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FILTER DESIGN WORKSHOP

Domingo Rodríguez

Automated Information Processing (AIP) Lab.

Electrical and Computer Engineering Department

INEL 4102, INEL 4095, INEL 4301, & INEL 5309

FILTER DESIGN WORKSHOP – January 17, 2017

Analog FiltersA Quick Basic Introduction

A More Detailed Introduction

Digital FiltersA Digital Filter Design Approach

Filter DesignFilter Design Using MATLAB

FDATool Design Technique

PSPICEFilter Simulation Using PSPICE

Workshop Outline

2

ANALOG FILTERS

3

Filters

Background:

. Filters may be classified as either digital or analog.

. are implemented using a digital computer

or special purpose digital hardware.

. may be classified as either passive or

active and are usually implemented with R, L, and C

components and operational amplifiers.

Courtesy of the University of Tennessee4

FiltersBackground:

. An is one that, along with R, L, and C

components, also contains an energy source, such

as that derived from an operational amplifier.

. A is one that contains only R, L, and

C components. It is not necessary that all three be

present. L is often omitted (on purpose) from

passive filter design because of the size and cost

of inductors – and they also carry along an R that

must be included in the design.

5

FiltersBackground:

. The (realization) of analog filters, that is, the

way one builds (topological layout) the filters, received

significant attention during 1940 thru 1960. Leading

the work were Cauer and Tuttle. Since that time,

very little effort has been directed to analog filter

realization.

. The of analog filters is well described in

filter text books. The most popular include Butterworth,

Chebyshev and elliptic methods.

6

FiltersBackground:

. Generally speaking, digital filters have become the focus

of attention in the last 40 years. The interest in digital

filters started with the advent of the digital computer,

especially the affordable PC and special purpose signal

processing boards. People who led the way in the work

(the analysis part) were Kaiser, Gold, and Rader.

. A digital filter is simply the implementation of an

equation(s) in computer software. There are no R, L,

C components as such. However, digital filters can also

be built directly into special purpose computers in

hardware form. But the execution is still in software.

7

FiltersBackground:

. In this course we will only be concerned with an

introduction to filters. We will look at both passive

and active filters; but, will concentrate on passive filters.

. We will not cover any particular design or realization

methods in detail; but, rather use our understanding of

poles and zeros in the s-plane, for basic designs.

. All EE and CE undergraduate students should take a

course in digital filter design, as an opinion.

8

Passive Analog Filters

Background: Four types of filters - “Ideal”

lowpass highpass

bandpass bandstop

9

Background: Realistic Filters:

lowpass highpass

bandpass bandstop


10


Background:

It will be shown later that the ideal

filter, sometimes called a “brickwall”

filter, can be approached by making the

order of the filter higher and higher.

The order here refers to the order of the

polynomial(s) that are used to define the

filter. MATLAB examples will be given later

to illustrate this.

11


Low Pass Filter Consider the circuit below.

R

CVI VO

+

_

+

_

1( ) 1

1( ) 1OV jw jwC

V jw jwRCRi

jwC

Low pass filter circuit

12


Low Pass Filter

0 dB

1

0

1/RC

1/RC

Bode

Linear Plot

.-3 dB

x0.707

Passes low frequencies

Attenuates high frequencies

13


High Pass Filter Consider the circuit below.

C

RVi VO

+

_

+

_

( )1( ) 1

OV jw jwRCRV jw jwRC

RijwC

High Pass Filter

14


High Pass Filter

0 dB

.

. -3 dB

0

1/RC

1/RC

1/RC

10.707

Bode

Linear

Passes high frequencies

Attenuates low frequencies

x

15


Bandpass Filter Consider the circuit shown below:

C L

RViVO

+

_

+

_

When studying series resonant circuit we showed that;

2

( )

1( )O

i

RsV s L

RV s s sL LC

16


Bandpass Filter

We can make a bandpass from the previous equation and select

the poles where we like. In a typical case we have the following shapes.

0

0 dB

-3 dB

lo

hi

.

. .

.10.707

Bode

Linear

lo

hi

17


Bandpass Filter Example

Suppose we use the previous series RLC circuit with output across R to

design a bandpass filter. We will place poles at –200 rad/sec and – 2000 rad/sec

hoping that our –3 dB points will be located there and hence have a bandwidth

of 1800 rad/sec. To match the RLC circuit form we use:

2

2200 2200 2200

( 200)( 2000)2200 400000 200 2000(1 )(1 )200 2000

s s ss ss ss s x

The last term on the right can be finally put in Bode form as;

0.0055

(1 )(1 )200 2000

jwjw jw

18



From this last expression we notice from the part involving the zero we

have in dB form;

20log(.0055) + 20logw

Evaluating at w = 200, the first pole break, we get a 0.828 dB

what this means is that our –3dB point will not be at 200 because

we do not have 0 dB at 200. If we could lower the gain by 0.829 dB

we would have – 3dB at 200 but with the RLC circuit we are stuck

with what we have. What this means is that the – 3 dB point will

be at a lower frequency. We can calculate this from

200log 20 0.828

low

dBx dB

w dec

19



This gives an wlow = 182 rad/sec. A similar thing occurs at whi where

the new calculated value for whi becomes 2200. These calculations

do no take into account a 0.1 dB that one pole induces on the other

pole. This will make wlo somewhat lower and whi somewhat higher.

