frequency

1. Variable-Frequency Response Analysis
Network performance as function of frequency.
Transfer function
Sinusoidal Frequency Analysis
Bode plots to display frequency response data
Resonant Circuits
The resonance phenomenon and its characterization
Scaling
Impedance and frequency scaling
Filter Networks
Networks with frequency selective characteristics:
low-pass, high-pass, band-pass
VARIABLE-FREQUENCY NETWORK
PERFORMANCE
LEARNING GOALS

2. °∠== 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 performance
varies with the frequency.
Variation in impedance of basic components

3. °∠== 90LLjZL ωωInductor

4. Capacitor °−∠== 90
11
CCj
Zc
ωω

5. Frequency dependent behavior of series RLC network
Cj
RCjLCj
Cj
LjRZeq
ω
ωω
ω
ω
1)(1 2
++
=++=
C
LCjRC
j
j
ω
ωω )1( 2
−+
=
−
−
×
C
LCRC
Zeq
ω
ωω 222
)1()(
||
−+
= 




 −
=∠ −
RC
LC
Zeq
ω
ω 1
tan
2
1
sC
sRCLCs
sZ
sj
eq
1
)(
2
++
=
≈ωnotation"intionSimplifica"

6. For all cases seen, and all cases to be studied, the impedance is of the form
01
1
1
01
1
1
...
...
)(
bsbsbsb
asasasa
sZ n
n
n
n
m
m
m
m
++++
++++
= −
−
−
−
sC
ZsLsZRsZ 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 V
sCsLR
R
sV
/1
)(
++
= SV
sRCLCs
sRC
12
++
=
So V
RCjLCj
RCj
V
js
1)( 2
++
=
≈
ωω
ω
ω
°∠
+××+××
××
= −−
−
010
1)1053.215()1053.21.0()(
)1053.215(
332
3
ωω
ω
jj
j
Vo
MATLAB can be effectively used to compute frequency response characteristics

7. USING MATLAB TO COMPUTE MAGNITUDE AND PHASE INFORMATION
01
1
1
01
1
1
...
...
)(
bsbsbsb
asasasa
sV n
n
n
n
m
m
m
m
o
++++
++++
= −
−
−
−
),(
];,,...,,[
];,,...,,[
011
011
dennumfreqs
bbbbden
aaaanum
nn
mm
>>
=>>
=>>
−
− MATLAB commands required to display magnitude
and phase as function of frequency
NOTE: Instead of comma (,) one can use space to
separate numbers in the array
1)1053.215()1053.21.0()(
)1053.215(
332
3
+××+××
××
= −−
−
ωω
ω
jj
j
Vo
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
2b 1b
0b
Missing coefficients must
be 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 also
work. Must be careful not
to insert blanks elsewhere

8. GRAPHIC OUTPUT PRODUCED BY MATLAB
Log-log
plot
Semi-log
plot

9. LEARNING EXAMPLE A possible stereo amplifier
Desired frequency characteristic
(flat between 50Hz and 15KHz)
Postulated amplifier
Log frequency scale

10. Frequency domain equivalent circuit
Frequency Analysis of Amplifier
)(
)(
)(
)(
sV
sV
sV
sV
in
o
S
in
=
)(
/1
)( sV
sCR
R
sV S
inin
in
in
+
=
]1000[
/1
/1
)( in
oo
o
o V
RsC
sC
sV
+
=






+





+
=
ooinin
inin
RsCRsC
RsC
sG
1
1
]1000[
1
)( 





+



+
=
π
π
π 000,40
000,40
]1000[
100 ss
s
( ) ( )
( ) ( ) π
π
000,401001058.79
100101018.3
191
1691
≈××=
≈××=
−−−
−−−
oo
inin
RC
RC
required
actual
π
π
ππ
000,40
000,40
]1000[)(000,40||100
s
s
sGs ≈⇒<<<<
Frequency dependent behavior is
caused by reactive elements
)(
)(
)(
sV
sV
sG
S
o
=
Voltage Gain
)50( Hz
)20( kHz

11. 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
EXAMPLE



=
admittanceTransfer
tanceTransadmit
)(
)(
)(
1
2
sV
sI
sYT
gainVoltage
)(
)(
)(
1
2
sV
sV
sGv =
To compute the transfer functions one must solve
the circuit. Any valid technique is acceptable

12. LEARNING EXAMPLE



=
admittanceTransfer
tanceTransadmit
)(
)(
)(
1
2
sV
sI
sYT
gainVoltage
)(
)(
)(
1
2
sV
sV
sGv =
The textbook uses mesh analysis. We will
use Thevenin’s theorem
sLR
sC
sZTH ||
1
)( 1+=
1
11
RsL
sLR
sC +
+=
)(
)(
1
11
2
RsLsC
RsLLCRs
sZTH
+
++
=
)()( 1
1
sV
RsL
sL
sVOC
+
=
−
+
)(sVOC
)(sZTH
−
+
)(2 sV
2R
)(2 sI
=
+
=
)(
)(
)(
2
2
sZR
sV
sI
TH
OC
)(
)(
1
11
2
2
1
1
RsLsC
RsLLCRs
R
sV
RsL
sL
+
++
+
+
12121
2
2
)()(
)(
RCRRLsLCRRs
LCs
sYT
++++
=
)(
)(
)(
)(
)(
)( 2
1
22
1
sYR
sV
sIR
sV
sV
sG T
s
v ===
)(
)(
1
1
RsLsC
RsLsC
+
+
×

13. POLES AND ZEROS (More nomenclature)
01
1
1
01
1
1
...
...
)(
bsbsbsb
asasasa
sH 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
21
0
n
m
pspsps
zszszs
KsH
−−−
−−−
=
functionnetworktheofpoles
functionnetworktheofzeros
=
=
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
s
KsH
84
1
20
++
+
ss
s
K
⇒== 1
8
1
)0( 0KH
84
1
8)( 2
++
+
=
ss
s
sH

14. LEARNING EXTENSION Find the driving point impedance at )(sVS
)(
)(
)(
sI
sV
sZ S
=
)(sI
)(
1
)()(: sI
sC
sIRsV
in
inS +=KVL
=+=
in
in
sC
RsZ
1
)(
Replace numerical values
Ω



+ M
s
π100
1

15. LEARNING EXTENSION
π)104(
000,20,50
0
7
0
21
1
×=
−=−=
=
K
HzpHzp
z
:poles
:zero
)(
)(
)(
sV
sV
sG
K
S
o
o
=gainvoltagethefor
ofvaluetheandlocationszeroandpoletheFind






