Transcript Active Filters

ENTC 3320
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)
0
H(f)
fc
Ideal “brick wall” filter
f
0
fc
Practical filter
f
Ideal Filters
Lowpass Filter
Highpass Filter
M(w)
Passband
Stopband
Stopband
wc
w
Passband
wc
Bandstop Filter
w
Bandpass Filter
M(w)
Passband
wc
1
Stopband
Passband
wc
2
Stopband
w
Passband
wc
1
Stopband
wc
2
w

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 a
unit of measure called the bel.
B  log 10
P2
P1
The decibel is defined as:
1 bel = 10 decibels (dB)
dB  10 log 10
P2
P1
A common dB term is the half power point
which is the dB value when the P2 is onehalf P1.
1
10 log 10  3.01 dB  3 dB
2
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).
N  (b)
x
The value power (x) can be determined by
taking the logarithm of the number (N) to
base (b).
x  log b N


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:
1
log q w 
log 10 w
log 10 q
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
R
vin



C
vout
vout
ZC
1 / sC


vin R  Z C R  1 / sC

1
1 / RC

sCR  1 s  1 / RC
First order low pass filter
Cut-off frequency = 1/RC rad/s
Problem : Any load (or source)
impedance will change frequency
response.
Single-Pole Active Filter
R
vin



C
vout
Same frequency response as passive
filter.
Buffer amplifier does not load RC
network.
Output impedance is now zero.
Low-Pass and High-Pass
Designs
High Pass
vout
1
1


1
1  sRC
vin
1
sCR
sCR
sRC
s


RC ( s  1 / RC ) ( s  1 / RC )
Low Pass
vout
1 / RC

vin s  1 / RC
To understand Bode plots, you need to
use Laplace transforms!
R
The transfer function
Vin(s)
of the circuit is:
Vo ( s )
1 / sC
1
Av 


Vin ( s) R  1 / sC sRC  1
Break Frequencies
Replace s with jw in the transfer function:
1
1
Av ( f ) 


jwRC  1 1  j 2RCf
1
 f 
1  j 
 fb 
where fc is called the break frequency, or corner
frequency, and is given by:
1
fc 
2RC
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.
One decade
10
20
30
40
50
60
70
80
90
100 200
where a decade is defined as a range of
frequencies where the highest and lowest
frequencies differ by a factor of 10.

Consider the magnitude of the transfer
function:
1
Av ( f ) 
1   f / fb 
2
Expressed in dB, the expression is
Av ( f ) dB  20 log 1  20 log
2


1  f / fb

 20 log 1   f / f b   10 log 1   f / f b 
2
 20 log  f / f b 
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
Low frequency asymptote
Note that the two
asymptotes intersect at fb
where
Magnitude
|Av(fb )|dB = -20log f/ fb
3 dB
0
20
Actual response curve
20 log( P ( w ) )
40
High frequency asymptote
60
0.1
1
10
w
rad
sec
100


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:
M (w )  H ( jw ) 
where n is the order of the filter.
1
1  w 
2 n 1/ 2
From the previous slide:
M (0)  1
1
M (1) 
2
for all values of n
For large w:
M (w ) 
1
w
n
And
20 log 10 M (w )  20 log 10 1  20 log 10 w n
 20n log 10 w
implying the M(w) falls off at 20n db/decade for large values
of w.
10
T1
i
1
20 db/decade
T2
i
T3
i
40 db/decade
0.1
60 db/decade
0.01
0.1
1
w
i
1000
10
To obtain the transfer function H(s) from the magnitude
response, note that
M (w )  H ( jw )  H ( jw ) H ( jw ) 
2
2
1
 
1 w
2 n
Because s  jw for the frequency response, we have s2   w2.
H ( s) H ( s) 
1
 
1  s
2 n
1

n 2n
1   1 s
The poles of this function are given by the roots of
1   1 s 2 n  1  e  j ( 2 k 1) , k  1,2,,2n
n
The 2n pole are:
e j[(2k-1)/2n] n even, k = 1,2,...,2n
sk =
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.
Recall from complex numbers that the rectangular form
of a complex can be represented as:
z  x  jy
Recalling that the previous equation is a phasor, we can
represent the previous equation in polar form:
z  r cos  j sin  
where
x  r cos  and y  r sin 
Definition: If z = x + jy, we define e z = e x+ jy to be the
complex number
e  e (cos y  j sin y)
z
x
Note: When z = 0 + jy, we have
e jy  (cos y  j sin y)
which we can represent by symbol:
e
j
The following equation is known as Euler’s law.
e
j
 (cos   j sin  )
Note that
cos    cos  
sin      sin  
even function
odd function
This implies that
e
 j
 (cos   j sin  )
This leads to two axioms:
j
e e
cos 
2
j
 j
and
e e
sin  
2j
 j

Observe that e j represents 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,
s0  e
j0
s1  e
j / 3
s2  e
j 2 / 3
s3  e
j
s4  e
j 4 / 3
s5  e
j 5 / 3
The roots are:
p1
1
p2
.5
p3
.5
p6
1
0.8668  j
p5
.5
0.866  j
0.866  j
p4
.5
0.866  j
2
Im p
i
2
0
2
2
Re p
i
Using the left half-plane poles for H(s), we get
1
H ( s) 
( s  1)( s  1 / 2  j 3 / 2)( s  1 / 2  j 3 / 2)
which can be expanded to:
1
H ( s) 
2
( s  1)( s  s  1)

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
w c  1 rad / sec
By means of a frequency transformation, we
can obtain a lowpass, bandpass, bandstop, or
highpass filter with specific cutoff frequencies.
Lowpass with Cutoff Frequency
wu

Transformation:
sn  s / w u
Highpass with Cutoff Frequency
wl

Transformation:
sn  w l / s
Bandpass with Cutoff
Frequencies wl and wu

Transformation:
F
G
H
s 2  w 20 w 0 s w 0
sn 


Bs
B w0
s
where
w 0  w uw l
B  wu wl
IJ
K
Bandstop with Cutoff
Frequencies wl and wu

Transformation:
Bs
sn  2

2
s w0
B
F
s w I
w G  J
Hw s K
0
0
0