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8/10/2019 Variable frequency networks - basic engineering circuit analysis Irwin
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Variable-Frequency Response Analysis
Network performance as function of frequency.
Transfer function
Sinusoidal Frequency AnalysisBode plots to display frequency response data
Resonant Circuits
The resonance phenomenon and its characterization
Filter NetworksNetworks with frequency selective characteristics:
low-pass, high-pass, band-pass
VARIABLE-FREQUENCY NETWORK
PERFORMANCE
LEARNING GOALS
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VARIABLE FREQUENCY-RESPONSE ANALYSIS
In AC steady state analysis the frequency is assumed constant (e.g., 60Hz).
Here we consider the frequency as a variable and examine how the performance
varies with the frequency.
Variation in impedance of basic components
0RRZRResistor
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90LLjZL Inductor
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Capacitor 9011
CCjZc
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Frequency dependent behavior of series RLC network
Cj
RCjLCj
CjLjRZeq
1)(1 2
C
LCjRC
j
j
)1( 2
C
LCRCZeq
222 )1()(||
RC
LCZeq
1tan
21
sC
sRCLCssZ
sj
eq
1)(
2
notation"intionSimplifica"
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For all cases seen, and all cases to be studied, the impedance is of the form
011
1
01
1
1
......)(
bsbsbsbasasasasZ n
n
n
n
m
m
m
m
sCZsLsZRsZ CLR
1,)(,)(
Simplified notation for basic components
Moreover, if the circuit elements (L,R,C, dependent sources) are real then the
expression for any voltage or current will also be a rational function in s
LEARNING EXAMPLE
sL
sC
1
R
So VsCsLR
RsV
/1)(
SV
sRCLCs
sRC
12
So VRCjLCj
RCjV
js
1)( 2
0101)1053.215()1053.21.0()(
)1053.215(332
3
jj
jVo
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LEARNING EXAMPLE A possible stereo amplifier
Desired frequency characteristic
(flat between 50Hz and 15KHz)
Postulated amplifier
Log frequency scale
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Frequency domain equivalent circuit
Frequency Analysis of Amplifier
)()(
)()(
sVsV
sVsV
i n
o
S
i n
)(/1
)( sVsCR
RsV S
i ni n
i ni n
]1000[/1
/1)( i n
oo
oo V
RsC
sCsV
ooi ni n
i ni n
RsCRsCRsCsG
11]1000[
1)(
000,40
000,40]1000[100 sss
000,401001058.79
100101018.3
191
1691
oo
i ni n
RC
RC
required
actual
000,40000,40]1000[)(000,40||100
sssGs
Frequency dependent behavior is
caused by reactive elements
)()()(
sVsVsG
S
o
Voltage Gain
)50( H z
)20( kH z
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NETWORK FUNCTIONS
INPUT OUTPUT TRANSFER FUNCTION SYMBOL
Voltage Voltage Voltage Gain Gv(s)
Current Voltage Transimpedance Z(s)
Current Current Current Gain Gi(s)
Voltage Current Transadmittance Y(s)
When voltages and currents are defined at different terminal pairs we
define the ratios as Transfer Functions
If voltage and current are defined at the same terminals we define
Driving Point Impedance/Admittance
Some nomenclature
admittanceTransfer
tanceTransadmit
)(
)()(
1
2
sV
sIsYT
gain Voltage)(
)(
)(1
2
sV
sV
sGv
To compute the transfer functions one must solve
the circuit. Any valid technique is acceptable
EXAMPLE
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LEARNING EXAMPLE
admittanceTransfer
tanceTransadmit
)(
)()(
1
2
sV
sIsYT
gain Voltage
)(
)()(
1
2
sV
sVsGv
The textbook uses mesh analysis. We will
use Thevenins theorem
sLRsC
sZT H ||1
)( 11
11
RsL
sLR
sC
)()(
1
112
RsLsC
RsLLCRssZT H
)()( 11
sVRsL
sLsVOC
)(sVOC
)(sZTH
)(2 sV2R
)(2 sI
)(
)()(
2
2sZR
sVsI
T H
OC
)(
)(
1
112
2
1
1
RsLsC
RsLLCRsR
sVRsL
sL
121212
2
)()()( RCRRLsLCRRs
LCssYT
2 2 22
1 1
( ) ( )( ) ( )
( ) ( )v T
V s R I s G s R Y s
V s V s
)(
)(
1
1
RsLsC
RsLsC
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POLES AND ZEROS (More nomenclature)
011
1
011
1
...
