Active Filters

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

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

Analog Filter Responses

H(f)

ffc

0

H(f)

ffc

0

Ideal “brick wall” filter Practical filter

Ideal Filters

Passband Stopband Stopband Passband

Passband PassbandStopband

Lowpass Filter Highpass Filter

Bandstop Filter

PassbandStopband Stopband

Bandpass Filter

M()

M()

c c

c1 c1

c2 c2

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.

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

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.

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

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.

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.

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

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

Logarithms

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

A logarithm is a one-to-one correspondence.

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 blog

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

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.

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.

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

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

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

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)

Break Frequencies

Replace s with j 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

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.

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

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

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

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

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.

Real Filters

The approximations to the ideal filter are the:• Butterworth filter

• Chebyshev filter

• Cauer (Elliptic) filter

• Bessel filter

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.

Butterworth Filter

The Butterworth filter magnitude is defined by:

where n is the order of the filter.

2/121

1)()(

njHM

From the previous slide:

for all values of n

For large :

1)0( M

2

1)1( M

nM

1

)(

And

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

10

101010

log20

log201log20)(log20

n

M n

T1i

T2i

T3i

wi

1000

0.1 1 100.01

0.1

1

10

20 db/decade

40 db/decade

60 db/decade

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

njHjHjHM2

22

1

1)()()()(

Because sj for the frequency response, we have s22.

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

The 2n pole 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 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.

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

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

)sin(cos je j

The following equation is known as Euler’s law.

Note that

functionodd

functioneven

sinsin

coscos

)sin(cos je j

This implies that

This leads to two axioms:

2cos

jj ee

andj

ee jj

2sin

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

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

Therefore,

3/55

3/44

3

3/22

3/1

00

j

j

j

j

j

j

es

es

es

es

es

es

p1

1 p6

1

p2

.5 0.8668j p5

.5 0.866j

p3

.5 0.866j p4

.5 0.866j

The roots are:

Im pi

Re pi

2 0 2

2

2

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

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

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)

Frequency Transformations

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.

Lowpass with Cutoff Frequency u

Transformation:

s sn u /

Highpass with Cutoff Frequency l

Transformation:

s sn l /

Bandpass with Cutoff Frequencies l and u

Transformation:

ss

Bs B

s

sn

FHG

IKJ

202

0

0

0

where

0

u l

u lB

Bandstop with Cutoff Frequencies l and u

Transformation:

sBs

s

B

s

s

n

FHG

IKJ

202

00

0

Lect23 EEE 202 61

Active Filters

To achieve a gain > 1, we need to utilize an op-amp based circuit to form an active filter

We can design all the common filter types using op-amp based active filters

Recall for an ideal op-amp that

V+ = V– I+ = I– = 0

Lect23 EEE 202 62

Op Amp Model

+

–

Inverting input

Non-inverting input

Rin

V+

V–

+–

A(V+ –V– )

VoRout

Lect23 EEE 202 63

Non-inverting Op-Amp Circuit+

– +

–

Vin+– Vout

Z1

Z2 1

2

in

out

1

)(

Z

Z

V

VH

j

Lect23 EEE 202 64

Inverting Op-Amp Circuit

–

+Vin+–

+

–

Vout

Z2

Z1

1

2

in

out)(Z

Z

V

VH j

Lect23 EEE 202 65

An Integrator

–

+Vin

+

–

Vout

R

C

+–

Earlier in the semester, we saw this op-amp based integrator circuit. What type of filter does it create? Does this help you to understand time and frequency domain interrelations?

Lect23 EEE 202 66

A Differentiator

–

+Vin

+

–

Vout

C

R

+–

We also observed this op-amp based differentiator circuit. What type of filtering does it produce? How could we move the zero away from the origin?

Lect23 EEE 202 67

Practical Example

You are shopping for a stereo system, and the following specifications are quoted:• Frequency range: –6 dB at 65 Hz and 22 kHz

• Frequency response: 75 Hz to 20 kHz ±3 dB What do these specs really mean?

• Note that the human hearing range is around 20 Hz to 20 kHz (audible frequencies)

Could you draw a rough Bode (magnitude) plot for the stereo system?

Lect23 EEE 202 68

MATLAB Filter Example

How can we filter a real measurement signal? One method involves using a numerical algorithm

called the Fast Fourier Transform (FFT) which converts a time domain signal into the frequency domain

Start MATLAB, then download and run the ‘EEE202Filter.m’ file• Can you observe the reduction in the high frequency

components in both the time and frequency domain plots of the output signal?

Lect23 EEE 202 69

MATLAB Filter Example (cont’d)

Can we extend the knowledge acquired about transfer functions and filtering?• How can we further reduce the high

frequency component magnitudes? Show me.

• How can we remove more high frequencies from the output signal? Show me.

InputSignal

FastFourier

Transform

Freq.DomainFiltering

InverseFFT

OutputSignal