Funções de rede H(s): pólos e zeros Diagramas de Bode de ... · Programa 1. Introdução aos circuitos eléctricos 2. Grafos e circuitos resistivos lineares 3. Circuitos dinâmicos

Programa

1. Introdução aos circuitos eléctricos2. Grafos e circuitos resistivos lineares3. Circuitos dinâmicos lineares4. Regime forçado sinusoidal5. Análise no domínio da frequência complexa

– Funções de rede H(s): pólos e zeros– Diagramas de Bode de amplitude e de fase– Traçado assimptótico

6. Circuitos resistivos não-lineares

1. Variable-Frequency Response AnalysisNetwork performance as function of frequency.Transfer function

2. Sinusoidal Frequency AnalysisBode plots to display frequency response data

3. Resonant CircuitsThe resonance phenomenon and its characterization

4. ScalingImpedance and frequency scaling

5. 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 ωωInductor

Capacitor °−∠== 9011CCj

Zc ωω

Frequency dependent behavior of series RLC network

CjRCjLCj

CjLjRZeq ω

ωωω

ω 1)(1 2 ++=++=

CLCjRC

jj

ωωω )1( 2 −+

=−−

×

CLCRCZeq ω

ωω 222 )1()(|| −+= ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=∠ −

RCLCZeq ω

ω 1tan2

1

sCsRCLCssZ

sj

eq1)(

notation"in tion Simplifica"2 ++

=

≈ω

For all cases seen, and all cases to be studied, the impedance is of the form

011

1

011

1

......)(

bsbsbsbasasasasZ 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

sC1

R

So VsCsLR

RsV/1

)(++

= SVsRCLCs

sRC12 ++

=

So VRCjLCj

RCjV

js

1)( 2 ++=

≈

ωωω

ω

°∠+××+××

××= −−

−

0101)1053.215()1053.21.0()(

)1053.215(332

3

ωωω

jjjVo

MATLAB can be effectively used to compute frequency response characteristics

USING MATLAB TO COMPUTE MAGNITUDE AND PHASE INFORMATION

011

1

011

1

......)(

bsbsbsbasasasasV n

nn

n

mm

mm

o ++++++++

= −−

−−

),(];,,...,,[

];,,...,,[

011

011

dennumfreqsbbbbden

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

+××+××××

= −−

−

ωωω

jjjVo

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 1b0b

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

)()(

)()(

sVsV

sVsV

in

o

S

in= )(/1

)( sVsCR

RsV Sinin

inin +

= ]1000[/1

/1)( inoo

oo V

RsCsCsV+

=

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡+

=ooinin

inin

RsCRsCRsCsG

11]1000[

1)( ⎥⎦

⎤⎢⎣⎡+⎥⎦

⎤⎢⎣⎡+

=π

ππ 000,40

000,40]1000[100 sss

( ) ( )( ) ( ) π

π

000,401001058.79

100101018.3191

1691

≈××=

≈××=−−−

−−−

oo

inin

RC

RC

required

actual

ππππ

000,40000,40]1000[)(000,40||100

sssGs ≈⇒<<<<

Frequency dependent behavior iscaused by reactive elements

)()()(

sVsVsG

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

sVsIsYT

gain Voltage)()()(

1

2

sVsVsGv =

To compute the transfer functions one must solve the circuit. Any valid technique is acceptable

LEARNING EXAMPLE

⎩⎨⎧

=admittanceTransfer

tanceTransadmit)()()(

1

2

sVsIsYT

gain Voltage)()()(

1

2

sVsVsGv =

The textbook uses mesh analysis. We willuse Thevenin’s theorem

sLRsC

sZTH ||1)( 1+=1

11RsL

sLRsC +

+=

)()(

1

112

RsLsCRsLLCRssZTH +

++=

)()( 11

sVRsL

sLsVOC +=

−+

)(sVOC

)(sZTH

−

+)(2 sV

2R)(2 sI

=+

=)(

)()(2

2 sZRsVsI

TH

OC

)(

)(

1

112

2

11

RsLsCRsLLCRsR

sVRsL

sL

++++

+

121212

2

)()()(

RCRRLsLCRRsLCssYT ++++

=

)()(

)()()()( 2

1

22

1sYR

sVsIR

sVsVsG T

sv ===

)()(

1

1

RsLsCRsLsC

++

×

POLES AND ZEROS (More nomenclature)

011

1

011

1

......)(

bsbsbsbasasasasH n

nn

n

mm

mm

++++++++

= −−

−− Arbitrary network function

Using the roots, every (monic) polynomial can be expressed as aproduct of first order terms

))...()(())...()(()(

21

21

n

m

pspspszszszsKsH

−−−−−−

=

function network the of polesfunctionnetworktheofzeros

==

n

m

pppzzz

,...,,,...,,

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

=−−=+−=

−=

Hjpjp

z :poles:zeros

=++−+

+=

)22)(22()1()(

jsjssKsH

841

2 +++

sssK

⇒== 181)0( KH

8418)( 2 ++

+=

ssssH

LEARNING EXTENSIONFind the driving point impedance at )(sVS

)()()(

sIsVsZ S=

)(sI

)(1)()(: sIsC

sIRsVin

inS +=KVL

=+=in

in sCRsZ 1)(

Replace numerical values

Ω⎥⎦⎤

⎢⎣⎡ + M

sπ1001

LEARNING EXTENSION

π)104(

000,20,50 :poles0 :zero

721

1

×=

−=−==

K

HzpHzpz

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡+

=ooinin

inin

RsCRsCRsCsG

11]1000[

1)( ⎥⎦

⎤⎢⎣⎡+⎥⎦

⎤⎢⎣⎡+

=π

ππ 000,40

000,40]1000[100 sss

For this case the gain was shown to be

))...()(())...()(()(

21

21

n

m

pspspszszszsKsH

−−−−−−

= Zeros = roots of numeratorPoles = roots of denominator

Find the pole and zero locations and value K for the voltage gain G(s) = Vo(s) / Vs(s)