One other thing that should have given us a hint that our w1 and w2

were not going to be correct is the following:

1 2

22 1 2 1 2

( )

1 ( ( ) )( )

Rs

w w sLR s w w s w w

s sL LC

What is the problem with this?

20



The problem is that we have

1 2 2 1( )R

w w BW w wL

Therein lies the problem. Obviously the above cannot be true and that

is why we have aproblem at the –3 dB points.

We can write a Matlab program and actually check all of this.

We will expect that w1 will be lower than 200 rad/sec and w2 will be

higher than 2000 rad/sec.

21

Frequency (rad/sec)

Pha

se

(d

eg

); M

ag

nitud

e (

dB

)

Bode Diagrams

-15

-10

-5

0From: U(1)

102

103

104

-100

-50

0

50

100

To:

Y(1

)

-3 dB

-5 dB


22

A Bandpass Digital Filter

Perhaps going in the direction to stimulate your interest in taking a course

on filtering, a 10 order analog bandpass butterworth filter will be

simulated using Matlab. The program is given below.

N = 10; %10th order butterworth analog prototype

[ZB, PB, KB] = buttap(N);

numzb = poly([ZB]);

denpb = poly([PB]);

wo = 600; bw = 200; % wo is the center freq

% bw is the bandwidth

[numbbs,denbbs] = lp2bs(numzb,denpb,wo,bw);

w = 1:1:1200;

Hbbs = freqs(numbbs,denbbs,w);

Hb = abs(Hbbs);

plot(w,Hb)

grid

xlabel('Amplitude')

ylabel('frequency (rad/sec)')

title('10th order Butterworth filter')

23

A Bandpass Filter

24

RLC Band Stop Filter

Consider the circuit below:

R

L

C

+

_

VO

+

_Vi

The transfer function for VO/Vi can be expressed as follows:

)(sGv

LCs

L

Rs

LCs

sGv 1

1

)(2

2

25

This is of the form of a band stop filter. We see we have complex

zeros on the jw axis located at


Comments

LCj

1

From the characteristic equation we see we have two poles. The poles

an essentially be placed anywhere in the left half of the s-plane. We

see that they will be to the left of the zeros on the jw axis.

We now consider an example on how to use this information.

26


Example

Design a band stop filter with a center frequency of 632.5 rad/sec

and having poles at –100 rad/sec and –3000 rad/sec.

The transfer function is:

3000003100

3000002

2

ss

s

We now write a Matlab program to simulate this transfer function.

27


Example

num = [1 0 300000];

den = [1 3100 300000];

w = 1 : 5 : 10000;

Bode(num,den,w)

28


Example

Bode

Matlab

29

VinVO

C

Rfb

+

_

+

_

Rin

Basic Active Filters

Low pass filter

30


RinC

Vin

Rfb

VO

+

_

+

_

High pass

31


Vin

R1

R1

C1

C2

R2

R2

Rfb

Ri

VO

+

+

_

_

Band pass filter

32


Vin

R1

R1

C1

C2

R2Ri

Rfb

VO

+

_

+

_

Band stop filter

33

ANALOG FILTERS

34

Filters

A filter is a system that processes a signal in some desired fashion.

A continuous-time signal or continuous signal of x(t) is a function of the continuous variable t. A continuous-time signal is often called an analog signal.

A discrete-time signal or discrete signal x(kT) is defined only at discrete instances t=kT, where k is an integer and T is the uniform sampling spacing or period between samples

35

Types of Filters

There are two broad categories of filters:

An analog filter processes continuous-time signals

A digital filter processes discrete-time signals.

The analog or digital filters can be subdivided into four categories:

Lowpass Filters

Highpass Filters

Bandstop Filters

Bandpass Filters36

Analog Filter Responses

H(f)

ffc

0

H(f)

ffc

0

Ideal “brick wall” filter Practical filter

37

Ideal Filters

Passband StopbandStopband Passband

Passband PassbandStopband

Lowpass Filter Highpass Filter

Bandstop Filter

PassbandStopband Stopband

Bandpass Filter

M()

M()

c

c

c1

c1

c2

c2

38

Passive Filters

Passive filters use resistors, capacitors, and inductors (RLC networks).

To minimize distortion in the filter characteristic, it is desirable to use inductors with high quality factors (remember the model of a practical inductor includes a series resistance), however these are difficult to implement at frequencies below 1 kHz.

They are particularly non-ideal (lossy)

They are bulky and expensive39

Active Filters

Active filters overcome these drawbacks and are realized using resistors, capacitors, and active devices (usually op-amps) which can all be integrated:

Active filters replace inductors using op-amp based equivalent circuits.

40

Op Amp Advantages

Advantages of active RC filters include:reduced size and weight, and therefore parasitics

increased reliability and improved performance

simpler design than for passive filters and can realize a wider range of functions as well as providing voltage gain

in large quantities, the cost of an IC is less than its passive counterpart

41

Op Amp Disadvantages

Active RC filters also have some disadvantages:limited bandwidth of active devices limits the highest attainable pole frequency and therefore applications above 100 kHz (passive RLC filters can be used up to 500 MHz)the achievable quality factor is also limitedrequire power supplies (unlike passive filters)increased sensitivity to variations in circuit parameters caused by environmental changes compared to passive filters

For many applications, particularly in voice and data communications, the economic and performance advantages of active RC filters far outweigh their disadvantages.

