Post on 02-Feb-2016
description
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
ScalingImpedance and frequency scaling
Filter NetworksNetworks with frequency selective characteristics:low-pass, high-pass, band-pass
VARIABLE-FREQUENCY NETWORKPERFORMANCE
LEARNING GOALS
0RRZRResistor
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
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
MATLAB can be effectively used to compute frequency response characteristics
USING MATLAB TO COMPUTE MAGNITUDE AND PHASE INFORMATION
011
1
011
1
...
...)(
bsbsbsb
asasasasV n
nn
n
mm
mm
o
),(
];,,...,,[
];,,...,,[
011
011
dennumfreqs
bbbbden
aaaanum
nn
mm
MATLAB commands required to display magnitudeand phase as function of frequency
NOTE: Instead of comma (,) one can use space toseparate numbers in the array
1)1053.215()1053.21.0()(
)1053.215(332
3
jj
jVo
EXAMPLE
» num=[15*2.53*1e-3,0];» den=[0.1*2.53*1e-3,15*2.53*1e-3,1];» freqs(num,den)
1a
2b1b
0b
Missing coefficients mustbe entered as zeros
» num=[15*2.53*1e-3 0];» den=[0.1*2.53*1e-3 15*2.53*1e-3 1];» freqs(num,den)
This sequence will alsowork. Must be careful notto insert blanks elsewhere
GRAPHIC OUTPUT PRODUCED BY MATLAB
Log-logplot
Semi-logplot
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
EXAMPLE
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
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
1
22
1
sYRsV
sIR
sV
sVsG T
sv
)(
)(
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 EXTENSIONFind the driving point impedance at )(sVS
)(
)()(
sI
sVsZ S
)(sI
)(1
)()(: sIsC
sIRsVin
inS KVL
in
in sCRsZ
1)(
Replace numerical values
M
s
1001
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 1802t
1 )2(log20|| 102 dBt 902tCorner/break frequency
221 2102 12log20|| dBt
2
12
21tan
t
2
2Resonance frequency
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
dB|25
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
LEARNING EXAMPLEA function with complex conjugate poles
1004)()5.0(
25)( 2
jjj
jjG
Put in standard form
125/)10/()15.0/(
5.0)( 2
jjj
jjG
40
20
0
20
dB
90
90
01.0 1.0 1 10 100 270
1 )2(log20|| 102 dBt
2.01.0
25/12
])()(21[ 22 jjt
dB8
Draw composite asymptote
Behavior close to corner of conjugate pole/zerois too dependent on damping ratio.Computer evaluation is better
Evaluation of frequency response using MATLAB
1004)()5.0(
25)( 2
jjj
jjG
» num=[25,0]; %define numerator polynomial» den=conv([1,0.5],[1,4,100]) %use CONV for polynomial multiplicationden = 1.0000 4.5000 102.0000 50.0000» freqs(num,den)
LEARNING EXTENSIONSketch the magnitude characteristic
]136/)12/[(
)1(2.0)( 2
jjj
jjG 6/136/12
12/1
])()(21[ 22 jjt
40
20
0
20
dB
90
270
90
1.0 110
100
1 )2(log20|| 102 dBt
decdB /20
decdB /40
decdB /0
12
dB5.9
]136/)12/[(
)1(2.0)( 2
jjj
jjG
» num=0.2*[1,1];» den=conv([1,0],[1/144,1/36,1]);» freqs(num,den)
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
CjLjRjZ
1
)(
circuit RLC Series
LjCjGjY
1
)(
circuit RLC Parallel
LCCL
110
when zero iscircuit each of reactance The
The frequency at which the circuit becomes purely resistive is calledthe resonance frequency
Properties of resonant circuits
At resonance the impedance/admittance is minimal
Current through the serial circuit/voltage across the parallel circuit canbecome very large (if resistance is small)
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
At resonance the power factor is unity
CIRCUIT BELOW RESONANCE ABOVE RESONANCESERIES CAPACITIVE INDUCTIVEPARALLEL INDUCTIVE CAPACITIVE
Phasor diagram for series circuit Phasor diagram for parallel circuit
RV
C
IjVC
Lj
1GV1CVj
L
Vj
1
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 EXAMPLEGiven L = 0.02H with a Q factor of 200, determine the capacitornecessary to form a circuit resonant at 1000Hz
R
L0200200Q withL
LC
10
C02.0
110002 FC 27.1
What is the rating for the capacitor if the circuit is tested with a 10V supply?