+





+
=
ooinin
inin
RsCRsC
RsC
sG
1
1
]1000[
1
)( 





+



+
=
π
π
π 000,40
000,40
]1000[
100 ss
s
For this case the gain was shown to be
))...()((
))...()((
)(
21
21
0
n
m
pspsps
zszszs
KsH
−−−
−−−
= Zeros = roots of numerator
Poles = roots of denominator
Variable
Frequency
Response

16. SINUSOIDAL FREQUENCY ANALYSIS
)(sH
Circuit represented by
network function



+
+
)cos(0
)(
0
θω
θω
tB
eA tj
( )


∠++
+
)(cos|)(|
)(
0
)(
0
ωθωω
ω θω
jHtjHB
ejHA tj
)(
)()(
)()(
|)(|)(
ωφ
ωω
ωωφ
ωω
j
eMjH
jH
jHM
=
∠=
=
Notation
stics.characteriphaseandmagnitude
calledgenerallyareoffunctionasofPlots ωωφω ),(),(M
)(log
)(
))(log20
10
10
ω
ωφ
ω
PLOTSBODE vs
(M



.offunctionaasfunctionnetworkthe
analyzewefrequencytheoffunctionaasnetworkaofbehaviorthestudyTo
ωω)( jH

17. HISTORY OF THE DECIBEL
Originated as a measure of relative (radio) power
1
2
)2 log10(|
P
P
P dB =1
Pover
2
1
2
2
2
1
2
2
)2
2
2
log10log10(|
I
I
V
V
P
R
V
RIP dB ==⇒== 1
Pover
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

18. ]...)()(21)[1(
]...)()(21)[1()(
)( 2
2
33310
bbba
N
jjj
jjjjK
jH
ωτωτςωτ
ωτωτςωτω
ω
+++
+++
=
±
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
2
1010
2
33310110
10010
−++−+−
++++++
±=
bbba jjj
jjj
jNK
ωτωτςωτ
ωτωτςωτ
ω
DN
D
N
BAAB
loglog)log(
loglog)log(
−=
+=
|)(|log20|)(| 10 ωω jHjH dB =
21
2
1
2121
zz
z
z
zzzz
∠−∠=∠
∠+∠=∠
...
)(1
2
tantan
...
)(1
2
tantan
900)(
2
11
2
3
331
1
1
−
−
−−
+
−
++
°±=∠
−−
−−
b
bb
a
NjH
ωτ
ωτς
ωτ
ωτ
ωτς
ωτ
ω
Display each basic term
separately and add the
results to obtain final
answer
Let’s examine each basic term

19. Constant Term
Poles/Zeros at the origin



°±=∠
×±=
→ ±
±
±
90)(
)(log20|)(|
)( 10
Nj
Nj
j N
dB
N
N
ω
ωω
ω
linestraightaisthis
logisaxis-xthe 10ω

20. Simple pole or zero ωτj+1




=+∠
+=+
−
ωτωτ
ωτωτ
1
2
10
tan)1(
)(1log20|1|
j
j dB
asymptotefrequencylow0|1| ≈+ dBjωτ
(20dB/dec)asymptotefrequencyhighωτωτ 10log20|1| ≈+ dBj
frequency)akcorner/bre1whenmeetasymptotestwoThe (=ωτ
Behavior in the neighborhood of the corner
FrequencyAsymptoteCurve
distance to
asymptote Argument
corner 0dB 3dB 3 45
octave above 6dB 7db 1 63.4
octave below 0dB 1dB 1 26.6
1=ωτ
2=ωτ
5.0=ωτ
°≈+∠ 0)1( ωτj
°≈+∠ 90)1( ωτj
⇒<<1ωτ
⇒>>1ωτ
Asymptote for phase
High freq. asymptote
Low freq. Asym.

21. Simple zero
Simple pole

22. Quadratic pole or zero ])()(21[ 2
2 ωτωτς jjt ++= ])()(21[ 2
ωτωτς −+= j
( ) ( )222
102 2)(1log20|| ςωτωτ +−=dBt 2
1
2
)(1
2
tan
ωτ
ςωτ
−
=∠ −
t
1<<ωτ asymptotefrequencylow0|| 2 ≈dBt °≈∠ 02t
1>>ωτ asymptotefreq.high2
102 )(log20|| ωτ≈dBt °≈∠ 1802t
1=ωτ )2(log20|| 102 ς=dBt °=∠ 902tCorner/break frequency
2
21 ςωτ −= 2
102 12log20|| ςς −=dBt
ς
ς 2
1
2
21
tan
−
=∠ −
t
2
2
≤ςResonance frequency
Magnitude for quadratic pole Phase for quadratic pole
dB/dec40
These graphs are inverted for a zero

23. LEARNING EXAMPLE Generate magnitude and phase plots
)102.0)(1(
)11.0(10
)(
++
+
=
ωω
ω
ω
jj
j
jGvDraw asymptotes
for each term
1,10,50: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

24. asymptotes

25. LEARNING EXAMPLE Generate magnitude and phase plots
)11.0()(
)1(25
)( 2
+
+
=
ωω
ω
ω
jj
j
jGv 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

26. ( )2
1
02
0
0
)(
K
j
K
dB
=⇔= ω
ω
Final results . . . And an extra hint on poles at the origin
dec
dB
40−
dec
dB
20−
dec
dB
40−

27. LEARNING EXTENSION Sketch the magnitude characteristic
)100)(10(
)2(10
)(
4
++
+
=
ωω
ω
ω
jj
j
jG
formstandardinNOTisfunctiontheBut
10010,2,:breaks
Put in standard form
)1100/)(110/(
)12/(20
)(
++
+
=
ωω
ω
ω
jj
j
jG We need to show about
4 decades
40
20
0
20−
dB
°−90
°90
1 10 100 1000
dB|25

28. LEARNING EXTENSION Sketch the magnitude characteristic
2
)(
102.0(100
)(
ω
ω
ω
j
j
jG
+
=
origintheatpoleDouble
50atbreak
formstandardinisIt
40
20
0
20−
dB
°−90
°−270
°90
1 10 100 1000
Once each term is drawn we form the composites

29. Put in standard form
)110/)(1(
)(
++
=
ωω
ω
ω
jj
j
jG
LEARNING EXTENSION Sketch the magnitude characteristic
)10)(1(
10
)(
++
=
ωω
ω
ω
jj
j
jG
101,:breaks
origintheatzero
formstandardinnot
40
20
0
20−
dB
°−90
°−270
°90
1.0 1
10
100
Once each term is drawn we form the composites
decdB /20
decdB /20−

