INTEGRATED CIRCUITS (EEC-501) UNIT-2 FILTERS (ACTIVE) By: Mr. RAJAN VERMA Email:[email protected]

INTEGRATED CIRCUITS(EEC-501)

UNIT-2

FILTERS (ACTIVE)

By: Mr. RAJAN VERMA Email:[email protected]

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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 spacing or period between samples

By: Mr. RAJAN VERMA Email:[email protected]: Mr. RAJAN VERMA Email:[email protected]

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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 Filters


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Analog Filter Responses

H(f)

ffc

0

H(f)

ffc

0

Ideal “brick wall” filter Practical filter


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Ideal Filters

Passband Stopband Stopband Passband

Passband PassbandStopband

Lowpass Filter Highpass Filter

Bandstop Filter

PassbandStopband Stopband

Bandpass Filter

M(w)

M(w)

w w

w w

w c w c

w c1w c1

w c2w c2


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There are a number of ways to build filters and of these passive and active filters are the most commonly used in voice and data communications.


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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 expensive


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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.


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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


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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 RLCfilters can be used up to 500 MHz)

• the achievable quality factor is also limited• require 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.


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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.


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The decibel (dB)

Two levels of power can be compared using aunit of measure called the bel.

The decibel is defined as:

1 bel = 10 decibels (dB)

1

210log

P

PB


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A common dB term is the half power pointwhich is the dB value when the P2 is one-half P1.

1

210log10

P

PdB

dBdB 301.32

1log10 10


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Logarithms

A logarithm is a linear transformation used to simplify mathematical and graphical operations.

A logarithm is a one-to-one correspondence.


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Any number (N) can be represented as a base number (b) raised to a power (x).

The value power (x) can be determined bytaking the logarithm of the number (N) tobase (b).

xbN )(

Nx blogBy: Mr. RAJAN VERMA Email:[email protected]

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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 1010

loglog

1log


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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.


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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.


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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


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Single-Pole Active Filter

Same frequency response as passive filter.

Buffer amplifier does not load RC network.

Output impedance is now zero.

vin vout

C

R


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Low-Pass and High-Pass Designs

High Pass Low Pass

)/1()/1(

11

11

1

RCs

s

RCsRC

sRCsCR

sRCsCR

v

v

in

out

RCs

RC

v

v

in

out

/1

/1


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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)

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Break Frequencies

Replace s with jw in the transfer function:

where fc is called the break frequency, or corner frequency, and is given by:

b

v

ff

jRCfjRCj

fA

1

1

21

1

1

1)(

RCfc 2

1

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

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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

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Consider the magnitude of the transfer function:

Expressed in dB, the expression is 2/1

1)(

b

vff

fA

b

bb

bdBv

ff

ffff

fffA

/log20

/1log10/1log20

/1log201log20)(

22

2

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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/ fb

• high frequency asymptote

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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 fb where

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

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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.

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Real Filters

The approximations to the ideal filter are the:• Butterworth filter• Chebyshev filter• Cauer (Elliptic) filter• Bessel filter

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Standard Transfer Functions Butterworth

• Flat Pass-band.• 20n dB per decade roll-off.

Chebyshev• Pass-band ripple.• Sharper cut-off than Butterworth.

Elliptic• Pass-band and stop-band ripple.• Even sharper cut-off.

Bessel• Linear phase response – i.e. no signal distortion in

pass-band.

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Butterworth Filter

The Butterworth filter magnitude is defined by:

where n is the order of the filter.

2/121

1)()(

njHM


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From the previous slide:

for all values of n

For large w:

1)0( M

2

1)1( M

nM

1

)(


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And

implying the M(w) falls off at 20n db/decade for large valuesof w.

10

101010

log20

log201log20)(log20

n

M n


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T1i

T2i

T3i

wi

1000

0.1 1 100.01

0.1

1

10

20 db/decade

40 db/decade

60 db/decade


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

njHjHjHM2

22

1

1)()()()(


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Because s = jw for the frequency response, we have s2 = - w2.

The poles of this function are given by the roots of

nnn sssHsH

22 11

1

1

1)()(

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


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The 2n pole are:

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

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

Note that for any n, the poles of the normalized Butterworthfilter lie on the unit circle in the s-plane. The left half-planepoles are identified with H(s). The poles associated with H(-s) are mirror images.


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Recall from complex numbers that the rectangular formof a complex can be represented as:

Recalling that the previous equation is a phasor, we canrepresent the previous equation in polar form:

jyxz

sincos jrz where

sincos ryandrx


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

Note: When z = 0 + jy, we have

which we can represent by symbol:

e j

)sin(cos yjyee xz

)sin(cos yjye jy


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)sin(cos je j

The following equation is known as Euler’s law.

Note that

functionodd

functioneven

sinsin

coscos


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)sin(cos je j

This implies that

This leads to two axioms:

2cos

jj ee

andj

ee jj

2sin


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Observe that e jq represents a unit vector which makes an angle q with the positivie x axis.


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Find the transfer function that corresponds to a third-order (n = 3) Butterworth filter.

Solution:

From the previous discussion:

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


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Therefore,

3/55

3/44

3

3/22

3/1

00

j

j

j

j

j

j

es

es

es

es

es

es


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p1

1 p6

1

p2

.5 0.8668j p5

.5 0.866j

p3

.5 0.866j p4

.5 0.866j

The roots are:


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Im pi

Re pi

2 0 2

2

2


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Using the left half-plane poles for H(s), we get

H ss s j s j

( )( )( / / )( / / )

1

1 1 2 3 2 1 2 3 2

which can be expanded to:

H ss s s

( )( )( )

1

1 12


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


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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)


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Frequency Transformations


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So far we have looked at the Butterworth filter with a normalized cutoff frequency

c rad1 / sec

By means of a frequency transformation, wecan obtain a lowpass, bandpass, bandstop, or highpass filter with specific cutoff frequencies.


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Lowpass with Cutoff Frequency wu

Transformation:

s sn u /


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Highpass with Cutoff Frequency wl

Transformation:

s sn l /


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Band pass Butterworth FiltersMultistage Wide Band pass

The center frequency fc or fo for the wide band pass filter is defined as,

fc or fo = √ fH fL

fL = 1 / 2π √ RC

fH = 1 / 2π √ R’ C’

and Q (Quality Factor) = fC / (fH - fL)

BW = (fH - fL)

Q = fC / BW

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Band reject KRC Filters :

• Vo / Vin = HON HN

Where HON = DC gain and HN is given by

HN = 1 – ( ω / ωO )2 / 1 – ( ω / ωO )2 + (j ω / ωO) / Q

• HON = K ωO = 1 / RC and Q = 1 / 4-2k

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Documents

INTEGRATED CIRCUITS (EEC-501) UNIT-2 FILTERS (ACTIVE) By: Mr. RAJAN VERMA Email:[email protected]