Formas da função de transferência:

1. Forma geral

2. Forma de ganho, pólos e zeros

3. Forma das constantes de tempo

Ganho estático (K0) é o ganho que se observa quando ω=0, i.e. a resposta a sinais DC

Obtém-se a partir de qualquer razão entre tensões e correntes (ex: Vo/Vi, Vo/Ii, Io/Ii, Io/Vi)

zi = zero

pi = pólo

τZi=1/zi constante de tempo associada ao zero zi

τPi=1/pi constante de tempo associada ao pólo pi

SINUSOIDAL FREQUENCY ANALYSIS

)(sH

Circuit represented bynetwork function

⎭⎬⎫

+

+

)cos(0

)(0

θω

θω

tBeA tj

( )⎩⎨⎧

∠++

+

)(cos|)(|)(

0

)(0

ωθωωω θω

jHtjHBejHA tj

)()()(

)()(|)(|)(

ωφωω

ωωφωω

jeMjH

jHjHM

=

∠==

Notation

stics.characteri phase and magnitudecalledgenerally are offunctionas of Plots ωωφω ),(),(M

)(log)(

))(log20PLOTS BODE 10

10 ωωφ

ωvs

(M

⎩⎨⎧

. of function a as function network theanalyzewefrequency theof functionaasnetworkaofbehavior thestudy To

ωω)( jH

HISTORY OF THE DECIBEL

Originated as a measure of relative (radio) power

1

2)2 log10(|PPP dB =1Pover

21

22

21

22)2

22 log10log10(|

II

VVP

RVRIP dB ==⇒== 1Pover

By extension

||log20|||log20|||log20|

10

10

10

GGIIVV

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

jjjjjjjKjH

ωτωτςωτωτωτςωτωω

++++++

=±

General form of a network function showing basic terms

Frequencyindependent

...|)()(21|log20|1|log20

...|)()(21|log20|1|log20

||log20log20

21010

233310110

10010

−++−+−

++++++

±=

bbba jjj

jjj

jNK

ωτωτςωτ

ωτωτςωτ

ω DNDN

BAAB

loglog)log(

loglog)log(

−=

+=|)(|log20|)(| 10 ωω jHjH dB=

212

1

2121

zzzz

zzzz

∠−∠=∠

∠+∠=∠

...)(1

2tantan

...)(1

2tantan

900)(

211

23

3311

1

−−

−−

+−

++

°±=∠

−−

−−

b

bba

NjH

ωτωτςωτ

ωτωτςωτ

ω

Idea: Display each basic termseparately and add the results to

obtain final answer.

Let’s examine each basic term >>

]...)()(21)[1(]...)()(21)[1()( 2

233310

bbba

N

sssssssKsH

ττςτττςτ

++++++

=±

Poles/zerosat the origin

First order terms Quadratic terms for complex conjugate poles/zeros

Constant Term

Poles/Zeros at the origin

⎩⎨⎧

°±=∠×±=

→±

±±

90)()(log20|)(|)( 10

NjNjj N

dBN

N

ωωωω

linestraight a is thislogisaxis-xthe 10ω

0

1a

1bSimple pole or zero ωττ

ω

jsjs

+=+=

11 ⎪⎩

⎪⎨⎧

=+∠+=+

− ωτωτωτωτ

1

210

tan)1()(1log20|1|

jj dB

asymptotefrequency low 0|1| ≈+ dBjωτ

(20dB/dec)asymptotefrequency high ωτωτ 10log20|1| ≈+ dBjfrequency)akcorner/bre1whenmeet asymptotes two The (=ωτ

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

1=ωτ2=ωτ5.0=ωτ

°≈+∠ 0)1( ωτj

°≈+∠ 90)1( ωτj

⇒<<1ωτ

⇒>>1ωτ

Asymptote for phase

High freq. asymptoteLow freq. Asym.

Simple zero

Simple pole

1b

1b

Quadratic pole or zero ])()(21[ 22 ωτωτς jjt ++= ])()(21[ 2ωτωτς −+= j

( ) ( )222102 2)(1log20|| ςωτωτ +−=dBt 2

12 )(1

2tanωτςωτ

−=∠ −t

1<<ωτ 0|| 2 ≈dBt °≈∠ 02t1>>ωτ 2

102 )(log20|| ωτ≈dBt °≈∠ 1802t1=ωτ )2(log20|| 102 ς=dBt °=∠ 902tCorner/break frequency

