variable-frequency response analysis network performance as function of frequency. transfer function...
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Variable-Frequency Response AnalysisNetwork performance as function of frequency.Transfer function
Sinusoidal Frequency AnalysisBode plots to display frequency response data
Resonant CircuitsThe resonance phenomenon and its characterization
Filter NetworksNetworks with frequency selective characteristics:low-pass, high-pass, band-pass
VARIABLE-FREQUENCY NETWORKPERFORMANCE
LEARNING GOALS
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 performancevaries with the frequency.
Variation in impedance of basic components
0RRZRResistor
90LLjZL Inductor
Capacitor 9011
CCjZc
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
eq1
)(2
notation" in tionSimplifica"
For all cases seen, and all cases to be studied, the impedance is of the form
011
1
011
1
...
...)(
bsbsbsb
asasasasZ n
nn
n
mm
mm
sCZsLsZRsZ CLR
1,)(,)(
Simplified notation for basic components
Moreover, if the circuit elements (L,R,C, dependent sources) are real then theexpression 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
LEARNING EXAMPLE A possible stereo amplifier
Desired frequency characteristic(flat between 50Hz and 15KHz)
Postulated amplifier
Log frequency scale
Frequency domain equivalent circuit
Frequency Analysis of Amplifier
)(
)(
)(
)(
sV
sV
sV
sV
in
o
S
in
)(/1
)( sVsCR
RsV S
inin
inin
]1000[/1
/1)( in
oo
oo V
RsC
sCsV
ooinin
inin
RsCRsC
RsCsG
1
1]1000[
1)(
000,40
000,40]1000[
100 ss
s
000,401001058.79
100101018.3191
1691
oo
inin
RC
RC
required
actual
000,40
000,40]1000[)(000,40||100
s
ssGs
Frequency dependent behavior iscaused by reactive elements
)(
)()(
sV
sVsG
S
o
Voltage Gain
)50( Hz
)20( kHz
NETWORK FUNCTIONS
INPUT OUTPUT TRANSFER FUNCTION SYMBOLVoltage 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 defineDriving Point Impedance/Admittance
Some nomenclature
admittanceTransfer
tanceTransadmit
)(
)()(
1
2
sV
sIsYT
gain Voltage)(
)()(
1
2
sV
sVsGv
To compute the transfer functions one must solvethe circuit. Any valid technique is acceptable
EXAMPLE
LEARNING EXAMPLE
admittanceTransfer
tanceTransadmit
)(
)()(
1
2
sV
sIsYT
gain Voltage)(
)()(
1
2
sV
sVsGv
The textbook uses mesh analysis. We willuse Thevenin’s theorem
sLRsC
sZTH ||1
)( 11
11
RsL
sLR
sC
)()(
1
112
RsLsC
RsLLCRssZTH
)()( 11
sVRsL
sLsVOC
)(sVOC
)(sZTH
)(2 sV
2R)(2 sI
)(
)()(
22 sZR
sVsI
TH
OC
)(
)(
1
112
2
11
RsLsCRsLLCRs
R
sVRsL
sL
121212
2
)()()(
RCRRLsLCRRs
LCssYT
2 2 22
1 1
( ) ( )( ) ( )
( ) ( )v T
V s R I sG s R Y s
V s V s
)(
)(
1
1
RsLsC
RsLsC
POLES AND ZEROS(More nomenclature)
011
1
011
1
...