...)(
bsbsbsb
asasasasH
n
n
n
n
m
m
m
m
Arbitrary network function
Using the roots, every (monic) polynomial can be expressed as a
product of first order terms
))...()((
))...()(()(
21
210
n
m
pspsps
zszszsKsH
functionnetworktheofpolesfunctionnetworktheofzeros
n
m
ppp
zzz
,...,,
,...,,
21
21
The network function is uniquely determined by its poles and zeros
and its value at some other value of s (to compute the gain)
EXAMPLE
1)0(
22,22
,1
21
1
H
jpjp
z
:poles
:zeros
)22)(22()1()( 0
jsjs
sKsH84
120
ss
sK
18
1)0( 0KH
84
18)(
2
ss
ssH
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LEARNING EXTENSION
)104(
000,20,50
0
7
0
21
1
K
H zpH zp
z
:poles
:zero
)(
)()(
sV
sVsG
K
S
o
o
gainvoltagethefor
ofvaluetheandlocationszeroandpoletheFind
ooi ni n
i ni n
RsCRsCRsCsG
11]1000[
1)(
000,40
000,40]1000[100 sss
For this case the gain was shown to be
))...()((
))...()(()(
21
210
n
m
pspsps
zszszsKsH
Zeros = roots of numerator
Poles = roots of denominator
Variable
Frequency
Response
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SINUSOIDAL FREQUENCY ANALYSIS
)(sH
Circuit represented by
network function
)cos(0
)(0
tB
eA tj
)(cos|)(|
)(
0
)(0
jHtjHB
ejHA tj
)()()(
)()(
|)(|)(
jeMjH
jH
jHM
Notation
stics.characteriphaseandmagnitudecalledgenerallyareoffunctionasofPlots ),(),(M
)(log)(
))(log2010
10
PLOTSBODE vs
(M
.offunctionaasfunctionnetworkthe
analyzewefrequencytheoffunctionaasnetworkaofbehaviorthestudyTo
)(jH
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HISTORY OF THE DECIBEL
Originated as a measure of relative (radio) power
1
2)2 log10(|
P
PP dB 1Pover
21
22
21
22
)2
22
log10log10(|I
I
V
VP
R
VRIP dB 1Pover
By extension
||log20|
||log20|
||log20|
10
10
10
GG
II
VV
dB
dB
dB
Using log scales the frequency characteristics of network functions
have simple asymptotic behavior.The asymptotes can be used as reasonable and efficient approximations
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]...)()(21)[1(
]...)()(21)[1()(
)( 2
233310
bbba
N
jjj
jjjjK
jH
General form of a network function showing basic terms
Frequency independent
Poles/zeros at the origin
First order terms Quadratic terms for
complex conjugate poles/zeros
..|)()(21|log20|1|log20
...|)()(21|log20|1|log20
||log20log20
21010
233310110
10010
bbba jjj
jjj
jNK
DND
N
BAAB
loglog)log(
loglog)log(
|)(|log20|)(| 10 jHjH dB
212
1
2121
zzz
z
zzzz
...)(1
2tantan
...)(1
2tantan
900)(
2
11
23
331
11
b
bba
NjH
Display each basic term
separately and add theresults to obtain final
answer
Lets examine each basic term
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Constant Term
Poles/Zeros at the origin
90)(
)(log20|)(|
)(
10
Nj
Nj
j NdB
NN
linestraightaisthis
logisaxis-xthe 10
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Simple pole or zero j1
1
210
tan)1(
)(1log20|1|
j
j dB
asymptotefrequencylow0|1| dBj
(20dB/dec)asymptotefrequencyhigh 10log20|1| dBj
frequency)akcorner/bre1whenmeetasymptotestwoThe (
Behavior in the neighborhood of the corner
Frequenc Asymptot Curve
distance to
asymptote Argument
corner 0dB 3dB 3 45
octave above 6dB 7db 1 63.4
octave below 0dB 1dB 1 26.6
12
5.0
0)1( j
90)1( j
1
1
Asymptote for phase
High freq. asymptoteLow freq. Asym.