30. LEARNING EXAMPLE A function with complex conjugate poles
[ ]1004)()5.0(
25
)( 2
+++
=
ωωω
ω
ω
jjj
j
jG
Put in standard form
[ ]125/)10/()15.0/(
5.0
)( 2
+++
=
ωωω
ω
ω
jjj
j
jG
40
20
0
20−
dB
°−90
°90
01.0 1.0 1 10 100
°−270
1=ωτ )2(log20|| 102 ς−=dBt
2.0
1.0
25/12
=⇒



=
=
ς
τ
ςτ
])()(21[ 2
2 ωτωτς jjt ++=
dB8
Draw composite asymptote
Behavior close to corner of conjugate pole/zero
is too dependent on damping ratio.
Computer evaluation is better

31. Evaluation of frequency response using MATLAB
[ ]1004)()5.0(
25
)( 2
+++
=
ωωω
ω
ω
jjj
j
jG
» num=[25,0]; %define numerator polynomial
» den=conv([1,0.5],[1,4,100]) %use CONV for polynomial multiplication
den =
1.0000 4.5000 102.0000 50.0000
» freqs(num,den)
Using default options

32. Evaluation of frequency response using MATLAB User controlled
>> clear all; close all %clear workspace and close any open figure
>> figure(1) %open one figure window (not STRICTLY necessary)
>> w=logspace(-1,3,200);%define x-axis, [10^{-1} - 10^3], 200pts total
[ ]1004)()5.0(
25
)( 2
+++
=
ωωω
ω
ω
jjj
j
jG
>> G=25*j*w./((j*w+0.5).*((j*w).^2+4*j*w+100)); %compute transfer function
>> subplot(211) %divide figure in two. This is top part
>> semilogx(w,20*log10(abs(G))); %put magnitude here
>> grid %put a grid and give proper title and labels
>> ylabel('|G(jomega)|(dB)'), title('Bode Plot: Magnitude response')

33. >> semilogx(w,unwrap(angle(G)*180/pi)) %unwrap avoids jumps from +180 to
-180
>> grid, ylabel('Angle H(jomega)(circ)'), xlabel('omega (rad/s)')
>> title('Bode Plot: Phase Response')
Evaluation of frequency response using MATLAB User controlled Continued
Repeat for phase
No xlabel here to avoid clutter
USE TO ZOOM IN A SPECIFIC REGION OF INTEREST
Compare with default!

34. LEARNING EXTENSION Sketch the magnitude characteristic
]136/)12/[(
)1(2.0
)( 2
++
+
=
ωωω
ω
ω
jjj
j
jG 6/136/12
12/1
=⇒=
=
ςςτ
τ
])()(21[ 2
2 ωτωτς jjt ++=
40
20
0
20−
dB
°−90
°−270
°90
1.0 1
10
100
1=ωτ )2(log20|| 102 ς−=dBt
decdB /20−
decdB /40−
decdB /0
12
dB5.9=

35. ]136/)12/[(
)1(2.0
)( 2
++
+
=
ωωω
ω
ω
jjj
j
jG
» num=0.2*[1,1];
» den=conv([1,0],[1/144,1/36,1]);
» freqs(num,den)

36. 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/( −
+ωj
C
C. Simple zero at 0.5
)15.0/( +ωj
D
D. Simple pole at 3
1
)13/( −
+ωj
E
E. Simple pole at 20
1
)120/( −
+ωj
)120/)(13/)(11.0/(
)15.0/(10
)(
+++
+
=
ωωω
ω
ω
jjj
j
jG
20
|
00
0
1020|
dBK
dB KK =⇒=
f the slope is -40dB we assume double real pole. Unless we are given more data

37. LEARNING EXTENSION
Determine a transfer function from the composite
magnitude asymptotes plot
A
A. Pole at the origin.
Crosses 0dB line at 5
ωj
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
)(
++
++
=
ωωω
ωω
ω
jjj
jj
jG
Sinusoidal

38. SONANT CIRCUITS - SERIES RESONANCE
Im{ } 0Z⇒ =
⇒
RESONANT FREQUENCY
PHASOR DIAGRAM
QUALITY FACTOR

39. RESONANT CIRCUITS
These are circuits with very special frequency characteristics…
And resonance is a very important physical phenomenon
Cj
LjRjZ
ω
ωω
1
)( ++=
circuitRLCSeries
Lj
CjGjY
ω
ωω
1
)( ++=
circuitRLCParallel
LCC
L
11
0 =⇒= ω
ω
ω
whenzeroiscircuiteachofreactanceThe
The frequency at which the circuit becomes purely resistive is called
the resonance frequency

40. 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 (if resistance is small)
CRR
L
Q
0
0 1
ω
ω
==:FactorQuality
222
)
1
(||
1
)(
C
LRZ
Cj
LjRjZ
ω
ω
ω
ωω
−+=
++=
222
)
1
(||
1
)(
L
CGY
Cj
Lj
GjY
ω
ω
ω
ω
ω
−+=
++=
Given the similarities between series and parallel resonant circuits,
we will focus on serial circuits

41. Properties of resonant circuits
At resonance the power factor is unity
CIRCUIT BELOW RESONANCE ABOVE RESONANCE
SERIES CAPACITIVE INDUCTIVE
PARALLEL INDUCTIVE CAPACITIVE
Phasor diagram for series circuit Phasor diagram for parallel circuit
−
+
RV
−
−=
+
C
I
jVC
ω
−
+
Ljω
1GV
1CVjω
L
V
j
ω
1
−

42. LEARNING EXAMPLE Determine the resonant frequency, the voltage across each
element at resonance and the value of the quality factor
LC
1
0 =ω sec/2000
)1010)(1025(
1
63
rad
FH
=
××
=
−−
I
A
Z
V
I S
5
2
010
=
°∠
==
Ω= 2ZresonanceAt
Ω=××= −
50)1025)(102( 33
0Lω
)(902505500 VjLIjVL °∠=×== ω
°−∠=×−==
Ω==
90250550
1
50
1
0
0
0
jI
Cj
V
L
C
C
ω
ω
ω
R
L
Q 0ω
= 25
2
50
==
||||
|||| 0
SC
S
S
L
VQV
VQ
R
V
LV
=
==ω
resonanceAt

43. LEARNING EXAMPLE Given L = 0.02H with a Q factor of 200, determine the capacitor
necessary to form a circuit resonant at 1000Hz
R
L0
200
ω
=⇒= 200QwithL
LC
1
0 =ω
C02.0
1
10002 =×⇒ π FC µ27.1=⇒
What is the rating for the capacitor if the
circuit is tested with a 10V supply?
VVC 2000|| =⇒||||
|||| 0
SC
S
S
L
VQV
VQ
R
V
LV
=
==ω
resonanceAt
Ω=
××
=⇒ 59.1
200
02.010002π
R
AI 28.6
59.1
10
==
The reactive power on the capacitor
exceeds 12kVA

