Download - Filter dengan-op-amp
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Active Filters
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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
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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()
M()
c c
c1 c1
c2 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 blog
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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 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
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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 :
1)0( M
2
1)1( M
nM
1
)(
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And
implying the M() falls off at 20n db/decade for large valuesof .
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 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
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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.
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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 jrepresents a unit vector which makes an angle 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 jk/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 u
Transformation:
s sn u /
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Highpass with Cutoff Frequency l
Transformation:
s sn l /
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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
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Bandstop with Cutoff Frequencies l and u
Transformation:
sBs
s
B
s
s
n
FHG
IKJ
202
00
0
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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
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Lect23 EEE 202 62
Op Amp Model
+
–
Inverting input
Non-inverting input
Rin
V+
V–
+–
A(V+ –V– )
VoRout
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Lect23 EEE 202 63
Non-inverting Op-Amp Circuit+
– +
–
Vin+– Vout
Z1
Z2 1
2
in
out
1
)(
Z
Z
V
VH
j
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Lect23 EEE 202 64
Inverting Op-Amp Circuit
–
+Vin+–
+
–
Vout
Z2
Z1
1
2
in
out)(Z
Z
V
VH j
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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?
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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?
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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?
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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?
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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