221 ςωτ −= 2102 12log20|| ςς −=dBt

ςς 2

12

21tan −=∠ −t

22

≤ςResonance frequency

2

Magnitude for quadratic pole

Phase for quadratic pole

↑↑ formulas for a complex zero ↓↓ plots for a complex pole (mirror vertically for the zero)

low freq. asymptote

high freq. asymptote 40dB/dec

LEARNING EXAMPLE Generate magnitude and phase plots

)102.0)(1()11.0(10)(++

+=

ωωωω

jjjjGvDraw 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

)102.0)(1()11.0(10)(++

+=

ssssGv

Ex1

asymptotes

LEARNING EXAMPLE Generate magnitude and phase plots

)11.0()()1(25)( 2 +

+=

ωωωω

jjjjGv 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

)11.0()1(25)( 2 +

+=

ssssGv

Ex2

( )21

020 0)(

KjK

dB

=⇔= ωω

Final results . . . And an extra hint on poles at the origin

decdB40−

decdB20−

decdB40−

LEARNING EXTENSION Sketch the magnitude characteristic

)100)(10()2(10)(

4

+++

=ωω

ωωjj

jjGformstandardinNOTisfunctiontheBut

10010,2,:breaks

Put in standard form)1100/)(110/(

)12/(20)(++

+=

ωωωω

jjjjG We need to show about

4 decades

40

20

0

20−

dB

°−90

°90

1 10 100 1000

dB|25

)1100/)(110/()12/(20)(++

+=

ssssG

Ex3


2)()102.0(100)(

ωωω

jjjG +

=

origin theat pole Double50at break

formstandardinisIt

40

20

0

20−

dB

°−90

°−270

°90

1 10 100 1000

Once each term is drawn we form the composites

2

)102.0(100)(s

ssG +=

Ex4

Put in standard form

)110/)(1()(

++=

ωωωωjjjjG


)10)(1(10)(

++=

ωωωωjjjjG

10 1, :breaksorigin theat zero

formstandardinnot

40

20

0

20−

dB

°−90

°−270

°90

1.0 110

100Once each term is drawn we form the composites

decdB /20decdB /20−

)10)(1(10)(

++=

ssssG

Ex5

LEARNING EXAMPLE A function with complex conjugate poles

[ ]1004)()5.0(25)( 2 +++

=ωωω

ωωjjj

jjG

Put in standard form

[ ]125/)10/()15.0/(5.0)( 2 +++

=ωωω

ωωjjj

jjG

40

20

0

20−

dB

°−90

°90

01.0 1.0 1 10 100°−270

1=ωτ )2(log20|| 102 ς−=dBt

2.01.025/12

=⇒⎭⎬⎫

==

ςτςτ

])()(21[ 22 ωτωτς jjt ++=

dB8

Draw composite asymptote

Behavior close to corner of conjugate pole/zerois too dependent on damping ratio.Computer evaluation is better

)1004)(5.0(25)( 2 +++

=sss

ssG

Ex6

Evaluation of frequency response using MATLAB

[ ]1004)()5.0(25)( 2 +++

=ωωω

ωωjjj

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)

Using default options

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

ωωjjj

jjG

>> 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(j\omega)|(dB)'), title('Bode Plot: Magnitude response')

>> semilogx(w,unwrap(angle(G)*180/pi)) %unwrap avoids jumps from +180 to -180>> grid, ylabel('Angle H(j\omega)(\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!


]136/)12/[()1(2.0)( 2 ++

+=

ωωωωω

jjjjjG 6/136/12

12/1=⇒=

=ςςτ

τ])()(21[ 2

2 ωτωτς jjt ++=

40

20

0

20−

dB

°−90

°−270

°90

1.0 110

100

1=ωτ )2(log20|| 102 ς−=dBt

decdB /20−

decdB /40−

decdB /0

12

dB5.9=

]136/)12/[()1(2.0)( 2 ++

+=

sssssG

Ex7

]136/)12/[()1(2.0)( 2 ++

+=

ωωωωω

jjjjjG

» 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/( −+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

ωωjjj

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

ωj5

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)(++

++=

ωωωωωωjjj

jjjG

)1100/)(120/()150/)(15/(5)(

++++

=ssssssG

Resumo blocos diagrama de Bode

2. Pólo na origem:

H(s)=1/s

1. Pólo simples:

H(s)=a/(s+a)3. Pólo complexo

conjugado (assimpt.)

5. Zero na origem:

H(s)= s

4. Zero simples:

H(s)= (s+a)/a6. Zero complexo

conjugado (assimpt.)

Sistemas de 2ª ordem, sobre-amortecimento vs sub-amortecimento / ressonância

s → jω

RESONANT CIRCUITS - SERIES RESONANCE

Im { } 0Z⇒ =

⇒RESONANT FREQUENCY

PHASOR DIAGRAM

QUALITY FACTOR

RESONANT CIRCUITS

These are circuits with very special frequency characteristics…And resonance is a very important physical phenomenon

CjLjRjZ

ωωω 1)( ++=

circuit RLC Series

LjCjGjY

ωωω 1)( ++=

circuitRLCParallel

LCCL 11

0 =⇒= ωω

ω

whenzeroiscircuit eachof reactanceThe

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)

ξωω

211 :FactorQuality

0

0 ===CRR

LQ

222 )1(||

1)(

CLRZ

CjLjRjZ

ωω

ωωω

−+=

++=

222 )1(||

1)(

LCGY

CjLj

GjY

ωω

ωω

ω

−+=

++=

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

−

−=

+

CIjVC ω

−

+Ljω

1GV1CVjω

LVjω

1−

LEARNING EXAMPLE Determine the resonant frequency, the voltage across eachelement at resonance and the value of the quality factor

LC1

0 =ω sec/2000)1010)(1025(

163

radFH

=××

=−−

I

AZ

VI S 52

010=

°∠==

Ω= 2Z resonanceAt

Ω=××= − 50)1025)(102( 330Lω

)(902505500 VjLIjVL °∠=×== ω

°−∠=×−==

Ω==

902505501

501

0

00

jICj

V

LC

C ω

ωω

RLQ 0ω

= 252

50==

||||

|||| 0

SC

SS

L

VQV

VQR

VLV

=

==ω

resonanceAt

LEARNING EXAMPLE Given L = 0.02H with a Q factor of 200, determine the capacitornecessary to form a circuit resonant at 1000Hz

