Introduction To analog Filters , low pass

Introduction To
Analog Filters
The University of Tennessee
Knoxville, Tennessee
wlg

Filters
Background:
. Filters may be classified as either digital or analog.
. Digital filters
Digital filters are implemented using a digital computer
or special purpose digital hardware.
. Analog filters
Analog filters may be classified as either passive or
active and are usually implemented with R, L, and C
components and operational amplifiers.

Filters
Background:
. An active filter
active filter is one that, along with R, L, and C
components, also contains an energy source, such
as that derived from an operational amplifier.
. A passive filter
passive filter is one that contains only R, L, and
C components. It is not necessary that all three be
present. L is often omitted (on purpose) from
passive filter design because of the size and cost
of inductors – and they also carry along an R that
must be included in the design.

Filters
Background:
. The synthesis
synthesis (realization) of analog filters, that is, the
way one builds (topological layout) the filters, received
significant attention during 1940 thru 1960. Leading
the work were Cauer and Tuttle. Since that time,
very little effort has been directed to analog filter
realization.
. The analysis
analysis of analog filters is well described in
filter text books. The most popular include Butterworth,
Chebyshev and elliptic methods.

Filters
Background:
. Generally speaking, digital filters have become the focus
of attention in the last 40 years. The interest in digital
filters started with the advent of the digital computer,
especially the affordable PC and special purpose signal
processing boards. People who led the way in the work
(the analysis part) were Kaiser, Gold and Radar.
. A digital filter is simply the implementation of an
equation(s) in computer software. There are no R, L,
C components as such. However, digital filters can also
be built directly into special purpose computers in
hardware form. But the execution is still in software.

Filters
Background:
. In this course we will only be concerned with an
introduction to filters. We will look at both passive
and active filters.
. We will not cover any particular design or realization
methods but rather use our understanding of poles and
zeros in the s-plane.
.All EE and CE undergraduate students should take a
course in digital filter design, in my opinion.

Passive Analog Filters
Background: Four types of filters
Four types of filters - “Ideal”
- “Ideal”
lowpass
lowpass highpass
highpass
bandpass
bandpass bandstop
bandstop

Background: Realistic Filters:
Realistic Filters:
lowpass
lowpass highpass
highpass
bandpass
bandpass bandstop
bandstop

Background:
It will be shown later that the ideal
filter, sometimes called a “brickwall”
filter, can be approached by making the
order of the filter higher and higher.
The order here refers to the order of the
polynomial(s) that are used to define the
filter. Matlab examples will be given later
to illustrate this.

Low Pass Filter Consider the circuit below.
R
C
VI VO
+
_
+
_
1
( ) 1
1
( ) 1
O
V jw jwC
V jw jwRC
R
i
jwC
 


Low pass filter circuit

Low Pass Filter
0 dB
1


0
1/RC
1/RC
Bode
Linear Plot
.
-3 dB
x
0.707
Passes low frequencies
Attenuates high frequencies

High Pass Filter Consider the circuit below.
C
R
Vi VO
+
_
+
_
( )
1
( ) 1
O
V jw jwRC
R
V jw jwRC
R
i
jwC
 


High Pass Filter

High Pass Filter
0 dB
.
. -3 dB
0 

1/RC
1/RC
1/RC
1
0.707
Bode
Linear
Passes high frequencies
Attenuates low frequencies
x

Bandpass Pass Filter Consider the circuit shown below:
C L
R
Vi
VO
+
_
+
_
When studying series resonant circuit we showed that;
2
( )
1
( )
O
i
R
s
V s L
R
V s s s
L LC

 

Bandpass Pass Filter
We can make a bandpass from the previous equation and select
the poles where we like. In a typical case we have the following shapes.


