ECE 342 1. Networks & Systems Jose E. Schutt Aine Electrical & Computer Engineering University of Illinois jschutt@emlab.uiuc.edu 1
What is Capacitance? 1 2 3 Voltage=0 No Charge No Current Voltage build up Charge build up Transient Current Voltage = 6V Charge=Q No Current 2
4 What is Capacitance? 5 6 Voltage=6V Charge=Q No Current Voltage decaying Charge decaying Transient Current Voltage=0 No Charge No Current 3
Relation: Q = Cv Q: charge stored by capacitor C: capacitance v: voltage across capacitor i: current into capacitor dv dq it () C dt dt 1 t () ( ) Capacitance vt i d C 0 4
What is Inductance? z Current in wire produces magnetic field Flux is magnetic field integrated over area Idl r R H Inductance Total Flux Linked Current 5
Inductance L L L d N di v I B Hdv 2 I 6
Inductance Relation: = Li : flux stored by inductor L: inductance i: current through inductor v: voltage across inductor di d vt () L dt dt 1 t () ( ) it v d L 0 7
Low-Pass Circuit dvo ir() t C ic() t dt v () t Ri () t v () t i R o dvo vi() t RC vo() t dt Need to solve for v o (t) 8
Low-Pass Circuit dvo ir() t C ic() t dt v () t Ri () t v () t i R o dvo vi() t RC vo() t dt Need to solve for v o (t) 9
Reactive Circuit i () t i () t i () t v C R L x di L dt d v L i x ic C dt v Need to solve for v x (t) 10
st L f() t f() t e dt F() s Laplace Transforms The Laplace transform F(s) of a function f(t) is defined as: 0 To a mathematician, this is very meaningful; however circuit engineers seldom use that integral The Laplace transform provides a conversion from the time domain into a new domain, the s domain 11
Laplace Transforms The Laplace transform of the derivative of a function f(t) is given by df () t df () t st L e dt sf() s dt dt 0 Differentiation in the time domain becomes a multiplication in the s domain. That is why Laplace transforms are useful to circuit engineers s = j 12
Low-Pass Circuit Time domain s domain v () t V () s o dvo () t dt o sv () s o Time domain dvo vi() t RC vo() t dt s domain V() s srcv () s V () s i o o 13
Low-Pass Circuit - Solution V () s o Vi() s Vo() s 1 1 src V ( s) 1sRC i Pole exists at s 1/ RC Assume that V i (s)=v d /s V () s o Vd A B s 1sRC s s 1/ RC Inversion v () t Au() t Be o t/ RC 14
Reactive Circuit v x di L dt L V () s x sli L i C C d v i dt v x I sc( V V ) C i x 15
Reactive Circuit - Solution Vx Vx sc( Vi Vx) R sl 1 1 scvi V xsc R sl V x scvi Vi 1 1 1 1 sc 1 R sl src s 2 LC Exercise: Use inversion to solve for v x (t) 16
Table of Laplace Transforms Function Transform ut ( ) 1 e t cos( t) s s 1 s s e s 2 2 17
Capacitance s Domain Assume time-harmonic behavior of voltage and current variables j = s it (), vt ()~ e jt it () I( s) vt () V() s d dt s I() s scv() s The capacitor is a reactive element 1 Reactance : X C sc 18
Inductance s Domain Assume time-harmonic behavior of voltage and current variables j = s it (), vt () e j t it () I( s) vt () V() s d dt s V() s sli() s The inductor is a reactive element Reactance : X L sl 19
Thevenin Equivalent Network V th + - Z th Principle Any linear two-terminal network consisting of current or voltage sources and impedances can be replaced by an equivalent circuit containing a single voltage source in series with a single impedance. Application To find the Thevenin equivalent voltage at a pair of terminals, the load is first removed leaving an open circuit. The open circuit voltage across this terminal pair is the Thevenin equivalent voltage. The equivalent resistance is found by replacing each independent voltage source with a short circuit (zeroing the voltage source), replacing each independent current source with an open circuit (zeroing the current source) and calculating the resistance between the terminals of interest. Dependent sources are not replaced and can have an effect on the value of the equivalent resistance. 20