44. LEARNING EXTENSION Find the value of C that will place the circuit in resonance
at 1800rad/sec
LC
1
0 =ω 2
18001.0
1
)(1.0
1
1800
×
=⇒
×
= C
CH
FC µ86.3=
Find the Q for the network and the magnitude of the voltage across the
capacitor
R
L
Q 0ω
= 60
3
1.01800
=
×
=Q
||||
|||| 0
SC
S
S
L
VQV
VQ
R
V
LV
=
==ω
resonanceAt
VVC 600|| =

45. Resonance for the series circuit
222
)
1
(||
1
)(
C
LRZ
Cj
LjRjZ
ω
ω
ω
ωω
−+=
++=
QR
CQRL
1
, 00 == ωω
:resonanceAt






−+=
−+=
)(1
)(
0
0
0
0
ω
ω
ω
ω
ω
ω
ω
ω
ω
jQR
QRjQRjRjZ
)(1
1
0
0
1
ω
ω
ω
ω
−+
==
jQV
V
G R
v
isgainvoltageThe:Claim
)(1 ω
ω
ω jZ
R
Cj
LjR
R
Gv =
++
=
vv GGM ∠== |)(|,|)( ωφω
2/1
20
0
2
)(1
1
)(






−+
=
ω
ω
ω
ω
ω
Q
M
)(tan)( 0
0
1
ω
ω
ω
ω
ωφ −−= −
Q
Q
BW 0ω
=








+





+−= 1
2
1
2
1
2
0
QQ
LO ωω
sfrequenciepowerHalf
Z
R
Gv =

46. The Q factor
CRR
L
Q
0
0 1
ω
ω
==
RLowQHigh:circuitseriesFor ⇔
G)(lowRHighQHigh:circuitparallelFor ⇔
M
BWSmallQHigh ⇔
dissipates
Stores as E
field
Stores as M
field
Capacitor and inductor exchange stored
energy. When one is at maximum the
other is at zero
D
S
W
W
Q π2=
cyclebydissipatedenergy
storedenergymaximum
π2=
Q can also be interpreted from an
energy point of view π
ω
π
ω
22
1
2
0202
×=×= mxeffD RIRIW
22
2
1
2
1
mxmxS CVLIW ==
ππ
ω
22
0 Q
R
L
W
W
D
s
=
×
=

47. ERGY TRANSFER IN RESONANT CIRCUITS
Normalization
factor
( ) cos [ ]m
O
V
i t t A
R
ω=

48. LEARNING EXAMPLE
Ω2
mH2
Fµ5
Determine the resonant frequency, quality factor and
bandwidth when R=2 and when R=0.2
CRR
L
Q
0
0 1
ω
ω
==
LC
1
0 =ω
Q
BW 0ω
=
sec/10
)105)(102(
1 4
630 rad=
××
=
−−
ω
R Q
2 10
0.2 100
R Q BW(rad/sec)
2 10 1000
0.2 100 100
Evaluated with EXCEL
R
Q
002.010000×
= QBW /10000=

49. LEARNING EXTENSION A series RLC circuit as the following properties:
sec/100sec,/4000,4 0 radBWradR ==Ω= ω
Determine the values of L,C.
CRR
L
Q
0
0 1
ω
ω
==
LC
1
0 =ω
Q
BW 0ω
=
1. Given resonant frequency and bandwidth determine Q.
2. Given R, resonant frequency and Q determine L, C.
40
100
40000
===
BW
Q
ω
H
QR
L 040.0
4000
440
0
=
×
==
ω
F
RQL
C 6
62
0
2
0
1056.1
1016104
111 −
−
×=
×××
===
ωω

50. LEARNING EXAMPLE Find R, L, C so that the circuit operates as a band-pass filter
with center frequency of 1000rad/s and bandwidth of 100rad/s
)(1 ω
ω
ω jZ
R
Cj
LjR
R
Gv =
++
=
CRR
L
Q
0
0 1
ω
ω
==
LC
1
0 =ω
Q
BW 0ω
=
dependent
Strategy:
1. Determine Q
2. Use value of resonant frequency and Q to set up two equations in the three
unknowns
3. Assign a value to one of the unknowns
10
100
10000
===
BW
Q
ω
R
L
R
L
Q
1000
100
=⇒=
ω
LCLC
1
)10(
1 23
0 =⇒=ω
For example FFC 6
101 −
== µ
HL 1=
Ω=100R

51. PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR
|||| 0 RVQV =
resonanceAt
But this is NOT the maximum value for the
voltage across the capacitor
CRjLC
Cj
LjR
Cj
V
V
S ωω
ω
ω
ω
+−
=
++
= 2
0
1
1
1
1
( )














+−
= 2
22
1
1
)(
Q
u
u
ug
2
0
0
;
SV
V
gu ==
ω
ω
( )
22
22
2
1
)/1)(/(2)2)(1(2
0














+−
+−−
==
Q
u
u
QQuuu
du
dg
CRR
L
Q
0
0 1
ω
ω
==
LC
1
0 =ω
2
2 1
)1(2
Q
u =−⇒
2
0
max
max
2
1
1
Q
u −==
ω
ω
2
2
424
max
4
1
1
2
11
4
1
1
Q
Q
QQQ
g
−
=






−+
=
2
0
4
1
1
||
||
Q
VQ
V S
−
=

52. LEARNING EXAMPLE
mH50
Fµ5
Ω=Ω= 150, RR andwhenDetermine max0 ωω
Natural frequency depends only on L, C.
Resonant frequency depends on Q.
srad
LC
/2000
)105)(105(
11
620 =
××
==
−−
ω
CRR
L
Q
0
0 1
ω
ω
==
LC
1
0 =ω
2
0
max
max
2
1
1
Q
u −==
ω
ω
R
Q
050.02000×
= 2max
2
112000
Q
−×=ω
R Q Wmax
50 2 1871
1 100 2000
Evaluated with EXCEL and rounded to zero decimals
Using MATLAB one can display the frequency response

53. R=50
Low Q
Poor selectivity
R=1
High Q
Good selectivity

54. LEARNING EXAMPLE The Tacoma Narrows Bridge Opened: July 1, 1940
Collapsed: Nov 7, 1940
Likely cause: wind
varying at frequency
similar to bridge
natural frequency
2.020 ×= πω

55. Tacoma Narrows Bridge Simulator
)11( ftV =
42
40
≈
+
=
BA
B
in RR
R
v
v
resonanceatmodeltheFor
.deflection4'causedwind42mphafailureAt
Ω1
Ω5.9H20
Fµ66.31
42=mxVin
Assume a low Q=2.39
0.44’
1.07’
'77.3