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Simple zero
Simple pole
2 2
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Quadratic pole or zero ])()(21[ 2
2 jjt ])()(21[ 2
j
222102 2)(1log20|| dBt 21
2)(1
2tan
t
1 asymptotefrequencylow0|| 2 dBt 02t
1 asymptotefreq.high2102 )(log20|| dBt 1802
t
1 )2(log20|| 102 dBt 902tCorner/break frequency
221 2102 12log20|| dBt
2
12
21tan
t
2
2
Magnitude for quadratic pole Phase for quadratic pole
dB/dec40
These graphs are inverted for a zero
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LEARNING EXAMPLE Generate magnitude and phase plots
)102.0)(1(
)11.0(10)(
jj
jjGvDraw asymptotes
for each term1,10,50:nersBreaks/cor
40
20
0
20
dB
90
90
1.0 1 10 100 1000
dB|10
decdB /20
dec/45
decdB/20
dec/45
Draw composites
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asymptotes
G t it d d h l t
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LEARNING EXAMPLE Generate magnitude and phase plots
)11.0()(
)1(25)(
2
jj
jjGv 101,:(corners)Breaks
40
20
0
20
dB
90
270
90
1.0 1 10 100
Draw asymptotes for each
dB28
decdB/40
180
dec/45
45
Form composites
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21
020 0)(
KjK
dB
Final results . . . And an extra hint on poles at the origin
dec
dB40
dec
dB20
dec
dB40
LEARNING EXTENSION Sk t h th it d h t i ti
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LEARNING EXTENSION Sketch the magnitude characteristic
)100)(10(
)2(10)(
4
jj
jjG
formstandardinNOTisfunctiontheBut
10010,2,:breaks
Put in standard form
)1100/)(110/(
)12/(20)(
jj
jjG
We need to show about
4 decades
40
20
0
20
dB
90
90
1 10 100 1000
26 |dB
LEARNING EXTENSION Sk t h th it d h t i ti
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LEARNING EXTENSION Sketch the magnitude characteristic
2)(
102.0(100)(
j
jjG
origintheatpoleDouble
50atbreak
formstandardinisIt
40
20
0
20
dB
90
270
90
1 10 100 1000
Once each term is drawn we form the composites
LEARNING EXTENSION Sketch the magnitude characteristic
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Put in standard form
)110/)(1()(
jj
jjG
LEARNING EXTENSION Sketch the magnitude characteristic
)10)(1(
10)(
jj
jjG
101,:breaks
origintheatzero
formstandardinnot
40
20
0
20
dB
90
270
90
1.0 110
100Once each term is drawn we form the composites
decdB/20
decdB/20
DETERMINING THE TRANSFER FUNCTION FROM THE BODE PLOT
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DETERMINING THE TRANSFER FUNCTION FROM THE BODE PLOT
This is the inverse problem of determining frequency characteristics.
We will use only the composite asymptotes plot of the magnitude to postulate
a transfer function. The slopes will provide information on the order
A
A. different from 0dB.
There is a constant Ko
B
B. Simple pole at 0.1
1
)11.0/(
j
C
C. Simple zero at 0.5
)15.0/( j
D
D. Simple pole at 3
1)13/( j
E
E. Simple pole at 20
1)120/( j
)120/)(13/)(11.0/(
)15.0/(10)(
jjj
jjG
20
|
00
0
1020|dBK
dB KK
If the slope is -40dB we assume double real pole. Unless we are given more data
Determine a transfer function from the composite
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LEARNING EXTENSIONDetermine a transfer function from the composite
