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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 v(t ) Ri (t ) V ( s) RI ( s ) • Inductor – The time domain relation ships are v(t ) L di (t ) dt t 1 i (t ) v( )d i (0 ) L 0 – In frequency domain they become V ( s ) sLI ( s ) Li (0 ) V ( s ) i (0 ) I ( s) sL s • 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 t 1 v(t ) i ( )d v(0 ) C 0 i (t ) C dv(t ) dt – In frequency domain they become I ( s ) v (0 ) V (s) sC s I ( s ) sCV ( s ) Cv(0 ) • A capacitor is represented in frequency domain as – An impedance 1/sC source in series with a voltage 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). i (0 ) 2amp L vC (0 ) 2V • The transformed circuit is • Writing the transformed equation 5 2 2 2 3 s I ( s) s s s • Solving for I(s) 2s 3 ( s 2)( s 1) 1 1 I (s) s 2 s 1 I (s) • Inverse transforming i(t ) e 2t e t Example 3 • At t=0, the switch is opened. Find the node voltages v1 and v2 1 h 2 G 1 mho L C 1f V 1v • 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 R( s) H ( s) E ( s) • 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. V0 ( s) H ( s) I g ( s) • When both the excitation and response is at the same port it is called driving point impedance. 1 sL H ( s ) R sC 1 sL sC Admittance • Transfer admittance is when the excitation is a voltage source and the response is a current. I 0 ( s) H ( s) Vg (s) 1 H (s) 1 sL R sC Voltage ratio transfer function • When the excitation is a voltage source and the response is a voltage. V0 (s) H ( s) Vg (s) Z 2 ( s) H ( s) Z1 ( s) Z 2 ( s) Current ratio transfer function • When the excitation is a current source and the response is a current. I 0 ( s) H ( s) I g ( s) 1 H ( s ) sL R 1 sC sL R 1 sC H (s) 1 R sL sC • 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. 1. ig (t ) Sinwot u (t ) 2. The square pulse on figure b 3. 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 1 s H ( s) 1 sC G C s 2 G s 1 sL C CL 1. ig (t ) Sinwot u (t ) Its transform is w0 I (S ) 2 s wo2 Hence, the response is wo s Vo ( s ) I g ( s ) H ( s ) 2 . 2 G 1 s wo 2 C s s C LC 2. The excitation is given as i (t ) u (t ) u (t a ) 1 1 as I (s) e s s Hence, the response is 1 e Vo ( s) I g ( s) H ( s) s as s . G 1 2 C s s C LC 3. The excitation is given as t t a ig (t ) u (t ) u (t ) u (t a) a a 1 1 e as I (s) 2 2 s as 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) si t • The terms Ai e are associated with the system H(s) and are called the free response terms. s jt • The terms B j e 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. 1 v g (t ) (cos t )u (t ) 2 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. • To find the steady-state sinusoidal response of a system function we replace ‘s’ by ‘jw’. • Hence, the system function becomes H ( jw) H ( s) |s jw H ( jw) M ( w)e j ( w) Where M(w) is the amplitude or magnitude response φ(w) is the phase response • 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 • 1 sC 1 V ( s) H ( s) 2 RC 1 V1 ( s ) R 1 s sC RC • To obtain H(jw) we substitute s by jw. H ( jw) 1 RC 1 jw RC • In polar form H(jw) becomes 1 RC H ( jw) 1 2 w 2 2 RC M ( w) 1 RC 1 2 w 2 2 R C ( w) tan 1 wRC 1 2 1 2 e j tan 1 wRC M ( w)e j ( w ) • 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. Half power point • As w increases to infinity M(w) goes to zero and the phase approaches -900. Amplitude and phase from pole-zero diagram • For the system function H ( s) A0 ( s z0 )( s z1 ) ( s p0 )( s p1 )( s p2 ) • H(jw) can be written as A0 ( jw z0 )( jw z1 ) H ( jw) ( jw p0 )( jw p1 )( jw p2 ) • Each one of the ( jw zi ) or ( jw p j ) represent a vector from zi or pj to the jw axis at w. • If we express j i jw zi Ni e , jw p j M j e j j • Then H(jw) can be given as A0 N1 N 2 j 0 1 2 0 1 2 H ( jw) e M 0 M 1M 2 • In general, Example 4s F ( s) 2 s 2s 2 • For phase for w=2. • Solution find the magnitude and – First let us find the zeros and poles 4 jw F ( jw) ( jw 1 j )( jw 1 j ) – Zero at jw=0 – Poles at ( jw 1 j ) and ( jw 1 j ) • Magnitude 2 4 M ( j 2) 4 * 2 * 10 5 • Phase ( j 2) 900 71.80 450 26.80 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 20 log M ( jw) • For system function H ( s) N ( s) D( s ) M ( jw) | H ( jw) | | N ( jw) | | D( jw) | • If we express the amplitude in terms of decibels we have 20 log M ( jw) 20 log | N ( jw) | 20 log | D( jw) | • In factored from both N(s) and D(s) are made up of 4 kinds of terms 1. 2. 3. 4. • • Constant K A root at origin, s A simple real root, s-a A complex set of roots, s 2 2s 2 2 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 20 log K K2 • 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 20 log | jw | 20 log w • 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 20 log | jw 1 | 20 log w 1 2 1 2 • The phase is arg( jw 1) tan 1 w • 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 20 log w 1 20 log 1 0dB 1 2 2 • For w>>1, the high frequency asymptote is 20 log w 1 20 log w 2 1 2 Which has a slope of 20 log w decibel/de cade • 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 1 2 w 20 log 2 1 20 log 1 0dB 2 • The high frequency asymptote is 1 2 w 20 log 2 1 20 log w 20 log 2 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 p1, 2 j • α and β are related to ω0 and ζ with 0 cos 0 0 sin 0 1 2 • Substituting these terms in the conjugate equation (s p1 )(s p2 ) ( jw j )( jw j ) jw 0 j0 1 2 jw 0 j0 1 2 w2 2 jw0 0 2 • For ω0=1 (for convenience), the magnitude of conjugate pairs can be expressed as 20 log 1 w2 j 2w 20 log 1 w • The phase is ( w) tan 1 2 2 1 w2 2 2 4 2 w 1 2 2 • The asymptotic behavior is – For low frequency, w<<1 20 log 1 w 2 2 4 2 w 20 log 1 0dB 1 2 2 – For high frequency, w>>1 20 log 1 w 2 2 4 2 w 40 log w 1 2 2 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 0.1s G ( s) 2 s s s 3 1 1 4 50 16 *10 10 Actual plot