Network Norton Equivalent I th Gth Principle Any linear two-terminal network consisting of current or voltage sources and impedances can be replaced by an equivalent circuit containing a single current source in parallel with a single impedance. Application To find the Norton equivalent current at a pair of terminals, the load is first removed and replaced with a short circuit. The short-circuit current through that branch is the Norton equivalent current. The equivalent resistance is found by replacing each independent voltage source with a short circuit (zeroing the voltage source), replacing each independent current source with an open circuit (zeroing the current source) and calculating the resistance between the terminals of interest. Dependent sources are not replaced and can have an effect on the value of the equivalent resistance. 21
Thevenin Equivalent - Example Circuit for calculating impedance Calculating Thevenin voltage KCL law at node A gives 1+I y =I x Also KVL gives 10=2I y +3I x Combining these equations gives I x = 2.4 ma From which we calculate V x = V th =3(2.4)=7.2 V (2)(3) Rth 1.2 2 3 k 22
Thevenin Example Find voltage across R 4 Define G1 1/ R 1,G2 1/ R 2, and G3 1/ R 3, G G v G v i 1 2 1 2 2 s Gv G G v 0 2 1 2 3 2 From which v i 5 s 2 Vth 23
Thevenin Equivalent Circuit for calculating impedance (current source is replaced with open circuit) R 2 R 1 R 3 Z th (R1 R 2 )R3 2 R th (R1 R 2 ) R3 R R R 5 1 2 3 24
Circuit with Thevenin Equivalent Z TH V TH v(t) o R i / 5 4s 2/5R 4 25
Thevenin Example Need to find the current i 1 (t) in resistor R 3 R 1 L 1 L 2 10 1H 1H V 100 V i R 1 (t) 2 10 R 3 10 26
Thevenin Example Thevenin voltage calculation 10 +s 100/s + - 10 10 +s V th Thevenin Network 10(100 / s ) 1,000 10 s 10 s s 20 27
Thevenin Example Calculating the impedance 10 +s 10 Z th 10( s 10 ) s 20 28
Thevenin Example Z th I 1 V th + - Z Load =10+s I(s) 1 I(s) V th(s) 1,000 Z th(s) ZLoad 10s 10 (s10) s 20 1,000 1 2 s s 40s 300 29
Thevenin Example Expanding using partial fraction I(s) 1 K1 K2 K3 s (s10) (s30) With K 1 = 3.33, K 2 = -5, K 3 = 1.67 The time-domain current i 1 (t) is: 1 10t 30t i ( t ) 3.33 5e 1.67e A 30
Low-Pass Circuit V o In frequency domain: Vi Vo 1 Av 1 jrc V 1 jrc A v 1 1 1 jrc 1 jf / f i o V o Vi 1 1 R jc j C 31
f o Low-Pass Circuit 1 1 2 RC 2 RC time constant At very high frequencies, the single time constant (STC) transfer function goes as G 20log 1 ( / o ) 2 G 20log( / ) 20X where X log( / o ) o At high frequencies, slope of curve is 20 db if X 1( 10 ), decrease is 20 db 20 db / decade o 32
High-Pass Circuit V o VR i Vi 1 1 R 1 jc jrc A v Vo 1 1 V 1 i 1 j 1 jfo/ f 2 frc 33
Octave & Decade If f 2 = 2f 1, then f 2 is one octave above f 1 If f 2 = 10f 1, then f 2 is one decade above f 1 f # of octaves log 3.32log # of decades log 2 2 2 10 f1 f1 2 GHz is one octave above 1 GHz 10 GHz is one decade above 1 GHz 10 f f 2 1 f 34