VVC 2000|| ||||
|||| 0
SC
SS
L
VQV
VQR
VLV
resonanceAt
59.1200
02.010002R
AI 28.659.1
10
The reactive power on the capacitorexceeds 12kVA
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
The Q factorCRR
LQ
0
0 1
R LowQ High :circuit seriesFor G) (low R HighQ High :circuit parallelFor
M
BW SmallQ High
dissipates
Stores as Efield
Stores as M field
Capacitor and inductor exchange storedenergy. When one is at maximum the other is at zero
D
S
W
WQ 2
cycleby dissipatedenergy storedenergy maximum2
Q can also be interpreted from anenergy point of view
22
1
20202 mxeffD RIRIW
22
2
1
2
1mxmxS CVLIW
220 Q
R
L
W
W
D
s
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
Evaluated with EXCEL
RQ
002.010000 QBW /10000
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
LEARNING EXAMPLEFind R, L, C so that the circuit operates as a band-pass filterwith center frequency of 1000rad/s and bandwidth of 100rad/s
)(1
jZ
R
CjLjR
RGv
CRR
LQ
0
0 1
LC
10
QBW 0
dependent
Strategy: 1. Determine Q2. Use value of resonant frequency and Q to set up two equations in the three unknowns3. Assign a value to one of the unknowns
10100
10000 BW
Q
R
L
R
LQ
1000100
LCLC
1)10(
1 230
For example FFC 6101
HL 1
100R
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(2
Qu
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
Natural frequency depends only on L, C.Resonant frequency depends on Q.
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
LEARNING EXAMPLE The Tacoma Narrows Bridge Opened: July 1, 1940Collapsed: Nov 7, 1940
Likely cause: windvarying at frequencysimilar to bridgenatural frequency
2.020
Tacoma Narrows Bridge Simulator
)11( ftV
42
40
BA
B
in RR
R
v
v
resonanceat model theFor
.deflection 4' caused wind 42mph a failureAt
1
5.9H20
F66.31
42mxVin
Assume a low Q=2.39
0.44’
1.07’
'77.3
PARALLEL RLC RESONANT CIRCUITS
222 )1
(||
1)(
CLRZ
CjLjRjZ
222 )1
(||
1)(
LCGY
CjLj
GjY
Impedance of series RLC Admittance of parallel RLC
IVYZ
LCCLGR
,
,,
esequivalenc Notice
SS YVI
SSL
SSC
SSG
IYLj
VLj
I
IY
CjCVjI
IY
GGVI
11
||1
||
||||
1
0
0
SL
SC
LC
SG
ILG
I
IG
CI
II
II
GYL
C
0
0
resonanceAt
CRR
LQ
0
0 1
LC
10
Series RLC
Parallel RLC
LGG
CQ
0
0 1
LC
10
|| SIQ
Series RLCQ
BW 0
Parallel RLCQ
BW 0
LEARNING EXAMPLE
mHLFC
SGVS120,600
01.0,0120
If the source operates at the resonant frequency of the network, compute all the branch currents
SSL
SSC
SSG
IYLj
VLj
I
IY
CjCVjI
IY
GGVI
11
||1
||
||||
1
0
0
SL
SC
LC
SG
ILG
I
IG
CI
II
II
GYL
C
0
0
resonanceAt
|| SIQ
SG IAI )(02.1012001.0
sradLC
/85.117)106(120.0
1140
)(9049.80120)10600()85.117()901( 6 AIC
)(9049.8 AIL
_______xI
LEARNING EXAMPLEDerive expressions for the resonant frequency, half powerfrequencies, bandwidth and quality factor for the transfercharacteristic
in
out
I
VH
LjCjGYT
1
Tin
out
T
inout YI
VH
Y
IV
1
22 1
11
1||
LCGLj
CjGH
LC
10 :frequencyResonant
22max
||5.0|)(| HjH h sfrequenciepower Half
22
2 21
GL
CGh
h
RG
H 1
|| max
GL
Ch
h
1
LCC
G
C
Gh
1
22
2
C
GBW LOHI
L
CR
L
C
GBWQ
10
LGG
CQ
0
0 1
12
1
2
12
0 QQLO
Replace and show
LEARNING EXAMPLE Increasing selectivity by cascading low Q circuits
Single stage tuned amplifier
MHzsradFHLC
9.99/10275.61054.210
11 8
1260
398.010
1054.2250 6
12
L
CR
L
C
GBWQ
10
LEARNING EXTENSIONDetermine the resonant frequency, Q factor and bandwidth
FCmHLkR 150,20,2
Parallel RLC
LGG
CQ
0
0 1
LC
10
QBW 0
srad /577)10150)(1020(
1630
1732000/1
10150577 6
Q
sradBW /33.3173
577
LEARNING EXTENSION 0 C, L, Determine
120,/1000,6 QsradBWkR
Parallel RLC
LGG
CQ
0
0 1
LC
10
QBW 0
sradBWQ /102.11000120 50
FR
QC
167.0
102.16000
1205
0
HQ
RL
417
102.1120
60005
0
Can be used to verify computations
PRACTICAL RESONANT CIRCUITThe resistance of the inductor coils cannot beneglected
LjRCjjY
1)(
LjR
LjR
22 )()(
LR
LjRCjjY
2222 )()()(
LR
LCj
LR
RjY
2
22
10
)(
L
R
LCLR
LCY R
real
R
LQ
LC0
00 ,1 2
00
11
QR
maxima are impedance and voltagethe resonanceAt .Y
IZIV
20
2
0
222
11R
LR
R
LR
R
LRZ RRRMAX
20RQZMAX
How do you define a quality factor for this circuit?