magnitude asymptotes plot
A
A. Pole at the origin.
Crosses 0dB line at 5
j
5
B
B. Zero at 5
C
C. Pole at 20
D
D. Zero at 50
E
E. Pole at 100
)1100/)(120/(
)150/)(15/(5)(
jjj
jjjG
Sinusoidal
RESONANT CIRCUITS
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RESONANT CIRCUITS
These are circuits with very special frequency characteristics
And resonance is a very important physical phenomenon
L CCL
110
whenzeroiscircuiteachofreactanceThe
The frequency at which the circuit becomes purely resistive is called
the resonance frequency
CjLjRjZ
1)(
circuitRLCSeries
LjCjGjY
1)(
circuitRLCParallel
P i f i i
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Properties of resonant circuits
At resonance the impedance/admittance is minimal
Current through the serial circuit/
voltage across the parallel circuit can
become very large
CRR
LQ
0
0 1
:FactorQuality
222)
1(||
1)(
CLRZ
Cj
LjRjZ
222 )1
(||
1)(
LCGY
Cj
Lj
GjY
Given the similarities between series and parallel resonant circuits,
we will focus on serial circuits
P ti f t i it
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Properties of resonant circuits
CIRCUIT BELOW RESONANCE ABOVE RESONANCE
SERIES CAPACITIVE INDUCTIVE
PARALLEL INDUCTIVE CAPACITIVE
Phasor diagram for series circuit Phasor diagram for parallel circuit
RV
C
IjVC
Lj
1GV 1CVj
L
Vj
1
LEARNING EXAMPLE Determine the resonant frequency the voltage across each
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LEARNING EXAMPLE Determine the resonant frequency, the voltage across each
element at resonance and the value of the quality factor
LC
10 sec/2000
)1010)(1025(
163
radFH
I
A
Z
VI S 5
2
010
2ZresonanceAt
50)1025)(102( 330 L
)(902505500 VjLIjVL
902505501
501
0
0
0
jICj
V
LC
C
R
LQ 0
25
2
50
||||
|||| 0
SC
SS
L
VQV
VQR
VLV
resonanceAt
LEARNING EXTENSION Fi d th l f C th t ill l th i it i
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LEARNING EXTENSION Find the value of C that will place the circuit in resonance
at 1800rad/sec
LC
10 218001.0
1
)(1.0
11800
C
CH
FC 86.3
Find the Q for the network and the magnitude of the voltage across the
capacitor
R
LQ 0
60
3
1.01800
Q
||||
|||| 0
SC
SS
L
VQV
VQR
VLV
resonanceAt
VVC 600||
Resonance for the series circuit
1)( M
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Resonance for the series circuit
222)
1(||
1)(
C
LRZ
CjLjRjZ
QR
CQRL 1
, 00
:resonanceAt
)(1
)(
0
0
0
0
j QR
QRjQRjRjZ
)(1
1
0
0
1
jQV
VG Rv
isgainvoltageThe:Claim
)(1
jZ
R
CjLjR
RGv
vv GGM |)(|,|)(
2/1
20
0
2 )(1
)(
Q
M
(tan)( 0
0
1
Q
QBW
0
12
1
2
1 2
0 QQL O
sfrequenciepowerHalf
Z
RGv
LEARNING EXAMPLE D t i th t f lit f t d
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LEARNING EXAMPLE
2
mH2
F5
Determine the resonant frequency, quality factor and
bandwidth when R=2 and when R=0.2
CRR
LQ
0
0 1
LC
10
QBW 0
sec/10)105)(102(
1 4630
rad
R Q
2 10
0.2 100
R Q BW(rad/sec)
2 10 1000
0.2 100 100
LEARNING EXTENSION A series RLC circuit as the following properties:
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LEARNING EXTENSION A series RLC circuit as the following properties:
sec/100sec,/4000,4 0 radBWr adR
Determine the values of L,C.