RL0200 ω

=⇒= 200Q withL

LC1

0 =ωC02.0

110002 =×⇒ π FC μ27.1=⇒

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

=

==ω

resonanceAt

Ω=××

=⇒ 59.1200

02.010002πR

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

LC1

0 =ω 218001.01

)(1.011800

×=⇒

×= C

CH

FC μ86.3=

Find the Q for the network and the magnitude of the voltage across thecapacitor

RLQ 0ω

=

603

1.01800=

×=Q

||||

|||| 0

SC

SS

L

VQV

VQR

VLV

=

==ω

resonanceAt

VVC 600|| =

Resonance for the series circuit

222 )1(||

1)(

CLRZ

CjLjRjZ

ωω

ωωω

−+=

++=

QRCQRL 1, 00 == ωω

:resonanceAt

⎥⎦

⎤⎢⎣

⎡−+=

−+=

)(1

)(

0

0

0

0

ωω

ωω

ωω

ωωω

jQR

QRjQRjRjZ

)(1

10

0

1ωω

ωω −+

==jQV

VG Rv

isgainvoltageThe:Claim

)(1 ωω

ω jZR

CjLjR

RGv =++

=

vv GGM ∠== |)(|,|)( ωφω

2/120

0

2 )(1

1)(

⎥⎦

⎤⎢⎣

⎡−+

=

ωω

ωω

ω

Q

M

)(tan)( 0

0

1

ωω

ωωωφ −−= − Q

QBW 0ω=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠

⎞⎜⎝

⎛+−= 121

21 2

0 QQLO ωω

sfrequenciepower Half

ZRGv =

The Q factorCRR

LQ0

0 1ω

ω==

RLowQHigh :circuit seriesFor ⇔G)(lowR HighQHigh :circuitparallelFor ⇔

M

BW SmallQHigh ⇔

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

WWQ π2=

cycleby dissipatedenergy storedenergy maximumπ2=

Q can also be interpreted from anenergy point of view π

ωπω

221

20202 ×=×= mxeffD RIRIW

22

21

21

mxmxS CVLIW ==

ππω

220 QR

LWW

D

s =×

=

ENERGY TRANSFER IN RESONANT CIRCUITS

Normalizationfactor

( ) cos [ ]mO

Vi t t AR

ω=

LEARNING EXAMPLE

Ω2

mH2

Fμ5

Determine the resonant frequency, quality factor andbandwidth when R=2 and when R=0.2

CRRLQ

0

0 1ω

ω==

LC1

0 =ωQ

BW 0ω=

sec/10)105)(102(

1 4630 rad=

××=

−−ω

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 EXTENSION

A series RLC circuit as the following properties:sec/100sec,/4000,4 0 radBWradR ==Ω= ω

Determine the values of L,C.

CRRLQ

0

0 1ω

ω==

LC1

0 =ωQ

BW 0ω=

1. Given resonant frequency and bandwidth determine Q.2. Given R, resonant frequency and Q determine L, C.

4010040000 ===

BWQ ω

HQRL 040.04000

4400

=×

==ω

FRQL

C 662

020

1056.11016104

111 −− ×=

×××===

ωω

LEARNING EXAMPLE Find R, L, C so that the circuit operates as a band-pass filterwith center frequency of 1000rad/s and bandwidth of 100rad/s

)(1 ωω

ω jZR

CjLjR

RGv =++

=

CRRLQ

0

0 1ω

ω==

LC1

0 =ωQ

BW 0ω=

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 ω

RL

RLQ 1000100 =⇒=

ωLCLC1)10(1 23

0 =⇒=ω

For example FFC 6101 −== μ

HL 1=

Ω=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

CjVV

S ωωω

ω

ω+−

=++

= 20

11

1

1

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛+−

= 2221

1)(

Quu

ug

20

0;

SVVgu ==

ωω

( )22

22

2

1

)/1)(/(2)2)(1(20

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛+−

+−−==

Quu

QQuuududg

CRRLQ

0

0 1ω

ω==

LC1

0 =ω

22 1)1(2

Qu =−⇒

20

maxmax 2

11Q

u −==ωω

2

2

424

max

411

211

41

1

Q

Q

QQQ

g−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=

2

0

411

||||

Q

VQV S

−=

LEARNING EXAMPLE

mH50

Fμ5

Ω=Ω= 150, RR andwhenDetermine max0 ωω

Natural frequency depends only on L, C.Resonant frequency depends on Q.

sradLC

/2000)105)(105(

11620 =

××==

−−ω

CRRLQ

0

0 1ω

ω==

LC1

0 =ω

20

maxmax 2

11Q

u −==ωω

RQ 050.02000×= 2max 2

112000Q

−×=ω

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 ×= πω

0.44’

1.07’