56. PARALLEL RLC RESONANT CIRCUITS
222
)
1
(||
1
)(
C
LRZ
Cj
LjRjZ
ω
ω
ω
ωω
−+=
++=
222
)
1
(||
1
)(
L
CGY
Cj
Lj
GjY
ω
ω
ω
ω
ω
−+=
++=
Impedance of series RLC Admittance of parallel RLC
IVYZ
LCCLGR
↔↔
↔↔↔
,
,,
esequivalencNotice
SS YVI =
SSL
SSC
SSG
I
Y
Lj
V
Lj
I
I
Y
Cj
CVjI
I
Y
G
GVI
ω
ω
ω
ω
1
1
==
==
==
||
1
||
||||
1
0
0
SL
SC
LC
SG
I
LG
I
I
G
C
I
II
II
GY
L
C
ω
ω
ω
ω
=
=
−=
=
=⇒=
0
0
resonanceAt
CRR
L
Q
0
0 1
ω
ω
==
LC
1
0 =ω
Series RLC
Parallel RLC
LGG
C
Q
0
0 1
ω
ω
==
LC
1
0 =ω
|| SIQ=
Series RLC
Q
BW 0ω
=
Parallel RLC
Q
BW 0ω
=

57. VARIATION OF IMPEDANCE AND PHASOR DIAGRAM – PARALLEL CIRCUIT

58. LEARNING EXAMPLE
mHLFC
SGVS
120,600
01.0,0120
==
=°∠=
µ
If the source operates at the resonant frequency of the
network, compute all the branch currents
SSL
SSC
SSG
I
Y
Lj
V
Lj
I
I
Y
Cj
CVjI
I
Y
G
GVI
ω
ω
ω
ω
1
1
==
==
==
||
1
||
||||
1
0
0
SL
SC
LC
SG
I
LG
I
I
G
C
I
II
II
GY
L
C
ω
ω
ω
ω
=
=
−=
=
=⇒=
0
0
resonanceAt
|| SIQ=
SG IAI =°∠=°∠×= )(02.1012001.0
srad
LC
/85.117
)106(120.0
11
40 =
××
==
−
ω
)(9049.80120)10600()85.117()901( 6
AIC °∠=°∠××××°∠= −
)(9049.8 AIL °−∠=
_______=xI

59. LEARNING EXAMPLE Derive expressions for the resonant frequency, half power
frequencies, bandwidth and quality factor for the transfer
characteristic
in
out
I
V
H =
Lj
CjGYT
ω
ω
1
++=
Tin
out
T
in
out
YI
V
H
Y
I
V
1
==⇒=
2
2 1
1
1
1
||






−+
=
++
=
L
CGLj
CjG
H
ω
ωω
ω
LC
1
0 =ω:frequencyResonant
22
max
||5.0|)(| HjH h =⇒ ωsfrequenciepowerHalf
2
2
2
2
1
G
L
CG
h
h =





−+
ω
ω
R
G
H ==
1
|| max
G
L
C
h
h ±=−⇒
ω
ω
1
LCC
G
C
G
h
1
22
2
+





+= ω
C
G
BW LOHI =−= ωω
L
C
R
L
C
GBW
Q ===
10ω
LGG
C
Q
0
0 1
ω
ω
==








+





+−= 1
2
1
2
1
2
0
QQ
LO ωω
Replace and show

60. LEARNING EXAMPLE Increasing selectivity by cascading low Q circuits
Single stage tuned amplifier
( )( )
MHzsrad
FHLC
9.99/10275.6
1054.210
11 8
1260 =×=
×
==
−−
ω
398.0
10
1054.2
250 6
12
=
×
×= −
−
L
C
R
L
C
GBW
Q ===
10ω

61. LEARNING EXTENSION Determine the resonant frequency, Q factor and bandwidth
FCmHLkR µ150,20,2 ==Ω=
Parallel RLC
LGG
C
Q
0
0 1
ω
ω
==
LC
1
0 =ω
Q
BW 0ω
=
srad /577
)10150)(1020(
1
630 =
××
=
−−
ω
( )
173
2000/1
10150577 6
=
××
=
−
Q
sradBW /33.3
173
577
==

62. LEARNING EXTENSION 0C,L,Determine ω
120,/1000,6 ==Ω= QsradBWkR
Parallel RLC
LGG
C
Q
0
0 1
ω
ω
==
LC
1
0 =ω
Q
BW 0ω
=
sradBWQ /102.11000120 5
0 ×=×=×=ω
F
R
Q
C µ
ω
167.0
102.16000
120
5
0
=
××
==
H
Q
R
L µ
ω
417
102.1120
6000
5
0
=
××
==
Can be used to verify computations

63. PRACTICAL RESONANT CIRCUIT The resistance of the inductor coils cannot be
neglected
LjR
CjjY
ω
ωω
+
+=
1
)(
LjR
LjR
ω
ω
−
−
×
22
)(
)(
LR
LjR
CjjY
ω
ω
ωω
+
−
+=






+
−+
+
= 2222
)()(
)(
LR
L
Cj
LR
R
jY
ω
ω
ω
ω
2
22
1
0
)(






−=⇒=
+
−⇒
L
R
LCLR
L
CY Rω
ω
ω
real
⇒==
R
L
Q
LC
0
00 ,
1 ω
ω 2
0
0
1
1
Q
R −=ωω
maximaareimpedanceandvoltagetheresonanceAt.
Y
I
ZIV ==
( )




















+=














+=
+
=
2
0
2
0
222
11
R
L
R
R
L
R
R
LR
Z RRR
MAX
ω
ω
ωωω
2
0RQZMAX =
How do you define a quality factor for
this circuit?

64. LEARNING EXAMPLE ΩΩ= 5,50, RR forbothDetermine 0 ωω
⇒==
R
L
Q
LC
0
00 ,
1 ω
ω 2
0
0
1
1
Q
R −=ωω
srad
FH
/2000
)105)(1050(
1
630 =
××
=
−−
ω
2
0
0
1
12000,
050.02000
QR
Q R −=
×
= ω
R Q0 Wr(rad/s) f(Hz)
50 2 1732 275.7
5 20 1997 317.8

65. RESONANCE IN A MORE GENERAL VIEW
222
)
1
(||
1
)(
C
LRZ
Cj
LjRjZ
ω
ω
ω
ωω
−+=
++=
222
)
1
(||
1
)(
L
CGY
Cj
Lj
GjY
ω
ω
ω
ω
ω
−+=
++=
For series connection the impedance reaches maximum at resonance. For parallel
connection the impedance reaches maximum
1)(1)( 22
++
=
++
=
LGjLCj
Lj
Z
CRjLCj
Cj
Y ps
ωω
ω
ωω
ω
12)( 2
++ ωτςωτ jj
aswrittenwastermquadratictheplotsBodeIn
0
1
ω
τ == LC
Q
CRCR
1
22 0 ==⇒= ωςςτ
series
Q
LGLG
1
22 0 ==⇒= ωςςτ
parallel
ς2
1
=QA high Q circuit is highly
under damped
Resonance

