Network Analysis and Synthesis

1. Network Analysis and Synthesis Chapter 4 Synthesis of deriving point functions (one port networks)

2. Elementary Synthesis procedures • The basic philosophy behind the synthesis of driving-point functions is to break up a positive real (p.r.) function Z(s) into a sum of simpler p.r. functions Z1(s), Z2(s) . . . Zn(s). • Then to synthesize these individual Zi(s) as elements of the overall network whose dp impedance is

3. Breaking up process • One important restriction is that all Zi(s) must be positive real. • If we were given all the Zi(s), we could synthesize a network whose driving point impendance is Z(s) by simply connecting the Zi(s) in series. • However, if we were to start from Z(s) alone, how do we decompose Z(s) into Zi(s)?

4. Removing a pole at s=0 • If there is a pole at s=0, we can write Q(s) as • Hence, Z(s) becomes • Z1(s) is a capacitor. • We know Z1(s) is positive real, is Z2(s) positive real?

5. Is Z2(s) positive real? • The poles of Z2(s) are also poles of Z1(s), hence, Z2(s) doesn’t’ have poles on the right hand side of the s plane and no multiple poles on the jw axis. • Satisfies the first 2 properties of p.r. functions. • What about Re(Z2(jw))? • Since Z(s) is p.r. Re(Z2(jw))=Re(Z(jw))>0. • Hence, Z2(s) is p.r.

6. Removing a pole at s=∞ • If Z(s) has a pole at s=∞, we can write Z(s) as • Using a similar argument as previous we can show that Z2(s) is p.r. • Z1(s) is an inductor.

7. Removing complex conjugate poles on the jw axis. • If Z(s) has complex conjugate poles on the jw axis, Z(s) can be expanded into • Note that • Hence, • Z2(s) is p.r.

8. Removing a constant K • If Re(Z(jw)) is minimum at some point wi and if Re(Z(jw)) = Ki as shown in the figure • We can remove that Ki as • Z2(s) is p.r. • This is essentially removing a resistor.

9. Constructing • Assume that using one of the removal processes discussed we expanded Z(s) into Z1(s) and Z2(s). • We connect Z1(s) and Z2(s) in series as shown on the figure.

10. Example 1 • Synthesize the following p.r. function • Solution: • Note that we have a pole at s=0. Lets remove it • Note that 2/s is a capacitor, while s/(s+3) is a parallel connection of a resistor and an inductor.

11. 2/s is a capacitor with C=1/2. • While s/(s+3) is a R=1 connected in parallel with an inductor L=1/3.

12. Example 2 • Synthesis the following p.r. function • Solution • Note that there are no poles on s=0 or s=∞ or jw axis. • Lets find the minimum of Re(Y(jw))

13. Note that minimum of Re(Y(jw))=1/2. • Lets remove it • ½ is a conductance in parallel with Y2(s)= • Note that Y2(s) is a conductance 1/3 in series with an inductor 3/2.

14. Exercise • Synthesize the following p.r. function.

15. Synthesis of one port networks with two kinds of elements • In this section we will focus on the synthesis of networks with only L-C, R-C or R-L elements. • The deriving point impedance/admittance of these kinds of networks have special properties that makes them easy to synthesize.

16. 1. L-C imittance functions • These networks have only inductors and capacitors. • Hence, the average power consumed in these kind of networks is zero. (Because an inductor and a capacitor don’t dissipate energy.) • If we have an L-C deriving point impedance Z(s) M1 and M2 even parts N1 and N2 odd parts

17. The average power dissipated by the network is

18. Properties of L-C function • The driving point impedance/admittance of an L-C network is even/odd or odd/even. • Both are Hurwitz, hence only simple imaginary zeros and poles on the jw axis. • Poles and zeros interlace on the jw axis. • Highest power of the numerator and denominator may only differ by 1. • Either a zero or a pole at origin or infinity.

19. Synthesis of L-C networks • There are two kinds of network realization types for two element only networks. • Foster and • Cauer

20. Foster synthesis • Uses decomposition of the given F(s) into simpler two element impedances/admittances. • For an L-C network with system function F(s), it can be written as • This is because F(s) has poles on the jw axis only.