CRR
L
Q 0
0 1
LC
1
0
QBW
0
1. Given resonant frequency and bandwidth determine Q.
2. Given R, resonant frequency and Q determine L, C.
40100
40000 BW
Q
HQR
L 040.04000
440
0
FRQL
C 6
620
20
1056.11016104
111
PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR
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PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR
|||| 0 RVQV
resonanceAt
But this is NOT the maximum value for the
voltage across the capacitor
CRjL C
CjLjR
Cj
V
V
S
20
1
1
1
1
2
221
1)(
Q
uu
ug
2
0
0
;SV
Vgu
22
22
2
1
)/1)(/(2)2)(1(20
Q
u
u
QQuuu
du
dg
CRR
LQ
0
0 1
LC
10
2
2 1)1(2Q
u
20
maxmax
2
11
Qu
2
2
424
max
4
11
2
11
4
1
1
Q
Q
QQQ
g
2
0
4
11
||||
Q
VQV S
LEARNING EXAMPLE 150, RR andwhenDetermine max0
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LEARNING EXAMPLE
mH50
F5
, max0
sr adLC
/2000)105)(105(
11620
CRR
L
Q 0
0 1
LC
1
0
20
maxmax
2
11
Qu
RQ
050.02000 2max 2
112000Q
R Q Wmax
50 2 18711 100 2000
Evaluated with EXCEL and rounded to zero decimals
Using MATLAB one can display the frequency response
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R=50Low Q
Poor selectivity
R=1
High Q
Good selectivity
FILTER NETWORKS
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FILTER NETWORKS
Networks designed to have frequency selective behavior
COMMON FILTERS
Low-pass filterHigh-pass filter
Band-pass filter
Band-reject filter
We focus first on
PASSIVE filters
Simple low-pass filter
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Simple low-pass filter
RCjCj
R
Cj
V
VGv
1
1
1
1
1
0
RCj
Gv
;1
1
1
2
tan)(
1
1||)(
v
v
G
GM
2
11,1max
MM
frequencypowerhalf 1
1BW
Simple high-pass filter
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Simple high-pass filter
CRj
CRj
CjR
R
V
VGv
111
0
RCj
jGv
;1
1
2
tan2)(
1||)(
v
v
G
GM
2
11,1max
MM
frequencypowerhalf
1
1L O
Simple band-pass filter
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p p
Band-pass
CLjR
R
V
VGv
11
0
222 1)(
LCRC
RCM
11
LC
M 0)()0( MM
2
4/)/( 202
LRLRL O
LC
10
2
4/)/( 202 LRLR
H I
L
RBW L OH I
)(2
1)( H IL O MM
Simple band reject filter
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Simple band-reject filter
0110
00
CLj
LC
10 VV circuitopenasactscapacitorthe0at
10 VV circuitopenasactsinductortheat
filterpass-band
theinasdeterminedareH IL O ,
LEARNING EXAMPLE Depending on where the output is taken, this circuit
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can produce low-pass, high-pass or band-pass or band-
reject filters
Band-pass
Band-reject filter
CLjR
Lj
V
V
S
L
1 1)(,00 S
L
S
L
V
V
V
VHigh-pass
CLjR
Cj
V
V
S
C
1
1
0)(,10 S
C
S
C
V
V
V
VLow-pass
FCHLR 159,159,10 forplotBode
ACTIVE FILTERS
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Passive filters have several limitations
1. Cannot generate gains greater than one
2. Loading effect makes them difficult to interconnect
3. Use of inductance makes them difficult to handle
Using operational amplifiers one can design all basic filters, and more,
with only resistors and capacitors
The linear models developed for operational amplifiers circuits are valid, in a
more general framework, if one replaces the resistors by impedances
Ideal Op-Amp
These currents are
zero
Basic Inverting Amplifier
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Basic Inverting Amplifier
0V
VVgainInfinite
0V
0 II-impedanceinputInfinite
1
1 2
0OVV
Z Z
21
1
O
ZV V
Z
Linear circuit equivalent
0
I
1
2
Z
ZG
1
11
Z
VI
EX MPLE
USING INVERTING MPLIFIER
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LOW P SS FILTER
Basic Non-inverting amplifier
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Basic Non-inverting amplifier
1V
1V
0I
1
1
2
10Z
V
Z
VV
1
1
120 V
Z
ZZV
1
21Z
ZG
01I
EX MPLE
USING NON INVERTING CONFIGUR TION
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EX MPLE SECON ORDER FILTER
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