (E.79) s s 0 (E.80) s 3 s 0 (E.8) s s 0 (E.8) s + s 3 (E.83) From Equations E.76 and E.77 we get (for s 3 0): s s (E.84) From Equation E.80 we then conclude: s s 0 (E.85) This result contradicts Equation E.75 since s + s 0 (Conflict for s 3 0) (E.86) The conflict can be solved for s 3 0. This implies (Equation E.79) and (Equation E.83) This result contradicts Equation E.75 since s (E.87) s (E.88) s + s (Conflict for s 3 0) (E.89) The set of equations shows that a three-port network cannot satisfy the following conditions simultaneously

18 APPENDIX E. SOLUTIONS TO PROBLEMS IN CHAPTER 5 8 matching at all ports (s ii 0), reciprocity (s ij s ji ) and losslessness. (Last modified: 0.0.0)