LEARNING EXAMPLE 5,50, RR for both Determine 0
R
LQ
LC0
00 ,1 2
00
11
QR
sradFH
/2000)105)(1050(
1630
20
01
12000,050.02000
QRQ R
R Q0 Wr(rad/s) f(Hz)50 2 1732 275.75 20 1997 317.8
RESONANCE IN A MORE GENERAL VIEW
222 )1
(||
1)(
CLRZ
CjLjRjZ
222 )1
(||
1)(
LCGY
CjLj
GjY
For series connection the impedance reaches maximum at resonance. For parallelconnection the impedance reaches maximum
1)(1)( 22
LGjLCj
LjZ
CRjLCj
CjY ps
12)( 2 jj
as written was term quadratic the plots Bode In
0
1
LC
QCRCR
122 0
series
QLGLG
122 0
parallel
2
1QA high Q circuit is highly
under damped
Resonance
SCALING
Scaling techniques are used to change an idealized network into a morerealistic one or to adjust the values of the components
M
M
M
K
CC
LKL
RKR
'
'
'
scaling impedanceor Magnitude
''
11'' 0 CLLC
CLLC
'
'00
R
L
R
LQ
Magnitude scaling does not change thefrequency characteristics nor the qualityof the network.
CCLL
ωKω' F
1
''
1,''
unchanged iscomponent each of Impedance
scaling timeor Frequency
F
F
K
CC
K
LL
RR
'
'
'
0'0 FK
)('
''0 BWKQ
BW F
QR
LQ
'
''
'0 Constant Q
networks
LEARNING EXAMPLE
2
H1
F2
1
Determine the value of the elements and the characterisitcsof the network if the circuit is magnitude scaled by 100 andfrequency scaled by 1,000,000
2,2
2,/20 BWQsrad
FC
HL
R
200
1'
100'
200'
M
M
M
K
CC
LKL
RKR
'
'
'
scaling impedanceor Magnitude F
F
K
CC
K
LL
RR
'
'
'
FC
mHL
R
200
1''
100''
200''
0'0 FK
)('
''0 BWKQ
BW F
srad /10414.1 6''0
unchanged are 0,Q
LEARNING EXTENSION
elementscircuit resulting the
Determine 10,000.by scaledfrequency and 100by scaled
magnitude is 2FC 1H,L ,10R with network RLC An
M
M
M
K
CC
LKL
RKR
'
'
'
scaling impedanceor Magnitude
FC
HL
R
02.0'
100'
1000'
F
F
K
CC
K
LL
RR
'
'
'
scalingFrequency
FC
HL
kR
2''
01.0''
1''
Scaling
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
CLj
LC
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
LEARNING EXAMPLEA simple notch filter to eliminate 60Hz interference
)1
(1
1
CLj
CL
CjLj
CjLj
ZR
inReq
eq VZR
RV
0
LCZR
1 01
0
LCV
tttvin 10002sin2.0602sin)( FCmHL 100,3.70
LEARNING EXTENSION )( jGvfor plot Bode the of sticcharacteri magnitude the Sketch
RCjCj
R
CjjGv
1
11
1
)(
sradFRC /2.0)1020)(1010( 63 20dB/dec- of asymptotefrequency High
0dB/dec of asymptotefrequency low
5rad/s :frequencyer Break/corn
LEARNING EXTENSION )( jGvfor plot Bode the of sticcharacteri magnitude the Sketch
RCj
RCj
CjR
RjGv
11
)(
sradFRC /5.0)1020)(1025( 63
srad /21
at 0dB Crosses 20dB/dec.
