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ENTC 3320
Active Filters
Filters
l 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
l There are two broad categories of filters:• An analog filter processes continuous-time signals• A digital filter processes discrete-time signals.
l The analog or digital filters can be subdivided into four categories:• Lowpass Filters• Highpass Filters• Bandstop Filters• Bandpass Filters
Analog Filter Responses
H(f)
ffc0
H(f)
ffc0
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(w)
M(w)
w w
w w
w c w c
w c1w c1
w c2w c2
l 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 filtersl Passive filters use resistors, capacitors, and
inductors (RLC networks). l 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
l 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
l 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 Disadvantagesl 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
l For many applications, particularly in voice and data communications, the economic and performance advantages of active RC filters far outweigh their disadvantages.
Bode Plots
l 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
PPB =
A common dB term is the half power pointwhich is the dB value when the P2 is one-half P1.
1
210log10
PPdB =
dBdB 301.321log10 10 -»-=
Logarithms
l A logarithm is a linear transformation used to simplify mathematical and graphical operations.
l 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=
l 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).
l Any base can be found in terms of the common logarithm by:
wq
wq 1010
loglog
1log =
Properties of Logarithms
l The common or natural logarithm of the number 1 is 0.
l The log of any number less than 1 is a negative number.
l The log of the product of two numbers is the sum of the logs of the numbers.
l The log of the quotient of two numbers is the log of the numerator minus the denominator.
l 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
l pole—value of s where the denominator goes to zero.
l zero—value of s where the numerator goes to zero.
Single-Pole Passive Filter
l First order low pass filterl Cut-off frequency = 1/RC rad/sl Problem : Any load (or source)
impedance will change frequency response.
vin voutC
R
RCsRC
sCR
sCRsC
ZRZ
vv
C
C
in
out
/1/1
11
/1/1
+=
+=
+=
+=
Single-Pole Active Filter
l Same frequency response as passive filter.
l Buffer amplifier does not load RC network.
l Output impedance is now zero.
vin vout
C
R
Low-Pass and High-Pass Designs
High Pass Low Pass
)/1()/1(
11
111
RCss
RCsRCsRC
sCRsRC
sCRvv
in
out
+=
+=
+=
+=
RCsRC
vv
in
out
/1/1+
=
To understand Bode plots, you need to use Laplace transforms!
The transfer function of the circuit is:
11
/1/1
)()(
+=
+==
sRCsCRsC
sVsVA
in
ov
R
Vin(s)
Break Frequencies
Replace s with jw in the transfer function:
where fc is called the break frequency, or corner frequency, and is given by:
÷÷ø
öççè
æ+
=+
=+
=
b
v
ffj
RCfjRCjfA
1
1211
11)(
pw
RCfc p2
1=
Corner Frequencyl The significance of the break frequency is that
it represents the frequency whereAv(f) = 0.707Ð -45°.
l 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.
l 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
l Consider the magnitude of the transfer function:
Expressed in dB, the expression is( )2/1
1)(b
vff
fA+
=
( )
( ) ( )[ ]( )b
bb
bdBv
ffffff
fffA
/log20/1log10/1log20
/1log201log20)(22
2
-=
+-=+-=
+-=
l 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 w( )( ).
w
radsec
0.1 1 10 10060
40
20
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
l The technique for approximating a filter function based on Bode plots is useful for low order, simple filter designs
l 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
l The approximations to the ideal filter are the:• Butterworth filter• Chebyshev filter• Cauer (Elliptic) filter• Bessel filter
Standard Transfer Functionsl Butterworth
• Flat Pass-band.• 20n dB per decade roll-off.
l Chebyshev• Pass-band ripple.• Sharper cut-off than Butterworth.
l Elliptic• Pass-band and stop-band ripple.• Even sharper cut-off.
l Bessel• Linear phase response – i.e. no signal distortion in
pass-band.
Butterworth FilterThe Butterworth filter magnitude is defined by:
where n is the order of the filter.
( ) 2/1211)()(
njHM
www
+==
From the previous slide:
for all values of n
For large w:
1)0( =M
21)1( =M
nMw
w1)( @
And
implying the M(w) falls off at 20n db/decade for large valuesof w.
www
10
101010
log20log201log20)(log20
nM n
-=-=
T1i
T2 i
T3 i
wi1000
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
11)()()()(w
wwww+
=-==
Because s = jw for the frequency response, we have s2 = - w2.
The poles of this function are given by the roots of
( ) ( ) nnn sssHsH
22 111
11)()(
-+=
-+=-
( ) nkes kjnn 2,,2,1,111 )12(2 K==-=-+ -- p
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.
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 +=
( )qq sincos jrz +=where
qq 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 jq
)sin(cos yjyee xz +=
)sin(cos yjye jy +=
)sin(cos qqq je j +=
The following equation is known as Euler’s law.
Note that
( ) ( )( ) ( ) functionodd
functionevenqqqq
sinsincoscos-=-
-=-
)sin(cos qqq je j -=-
This implies that
This leads to two axioms:
2cos
qjj ee -+
= and jee jj
2sin
q--
=
l Observe that e jq represents a unit vector which makes an angle qwith 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 jkp/3, k=0,1,2,3,4,5
Therefore,
3/55
3/44
3
3/22
3/1
00
p
p
p
p
p
j
j
j
j
j
j
eseseseseses
=
=
=
=
=
=
p1 1 p6 1
p2 .5 0.8668j. p5 .5 0.866 j.
p3 .5 0.866 j. p4 .5 0.866 j.
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
( )( )( / / )( / / )
=+ + - + +
11 1 2 3 2 1 2 3 2
which can be expanded to:
H ss s s
( )( )( )
=+ + +
11 12
l 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 rad= 1 / sec
By means of a frequency transformation, wecan obtain a lowpass, bandpass, bandstop, or highpass filter with specific cutoff frequencies.
Lowpass with Cutoff Frequency wu
l Transformation:
s sn u= /w
Highpass with Cutoff Frequency wl
l Transformation:
s sn l=w /
Bandpass with Cutoff Frequencies wl and wu
l Transformation:
ss
Bs Bs
sn =+
= +FHG
IKJ
202
0
0
0w ww
w
where
w w w
w w0 =
= -u l
u lB
Bandstop with Cutoff Frequencies wl and wu
l Transformation:
s Bss
Bs
s
n = +=
+FHG
IKJ
202
00
0ww
ww