0
0 dB
-3 dB
lo
hi
.
. .
.
1
0.707
Bode
Linear
lo
hi

Bandpass Pass Filter Example
Suppose we use the previous series RLC circuit with output across R to
design a bandpass filter. We will place poles at –200 rad/sec and – 2000 rad/sec
hoping that our –3 dB points will be located there and hence have a bandwidth
of 1800 rad/sec. To match the RLC circuit form we use:
2
2200 2200 2200
( 200)( 2000)
2200 400000 200 2000(1 )(1 )
200 2000
s s s
s s
s s
s s x
 
 
   
The last term on the right can be finally put in Bode form as;
0.0055
(1 )(1 )
200 2000
jw
jw jw
 

From this last expression we notice from the part involving the zero we
have in dB form;
20log(.0055) + 20logw
Evaluating at w = 200, the first pole break, we get a 0.828 dB
what this means is that our –3dB point will not be at 200 because
we do not have 0 dB at 200. If we could lower the gain by 0.829 dB
we would have – 3dB at 200 but with the RLC circuit we are stuck
with what we have. What this means is that the – 3 dB point will
be at a lower frequency. We can calculate this from
200
log 20 0.828
low
dB
x dB
w dec


This gives an wlow = 182 rad/sec. A similar thing occurs at whi where
the new calculated value for whi becomes 2200. These calculations
do no take into account a 0.1 dB that one pole induces on the other
pole. This will make wlo somewhat lower and whi somewhat higher.
One other thing that should have given us a hint that our w1 and w2
were not going to be correct is the following:
1 2
2
2 1 2 1 2
( )
1 ( ( ) )
( )
R
s
w w s
L
R s w w s w w
s s
L LC


  
 
What is the problem with this?

The problem is that we have
 
1 2 2 1
( )
R
w w BW w w
L
    
Therein lies the problem. Obviously the above cannot be true and that
is why we have aproblem at the –3 dB points.
We can write a Matlab program and actually check all of this.
We will expect that w1 will be lower than 200 rad/sec and w2 will be
higher than 2000 rad/sec.

Phase
(deg);
Magnitude
(dB)
Bode Diagrams
-15
-10
-5
0
From: U(1)
10
2
10
3
10
4
-100
-50
0
50
100
To:
Y(1)
-3 dB
-5 dB

A Bandpass Digital Filter
Perhaps going in the direction to stimulate your interest in taking a course
on filtering, a 10 order analog bandpass butterworth filter will be
simulated using Matlab. The program is given below.
N = 10; %10th order butterworth analog prototype
[ZB, PB, KB] = buttap(N);
numzb = poly([ZB]);
denpb = poly([PB]);
wo = 600; bw = 200; % wo is the center freq
% bw is the bandwidth
[numbbs,denbbs] = lp2bs(numzb,denpb,wo,bw);
w = 1:1:1200;
Hbbs = freqs(numbbs,denbbs,w);
Hb = abs(Hbbs);
plot(w,Hb)
grid
xlabel('Amplitude')
ylabel('frequency (rad/sec)')
title('10th order Butterworth filter')

RLC Band stop Filter
Consider the circuit below:
R
L
C
+
_
VO
+
_
Vi
The transfer function for VO/Vi can be expressed as follows:

)
(s
Gv
LC
s
L
R
s
LC
s
s
Gv
1
1
)
(
2
2





This is of the form of a band stop filter. We see we have complex zeros
on the jw axis located
RLC Band Stop Filter
Comments
LC
j
1

From the characteristic equation we see we have two poles. The poles
an essentially be placed anywhere in the left half of the s-plane. We
see that they will be to the left of the zeros on the jw axis.
We now consider an example on how to use this information.

Example
Design a band stop filter with a center frequency of 632.5 rad/sec
and having poles at –100 rad/sec and –3000 rad/sec.
The transfer function is:
300000
3100
300000
2
2



s
s
s
We now write a Matlab program to simulate this transfer function.

Example
num = [1 0 300000];
den = [1 3100 300000];
w = 1 : 5 : 10000;
Bode(num,den,w)

Example
Bode
Matlab

Vin
VO
C
R fb
+
_
+
_
R in
Basic Active Filters
Low pass filter

Rin
C
Vin
Rfb
VO
+
_
+
_
High pass

Vin
R 1
R 1
C 1
C 2
R 2
R 2
R fb
R i
VO
+
+
_
_
Band pass filter

Vin
R 1
R 1
C 1
C 2
R 2
R i
R fb
VO
+
_
+
_
Band stop filter

Introduction To analog Filters , low pass

Introduction To analog Filters , low pass

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