Network Analysis and Synthesis

1. Network Analysis and Synthesis Chapter 2 Network transform representation and analysis

2. 2.1 The transformed circuit • When analyzing a network in time domain we will be dealing with • Derivation and • Integration • However, when transformed to complex frequency domain these become • Derivation -> multiplication by ‘s’ • Integration -> division by ‘s’ • Hence, it is easier to do network analysis in complex frequency domain.

3. The voltage current relationships of network elements in time domain and complex frequency domain are given as: • Resistor

4. Inductor • The time domain relation ships are • In frequency domain they become

6. An inductor is represented in frequency domain as • An impedance sL in series with a voltage source Used in mesh analysis. or • An admittance 1/sL in parallel with a current source Used in nodal analysis.

7. Capacitor • The time domain relation ships are • In frequency domain they become

9. A capacitor is represented in frequency domain as • An impedance 1/sCin series with a voltage source Used in mesh analysis. or • An admittance sC in parallel with a current source Used in nodal analysis.

10. Example 1 • In the figure below, the switch is switched from postion 1 to 2 at t=0. Draw its transformed circuit and write the transformed equations using mesh analysis.

11. The transformed circuit is

12. The transformed equations become

13. Example 2 • The switch is thrown to position 2 at t=0. Find i(t).

14. The transformed circuit is

15. Writing the transformed equation • Solving for I(s) • Inverse transforming

16. Example 3 • At t=0, the switch is opened. Find the node voltages v1 and v2

17. The transformed circuit becomes

18. The transformed equations become • Solving these 2 equations

19. 2.2 System function • The excitation , e(t), and response, r(t), of a linear system are related by a linear differential equation. • When transformed to complex frequency domain the relationship between excitation and response is algebraic one. • When the system is initially inert, the excitation and response are related by the system function H(s) given by

20. The system function may have many different forms and may have special names. Such as: • Driving point admittance • Transfer impedance • Voltage or current ratio transfer function • This is because the excitation and response may be taken from the same port or different ports and the excitation and response can be either voltage or current.

21. Impedance • Transfer impedance is when the excitation is a current source and the response is a voltage. • When both the excitation and response is at the same port it is called driving point impedance.

22. Admittance • Transfer admittance is when the excitation is a voltage source and the response is a current.

23. Voltage ratio transfer function • When the excitation is a voltage source and the response is a voltage.

24. Current ratio transfer function • When the excitation is a current source and the response is a current.

25. Note that, the system function is a function of the system elements only. • It is obtained from the network by using the standard circuit laws. Such as: • Kirchhoffs law • Nodal analysis • Mesh analysis

26. Example 4 • Obtain the driving point impedance of the network. Then using the following excitations determine the response. • The square pulse on figure b • The waveform on figure c b c a

27. First lets find the driving point impedance • Note that it is the equivalent impedance of the 3 elements

28. Its transform is Hence, the response is

29. The excitation is given as Hence, the response is

30. The excitation is given as

31. Consider the partial fraction expansion of R(s) where si are the poles of H(s) and sj are the poles of E(s). • Taking the inverse Laplace transform of R(s) • The terms are associated with the system H(s) and are called the free response terms.

32. The terms are due to the excitation E(s) and are called the forced response terms. • The frequencies si are the natural frequency of the system, while the frequencies sj are the frequencies of the excitation.

33. Problem • Find the free response and the forced response for the circuit below. The system is inert before applying the source.

34. 2.3 Poles and zeros of system • We will discuss the relationship between the poles and zeros of a system function and its steady state sinusoidal response. • In other words, we will investigate the effect of positions of poles and zeros upon H(s) on thejwaxis.

35. To find the steady-state sinusoidal response of a system function we replace ‘s’ by ‘jw’. • Hence, the system function becomes Where • M(w) is the amplitude or magnitude response • φ(w) is the phase response

36. The amplitude and phase response of a system provide valuable information in the analysis and design of transmission circuits. • Consider the low pass filter • Observe that • It passes only frequency below wc • The phase response is almost linear till wc

37. Hence, if all the significant harmonic terms are less than wc , then the system will produce minimum phase distortion. • In the rest of this section, we will concentrate on methods to obtain amplitude and phase response curves.

38. R-C network • To obtain H(jw) we substitute s by jw.

39. In polar form H(jw) becomes

40. The amplitude is unity and the phase is zero degrees at w=0. • The amplitude and phase decrease monotonically as we increase w. • When w=1/RC, the amplitude is 0.707 and phase is -450. • As w increases to infinity M(w) goes to zero and the phase approaches -900. Half power point

41. Amplitude and phase from pole-zero diagram • For the system function • H(jw) can be written as • Each one of the or represent a vector from zi or pjto the jw axis at w.

42. If we express • Then H(jw) can be given as

44. In general,

45. Example • For find the magnitude and phase for w=2. • Solution • First let us find the zeros and poles • Zero at jw=0 • Poles at

47. Magnitude • Phase

48. Exercise • Examine the property of F(s) around the poles and zeroes.

49. Bode plots • In this section we turn our attention to semi logarithmic plots of system function, called Bode plots. • In these plots we take the logarithm of the amplitude and plot it on linear frequency scale. • For amplitude M(jw), if we express in terms of decibel it becomes

50. For system function • If we express the amplitude in terms of decibels we have