21. Using the above decomposition, we can realize F(s) as For a driving point impedance For a driving point admittance

22. Example • Synthesize as driving point impedance and admittance. Solution: • Decompose F(s) into simpler forms

23. For driving point impedance • For driving point admittance

24. Cauer synthesis • Uses partial fraction expansion method. • It is based on removing pole at s=∞. • Since the degree of the numerator and denominator differ by only 1, there is either a pole at s=∞ or a zero at s=∞. • If a pole at s=∞, then we remove it. • If a zero at s=∞, first we inverse it and remove the pole at s=∞.

25. Case 1: pole at s=∞ • In this case, F(s) can be written as

26. This expansion can easily be realized as

27. Case 2: zero at s=∞ • In this case will have a pole at s=∞. • We synthesize G(s) using the procedure in the previous step. • Remember that if F(s) is an impedance function, G(s) will be an admittance function and vice versa.

28. Example • Using Cauer realization synthesize Solution: • This is an impedance function. • We have a pole at s=∞, hence, we should remove it.

30. If we were given Y(s) instead our realization would be

31. R-C driving point impedance/ R-L admittance • R-C impedance and R-L admittance driving point functions have the same properties. • By replacing the inductor in LC by a resistor an R-C driving point impedance or R-L driving point admittance, it can be written as • Where

32. Properties of R-C impedance or R-L admittance functions • Poles and zeros lie on the negative real axis. • The singularity nearest origin must be a pole and a zero near infinity. • The residues of the poles must be positive and real. • Poles and zeros must alternate on the negative real axis.

33. Synthesis of R-C impedance or R-L admittance • Foster • In foster realization we decompose the function into simple imittances according to the poles. That is we write F(s) as • For R-C impedance

34. For R-L admittance

35. Example • Synthesize as R-C impedance and R-L admittance in foster realization. Solution: • Note that the singularity near origin is a pole and a zero near infinity. • The poles and zeros alternate • We can expand F(s) as • R-C impedance

36. R-L admittance

37. Cauer realization • Cauer realization uses continued fraction expansion. • For R-C impedance and R-L admittance we remove a resistor first. • Then invert and remove a capacitor • Then invert and remove a resistor . . .

38. Example • Synthesize using Cauer realization as R-C impedance and R-L admittance. Solution: • Note that the singularity near origin is a pole. • The singularity near infinity is a zero. • The zeros and the poles alternate. • Note that the power of the numerator and denominator is equal, hence, we remove the resistor first. F(s) is R-L impedance or R-C admittance

40. For R-L admittance For R-C impedance

41. R-L impedance/R-C admittance • R-L impedance deriving point function and R-C admittance deriving point function have the same property. • If F(s) is R-L impedance or R-C admittance, it can be written as

42. Properties of R-L impedance/R-C admittance • Poles and zeros are located on the negative real axis and they alternate. • The nearest singularity near origin is zero. The singularity near infinity is a pole. • The residues of the poles must be real and negative. • Because the residues are negative, we can’t use standard decomposition method to synthesize.

43. Synthesis of R-L impedance and R-C admittance • Foster • If F(s) is R-L impedance d.p or R-C admittance d.p function. We can write it as • Because of the third property of R-L impedance/R-C admittance d.p. functions, we can’t decompose F(s) into synthesizable components with the way we were using till now. • We have to find a new way where the residues wont be negative.

44. If we divide F(s) by s, we get • Note that this is a standard R-C impedance d.p. function, hence, the residues of the poles of F(s)/s will be positive. • Once we find Ki and σi we multiply by s and draw the foster realization.

45. Example • Synthesize as R-L impedance and R-C admittance using Foster realization. Solution: • Note that the singularity near origin is a zero. • The singularity near infinity is a pole. • The zeros and the poles alternate. F(s) is R-L impedance or R-C admittance

46. We divide F(s) by s.

47. R-L impedance • R-C admittance

48. Cauer realization • Using continued fractional expansion • We first remove R0. To do this we use fractional expansion method by focusing on removing the lowest s term first. • We write N(s) and M(s) starting with the lowest term first.

49. Example • Synthesize as R-L impedance and R-C admittance using Cauer realization. • Solution: • We write P(s) and M(s) as