S-Domain Analysis in Network Function Lectures
Explore complex frequency analysis, impedance inductors for AC circuits, complete response concepts, and more in this lecture on network functions and S-domain analysis by Hung-yi Lee.
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Lecture 21 Network Function and s-domain Analysis Hung-yi Lee
Outline Chapter 10 (Out of the scope) Frequency (chapter 6) Complex Frequency (s- domain) Impedance (chapter 6) Generalized Impedance Network function
What are we considering? Chapter 5 and 9 Complete Response What really observed Final target Zero State Response Zero Input Response Natural Response Forced Response Chapter 6 and 7 Transient Response Steady State Response = cos A ( ) + t ( ) x t
What are we considering? Chapter 5 and 9 Complete Response What really observed Final target Zero State Response Zero Input Response Natural Response Forced Response Transient Response Steady State Response = Ae This lecture ) ( + t ( ) cos x t t
Complex Frequency In Chapter 6 Current or Voltage Sources = t x A ) ( Currents or Voltages in the circuit = t x cos A ) ( ( ) ( ) + t + cos t The same frequency Different magnitude and phase In Chapter 10 Current or Voltage Sources = t x Ae ) ( Currents or Voltages in the circuit = t x e A ) ( ( ) ( ) + t + cos t t cos t The same frequency and exponential term Different magnitude and phase You can observe the results from differential equation.
Complex Frequency s plane ( ) = + t ( ) Ae cos x t t Phasor: = A X Complex Frequency: = + j s
Complex Frequency - Inductor For AC Analysis (Chapter 6) ( ) = I L I = + ( ) I cos i t t m i L m i ( ) ( ) Lt i = + ( ) I sin v t L t L m i ( ) ( ) = + ( ) v Lt VL LI 90 = + + I cos 90 L t m i m i Impedance of inductor V Z L ( ) + LI 90 ( ) di t L = = = = j L 90 L m i = ( ) L v t L L I I L dt m i
Complex Frequency - Inductor ( ) = cos ( ) x Ae t = + t ( ) I cos i t e t L m i ( ) + t t ( ) ( ) = ( + ) + ) t t ( ) I cos I sin v t L e t L e t ) L m i m i ( ( ( ) Lt i = + + t I cos sin L e t t m i i ( ) v Lt ( ) = + + 2 2 t I cos L e t m i + 2 2 ( ) + sin t i + 2 2 ( ) di t = ( ) L v t L L dt = + + + 2 2 1 t I cos tan L e t m i
Complex Frequency - Inductor ( ) = I L I = + t ( ) I cos i t e t m i L m i ( ) = + = + + 2 2 t 2 2 1 I V LI tan v t L e L L m m i + + 1 cos tan t i Generalized Impedance of inductor + + 2 2 1 LI tan V m i = + 2 2 1 L tan L = = Z L I I L m i ( )L = = + + 2 2 L j = = + j s L j s + + 2 2 2 2 +
Generalized Impedance Generalized Impedance (Table 10.1) Generalized Impedance Element Impedance Resistor R R sL j L Inductor j / 1 C / 1 s C Capacitor s = j Special case: The circuit analysis for DC circuits can be used.
Example 10.2 s domain diagram 1 = = = = H, 1 5 , F L R C 20 90 V V 10 = + 2 4 s j = + 2 t ( ) 20 cos( 4 90 ) V v t e t 1 V 20 90 R = = = = + = + I 8 53 ( ) // Z s sL R sL 143 + ( ) 5 . 2 1 Z s sC sCR = 2 t ( ) 8 cos( 4 53 ) A i t e t = 5 . 2 143
Network Function / Transfer Function Given the phasors of two branch variables, the ratio of the two phasors is the network function/transfer function Y ( ) s = H phasors of current or voltage X Complex number The ratio depends on complex frequency ( )X s = H X : Y : Y input output
Network Function / Transfer Function Network Function/Transfer Function is not new idea I V ( ) s = Z Impedance: I V 1 I ( ) s = = Y Admittance: ( ) s V Z Y ( ) s = Impedance and admittance are special cases for network function H X
Network Function / Transfer Function Y ( ) s = Current or voltage H X Current or voltage In general, network function can have four meaning V V V ( ) s ( ) s = = 2 H H Voltage Gain Impedance I 1 2I I ( ) s = I ( ) s H = H Current Gain Admittance V 1
Example 10.5 V V L R C I sL LI R s V V 1 C sC I 1 L = = + + ( ) H s V I sL R 1 V s L sC s Polynomial of s I 1 sC ( ) s L = = = H 1 1 + + 2 1 V s LC sRC + + sL R s sC Polynomial of s