66. SCALING
Scaling techniques are used to change an idealized network into a more
realistic one or to adjust the values of the components
M
M
M
K
C
C
LKL
RKR
→
→
→
'
'
'
scalingimpedanceorMagnitude
''
11
'' 0
CLLC
CLLC ==⇒= ω
'
'00
R
L
R
L
Q
ωω
==
Magnitude scaling does not change the
frequency characteristics nor the quality
of the network.
CC
LL
ωKω' F
ωω
ωω
1
''
1
,'' ==
→
unchangediscomponenteachofImpedance
scalingtimeorFrequency
F
F
K
C
C
K
L
L
RR
→
→
→
'
'
'
0
'
0 ωω FK=
)(
'
'
'
0
BWK
Q
BW F==
ω
Q
R
L
Q ==
'
'
'
'
0ω Constant Q
networks

67. LEARNING EXAMPLE
Ω= 2
H1=
F
2
1
Determine the value of the elements and the characterisitcs
of the network if the circuit is magnitude scaled by 100 and
frequency scaled by 1,000,000
2,
2
2
,/20 === BWQsradω
FC
HL
R
200
1
'
100'
200'
=
=
Ω=
M
M
M
K
C
C
LKL
RKR
→
→
→
'
'
'
scalingimpedanceorMagnitude F
F
K
C
C
K
L
L
RR
→
→
→
'
'
'
FC
mHL
R
µ
200
1
''
100''
200''
=
=
Ω=
0
'
0 ωω FK=
)(
'
'
'
0
BWK
Q
BW F==
ω
srad /10414.1 6''
0 ×=ω
unchangedare0,ωQ

68. LEARNING EXTENSION
elementscircuitresultingthe
Determine10,000.byscaledfrequencyand100byscaled
magnitudeis2FC1H,L,10RwithnetworkRLCAn ==Ω=
M
M
M
K
C
C
LKL
RKR
→
→
→
'
'
'
scalingimpedanceorMagnitude
FC
HL
R
02.0'
100'
1000'
=
=
Ω=
F
F
K
C
C
K
L
L
RR
→
→
→
'
'
'
scalingFrequency
FC
HL
kR
µ2''
01.0''
1''
=
=
Ω=
Scaling

69. FILTER NETWORKS
Networks designed to have frequency selective behavior
COMMON FILTERS
Low-pass filter
High-pass filter
Band-pass filter
Band-reject filter
We focus first on
PASSIVE filters

70. Simple low-pass filter
RCj
Cj
R
Cj
V
V
Gv
ω
ω
ω
+
=
+
==
1
1
1
1
1
0
RC
j
Gv =
+
= τ
ωτ
;
1
1
( )
ωτωφ
ωτ
ω
1
2
tan)(
1
1
||)(
−
−==∠
+
==
v
v
G
GM
2
11
,1max =





==
τ
ωMM
frequencypowerhalf==
τ
ω
1
τ
1
=BW

71. Simple high-pass filter
CRj
CRj
Cj
R
R
V
V
Gv
ω
ω
ω
+
=
+
==
111
0
RC
j
j
Gv =
+
= τ
ωτ
ωτ
;
1
( )
ωτ
π
ωφ
ωτ
ωτ
ω
1
2
tan
2
)(
1
||)(
−
−==∠
+
==
v
v
G
GM
2
11
,1max =





==
τ
ωMM
frequencypowerhalf==
τ
ω
1
τ
ω
1
=LO

72. Simple band-pass filter
Band-pass






−+
==
C
LjR
R
V
V
Gv
ω
ω
11
0
( ) ( )222
1
)(
−+
=
LCRC
RC
M
ωω
ω
ω
1
1
=





=
LC
M ω 0)()0( =∞=== ωω MM
( )
2
4/)/( 2
0
2
ω
ω
++−
=
LRLR
LO
LC
1
0 =ω
( )
2
4/)/( 2
0
2
ω
ω
++
=
LRLR
HI
L
R
BW LOHI =−= ωω
)(
2
1
)( HILO MM ωω ==

73. Simple band-reject filter
0
11
0
00 =





−⇒=
C
Lj
LC ω
ωω
10 VV =⇒= circuitopenasactscapacitorthe0at ω
10 VV =⇒∞= circuitopenasactsinductortheat ω
filterpass-band
theinasdeterminedareHILO ωω ,

74. LEARNING EXAMPLE Depending on where the output is taken, this circuit
can produce low-pass, high-pass or band-pass or band-
reject filters
Band-pass
Band-reject filter






−+
=
C
LjR
Lj
V
V
S
L
ω
ω
ω
1 ( ) 1)(,00 =∞=== ωω
S
L
S
L
V
V
V
V
High-pass






−+
=
C
LjR
Cj
V
V
S
C
ω
ω
ω
1
1
( ) 0)(,10 =∞=== ωω
S
C
S
C
V
V
V
V
Low-pass
FCHLR µµ 159,159,10 ==Ω=forplotBode

75. LEARNING EXAMPLE A simple notch filter to eliminate 60Hz interference
)
1
(
1
1
C
Lj
C
L
Cj
Lj
Cj
Lj
ZR
ω
ω
ω
ω
ω
ω
−
=
+
=
in
Req
eq
V
ZR
R
V
+
=0
∞=





=
LC
ZR
1
ω 0
1
0 =





=∴
LC
V ω
( ) ( )tttvin 10002sin2.0602sin)( ×+×= ππ
FCmHL µ100,3.70 ==

76. LEARNING EXTENSION )( ωjGvforplotBodetheofsticcharacterimagnitudetheSketch
RCj
Cj
R
Cj
jGv
ω
ω
ω
ω
+
=
+
=
1
1
1
1
)(
sradFRC /2.0)1020)(1010( 63
=×Ω×== −
τ 20dB/dec-ofasymptotefrequencyHigh
0dB/decofasymptotefrequencylow
5rad/s:frequencyerBreak/corn

77. LEARNING EXTENSION )( ωjGvforplotBodetheofsticcharacterimagnitudetheSketch
RCj
RCj
Cj
R
R
jGv
ω
ω
ω
ω
+
=
+
=
11
)(
sradFRC /5.0)1020)(1025( 63
=×Ω×== −
τ
srad /2
1
==
τ
ωat0dBCrosses20dB/dec.
20dB/dec-ofasymptotefrequencyHigh
0dB/decofasymptotefrequencylow
2rad/s:frequencyerBreak/corn