3-dB points are points where the magnitude is divided by 2 1/2 (power is halved) 1+j = 2 1/2 A db = 20log1.414 = 3 db Frequency Response Amplifier has intrinsic gain A o Low-pass characteristics is: 1 1 jf / fhi High-pass characteristics is: Overall gain A(f) is A o jf / f 1 jf / lo f lo jf / flo 1 1 jf / f 1 jf / f lo hi 35
Octave & Decade Overall gain A(f) is A( f) jf / flo 1 Ao 1 jf / f 1 jf / f lo hi 36
Model for general Amplifying Element C c1 and C c2 are coupling capacitors (large) F C in and C out are parasitic capacitors (small) pf 37
Midband Frequencies - Coupling capacitors are short circuits - Parasitic capacitors are open circuits A MB vout Rin RL A v R R R R in g in out L 38
Low Frequency Model - Coupling capacitors are present - Parasitic capacitors are open circuits v vinrin vin jcc 1Rin 1 R 1 j Cc 1( Rg Rin) g Rin jcc 1 R jcc 1( Rg Rin) in vab vin R g R in 1 jcc 1( Rg Rin) ab 39
Low Frequency Model 1 1 define fl1 and f 2 2 R R C l R 2 g Rin C c1 L out c2 R jf / f in l1 vab vin R g R in 1 jf / f l1 Similarly, v out Av ab RL jf / fl 2 R R 1 jf / f L out l 2 40
Low Frequency Model Overall gain vout Rin RL jf / fl1 jf / fl 2 A v R R R R 1 jf / f 1 jf / f in g in L out l1 l 2 v jf / f jf / f A v jf f jf f out l1 l 2 MB in 1 / l1 1 / l 2 41
Example R out = 3 k, R g =200, R in =12 k, R L =10 k C c1 =5 F and C c2 =1 F fl 1 2 (12, 200510 ) 1 6 2.61 Hz fl 1 2 6 2 (13,00010 ) 12.2 Hz 42
High Frequency Model - Assume coupling capacitors are short - Account for parasitic capacitors Potential Thevenin equivalent for input as seen by C in V th1 vinrin R R g in R R R th1 g in 43
v v ab ab High Frequency Model vinrin 1 R R 1 jc R g in in th1 v R 1 1 where f R R jf f R C in in h1 g in 1 / h1 2 th1 in Likewise v out AvabRL 1 R R 1 jc R out L out th2 v out with R R R th2 out L Av R 1 1 where f R R jf f R C ab L h2 L out 1 / h2 2 th2 out 44
Overall gain is: High Frequency vo Rin RL 1 1 A v R R R R 1 jf / f 1 jf / f i in g L out h1 h2 or v 1 1 A v jf f jf f o MB i 1 / h1 1 / h2 45
Important Remarks An arbitrary network s transfer function can be described in terms of its s-domain representation s is a complex number s = + j The impedance (or admittance) or networks can be described in the s domain 46
T() s Transfer Function Representation m as m a s... asa n s b s bsb m1 m1 1 0 n1 n1... 1 0 The coefficients a and b are real and the order m of the numerator is smaller than or equal to the order n of the denominator A stable system is one that does not generate signal on its own. For a stable network, the roots of the denominator should have negative real parts 47
Transfer Function Representation The transfer function can also be written in the form T() s a m sz sz... sz 1 2 sp sp... sp 1 2 m m Z 1, Z 2, Z m are the zeros of the transfer function P 1, P 2, P m are the poles of the transfer function 48
Transfer Function Representation Use a two terminal representation of system for input and output 49
Y-parameter Representation I yvyv 1 11 1 12 2 I yv yv 2 21 1 22 2 50
Y Parameter Calculations y I 1 2 11 21 V1 V V 0 1 V 0 y 2 2 I To make V 2 = 0, place a short at port 2 51
Z Parameters V z I z I 1 11 1 12 2 V z I z I 2 21 1 22 2 52
Z-parameter Calculations z V V 1 2 11 21 I1 I I 0 1 I 0 z 2 2 To make I 2 = 0, place an open at port 2 53
H Parameters V h I h V 1 11 1 12 2 I h I h V 2 21 1 22 2 54
H Parameter Calculations h V 1 2 11 21 I1 I V 0 1 V 0 h 2 2 I To make V 2 = 0, place a short at port 2 55
G Parameters I gvg I 1 11 1 12 2 V g V g I 2 21 1 22 2 56
G-Parameter Calculations g I V 1 2 11 21 V1 V I 0 1 I 0 g 2 2 To make I 2 = 0, place an open at port 2 57