Example 10.5 V V L R C I sL LI R s V V 1 C sC V sL L = = ( ) H s 2 1 s V + + sL R sC Polynomial of s 2 s LC = + + 2 1 s LC sRC Polynomial of s
Network Function / Transfer Function ( )X s Y H = H | | | Y = ( ) Y ( ( ) s ) ( ) X ( ) s = + H || | X + 4 7 j |H(s)| is complex frequency dependent = + j s = represents 0 dc s gain dc the is ) 0 ( H Output will be very large when = t x Ae ) ( ( ) + t 4 cos 7 t
Example 10.5 Check your results by DC Gain V V L R C I sL LI R s V V 1 C sC For DC I sC Capacitor = open circuit ( ) 0 L = = ( ) H s = H1 0 1 + + V 2 1 s LC sRC s V 2 s LC Inductor = short circuit ( ) 0 L = = ( ) H s = H2 0 2 s V + + 2 1 s LC sRC
Example 10.5 Check your results by Units V V L R C I sL LI R s V V 1 C sC 1 V A V = = = C : F L : H t t R : ? : s t A V A 2 1 V A 1 A t t t t A V V A t V I V sC 2 s LC L L = = = ( ) H s = ( ) H s 1 2 + + V 2 s V 1 s LC sRC + + 2 1 s LC sRC s V 2 1 V A 1 V A 2 1 V A V 1 V A t t t t t t t A V t A V t A V t A V
Poles/Zeros General form of network function ( ) ( ) s + + + + 1 m m b s b s b s b N s = 1 1 s 0 ( ) m m H s = + + + + 1 n n a s a s a a D 1 1 0 n n The zeros is the root of N(s). If z is a zero, H(z) is zero. The poles is the root of D(s). If p is a pole, H(s) is infinite.
Poles/Zeros General form of network function + + + + 1 m m b s b s b s b = 1 1 s 0 ( ) m m H s + + + + 1 n n a s a s a a 1 1 0 n n b b b + + + + 1 m m s s s 1 0 1 m ( s )( )( ) ( ) ( 2 ) s z s z s z b b b b = = m m m m 1 2 K m p ( ) a a a a p s p s + + + + 1 n n s s s 1 0 1 n n 1 n a a a 1 n n b m zeros: z1, z2, ,zm n poles: p1, p2, ,pn = K m a n
Example 10.8 + + s 4 3 2 16 + 164 2 + s s s = ( ) 5 H s )( )( ) Find its zeros and poles ( + + 2 32 36 40 400 s s s Denominator: Numerator: + = 32 32 s p + s + 16 0 4 3 2 16 ( s )( 0 164 + s )( s s s ( z3 and z4are the two roots of s2+16s+164 s + s s 1 ) )( + = 2 36 , 6 s p p j = = 2 164 z + 2 3 ) + + = = 2 + 40 400 20 s s p p s s z 4 5 3 4 Poles: p1=-32, p2=j6, p3=-6j, p4=-20, p5=-20 Zeros: z1=0, z2=0, z3=-8+j10, z4=-8-j10
Example 10.8 + + s 4 3 2 16 + 164 2 + s s s = ( ) 5 H s )( )( ) Find its zeros and poles ( + + 2 32 36 40 400 s s s Zeros: z1=0, z2=0, z3=-8+j10, z4=-8-j10 Poles: p1=-32, p2=j6, p3=-6j, p4=-20, p5=-20 Pole and Zero Diagram Zero (O), pole (X) We can read the characteristics of the network function from this diagram For example, stability of network
Stability A network is stable when all of its poles fall within the left half of the s plane ( )X s H Y = If p = p + j pis a pole H(p)= ( cos ) Complex frequency is p ( ) t = + y Ae p If the output is p Y Y = = = X 0 No input ( ) p H The waveforms corresponding to the complex frequencies of the poles can appear without input.
Stability A network is stable when all of its poles fall within the left half of the s plane The poles are at the right plane. Appear automatically Appear automatically The poles are at the left plane. Stable Unstable
Stability A network is stable when all of its poles fall within the left half of the s plane p= 0 The poles are on the j axis. Appear automatically Marginally stable oscillator
Acknowledgement (b02)
What is Network/Transfer Function considered? Input Network Function H(s) Output Natural Response Forced Response
Natural Response It is also possible to observe natural response from network function. ( )X s Y = H
Fix 0, decrease The position of the two roots 1and 2. = 2 2 0 1, 2 =0 Undamped
Time-Domain Response of a System Versus Position of Poles (constant magnitude Oscillation) (unstable) (exponential decay) The location of the poles of a closed Loop system is shown.
Example 10.3 - Miller Effect V V in in = = Z in V V I in out 1 1 sC V in =sC ( ) V V in out V in = ( ) ( ) V V sC A in in 1 = ( ) sC + 1 A Capacitor with capacitance C(1+A)
Example 10.3 - Miller Effect V V out out = = Z out V V I out in 2 1 sC ) in V V out =sC ( V V ) V out ( A + 1 A in = ( ) ( ) V sC A in in 1 = = ( ) sC A + 1 1 +A sC A 1 Capacitor with capacitance 1 C