42

Bode Plots

Bode plots are important when considering the frequency response characteristics of amplifiers. They plot the magnitude or phase of a transfer function in dB versus frequency.

43

The decibel (dB)

Two levels of power can be compared using a

unit of measure called the bel.

The decibel is defined as:

1 bel = 10 decibels (dB)

1

210log

P

PB

44

Half Power Point

A common dB term is the half power point

which is the dB value when the P2 is one-

half P1.

1

210log10

P

PdB

dBdB 301.32

1log10 10

45

Logarithms

A logarithm is a linear transformation used to

simplify mathematical and graphical operations.

A logarithm is a one-to-one correspondence.

46

Base Number Representation

Any number (N) can be represented as a

base number (b) raised to a power (x).

The value power (x) can be determined by

taking the logarithm of the number (N) to

base (b).

xbN )(

Nx blog

47

Change of Base Logarithm

Although there is no limitation on the numerical value of the base, calculators are designed to handle either base 10 (the common logarithm) or base e (the natural logarithm).

Any base can be found in terms of the common logarithm by:

wq

wq 10

10

loglog

1log

48

Properties of Logarithms

The common or natural

logarithm of the number 1 is 0.

The log of any number less

than 1 is a negative number.

The log of the product of two

numbers is the sum of the logs

of the numbers.

The log of the quotient

of two numbers is the

log of the numerator

minus the denominator.

The log a number taken

to a power is equal to the

product of the power and

the log of the number.

49

Poles & Zeros of the Transfer Function

pole—value of s where the denominator goes to zero.

zero—value of s where the numerator goes to zero.

50

Single-Pole Passive Filter

First order low pass filter

Cut-off frequency = 1/RC rad/s

Problem : Any load (or source) impedance will change frequency response.

vin voutC

R

RCs

RC

sCR

sCR

sC

ZR

Z

v

v

C

C

in

out

/1

/1

1

1

/1

/1

51

Single-Pole Active Filter

Same frequency response as passive filter.

Buffer amplifier does not load RC network.

Output impedance is now zero.

vinvout

C

R

52

Low-Pass and High-Pass Active Filter Designs

High Pass Low Pass

)/1()/1(

1

1

11

1

RCs

s

RCsRC

sRC

sCR

sRC

sCR

v

v

in

out

RCs

RC

v

v

in

out

/1

/1

53

What are Body plots?

To understand Bode plots, you need to

use Laplace transforms!

The transfer function

of the circuit is:

1

1

/1

/1

)(

)(

sRCsCR

sC

sV

sVA

in

ov

R

Vin(s)

54

Break Frequency

Replace s with j in the transfer function:

Where is called the break frequency, or corner

frequency, and is given by:

b

v

f

fj

RCfjRCjfA

1

1

21

1

1

1)(

RCfb

2

1

bf

55

Corner Frequency

The significance of the break frequency is that it represents the frequency where

Av(f) = 0.707-45.

This is where the output of the transfer function has an amplitude 3-dB below the input amplitude, and the output phase is shifted by -45 relative to the input.

Therefore, is also known as the 3-dB frequency or the corner frequency.

bf

56

Body Plot Logarithmic Scale

Bode plots use a logarithmic scale for

frequency.

where a decade is defined as a range of

frequencies where the highest and lowest

frequencies differ by a factor of 10.

10 20 30 40 50 60 70 80 90 100 200

One decade

57

Gain or Magnitude of the Transfer Function

Consider the magnitude of the transfer function:

Expressed in dB, the expression is

2/1

1)(

b

v

fffA

b

bb

bdBv

ff

ffff

fffA

/log20

/1log10/1log20

/1log201log20)(

22

2

58

Asymptotes

Look how the previous expression changes with frequency:

at low frequencies f<< fb,

|Av|dB = 0 dB

low frequency asymptote

at high frequencies f>>fb,

|Av(f)|dB = -20log f/ fbhigh frequency asymptote

59

Magnitude

20 log P ( )( )

rad

sec

0.1 1 10 10060

40

20

0

arg P ( )( )

deg

rad

sec

0.1 1 10 100100

50

0

Actual response curve

High frequency asymptote

3 dB

Low frequency asymptote

Note that the two

asymptotes intersect at fbwhere

|Av(fb )|dB = -20log f/ fb

60

Bode Plot Filter Approximation

The technique for approximating a filter function based on Bode plots is useful for low order, simple filter designs

More complex filter characteristics are more easily approximated by using some well-described rational functions, the roots of which have already been tabulated and are well-known.

61

Real Filters

The approximations to the ideal filter are the:

Butterworth filter

Chebyshev filter

Cauer (Elliptic) filter

Bessel filter

62

Standard Transfer Functions

ButterworthFlat Pass-band.

20n dB per decade roll-off.

ChebyshevPass-band ripple.

Sharper cut-off than Butterworth.

EllipticPass-band and stop-band ripple.

Even sharper cut-off.

BesselLinear phase response – i.e. no signal distortion in pass-band.