20dB/dec- of asymptotefrequency High
0dB/dec of asymptotefrequency low
2rad/s :frequencyer Break/corn
LEARNING EXTENSION )( jGvfor plot Bode the of sticcharacteri magnitude the Sketch
Band-pass
LCjRCj
RCj
LjCj
R
RjGv 2)(11
)(
5.0102
1010102
,1010
3
363
362
RC
LC
sradRC
/10001
at 0dB Crosses 20dB/dec.
40dB/dec- of asymptotefrequency High
0dB/dec of asymptotefrequency low
rad/s 1000 :frequencyer Break/corn
2
4/)/( 20
2
LRLRLO
2
4/)/( 20
2
LRLRHI
10001
0 LC
srad /618
srad /1618
decdB /40
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
02
2
1
1 Z
V
Z
V
11
22 V
Z
ZV
Linear circuit equivalent
0I
1
2
Z
ZG
1
11 Z
VI
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
Basic Non-inverting Amplifier
Due to the internal op-amp circuitry, it haslimitations, e.g., for high frequency and/orlow voltage situations. The OperationalTransductance Amplifier (OTA) performswell in those situations
Operational Transductance Amplifier
0RRin :OTA Ideal
Comparison of Op-Amp and OTA
Amplifier Type Ideal Rin Ideal Ro Ideal Gain Input Current input VoltageOp-Amp 0 0 0
OTA gm 0 nonzero
Basic Op-Amp Circuit Basic OTA Circuit
inS
inv
L
L
S
SinS
inin
invL
L
RR
RA
RR
R
V
vA
VRR
Rv
vARR
Rv
0
0
00
inS
inm
Linm
SinS
inin
inmL
RR
Rg
RR
R
v
iG
VRR
Rv
vgRR
Ri
0
00
0
00
vAA
Amp-Op Idealmm gG
OTA Ideal
Basic OTA Circuits
)0()(1
00
00
10
vdxxiC
v
vgit
m
)0()()( 00
10 vdxxvC
gtv
tm
Integrator
In the frequency domain
10 VCj
gV m
00
0
ii
vgi
in
inm polarity) (notice
meq
in
in
gR
i
v 1
Resistor Simulated
Basic OTA Adder
11vgm
22vgm
21 21vgvg mm
Resistor Simulated)(1
213
0 21vgvg
gv mm
m
Equivalent representation
mg ofility Programmab
)10
10.,.(
10
7
mSgge
g
mSg
m
m
m
decades 7 - 3 :range
valuesTypical
ABCm IA
Sg
201
LEARNING EXAMPLE
ABCm
m
m
Ig
SmS
g
mSg
20
10410
4
4
74
resistor a Produce k25
SSgg mm
753 1041041
1025
meq
in
in
gR
i
v 1
Resistor Simulated
)(20104 5 AIA
SS ABC
AAIABC 2102 6
LEARNING EXAMPLE Floating simulated resistor
1101 vgi m 1202 vgi m
inmvgi 0
One grounded terminal
011 ii 102 ii
21 mm gg
operationproper For
ABCm
m
m
Ig
SmS
g
mSg
20
10410
4
4
74
resistor 10M a Produce
Sgm7
6 101010
1
S7104
The resistor cannot be producedwith this OTA!
LEARNING EXAMPLE
210
210
321
210
210
,,,
vvv
vvv
ggg mmm
b)
a)
produce to Select
ABCm
m
m
Ig
SmS
g
mSg
20
10410
4
4
74
)(1
213
0 21vgvg
gv mm
m
Case a
2;103
2
3
1 m
m
m
m
g
g
g
g
Two equations in three unknowns.Select one transductance
)(1020
11.0 4
33 AImSg ABCm A5
AImSg ABCm 102.0 22
AImSg ABCm 501 11
Case b
Reverse polarity of v2!