'77.3

Tacoma Narrows Bridge Simulator

)11( ftV =

424

resonanceat model For the .deflection 4'

caused 42mph wind a failureAt

0 ≈+

=BA

B

in RRR

vv

Ω1

Ω5.9H20

Fμ66.31

42=mxVin

Assume a low Q=2.39

PARALLEL RLC RESONANT CIRCUITS

222 )1(||

1)(

CLRZ

CjLjRjZ

ωω

ωωω

−+=

++=

222 )1(||

1)(

LCGY

CjLj

GjY

ωω

ωω

ω

−+=

++=

Impedance of series RLC Admittance of parallel RLC

IVYZLCCLGR

↔↔↔↔↔

,,,

esequivalencNotice

SS YVI =

SSL

SSC

SSG

IY

LjVLj

I

IY

CjCVjI

IYGGVI

ωω

ωω

11

==

==

==

||1||

||||

1

0

0

SL

SC

LC

SG

ILG

I

IG

CI

IIII

GYL

C

ω

ω

ωω

=

=

−==

=⇒=0

0

resonanceAt

CRRLQ

0

0 1ω

ω==

LC1

0 =ω

Series RLC

Parallel RLC

LGGCQ

0

0 1ω

ω==

LC1

0 =ω

|| SIQ=

Series RLC QBW 0ω=

Parallel RLCQ

BW 0ω=

VARIATION OF IMPEDANCE AND PHASOR DIAGRAM – PARALLEL CIRCUIT

LEARNING EXAMPLE

mHLFCSGVS

120,60001.0,0120

===°∠=

μ

If the source operates at the resonant frequency of the network, compute all the branch currents

SSL

SSC

SSG

IY

LjVLj

I

IY

CjCVjI

IYGGVI

ωω

ωω

11

==

==

==

||1||

||||

1

0

0

SL

SC

LC

SG

ILG

I

IG

CI

IIII

GYL

C

ω

ω

ωω

=

=

−==

=⇒=0

0

resonanceAt

|| SIQ=

SG IAI =°∠=°∠×= )(02.1012001.0

sradLC

/85.117)106(120.0

1140 =

××==

−ω

)(9049.80120)10600()85.117()901( 6 AIC °∠=°∠××××°∠= −

)(9049.8 AIL °−∠=

_______=xI

LEARNING EXAMPLE Derive expressions for the resonant frequency, half powerfrequencies, bandwidth and quality factor for the transfercharacteristic

in

out

IVH =

LjCjGYT ω

ω 1++=

Tin

out

T

inout YI

VHYIV 1

==⇒=

22 1

11

1||

⎟⎠⎞

⎜⎝⎛ −+

=++

=

LCGLj

CjGH

ωωω

ω

LC1

0 =ω :frequencyResonant

22max

||5.0|)(| HjH h =⇒ ωsfrequenciepower Half

22

2 21 GL

CGh

h =⎟⎟⎠

⎞⎜⎜⎝

⎛−+ω

ω

RG

H ==1|| max

GL

Ch

h ±=−⇒ω

ω 1

LCCG

CG

h1

22

2

+⎟⎠⎞

⎜⎝⎛+= mω

CGBW LOHI =−= ωω

LCR

LC

GBWQ ===

10ω

LGGCQ

0

0 1ω

ω==

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠

⎞⎜⎝

⎛+−= 121

21 2

0 QQLO ωω

Replace and show

LEARNING EXAMPLE Increasing selectivity by cascading low Q circuits

Single stage tuned amplifier

( )( ) MHzsradFHLC

9.99/10275.61054.210

11 81260 =×=

×==

−−ω

398.010

1054.2250 6

12=

××= −

−LCR

LC

GBWQ ===

10ω

LEARNING EXTENSION Determine the resonant frequency, Q factor and bandwidth

FCmHLkR μ150,20,2 ==Ω=

Parallel RLC

LGGCQ

0

0 1ω

ω==

LC1

0 =ωQ

BW 0ω=

srad /577)10150)(1020(

1630 =

××=

−−ω

( ) 1732000/1

10150577 6=

××=

−

Q

sradBW /33.3173577

==

LEARNING EXTENSION 0 C, L, Determine ω

120,/1000,6 ==Ω= QsradBWkR

Parallel RLC

LGGCQ

0

0 1ω

ω==

LC1

0 =ωQ

BW 0ω=

sradBWQ /102.11000120 50 ×=×=×=ω

FRQC μω

167.0102.16000

1205

0=

××==

HQ

RL μω

417102.1120

60005

0=

××==

Can be used to verify computations

PRACTICAL RESONANT CIRCUIT The resistance of the inductor coils cannot beneglected

LjRCjjY

ωωω

++=

1)(LjRLjR

ωω

−−

×

22 )()(

LRLjRCjjY

ωωωω

+−

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−++

= 2222 )()()(

LRLCj

LRRjY

ωω

ωω

2

2210

)(⎟⎠⎞

⎜⎝⎛−=⇒=

+−⇒

LR

LCLRLCY Rωω

ω real

⇒==R

LQLC

000 ,1 ωω 2

00

11QR −=ωω

maxima are impedance and voltagethe resonanceAt .YIZIV ==

( )⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+=

+=

20

2

0

22211

RLR

RLR

RLRZ RRR

MAXω

ωωωω

20RQZMAX =

How do you define a quality factor for this circuit?