78. LEARNING EXTENSION )( ωjGvforplotBodetheofsticcharacterimagnitudetheSketch
Band-pass
LCjRCj
RCj
Lj
Cj
R
R
jGv 2
)(11
)(
ωω
ω
ω
ω
ω
++
=
++
=
5.0
102
10
10102
,1010
3
3
63
362
=
×
=⇒×==
==⇒=
−
−
−
−−
ςςτ
ττ
RC
LC
srad
RC
/1000
1
==ωat0dBCrosses20dB/dec.
40dB/dec-ofasymptotefrequencyHigh
0dB/decofasymptotefrequencylow
rad/s1000:frequencyerBreak/corn
( )
2
4/)/( 2
0
2
ω
ω
++−
=
LRLR
LO
( )
2
4/)/( 2
0
2
ω
ω
++
=
LRLR
HI
1000
1
0 ==
LC
ω
srad /618=
srad /1618=
decdB /40−

79. 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 a
more general framework, if one replaces the resistors by impedances
Ideal Op-Amp
These currents are
zero

80. Basic Inverting Amplifier
0=+V
+− =⇒ VVgainInfinite
0=−V
0==⇒ +II-impedanceinputInfinite
0
2
2
1
1
=+
Z
V
Z
V
1
1
2
2 V
Z
Z
V −=
Linear circuit equivalent
0=−I
1
2
Z
Z
G −=
1
1
1
Z
V
I =

81. Basic Non-inverting amplifier
1V
1V
0=+I
1
1
2
10
Z
V
Z
VV
=
−
1
1
12
0 V
Z
ZZ
V
+
=
1
2
1
Z
Z
G +=
01 =I
Basic Non-inverting Amplifier
Due to the internal op-amp circuitry, it has
limitations, e.g., for high frequency and/or
low voltage situations. The Operational
Transductance Amplifier (OTA) performs
well in those situations

82. Operational Transductance Amplifier (OTA)
∞== 0RRin:OTAIdeal
MPARISON BETWEEN OP-AMPS AND OTAs – PHYSICAL CONSTRUCTION
Comparison of Op-Amp and OTA - Parameters
Amplifier Type Ideal Rin Ideal Ro Ideal Gain Input Current input Voltage
Op-Amp 0 0 0
OTA gm 0 nonzero
∞
∞ ∞
∞

83. Basic Op-Amp Circuit Basic OTA Circuit






+





+
==
+
=
+
=
inS
in
v
L
L
S
S
inS
in
in
inv
L
L
RR
R
A
RR
R
V
v
A
V
RR
R
v
vA
RR
R
v
0
0
0
0






+





+
==
+
=
+
=
inS
in
m
Lin
m
S
inS
in
in
inm
L
RR
R
g
RR
R
v
i
G
V
RR
R
v
vg
RR
R
i
0
00
0
0
0
∞== vAA
Amp-OpIdeal mm gG =
OTAIdeal

84. Basic OTA Circuits
)0()(
1
0
0
00
10
vdxxi
C
v
vgi
t
m
+=
=
∫
)0()()( 0
0
10 vdxxv
C
g
tv
t
m
+= ∫
Integrator
In the frequency domain
10 V
Cj
g
V m
ω
=
00
0
=+
−=
ii
vgi
in
inm polarity)(notice
m
eq
in
in
g
R
i
v 1
==
ResistorSimulated

85. Basic OTA Adder
11vgm
22vgm
21 21
vgvg mm +
ResistorSimulated
Equivalent representation
OTA APPLICATION
)(
1
21
3
0 21
vgvg
g
v mm
m
+=

86. mgofilityProgrammab
)
10
10
.,.(
10
7
mS
gge
g
mSg
m
m
m
≥
≤
decades7-3:range
valuesTypical
ABCm I
A
S
g 



= 201
Controlling transconductance

87. LEARNING EXAMPLE
ABCm
m
m
Ig
S
mS
g
mSg
20
104
10
4
4
7
4
=
×=≥
≤
−
resistoraProduce Ωk25
SSg
g
m
m
753
104104
1
1025 −−
×>×=⇒=×
m
eq
in
in
g
R
i
v 1
==
ResistorSimulated
)(20104 5
AI
A
S
S ABC



=× −
AAIABC µ2102 6
=×= −

88. LEARNING EXAMPLE Floating simulated resistor
1101 vgi m−= 1202 vgi m=
inmvgi −=0
One grounded terminal
011 ii −= 102 ii =
21 mm gg =
operationproperFor
ABCm
m
m
Ig
S
mS
g
mSg
20
104
10
4
4
7
4
=
×=≥
≤
−
resistor10MaProduce Ω
Sgm
7
6
10
1010
1 −
=
×
= S7
104 −
×<
The resistor cannot be produced
with this OTA!

89. LEARNING EXAMPLE
210
210
321
210
210
,,,
vvv
vvv
ggg mmm
−=
+=
b)
a)
producetoSelect
ABCm
m
m
Ig
S
mS
g
mSg
20
104
10
4
4
7
4
=
×=≥
≤
−
)(
1
21
3
0 21
vgvg
g
v mm
m
+=
Case a
2;10
3
2
3
1
==
m
m
m
m
g
g
g
g
Two equations in three unknowns.
Select one transductance
)(10
20
1
1.0 4
33 AImSg ABCm
−
×=⇒= Aµ5=
AImSg ABCm µ102.0 22 =⇒=
AImSg ABCm µ501 11 =⇒=
Case b
Reverse polarity of v2!

90. ANALOG MULTIPLIER
ASSUMES VG IS
ZERO
Based on ‘modulating the control current

91. AUTOMATIC GAIN CONTROL
For simplicity of analysis
we drop the absolute value
IN O IN
IN O
v small v Av
A
v big v
B
⇒ ≈
⇒ ≈

92. OTA-C CIRCUITS
Circuits created using capacitors, simulated resistors, adders and integrators
integrator
resistor
Frequency domain analysis assuming
ideal OTAs
1101 im VgI = 0202 VgI m−=
0201 IIIC +=CI
Cj
V
ω
1
0 =
[ ]02110
1
VgVg
Cj
V mim −=
ω
1
2
2
1
0
1
i
m
m
m
V
g
Cj
g
g
V
ω+
=
Magnitude Bode plot
1
0
i
v
V
V
G =
2
1
m
m
dc
g
g
A =
C
g
f
C
g
m
C
m
C
2
2
2 =
=
π
ω