63

Magnitude Plots of Standard Transfer Functions

64

65

Types of Filters

Butterworth – flat response in the passband and acceptable roll-off

Chebyshev – steeper roll-off but exhibits passband ripple (making it unsuitable for audio systems)

Bessel – yields a constant propagation delay

Elliptical – much more complicated

Butterworth Filter

The Butterworth filter magnitude is defined by:

where n is the order of the filter.

,

1

1)()(

2/12njHM

66

Filter Magnitude Approximation

From the previous slide:

for all values of n

For large :

1)0( M

2

1)1( M

nM

1)(

67

dB/decade Roll Off

And

implying the M() falls off at 20n dB/decade for

large values of .

10

101010

log20

log201log20)(log20

n

M n

68

dB/decade Roll Off Examples

T1i

T2i

T3i

wi

1000

0.1 1 100.01

0.1

1

10

20 db/decade

40 db/decade

60 db/decade

69

Relating Transfer Function and Magnitude Response

To obtain the transfer function H(s) from the magnitude response, note that

njHjHjHM

2

22

1

1)()()()(

70

Complex Conjugation and Frequency Response

Because s j for the frequency response, we have s2 2.

The poles of this function are given by the roots of

nnnss

sHsH22 11

1

1

1)()(

nkes kjnn2,,2,1,111 )12(2

71

Pole Calculation

The 2n poles are:

sk =e j[(2k-1)/2n] n even, k = 1,2,...,2n

e j(k/n) n odd, k = 0,1,2,...,2n-1

Note that for any n, the poles of the normalized Butterworth

filter lie on the unit circle in the s-plane. The left half-plane

poles are identified with H(s). The poles associated with H(-s)

are mirror images.

72

S-Plane Pole Placement

73

Frequency Response

74

Phasor’s Rectangular and Polar Representations

Recall from complex numbers that the rectangular form

of a complex can be represented as:

Recalling that the previous equation is a phasor, we can

represent the previous equation in polar form:

jyxz

sincos jrz where

sincos ryandrx

75

Complex Number Definition

Definition: If z = x + jy, we define e z = e x+ jy to be the

complex number

Note: When z = 0 + jy, we have

which we can represent by symbol:

e j

)sin(cos yjyee xz

)sin(cos yjye jy

76

Euler’s Law

)sin(cos je j

The following equation is known as Euler’s law.

Note that

functionodd

functioneven

sinsin

coscos

77

Sine & Cosine Axioms

)sin(cos je j

This implies that

This leads to two axioms:

2cos

jj ee

andj

ee jj

2sin

78

Complex Unit Vector

Observe that e j represents a unit vector which makes an angle with the positive x axis.

79

Third Order Butterworth Filter

Find the transfer function that corresponds to a third-order

(n = 3) Butterworth filter.

Solution:

From the previous discussion:

sk = e jk/3, k=0,1,2,3,4,5

80

S Values

Therefore,

3/5

5

3/4

4

3

3/2

2

3/

1

0

0

j

j

j

j

j

j

es

es

es

es

es

es

81

Roots

p1

1 p6

1

p2

.5 0.8668j p5

.5 0.866j

p3

.5 0.866 j p4

.5 0.866j

The roots are:

82

Filter Design Table

The factored form of the normalized Butterworth polynomials for various order n are tabulated in filter design tables.

83

Butterworth Filter Design Table

n Denominator of H(s) for Butterworth Filter

1 s + 1

2 s2 + 1.414s + 1

3 (s2 + s + 1)(s + 1)

4 (s2 + 0.765 + 1)(s2 + 1.848s + 1)

5 (s + 1) (s2 + 0.618s + 1)(s2 + 1.618s + 1)

6 (s2 + 0.517s + 1)(s2 + 1.414s + 1 )(s2 + 1.932s + 1)

7 (s + 1)(s2 + 0.445s + 1)(s2 + 1.247s + 1 )(s2 + 1.802s + 1)

8 (s2 + 0.390s + 1)(s2 + 1.111s + 1 )(s2 + 1.663s + 1 )(s2 + 1.962s + 1)

84

DIGITAL FILTERS

85

Digital Filter Design Parameters

86

http://cnx.org/contents/2YEHwQYh@1/FIR-FILTER-DESIGN-THE-WINDOW-D


Digital Filter Design Parameters

87


T

S

T

S

T

P

T

P

S

P

F

F

FS

F

F

FP

S

P

sS

PP

dB

dB

2

2

20/

20/

10

10

10

101

)(log20

)1(log20


Analog-Based Digital Filters

88

Filter Order Estimation

89

Estimation of the Filter OrderFIR(lowpass digital filter):

For narrowband filter:

For wideband filter:

2/)(27

94.5)(log20

2/)(

22.0)(log20

2/)(6.14

13)(log20

10

10

10

ps

p

ps

s

ps

sp

N

N

N

90

Digital Signal Processing

Filter Design: IIR

1. Filter Design Specifications

2. Analog Filter Design

3. Digital Filters from Analog Prototypes

Filter Order Estimation

91

1. Filter Design Specifications

Digital Filter Design

92

Performance Constraints


93

Performance Constraints


94

Passband Ripple


95

Stopband Ripple


96

Filter Type Choice : FIR vs. IIR


97

FIR vs. IIR


98

IIR Filter Design


99

2. Analog Filter Design


100

CT Transfer Functions


101

Butterworth Filters


102

Butterworth Filters


103

Digital Filter DesignButterworth Filters

104

Butterworth Filters


105

Butterworth Example


106

Butterworth Example


107

Chebyshev I Filter


108

Chebyshev I Filter


109

Chebyshev I Filter


110

Chebyshev II Filter


111

Elliptical (Cauer) Filters


112

Analog Filter Types Summary


FILTER DESIGN

113

MATLAB for Signal Processing

The MathWorks Inc.