LEARNING EXAMPLE ΩΩ= 5,50, RR for both Determine 0 ωω

⇒==R

LQLC

000 ,1 ωω 2

00

11QR −=ωω

sradFH

/2000)105)(1050(

1630 =

××=

−−ω

20

0112000,050.02000

QRQ R −=

×= ω

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

ωω

ωωω

−+=

++=

222 )1(||

1)(

LCGY

CjLj

GjY

ωω

ωω

ω

−+=

++=

For series connection the impedance reaches maximum at resonance. For parallelconnection the impedance reaches maximum

1)(1)( 22 ++=

++=

LGjLCjLjZ

CRjLCjCjY ps ωω

ωωω

ω

12)( 2 ++ ωτςωτ jj

aswrittenwastermquadratictheplotsBodeIn

0

1ω

τ == LC

QCRCR 122 0 ==⇒= ωςςτ

series

QLGLG 122 0 ==⇒= ωςςτ

parallel

ς21

=QA high Q circuit is highlyunder damped

Freq. Natural e Ressonância RLC série (Vs -> I) ou paralelo (Is -> V)

Em todas estas funções de transferência a frequência natural tem a expressão:

LCn1

=ω

Ressonância RLC série (Vs -> I) ou paralelo (Is -> V)

Ressonância nestes casos corresponde a Z ou Y reais (resistivos puros)

LCnr1

=≡ωω

Nota: definição mais geral: ωr é o máximizante da função de transferência

Nas funções de transferência I/Vsou Vr/Vs (RLC série), V/Is ou Ir/Is(RLC paralelo) temos:

Ressonância RLC série ou paralelo – outras funções

221 ξωω −= nr

Definição mais geral de ressonância: ωr é a frequência máximizante da função de transferência

Usando esta definição então os circuitos VC/Vs e IL/IS têm a seguinte frequência de ressonância:

221 ξωω−

= nr

Usando esta definição então os circuitos VL/VS e IC/IS têm a seguinte frequência de ressonância:

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

KCC

LKLRKR

→

→→

'

''

scalingimpedanceor Magnitude

''11'' 0 CLLC

CLLC ==⇒= ω

''00

RL

RLQ ωω==

Magnitude scaling does not change thefrequency characteristics nor the qualityof the network.

CCLL

ωKω' F

ωωωω 1

''1,'' ==

→unchanged iscomponent each of Impedance

scalingtimeor Frequency

F

F

KCC

KLL

RR

→

→

→

'

'

'

0'0 ωω FK=

)('

''0 BWK

QBW F==

ω

QR

LQ =='''

'0ω Constant Q

networks

LEARNING EXAMPLE

Ω= 2

H1=

F21

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

2,22,/20 === BWQsradω

FC

HLR

2001'

100'200'

=

=Ω=

M

M

M

KCC

LKLRKR

→

→→

'

''

scalingimpedanceor Magnitude F

F

KCC

KLL

RR

→

→

→

'

'

'

FC

mHLR

μ2001''

100''200''

=

=Ω=

0'0 ωω FK=

)('

''0 BWK

QBW F==

ω

srad /10414.1 6''0 ×=ω

unchangedare 0,ωQ

LEARNING EXTENSION

elementscircuit resulting theDetermine 10,000.by scaledfrequency and 100by scaledmagnitudeis2FC1H,L,10RwithnetworkRLCAn ==Ω=

M

M

M

KCC

LKLRKR

→

→→

'

''

scalingimpedanceor Magnitude

FCHL

R

02.0'100'1000'

==

Ω=

F

F

KCC

KLL

RR

→

→

→

'

'

'scalingFrequency

FCHL

kR

μ2''01.0''

1''

==

Ω=

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

CjVVGv ω

ω

ω+

=+

==1

11

1

1

0

RCj

Gv =+

= τωτ

;1

1

( )ωτωφ

ωτω

1

2

tan)(

11||)(

−−==∠

+==

v

v

G

GM

211,1max =⎟

⎠⎞

⎜⎝⎛ ==

τωMM

frequencypower half==τ

ω 1τ1

=BW

Simple high-pass filter

CRjCRj

CjR

RVVGv ω

ω

ω+

=+

==111

0

RCj

jGv =+

= τωτ

ωτ ;1

( )

ωτπωφ

ωτ

ωτω

1

2

tan2

)(

1||)(

−−==∠

+==

v

v

G

GM

211,1max =⎟

⎠⎞

⎜⎝⎛ ==

τωMM

frequencypower half==τ

ω 1

τω 1

=LO

Simple band-pass filter

Band-pass

⎟⎠⎞

⎜⎝⎛ −+

==

CLjR

RVVGv

ωω 11

0

( ) ( )222 1)(

−+=

LCRC

RCMωω

ωω

11=⎟

⎠⎞

⎜⎝⎛ =

LCM ω 0)()0( =∞=== ωω MM

( )2

4/)/( 20

2 ωω

++−=

LRLRLO

LC1

0 =ω

( )2

4/)/( 20

2 ωω

++=

LRLRHI

LRBW LOHI =−= ωω

)(2

1)( HILO MM ωω ==

Simple band-reject filter

0110

00 =⎟⎟⎠

⎞⎜⎜⎝

⎛−⇒=

CLj

LC ωωω

10 VV =⇒= circuitopenasactscapacitor the0at ω

10 VV =⇒∞= circuitopenasactsinductor theat ω

filter pass-bandtheinasdeterminedare HILO ωω ,

LEARNING EXAMPLE Depending 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

LjVV

S

L

ωω

ω1 ( ) 1)(,00 =∞=== ωω

S

L

S

L

VV

VV

High-pass

⎟⎠⎞

⎜⎝⎛ −+

=

CLjR

CjVV

S

C

ωω

ω1

1

( ) 0)(,10 =∞=== ωωS

C

S

C

VV

VV

Low-pass

FCHLR μμ 159,159,10 ==Ω=for plot Bode

LEARNING EXAMPLE A simple notch filter to eliminate 60Hz interference

)1(1

1

CLj

CL

CjLj

CjLj

ZR

ωω

ωω

ωω

−=

+=

inReq

eq VZR

RV

+=0

∞=⎟⎠⎞

⎜⎝⎛ =

LCZR

1ω 010 =⎟

⎠⎞

⎜⎝⎛ =∴

LCV ω

( ) ( )tttvin 10002sin2.0602sin)( ×+×= ππ

FCmHL μ100,3.70 ==

LEARNING EXTENSION )( ωjGvfor plot BodetheofsticcharacterimagnitudetheSketch

RCjCj

R

CjjGv ωω

ωω+

=+

=1

11

1

)(

sradFRC /2.0)1020)(1010( 63 =×Ω×== −τ 20dB/dec- of asymptotefrequency High0dB/dec of asymptotefrequency low

5rad/s :frequencyer Break/corn


RCjRCj

CjR

RjGv ωω

ω

ω+

=+

=11)(

sradFRC /5.0)1020)(1025( 63 =×Ω×== −τ

srad /21==

τωat 0dB Crosses 20dB/dec.