...)(
bsbsbsb
asasasasH n
nn
n
mm
mm
Arbitrary network function
Using the roots, every (monic) polynomial can be expressed as aproduct of first order terms
))...()((
))...()(()(
21
210
n
m
pspsps
zszszsKsH
function network the of poles
function network the of zeros
n
m
ppp
zzz
,...,,
,...,,
21
21
The network function is uniquely determined by its poles and zerosand 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
sKsH
84
120
ss
sK
18
1)0( 0KH
84
18)( 2
ss
ssH
LEARNING EXTENSION
)104(
000,20,50
0
70
21
1
K
HzpHzp
z
:poles
:zero
)(
)()(
sV
sVsG
K
S
o
o
gain voltagethefor
of valuethe and locations zero and pole the Find
ooinin
inin
RsCRsC
RsCsG
1
1]1000[
1)(
000,40
000,40]1000[
100 ss
sFor this case the gain was shown to be
))...()((
))...()(()(
21
210
n
m
pspsps
zszszsKsH
Zeros = roots of numeratorPoles = roots of denominator
VariableFrequencyResponse
SINUSOIDAL FREQUENCY ANALYSIS
)(sH
Circuit represented bynetwork function
)cos(0
)(0
tB
eA tj
)(cos|)(|
)(
0
)(0
jHtjHB
ejHA tj
)()()(
)()(
|)(|)(
jeMjH
jH
jHM
Notation
stics.characteri phase and magnitude
calledgenerally are of function as of Plots ),(),(M
)(log)(
))(log2010
10
PLOTS BODE vs
(M
. of function a as function network the
analyze wefrequency the of function a as network a ofbehavior thestudy To
)( jH
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 functionshave simple asymptotic behavior.The asymptotes can be used as reasonable and efficient approximations
]...)()(21)[1(
]...)()(21)[1()()( 2
233310
bbba
N
jjj
jjjjKjH
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)(
211
23
3311
1
b
bba
NjH
Display each basic termseparately and add theresults to obtain final answer
Let’s examine each basic term
Constant Term
Poles/Zeros at the origin
90)(
)(log20|)(|)( 10
Nj
Njj
NdB
NN
linestraight a is this
log is axis- xthe 10
Simple pole or zero j1
1
210
tan)1(
)(1log20|1|
j
j dB
asymptotefrequency low 0|1| dBj
(20dB/dec) asymptotefrequency high 10log20|1| dBj
frequency) akcorner/bre1 whenmeet asymptotes two The (
Behavior in the neighborhood of the corner
FrequencyAsymptoteCurvedistance to asymptote Argument
corner 0dB 3dB 3 45
octave above 6dB 7db 1 63.4
octave below 0dB 1dB 1 26.6
125.0
0)1( j
90)1( j
1
1
Asymptote for phase
High freq. asymptoteLow freq. Asym.
Simple zero
Simple pole
Quadratic pole or zero ])()(21[ 22 jjt ])()(21[ 2 j
222102 2)(1log20|| dBt 2
12
)(1
2tan
t
1 asymptotefrequency low 0|| 2 dBt 02t
1 asymptote freq. high 2102 )(log20|| dBt 1802t1 )2(log20|| 102 dBt 902tCorner/break frequency
221 2102 12log20|| dBt
21
221
tan
t2
2
Magnitude for quadratic pole Phase for quadratic pole
dB/dec40
These graphs are inverted for a zero
LEARNING EXAMPLEGenerate 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
asymptotes
LEARNING EXAMPLEGenerate magnitude and phase plots
)11.0()(
)1(25)( 2
jj
jjGv 10 1, :(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
21
020 0)(
Kj
K
dB
Final results . . . And an extra hint on poles at the origin
dec
dB40
dec
dB20
dec
dB40
LEARNING EXTENSION Sketch the magnitude characteristic
)100)(10(
)2(10)(
4
jj
jjG
form standard in NOT is function theBut
100 10, 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 Sketch the magnitude characteristic
2)(
102.0(100)(
j
jjG
origin theat pole Double
50at break
form standard in isIt
40
20
0
20
dB
90
270
90
1 10 100 1000
Once each term is drawn we form the composites
Put in standard form
)110/)(1()(
jj
jjG
LEARNING EXTENSION Sketch the magnitude characteristic
)10)(1(
10)(
jj
jjG
10 1, :breaks
origin theat zero
form standard innot
40
20
0
20
dB
90
270
90
1.0 110
100Once each term is drawn we form the composites
decdB /20decdB /20
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 postulatea 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 0.11)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
LEARNING EXTENSIONDetermine a transfer function from the composite magnitude asymptotes plot
A
A. Pole at the origin. Crosses 0dB line at 5
j5
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
These are circuits with very special frequency characteristics…And resonance is a very important physical phenomenon
LCCL
110
when zero iscircuit each of reactance The