Natick, MA

USA

Filter Design

114

Copyright 1984 - 2001 by The MathWorks, Inc.

115MATLAB for Signal Processing

Filter Design

• Frequency selective system

• Analog

• Digital

– Infinite impulse response (IIR)

– Finite impulse response (FIR) a=1

N

k

k

M

m

m knyamnxbny10

)()()(

M

m

m mnxbny0

)()(



Filters

DiscreteContinuous

IIR FIR

Chebyshev I

Chebyshev II

Eliptic

Yulewalk

Butterworth

Arbitrary Response

Equiripple

Least Squares

Raised Cosine

Windowing

Bessel

Analog

prototyped

filters

Bilinear

Transformation

Filter Design Methods



Transition Band

Stopband

Attenuation

Passband Ripple

Passband Stopband

Fs/2

Filter Specification

• Wp, Ws, Rp, Rs



FIR Filter Design

• FIR

–Features

–Windowing

Inverse FFT to get h(n)

Arbitrary response

–Other methods

• Example

b=fir1(20,0.5,hanning(21));

freqz(b,1)



Windowing• Finite data length

• Functions

• bartlett, blackman, boxcar, chebwin, hamming, hanning, kaiser, triang

•

• Example

x=hanning(N);

plot(hamming(32)

stem(kaiser(16,30)

0 5 10 15 20 25 30 350

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 2 4 6 8 10 12 14 160

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1



Filter Design Tradeoffs

Reduced width of

transition band

FIR

Higher Order

Always Stable

Passband Phase

Linear

IIR

Lower Order

Can be Unstable

Non-linear Phase

Increased complexity

(higher order)

<==>

Lower Rp, Higher Rs Increased complexity

(higher order)

<==>



Filter Design with FDATool

Import filter

coefficients or

design filter

Set quantization

parameters for

Filter Design

Toolbox

Analysis method for

analyzing filter design

Quantize

current filter

using Filter

Design Toolbox

Filter

specifications

Type of filter to

design and

method to use

»fdatool



Importing Existing Designs



Analyze Filters with FDATool



Print Preview and Annotation



Convenient Exporting from FDATool

126


127

Starting FDATool

Start MATLAB. At the command line, type in “fdatool”.

The following screen will appear:

128

Designing an FIR Filter Design

From the “Design Method” part of the screen, click on the “FIR” box and select “Window”.Setting the Filter CoefficientsClick on the “Design Filter” box when done.

129

Designing an FIR Filter

The frequency response curve of the filter will appear

It can be seen that for frequencies above 4 kHz, the output amplitude is reduced by more than 40 dB (100 times).

130

Designing an FIR Filter

Select File -> Save Session As…Give the filter a unique name such as “FIR_lowpass_800Hz”.

131

Designing a High Pass Filter

The next step is to design a highpass filter.At the FDATool window, Select File -> New Session to start the new design.

132

The Highpass Filter Design

Set the “Response Type” to “Highpass”.Notice that the “Filter Specifications” part of the screen changes.

133

Entering the Filter Parameters

Click on the “Design Filter” box when done.

134

Viewing the Highpass Filter Frequency Response

Click on the “Design Filter” box when done.

135

Saving the Finished Highpass Filter Design

Select File -> Save Session As… and give it a unique name.“FIR Highpass 800Hz Hamming Window”.

136

PSPICE

What is Spice?

Spice is the short form of:

Simulated

Program with

Integrated

Circuit

Emphasis

137

Why PSPICE Programming?

Don’t have to draw the circuit

More control over the parts

More control over the analysis

Don’t have to search for parts

Some SPICE software applications (HSPICE, etc.) don’t have GUI at all

Quick and efficient138

Steps of PSPICE Programming

Draw the circuit and label the nodes

Create netlist (*.cir) file

Add in control statements

Add in title, comment & end statements

Run PSPICE

Evaluate the results of the output

139

MATLAB and SPICE Features

MATLABSPICEFeatures

small circuitssmall to very large

circuits

Type of Circuit For

Analysis

YesYesFrequency

Response

NoYes

Inclusion of Device

Model in Software

Package

140

MATLAB and SPICE Features

YesNoDetermination and

plot of poles and

Zeros

YesNoBulk Semiconductor

Characteristics

YesNopn junction

characteristics -

excluding I-v

characteristics

141

142

PHASOR ANALYSIS

Continuity Conditions of Circuit Elements

The voltage across a capacitor cannot be changed instantaneously.

)0()0( CC VV

)0()0( LL II

The current across an inductor cannot be changed instantaneously.