93. LEARNING EXAMPLE
)10(2
1
4
5
1
0
π
ωjV
V
G
i
v
+
==:Desired
ABCm
m
m
Ig
S
mS
g
mSg
20
10
10
1
1
6
3
=
=≥
≤
−
4=dcA kHzfCC 100)10(2 5
=⇒= πω
2
1
m
m
dc
g
g
A =
C
g
f
C
g
m
C
m
C
2
2
2 =
=
π
ω
Two equations in three unknowns.
Select the capacitor value
pFC 25= Sgm
6125
2 107.15)1025)(10(2 −−
×=×= π
OK
AISg ABCm µµ 14.38.62 11 =⇒=
AIABC µ785.0
20
7.15
2 ==
biasesandncestransductatheFind

94. TOW-THOMAS OTA-C BIQUAD FILTER biquad ~ biquadratic
2
0
02
2
0
)()(
)()(
ωω
ω
ω
ωω
++
++
=
j
Q
j
CjBjA
V
V
i
Cj
VV
gV i
m
ω
021
101
−
=
)( 201202 im VVgI −= )( 23303 oim VVgI −=
)(
1
0302
2
02 II
Cj
V +=
ω
03i
)(unknownsfourandequationsFour 02010201 ,,, IIVV
1)()(
12
132
21
21
3
2
3
21
2
3
2
2
01
+

















−+





+
=
+
ωω
ω
j
gg
Cg
j
gg
CC
V
g
g
VV
g
g
g
Cj
V
mm
m
mm
i
m
m
ii
m
m
m
1)()(
12
132
21
21
3
21
31
2
1
1
1
02
+

















+





−
=
+
ωω
ωω
j
gg
Cg
j
gg
CC
V
gg
gCj
V
g
Cj
V
V
mm
m
mm
i
mm
m
i
m
i
1
2
2
3
21
2
30
21
21
0 ,,
C
C
g
gg
Q
C
g
QCC
gg
m
mmmmm
===
ω
ω
Filter Type A B C
Low-pass 0 0 nonzero
Band-pass 0 nonzero 0
High-pass nonzero 0 0








=
=
=
⇒



=
=
C
g
BW
g
g
Q
C
g
CC
gg
m
m
m
m
mm
3
3
0
21
21
ω

95. LEARNING EXAMPLE
ABCm
m
m
Ig
S
mS
g
mSg
20
104
10
4
4
7
4
=
×=≥
≤
−
5.-gainfrequencycenterand75kHz,ofbandwidth
500kHz,offrequencycenterwithfilterpass-bandaDesign
1)()(
12
132
21
21
3
21
31
2
1
1
1
02
+

















+





−
=
+
ωω
ωω
j
gg
Cg
j
gg
CC
V
gg
gCj
V
g
Cj
V
V
mm
m
mm
i
mm
m
i
m
i
1
2
2
3
21
2
30
21
21
0 ,,
C
C
g
gg
Q
C
g
QCC
gg
m
mmmmm
===
ω
ω
capacitorspF-50andionconfiguratThomas-TowtheUse
BW
03 =iV
3
2
0 |)(|
m
m
v
g
g
jG =ω
0)(1 2
21
21
0 =





+⇒= ωωω j
gg
CC
mm
21 CC =
S
g
BW m
µπ 56.2310752
1050
3
12
3
=××=
×
= −
Sgm µ8.1175 2 =⇒=
( ) ( ) 213
6
1252
0
)105(
108.117
1052 −
−
×
××
=××= mg
πω Sgm µ5.2091 =⇒
AI
AI
AI
ABC
ABC
ABC
µ
µ
µ
18.1
89.5
47.10
3
2
1
=
=
=

96. Bode plots for resulting amplifier

97. LEARNING BY APPLICATION Using a low-pass filter to reduce 60Hz ripple
Thevenin equivalent for AC/DC
converter
Using a capacitor to create a low-
pass filter
Design criterion: place the corner frequency
at least a decade lower
TH
TH
OF V
CRj
V
ω+
=
1
1
( )2
1
||
||
CR
V
V
TH
TH
OF
ω+
=
CRTH
C
1
=ω
||1.0|| THOF VV ≈
FCC µ
π
05.53
62
1
500 =⇒
×
=

98. Filtered output

99. LEARNING EXAMPLE Single stage tuned transistor amplifier
Select the capacitor for maximum
gain at 91.1MHz
Antenna
Voltage
Transistor Parallel resonant circuit



−=
Cj
LjR
V
V
A
ωω 1||||
1000
40
LCRC
j
j
Cj
V
V
Cj
Cj
Cj
LjR
A
1
)(
/
1000
4
/
/
11
1
1000
4
2
0
++
−=
×
++
−=
ω
ω
ω
ω
ω
ω
ω
LC/1frequencycenterwithpass-Band
( ) ⇒=×
−
C6
6
10
1
101.912π pFC 05.3=
100
1000
410
==





= R
LCV
V
A
ω
AV
V0forplotBodeMagnitude

100. LEARNING BY DESIGN Anti-aliasing filter
Nyquist Criterion
When 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 higher
than 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 ω+
=
1
101
050,222
1
×== πω
RC
C
Ω=⇒= kRnFC 18.721
Resulting magnitude Bode plot
Attenuation
in audio range

101. Improved anti-aliasing filter Two-stage buffered filter
−
+
01v
RCjV
V
ω+
=
1
1
01
02
RCjV
V
in ω+
=
1
101
One-stage
Two-stage
Four-stage
( )n
in
n
RCjV
V
ω+
=
1
10
stage-n

102. mHL
FC
704.0
10
=
= µ
Magnitude Bode plot
( )sCsLRR
R
V
V
tapeamp
amp
tape
amp
/1||++
=














+







+
+
+
+
=
1
1
2
2
tapeamp
tapeamp
amp
tape
amp
RR
L
sLCs
LCs
RR
R
V
V
LC
1
=frequencynotch To design, pick one, e.g., C and determine the other
LEARNING BY DESIGN Notch filter to eliminate 60Hz hum
Notch filter characteristic

103. DESIGN EXAMPLE ANTI ALIASING FILTER FOR MIXED MODE CIRCUITS
Visualization of aliasing
Signals of different
frequency and the same
samples
Ideally one wants to eliminate frequency
components higher than twice the sampling
frequency and make sure that all useful
frequencies as properly sampled
Design specification
Simplifying assumption
Infinite input resistance (no load on RC circuit)
Design equation
15.9R k∴ = Ω

104. (non-inverting op-amp)
DESIGN EXAMPLE“BASS-BOOST” AMPLIFIER
DESIRED BODE PLOT
OPEN SWITCH
(6dB)
500
2P
f =
Switch closed??

105. DESIGN EXAMPLE TREBLE BOOST
Original player response Desired boost
Proposed boost circuit
Non-inverting amplifier
Design equations
Filters