The frequency at which the circuit becomes purely resistive is calledthe resonance frequency
CjLjRjZ
1
)(
circuit RLC Series
LjCjGjY
1
)(
circuit RLC Parallel
Properties of resonant circuits
At resonance the impedance/admittance is minimal
Current through the serial circuit/voltage across the parallel circuit canbecome very large
CRR
LQ
0
0 1
:FactorQuality
222 )1
(||
1)(
CLRZ
CjLjRjZ
222 )1
(||
1)(
LCGY
CjLj
GjY
Given the similarities between series and parallel resonant circuits, we will focus on serial circuits
Properties of resonant circuits
CIRCUIT BELOW RESONANCE ABOVE RESONANCESERIES CAPACITIVE INDUCTIVEPARALLEL INDUCTIVE CAPACITIVE
Phasor diagram for series circuit Phasor diagram for parallel circuit
RV
C
IjVC
Lj
1GV 1CVj
L
Vj1
LEARNING EXAMPLEDetermine the resonant frequency, the voltage across eachelement at resonance and the value of the quality factor
LC
10 sec/2000
)1010)(1025(
163
radFH
I
AZ
VI S 5
2
010
2Z resonanceAt
50)1025)(102( 330L
)(902505500 VjLIjVL
902505501
501
0
00
jICj
V
LC
C
R
LQ 0 25
2
50
||||
|||| 0
SC
SS
L
VQV
VQR
VLV
resonanceAt
LEARNING EXTENSION Find the value of C that will place the circuit in resonanceat 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 thecapacitor
R
LQ 0 60
3
1.01800
Q
||||
|||| 0
SC
SS
L
VQV
VQR
VLV
resonanceAt
VVC 600||
Resonance for the series circuit
222 )1
(||
1)(
CLRZ
CjLjRjZ
QRCQRL
1, 00
:resonanceAt
)(1
)(
0
0
0
0
jQR
QRjQRjRjZ
)(1
1
0
0
1
jQV
VG Rv
is gain voltageThe :Claim
)(1
jZ
R
CjLjR
RGv
vv GGM |)(|,|)(
2/120
0
2 )(1
1)(
Q
M
)(tan)( 0
0
1
Q
QBW 0
12
1
2
12
0 QQLO
sfrequenciepower Half
Z
RGv
LEARNING EXAMPLE
2
mH2
F5
Determine the resonant frequency, quality factor andbandwidth when R=2 and when R=0.2
CRR
LQ
0
0 1
LC
10
QBW 0
sec/10)105)(102(
1 4
630 rad
R Q2 10
0.2 100
R Q BW(rad/sec)2 10 1000
0.2 100 100
LEARNING EXTENSIONA series RLC circuit as the following properties:sec/100sec,/4000,4 0 radBWradR
Determine the values of L,C.
CRR
LQ
0
0 1
LC
10
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 662
020
1056.11016104
111
PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR
|||| 0 RVQV resonanceAt
But this is NOT the maximum value for thevoltage across the capacitor
CRjLCCj
LjR
CjV
V
S
20
1
11
1
2
221
1)(
Qu
u
ug
2
0
0
;SV
Vgu
22
22
2
1
)/1)(/(2)2)(1(20
Qu
u
QQuuu
du
dg
CRR
LQ
0
0 1
LC
10
22 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
mH50
F5
150, RR and when Determine max0
sradLC
/2000)105)(105(
11620
CRR
LQ
0
0 1
LC
10
20
maxmax
2
11
Qu
RQ
050.02000 2max 2
112000Q
R Q Wmax50 2 18711 100 2000
Evaluated with EXCEL and rounded to zero decimals
Using MATLAB one can display the frequency response
R=50Low QPoor selectivity
R=1High QGood selectivity
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 onPASSIVE filters
Simple low-pass filter
RCjCj
R
CjV
VGv
1
11
1
1
0
RCj
Gv
;
1
1
1
2
tan)(
1
1||)(
v
v
G
GM
2
11,1max
MM
frequencypower half
1
1
BW
Simple high-pass filter
CRj
CRj
CjR
R
V
VGv
11
1
0
RCj
jGv
;1
1
2
tan2
)(
1||)(
v
v
G
GM
2
11,1max
MM
frequencypower half
1
1
LO
Simple band-pass filter
Band-pass
CLjR
R
V
VGv
11
0
222 1)(
LCRC
RCM
11
LCM 0)()0( MM
2
4/)/( 20
2
LRLRLO
LC
10
2
4/)/( 20
2
LRLRHI
L
RBW LOHI
)(2
1)( HILO MM
Simple band-reject filter
011
000
C
LjLC
10 VV circuit open as actscapacitor the 0at
10 VV circuit open as actsinductor the at
filter pass-band
the in as determined are HILO ,
LEARNING EXAMPLEDepending on where the output is taken, this circuitcan 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
CjV
V
S
C
1
1
0)(,10 S
C
S
C
V
V
V
VLow-pass
FCHLR 159,159,10 for plot Bode
ACTIVE FILTERS
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 amore general framework, if one replaces the resistors by impedances
Ideal Op-Amp
These currents arezero
Basic Inverting Amplifier
0V
VV gain Infinite
0V
0 II -impedanceinput Infinite
1
1 2
0OVV
Z Z
21
1O
ZV V
Z
Linear circuit equivalent
0I
1
2
Z
ZG
1
11 Z
VI
EXAMPLE USING INVERTING AMPLIFIER
LOW PASS FILTER
Basic Non-inverting amplifier
1V
1V
0I
1
1
2
10
Z
V
Z
VV
11
120 V
Z
ZZV
1
21Z
ZG
01 I
EXAMPLEUSING NON INVERTING CONFIGURATION
EXAMPLE SECON ORDER FILTER
2( )V s
22 2 2
1 1 2 3
01/
OINV VV V V V
R C s R R
2
2 2
01/
OVV
R C s
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