143

Time Phasor

Current in phase

Current lagging

Current leading

V = IR

V = jLI

I = jCV

144

RI

V

Resistor

145

Instantaneous Response

146

LjI

V

Inductor

147

Current Lags Voltage in an Inductor

148

CjI

V

1

Capacitor

149

Current Leads Voltage in an Capacitor

150

...321 ZZZZeqComplex Impedance

Resistance, Reactance

Impedance in Series

151

jXRY

Z

ZZZZab

1

...321

...1111

321

ZZZZab

jBGZ

Y

YYYYab

1

...321

Complex Admittance

Conductance, Susceptance

Impedance in Parallel

152

Thevenin and Norton Transformation

153

Thevenin Equivalent Circuit

154

Norton Equivalent Circuit

155

2221

2111

)(0

)(

IZLjRMIj

MIjILjRZV

L

ss

L

r

rab

ZLjR

MZ

ZLjRZ

22

22

11

Time differentiation replaced by j

Transformer

156

1

2

1

2

N

N

v

v

1

2

2

1

N

N

i

i

Power Conserved

Sinusoidal Ideal Transformer

158

1ST ORDER FILTER ANALYSIS

Figure 14.2 A circuit with voltage

input and output.

Filtering Circuit

159

response plots of the four types of filter

circuits. (a) An ideal low-pass filter. (b)

An ideal high-pass filter. (c) An ideal

bandpass filter. (d) An ideal bandreject

filter.

Filtering Types

160

Figure 14.4 (a) A series RL low-pass

filter. (b) The equivalent circuit at = 0

and (c) The equivalent circuit at = .

Low-Pass Filtering Analysis

161

Figure 14.5 The frequency response

plot for the series RL circuit in Fig.

14.4(a).

Magnitude & Phase Responses

162

Figure 14.6 The s-domain equivalent

for the circuit in Fig. 14.4(a).

LR Transfer Function Analysis

163

Figure 14.7 A series RC low-pass

filter.

Low-Pass Passive RC Circuit

164

Figure 14.8 The s-domain equivalent

for the circuit in Fig. 14.7.

RC Transfer Function Analysis

165

Figure 14.9 Two low-pass filters, the

series RL and the series RC, together

with their transfer functions and

cutoff frequencies.

Low-Pass Transfer Functions

166

Passive low-pass filter with cut-off frequency

First Order Low-Pass RC Passive Filter

167

Figure 14.10 A series RC high-pass

filter; (b) the equivalent circuit at = 0;

and (c) the equivalent circuit at = .

High-Pass Filtering Analysis

168

Figure 14.11 The frequency response

plot for the series RC circuit in Fig.

14.10(a).

Magnitude & Phase Responses

169

Figure 14.12 The s-domain

equivalent of the circuit in Fig. 14.10(a).

CR Transfer Function Analysis

170

Figure 14.13 The circuit for Example

14.3.

High-Pass Passive RL Circuit

171

Figure 14.14 The s-domain

equivalent of the circuit in Fig. 14.13.

RL Transfer Function Analysis

172

173

First-Order Filter Circuits

L+

–VS

C

R

Low Pass

High

Pass

HR = R / (R + sL)

HL = sL / (R + sL)

+

–VS

R

High Pass

Low

Pass

GR = R / (R + 1/sC)

GC = (1/sC) / (R + 1/sC)

174

2nd ORDER FILTER ANALYSIS

Mechanical Analogy

Automobile: Mass inductorSuspension: Spring capacitorShock Absorber: Damper resistor

Force voltageVelocity current

Wik

ipedia

.com

175

Series RLC Circuit

+

-VS(t)

L

C

Ri(t)

KVL:

VL + VR + VC = Vs

176

dt

dv

LLC

ti

dt

di

L

R

dt

id

dt

dv

C

ti

dt

diR

dt

idL

tvvdttiC

iRdt

diL

tvtvtvtv

S

S

t

SC

SCRL

1)(

0)(

)()0()(1

)()()()(

2

2

2

2

0

177

Series RLC Circuit Equations

0)(2

0:

.1

2...

1)(2

1)(

202

2

0

202

2

2

2

tidt

di

dt

id

vresponsenaturaltheFind

LCand

L

Rwhere

dt

dv

Lti

dt

di

dt

id

dt

dv

LLC

ti

dt

di

L

R

dt

id

S

S

S

178


2

0

2

2

0

2

2

0

2

2

0

2

2

02

2

)1(2

)1(4)2(2

:

02

:

02

:

)(

:

0)(2

s

equationquadratictheUse

ss

giving

keskekes

ngsubstituti

keti

formtheinsolutionatry

tidt

di

dt

id

ststst

st

179


Natural Overdamped Case

180

Case 1 - Overdamped: >o large R:LCL

R 1

2>

tsts

n eKeKti

s

s

21

21

2

0

2

2

2

0

2

1

)(

:lsexponentia decaying twois response natural The

0

0

Natural Underdamped Case

181

LCL

R 1

2

22

0

)(

2

)(

1

21

22

0

2

0

2

2

22

0

22

0

2

0

2

1

)...cos(

)(

phasorsdecayingtwoisresponsenaturalThe

1

220

220

21

nn

t

tjttjt

tsts

n

wheretKe

eeKeeK

eKeKti

js

js

Case 2 - Underdamped: o small R:

