Network Analysis and Synthesis Chapter 2 Network transform representation and analysis

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.

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

Inductor The time domain relation ships are In frequency domain they become

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.

Capacitor The time domain relation ships are In frequency domain they become

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

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.

The transformed circuit is

The transformed equations become

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

The transformed circuit is

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

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

The transformed circuit becomes

The transformed equations become Solving these 2 equations

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

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.

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.

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

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

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

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

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 a b c

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

Its transform is Hence, the response is

The excitation is given as Hence, the response is

The excitation is given as

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.

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.

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

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 the jw axis.

Hence, the system function becomes 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

Consider the low pass filter Observe that 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

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.

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

In polar form H(jw) becomes

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

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 pj to the jw axis at w.

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

In general,

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

Magnitude Phase

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

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

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

In factored from both N(s) and D(s) are made up of 4 kinds of terms Constant K A root at origin, s A simple real root, s-a A complex set of roots, To understand the nature of log-amplitude plots, we only need to discuss the amplitude response of these 4 terms. If the term is on the numerator it carries positive sign, if on denominator negative sign.

1. Constant K The dB gain or loss is K2 is either positive |K|>1 or negative |K|<1. The phase is either 00 for K>0, or 1800 for K<0.

Single root at origin, s The loss or gain of a single root at origin is Thus the plot of magnitude in dB vs frequency is a straight line with slope of 20 or -20. 20 when s is in the numerator. -20 when s is in the denominator. The phase is either 900 or -900. 900 when s is in the numerator. -900 when s is in the denominator.

The factor s+α For convenience lets set α=1. Then the magnitude is The phase is A straight line approximation can be obtained by examining the asymptotic behavior of the factor jw+1.

For w<<1, the low frequency asymptote is For w>>1, the high frequency asymptote is Which has a slope of These 2 asymptotic approximations meet at w=1.

Note that the maximum error is for w=1 or for the non normalized one w=α. For the general case α different from 1, we normalize the term by dividing by α. The low frequency asymptote is The high frequency asymptote is

For complex conjugates For complex conjugates it is convenient to adopt a standard symbol. We describe the pole (zero) in terms of magnitude ω0 and angle θ measured from the negative real axis. These parameters that describe the pole (zero) are ω0, the undamped frequency of oscillation, and ζ, the damping factor.

If the pole (zero) pair is given as α and β are related to ω0 and ζ with Substituting these terms in the conjugate equation

For ω0=1 (for convenience), the magnitude of conjugate pairs can be expressed as The phase is

The asymptotic behavior is For low frequency, w<<1 For high frequency, w>>1 which is a straight line with slope of 40dB/decade. These 2 asymptotes meet at w=1.

Example Using Bode plot asymptotes, draw the magnitude vs. frequency for the following system function

Actual plot