OTA-C CIRCUITS
Circuits created using capacitors, simulated resistors, adders and integrators
integrator
resistor
Frequency domain analysis assumingideal OTAs
1101 im VgI 0202 VgI m
0201 IIIC CICjV
1
0
021101
VgVgCj
V mim
1
2
2
1
01
i
m
m
m
V
gCj
gg
V
Magnitude Bode plot
1
0
iv V
VG
2
1
m
mdc g
gA
C
gf
C
g
mC
mC
2
2
2
LEARNING EXAMPLE
)10(21
4
51
0
jV
VG
iv
:Desired
ABCm
m
m
Ig
SmS
g
mSg
20
1010
1
1
63
4dcA kHzfCC 100)10(2 5
2
1
m
mdc g
gA
C
gf
C
g
mC
mC
2
2
2
Two equations in three unknowns.Select the capacitor value
pFC 25 Sgm6125
2 107.15)1025)(10(2
OK
AISg ABCm 14.38.62 11
AIABC 785.020
7.152
biases and ncestransducta the Find
TOW-THOMAS OTA-C BIQUAD FILTER biquad ~ biquadratic
20
02
20
)()(
)()(
jQ
j
CjBjA
V
V
i
Cj
VVgV im
021101
)( 201202 im VVgI )( 23303 oim VVgI
)(1
03022
02 IICj
V
03i
)( unknownsfour and equationsFour 02010201 ,,, IIVV
1)()(12
132
21
21
32
321
2
3
2
2
01
jggCg
jggCC
Vgg
VVgg
gCj
V
mm
m
mm
im
mii
m
m
m
1)()(12
132
21
21
321
312
1
11
02
jggCg
jggCC
VgggCj
VgCj
V
V
mm
m
mm
imm
mi
mi
1
22
3
21
2
30
21
210 ,,
C
C
g
ggQ
C
g
QCC
gg
m
mmmmm
Filter Type A B CLow-pass 0 0 nonzeroBand-pass 0 nonzero 0High-pass nonzero 0 0
C
gBW
g
gQ
Cg
CC
gg
m
m
m
m
mm
3
3
0
21
21
LEARNING EXAMPLE
ABCm
m
m
Ig
SmS
g
mSg
20
10410
4
4
74
5.- gainfrequency center and 75kHz, of bandwidth
500kHz, offrequency center withfilter pass-band a Design
1)()(12
132
21
21
321
312
1
11
02
jggCg
jggCC
VgggCj
VgCj
V
V
mm
m
mm
imm
mi
mi
1
22
3
21
2
30
21
210 ,,
C
C
g
ggQ
C
g
QCC
gg
m
mmmmm
capacitors pF-50 and ionconfigurat Thomas-Tow the Use
BW
03 iV
3
20 |)(|
m
mv g
gjG
0)(1 2
21
210
j
gg
CC
mm
21 CC
Sg
BW m 56.23107521050
312
3
Sgm 8.1175 2
213
61252
0)105(
108.1171052
mg Sgm 5.2091 AI
AI
AI
ABC
ABC
ABC
18.1
89.5
47.10
3
2
1
Bode plots for resulting amplifier
LEARNING BY APPLICATIONUsing a low-pass filter to reduce 60Hz ripple
Thevenin equivalent for AC/DCconverter
Using a capacitor to create a low-pass filter
Design criterion: place the corner frequencyat least a decade lower
THTH
OF VCRj
V
1
1
21
||||
CR
VV
TH
THOF
CRTHC
1
||1.0|| THOF VV
FCC
05.5362
1500
Filtered output
LEARNING EXAMPLESingle stage tuned transistor amplifier
Select the capacitor for maximumgain at 91.1MHz
AntennaVoltage
Transistor Parallel resonant circuit
CjLjR
V
V
A 1||||
1000
40
LCRCj
j
Cj
V
V
Cj
Cj
CjLjR
A1
)(
/
1000
4
/
/11
1
1000
4
2
0
LC/1frequency center with pass-Band
C6
6
10
1101.912 pFC 05.3
1001000
410
R
LCV
V
A
AVV0for plot Bode Magnitude
LEARNING BY DESIGN Anti-aliasing filter
Nyquist CriterionWhen digitizing an analog signal, such as music, any frequency components
greater than half the sampling rate will be distorted
In fact they may appear as spurious components. The phenomenon is known as aliasing.
SOLUTION: Filter the signal before digitizing, and remove all components higherthan half the sampling rate. Such a filter is an anti-aliasing filter
For CD recording the industry standard is to sample at 44.1kHz.An anti-aliasing filter will be a low-pass with cutoff frequency of 22.05kHz
Single-pole low-pass filter
RCjV
V
in
1
101
050,2221
RCC
kRnFC 18.721
Resulting magnitude Bode plot
Attenuationin audio range
Improved anti-aliasing filter Two-stage buffered filter
01v
RCjV
V
1
1
01
02
RCjV
V
in
1
101
One-stage
Two-stage
Four-stage
nin
n
RCjV
V
1
10
stage-n
mHL
FC
704.0
10
Magnitude Bode plot
sCsLRR
R
V
V
tapeamp
amp
tape
amp
/1||
1
1
2
2
tapeamp
tapeamp
amp
tape
amp
RRL
sLCs
LCs
RR
R
V
V
LC
1frequency notch To design, pick one, e.g., C and determine the other
LEARNING BY DESIGNNotch filter to eliminate 60Hz hum
Notch filter characteristic
Filters