Natural Critically Damped Case

182

LCL

R 1

2

tt

n teKeKti

s

s

21

2

0

2

2

2

0

2

1

)(

isresponsenaturalThe

00

0

Case 3 - Critically damped: o :

)4( 1)0(

0)0(

)0(

gives (2) and (1)

(3) 0

(2) 01

gives KVL Applying

(1b)

10

(1a) 0

: conditions initial Assumed

00

0

2

2

0

0

0

VRILdt

di

Vdt

diLRi

LC

i

dt

di

L

R

dt

id

idtCdt

diLRi

VidtC

v

Ii

t

C

00

0

2

2

1)0(

0

: conditions Initial

0

VRILdt

di

Ii

LC

i

dt

di

L

R

dt

id

Overdamped CaseNatural Response: Series RLC

183

01

01

0

constants are and : Let

1)0(

0

: conditions Initial

0

2

2

2

00

0

2

2

LCs

L

Rs

LCs

L

RsAe

eLC

Ase

L

AReAs

sAAei

VRILdt

di

Ii

LC

i

dt

di

L

R

dt

id

st

ststst

st

LC

L

R

s

s

LCL

R

L

Rs

LCL

R

L

Rs

1

2 where

1

22

1

22

0

2

0

2

2

2

0

2

1

2

2

2

1

Characteristic

equation

Natural

frequencies

Damping

factor

Resonant

frequency

Natural Response: Series RLC

184

tsts

tsts

eAeAti

eAieAi

ss

21

21

21

2211

21

)(

:solution generalA

,

:) (if solutions Two

LC

L

R

s

s

1

2 where

0

2

0

2

2

2

0

2

1

• Three cases discussed

– Overdamped case : > 0

– Critically damped case : = 0

– Underdamped case : < 0

Natural Solution: Series RLC

185

011

becomesequation sticcharacteri the

,Let

(3) 01

(2) 01

gives KCL Applying

(1b) 0

(1a) 1

0

: conditions initial Assumed

2

2

2

0

0

0

LCs

RCs

Aev(t)

LC

v

dt

dv

RCdt

vd

dt

dvCvdt

LR

v

Vv

v(t)dtL

Ii

st

t

LC

RC

s

1

2

1

where

0

2

0

2

2,1

Natural Response: Parallel RLC

186

• Overdamped case : > 0

• Critically damped case : = 0

• Underdamped case : < 0

tstseAeAtv 21

21)(

tetAAtv 21)(

tAtAetv

js

dd

t

d

d

sincos)(

where

21

22

0

2,1

Natural Response: Parallel RLC

187

L

RIV

dt

di

Ii

LC

L

R

s

00

0

0

2

0

2

2,1

)0(

)0(

:conditions Initial

1

2 where

• Series RLC Circuit • Parallel RLC Circuit

RC

RIV

dt

dv

Vv

LC

RC

s

00

0

0

2

0

2

2,1

)0(

)0(

:conditions Initial

1

2

1

where

Summary: Series & Parallel RLC

188

case. free-source in the

as form same thehas (2)

(2)

But

(1)

,0for KVL Applying

2

2

LC

V

LC

v

dt

dv

L

R

dt

vd

dt

dvCi

Vvdt

diLRi

t

S

S

>

response state-steady the:

response transientthe:

where

)()()(

ss

t

sst

v

v

tvtvtv

Step Response: Series RLC

189

case. free-source in the as Same

01

becomesequation sticcharacteri The

0

, Let

0

2

''

2

'2

'

2

2

LCs

L

Rs

LC

v

dt

dv

L

R

dt

vd

Vvv

LC

Vv

dt

dv

L

R

dt

vd

S

S


190

.)0( and )0( from obtained are where

sincos

)(

)()(

)()()(

2,1

21

21

2121

/dtdvvA

etAtA

etAA

eAeA

tv

Vvtv

tvtvtv

t

dd

t

tsts

t

Sss

sst

(Overdamped)

(Critically damped)

(Underdamped)


191

General Response

192

case. free-source in the

as form same thehas (2)

(2) 1

But

(1)

,0for KCL Applying

2

2

LC

I

LC

i

dt

di

RCdt

id

dt

diLv

Idt

dvCi

R

v

t

S

S

>

response state-steady the:

response transient the:

where

)()()(

ss

t

sst

i

i

tititi

Step Response: Parallel RLC

193

case. free-source in the as Same

011

becomesequation sticcharacteri The

01

, Let

01

2

''

2

'2

'

2

2

LCs

RCs

LC

i

dt

di

RCdt

id

Iii

LC

Ii

dt

di

RCdt

id

S

S


194

.)0( and )0( from obtained are where

sincos

)(

)()(

)()()(

2,1

21

21

2121

/dtdiiA

etAtA

etAA

eAeA

ti

Iiti

tititi

t

dd

t

tsts

t

Sss

sst

(Overdamped)

(Critically damped)

(Underdamped)