20dB/dec- of asymptotefrequency High0dB/dec of asymptotefrequency low

2rad/s :frequencyer Break/corn


Band-pass

LCjRCjRCj

LjCj

R

RjGv 2)(11)(ωω

ω

ωω

ω++

=++

=

5.0102

1010102

,1010

3

363

362

=×

=⇒×==

==⇒=

−

−−

−−

ςςτ

ττ

RC

LC

sradRC

/10001==ωat 0dB Crosses 20dB/dec.

40dB/dec- of asymptotefrequency High0dB/dec of asymptotefrequency low

rad/s1000 :frequencyer Break/corn

( )2

4/)/( 20

2 ωω

++−=

LRLRLO

( )2

4/)/( 20

2 ωω

++=

LRLRHI

100010 ==

LCω

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

0=+V

+− =⇒ VV gain Infinite

0=−V

0==⇒ +II-impedanceinput Infinite

02

2

1

1 =+ZV

ZV

11

22 V

ZZV −=

Linear circuit equivalent

0=−I

1

2

ZZG −=

1

11 Z

VI =

Basic Non-inverting amplifier

1V

1V

0=+I

1

1

2

10

ZV

ZVV

=−

11

120 V

ZZZV +

=

1

21ZZG +=

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 (OTA)

∞== 0RRin :OTAIdeal

COMPARISON 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 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

RRRA

RRR

VvA

VRR

Rv

vARR

Rv

0

0

00

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡+

==

+=

+=

inS

inm

Linm

SinS

inin

inmL

RRRg

RRR

viG

VRR

Rv

vgRR

Ri

0

00

0

00

∞== vAAAmp-Op Ideal mm gG =

OTAIdeal

Basic OTA Circuits

)0()(10

000

10

vdxxiC

v

vgit

m

+=

=

∫

)0()()( 00

10 vdxxvCgtv

tm += ∫

Integrator

In the frequency domain

10 VCj

gV m

ω=

00

0

=+−=ii

vgi

in

inm polarity)(notice

meq

in

in

gR

iv 1

==

ResistorSimulated

Basic OTA Adder

11vgm

22vgm

21 21vgvg mm +

ResistorSimulated

Equivalent representation

OTA APPLICATION

)(121

30 21

vgvgg

v mmm

+=

mg ofility Programmab

)10

10.,.(

10

7mSgge

gmSg

m

m

m

≥

≤decades 7 - 3 :range

valuesTypical

ABCm IASg ⎥⎦⎤

⎢⎣⎡= 201

Controlling transconductance

LEARNING EXAMPLE

ABCm

m

m

Ig

SmSg

mSg

20

1041044

74

=

×=≥

≤

−

resistoraProduce Ωk25

SSgg m

m

753 10410411025 −− ×>×=⇒=×

meq

in

in

gR

iv 1

==

Resistor Simulated

)(20104 5 AIASS ABC⎥⎦⎤

⎢⎣⎡=× −

AAI ABC μ2102 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

SmSg

mSg

20

1041044

74

=

×=≥

≤

−

resistor10MaProduce Ω

Sgm7

6 101010

1 −=×

= S7104 −×<

The resistor cannot be producedwith this OTA!

LEARNING EXAMPLE

210

210

321

210210

,,,

vvvvvv

ggg mmm

−=+=

b) a)

producetoSelect

ABCm

m

m

Ig

SmSg

mSg

20

1041044

74

=

×=≥

≤

−

)(121

30 21

vgvgg

v mmm

+=

Case a

2;103

2

3

1 ==m

m

m

m

gg

gg

Two equations in three unknowns.Select one transductance

)(102011.0 4

33 AImSg ABCm−×=⇒= Aμ5=

AImSg ABCm μ102.0 22 =⇒=

AImSg ABCm μ501 11 =⇒=

Case b

Reverse polarity of v2!

ANALOG MULTIPLIER

ASSUMES VG IS ZERO

Based on ‘modulating the control current

AUTOMATIC GAIN CONTROL

For simplicity of analysiswe drop the absolute value

IN O IN

IN O

v small v AvAv big vB

⇒ ≈

⇒ ≈

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 +=CICj

Vω1

0 =

[ ]021101 VgVgCj

V mim −=ω

1

2

21

0 1 i

m

mm

Vg

Cjg

gV

ω+=

Magnitude Bode plot

1

0

iv V

VG =

2

1

m

mdc g

gA =

Cgf

Cg

mC

mC

2

2

2 =

=

π

ω

LEARNING EXAMPLE

)10(21

4

51

0

πωjV

VGi

v+

== :Desired

ABCm

m

m

Ig

SmSg

mSg

20

101011

63

=

=≥

≤

−

4=dcA kHzfCC 100)10(2 5 =⇒= πω

2

1

m

mdc g

gA =

Cgf

Cg

mC

mC

2

2

2 =

=

π

ω

Two equations in three unknowns.Select the capacitor value

pFC 25= Sgm6125

2 107.15)1025)(10(2 −− ×=×= π

OK

AISg ABCm μμ 14.38.62 11 =⇒=

AI ABC μ785.020

7.152 ==

biasesandncestransductathe Find

TOW-THOMAS OTA-C BIQUAD FILTER biquad ~ biquadratic

20

02

20

)()(

)()(

ωωωω

ωω

++

++=

jQ

j

CjBjAVV

i

CjVVgV i

m ω021

101−

=)( 201202 im VVgI −= )( 23303 oim VVgI −=

)(10302

202 II

CjV +=

ω

03i

)( unknownsfour andequationsFour 02010201 ,,, IIVV

1)()(12

132

21

21

32

321

2

3

2

2

01

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡−+⎥

⎦

⎤⎢⎣

⎡+

=+ ωω

ω

jggCgj

ggCC

VggVV

gg

gCj

V

mm

m

mm

im

mii

m

m

m

1)()(12

132

21

21

321

312

1

11

02

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

=+ ωω

ωω

jggCgj

ggCC

VgggCjV

gCjV

V

mm

m

mm

imm

mi

mi

1

22

3

21

2

30

21

210 ,,

CC

gggQ

Cg

QCCgg

m

mmmmm ===ωω

Filter Type A B CLow-pass 0 0 nonzeroBand-pass 0 nonzero 0High-pass nonzero 0 0 ⎪

⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=

=

=

⇒⎭⎬⎫

==

CgBW

ggQ

Cg

CCgg

m

m

m

m

mm

3

3

0

21

21

ω

LEARNING EXAMPLE

ABCm

m

m

Ig

SmSg

mSg

20

1041044

74

=

×=≥

≤

−

5.- gainfrequency center and 75kHz, of bandwidth500kHz,offrequency center withfilter pass-bandaDesign

1)()(12

132

21

21

321

312

1

11

02

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

=+ ωω

ωω

jggCgj

ggCC

VgggCjV

gCjV

V

mm

m

mm

imm

mi

mi

1

22

3

21

2

30

21

210 ,,

CC

gggQ

Cg

QCCgg

m

mmmmm ===ωω

capacitorspF-50and ionconfiguratThomas-Towthe Use

BW03 =iV

3

20 |)(|

m

mv g

gjG =ω0)(1 2

21

210 =⎥

⎦

⎤⎢⎣

⎡+⇒= ωωω j

ggCC

mm

21 CC =

SgBW m μπ 56.23107521050

312

3 =××=×

= −

Sgm μ8.1175 2 =⇒=

( ) ( ) 213

61252

0 )105(108.1171052 −

−

×××

=××= mgπω Sgm μ5.2091 =⇒

AIAIAI

ABC

ABC

ABC

μμμ

18.189.547.10

3

2

1

===

Bode plots for resulting amplifier

LEARNING BY APPLICATION Using 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

Vω+

=1

1

( )21||||CR

VVTH

THOF

ω+=

CRTHC

1=ω

||1.0|| THOF VV ≈

FCC μπ

05.5362

1500 =⇒×

=

Filtered output

LEARNING EXAMPLE Single stage tuned transistor amplifier

Select the capacitor for maximumgain at 91.1MHz

AntennaVoltage

Transistor Parallel resonant circuit

⎥⎦⎤

⎢⎣⎡−= CjLjR

VV

Aωω 1||||

100040

LCRCjj

CjVV

CjCj

CjLjR

A 1)(

/1000

4

//

111

10004

2

0

++−=

×++

−=

ωω

ω

ωω

ωω

LC/1frequency center with pass-Band

( ) ⇒=×− C6

6

101101.912π pFC 05.3=

1001000

410 ==⎟⎠⎞

⎜⎝⎛ = R

LCVV

Aω

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

RCjVV

in ω+=

1101

050,2221×== πω

RCC

Ω=⇒= kRnFC 18.721

Resulting magnitude Bode plot

Attenuationin audio range

Improved anti-aliasing filter Two-stage buffered filter

−

+

01v

RCjVV

ω+=

11

01

02

RCjVV

in ω+=

1101

One-stage

Two-stage

Four-stage

( )nin

n

RCjVV

ω+=

110

stage-n

mHLFC

704.010

== μ

Magnitude Bode plot

( )sCsLRRR

VV

tapeamp

amp

tape

amp

/1||++=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+⎟⎟⎠

⎞⎜⎜⎝

⎛

++

++

=

1

1

2

2

tapeamp

tapeamp

amp

tape

amp

RRLsLCs

LCsRR

RVV

LC1

=frequency notch To design, pick one, e.g., C and determine the other

LEARNING BY DESIGN Notch filter to eliminate 60Hz hum

Notch filter characteristic

DESIGN EXAMPLE ANTI ALIASING FILTER FOR MIXED MODE CIRCUITS

Visualization of aliasing

Signals of differentfrequency and the samesamples

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

(non-inverting op-amp)DESIGN EXAMPLE “BASS-BOOST” AMPLIFIER

DESIRED BODE PLOT

OPEN SWITCH

(6dB)

5002P

f =

Switch closed??

DESIGN EXAMPLE TREBLE BOOST

Original player response Desired boost

Proposed boost circuit

Non-inverting amplifier

Design equations

Filters

Funções de rede H(s): pólos e zeros Diagramas de Bode de ... · Programa 1. Introdução aos circuitos eléctricos 2. Grafos e circuitos resistivos lineares 3. Circuitos dinâmicos

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