195

196

Second-Order Filter Circuits

C

+

–VS

R

Band Pass

Low

Pass

LHigh

Pass

Band

Reject

Z = R + 1/sC + sL

HBP = R / Z

HLP = (1/sC) / Z

HHP = sL / Z

HBR = HLP + HHP

f

f

f

fjQ

ffjQ

LCfff

f

R

Lfj

fRC

j

fH

LCfff

f

R

Lfj

fRC

j

fC

jfLjR

fC

j

ZZZ

Z

S

S

in

out

inininCLR

Cout

0

0

0

00

0

00

0

1

)/(

2

121

2)(

2

121

2

22

2

V

V

VVVV

First-Order Filter CircuitsSecond Order Low-Pass Filter

197

Second-Order Low-Pass Filter

200

2

0

0012

002

0

00

0

1

1

90

1

ffffQ

ffQfH

ffffQTanffffQ

ffQ

ffffjQ

ffjQfH

S

s

sS

s

s

s

in

out

V

V

Low-Pass Filter: Natural Frequency

198

First-Order Filter CircuitsLow-Pass Filter: Natural Frequency

199

Second Order High-Pass Filter

At low frequency the

capacitor is an open circuit

At high frequency the

capacitor is a short and the

inductor is open

High-Pass Filter: Natural Frequency

200

Second Order Band-Pass Filter


capacitor is an open circuit


inductor is an open circuit

Band-Pass Filter: Natural Frequency

201

Second Order Band-Reject Filter


capacitor is an open

circuit


inductor is an open

circuit

Band-Reject Filter: Natural Frequency

202

203

Series RLC Bandpass Filter DesignDesign a series RLC bandpass filter with cutoff frequencies f1=1kHz and f2 = 10 kHz.

Cutoff frequencies give us two equations but we have 3 parameters to choose.

Thus, we need to select a value for either R, L, or C and use the equations to find

other values. Here, we choose C=1μF.

0 1 2

0

22 6

0

2 1

3

2 6 2

(6280)(62800) 19,867rad/s

3162.28Hz2

1 12.533 mH

2 (3162.28) (10 )

19,867rad/s0.3514

(2 *10000 2 *1000)rad/s

2.533(10 )143.24

(10 )(0.3514)

o

o

f

LC

Q

LR

CQ

f1=1kHz 1 = 2f1 = 6280 rad/s

f2 = 10 kHz 2 = 2f2 = 62,800 rad/s

0 1 2

0

2 1

Q

0

1

LC

0

2

1L L L LR

Q Q C Q CQ LC

204

ACTIVE FILTER DESIGN

ffB

Bi

f

ffi

f

i

f

ff

f

f

f

ff

f

f

ff

i

f

i

o

CRf

ffjR

R

CfRjR

R

Z

ZfH

CfRj

RZ

R

CfRj

R

fCj

RZ

Z

Z

V

VfH

2

1

)/(1

1

21

1)(

21

21

2

1

111

)(

A low-pass filter with a dc gain of -Rf/Ri

First Order Low-Pass Active RC Filter

205

dttvRC

tv

t

o in

0

1

Integrator Circuit

206

iiB

B

B

i

f

ii

if

i

i

ii

if

ii

f

i

f

ffi

ii

i

f

i

o

CRf

ffj

ffj

R

R

CRfj

CRfj

R

R

CRfj

CRfj

CfjR

R

Z

ZfH

RZCfj

RZ

Z

Z

v

vfH

2

1

)/(1

)/(

21

2

21

2

2

1)(

2

1

)(

A high-pass filter with a high frequency gain of -Rf/Ri

First Order High-Pass RC Passive Filter

207

dt

dvRCtvo

in

Differentiator Circuit

208

n

B

n

i

fn

n

ffjR

R

fHfHfHfH

)/(1

1)1(

)()()()( 21

Higher Order Active Filters

209

n

Bff

HfH

2

0

1

Butterworth filters are

characterized by having a

particularly flat pass-band.

Butterworth Transfer Function

210

211

First Order Low-Pass RC Active Filter

+

+

+

vovi

R1R2

C

+

+

+

vovi

Zi

Zf

c

csC

i

f

sK

R

R

Z

ZsH

1

12 ||

)(

CRR

RK c

21

2 1

212

Useful Active Circuits

Non-Inverting Op Amp: gain = (1 + Rf / R1) Inverting Op Amp: gain = (- Rf / R1)

Voltage Follower: gain = 1

vi

+

vo

-Rf

R1 vi

+

vo

-

RfR1

vi

+

vo

-

213

Other Useful Active Circuits

The summing amp: The differential amp:

-

+

vi1

vi2

viN

+

vo

-

R1

R2

RN

Rf

vi1

vi2

+

vo

-

R1

R2

R1

R2

vR

Rv

R

Rv

R

Rvo

f

i

f

i

f

N

iN

1

1

2

2 vR

Rv vo i i 2

1

2 1

214

First Order Filter Functions

215

Second Order Filter Functions

216

Second Order Filter Functions

AC Analysis

Second-Order High-Pass Filter

Vin 1 0 AC 10V

Rf 1 2 4.0

Cf 2 3 2.0uF

Lf 3 0 127uH

.AC DEC 20 100Hz 1MEG

.PROBE

.END

217

THANK YOU

220

PART of this document’s CONTENT is NOT ORIGINAL.Its ORIGINATION is from many FRIENDLY SOURCES.THANKS to ALL the sources used for the compositionof this document. WE are TRULY INDEBTED to YOU.

http://www.ece.uprm.edu/~domingo/

http://www.ece.uprm.edu/~domingo/

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