Noty an27af Linear Technology

Application Note 27A
June 1988
A Simple Method of Designing Multiple Order All Pole Bandpass Filters by Cascading 2nd Order Sections
Nello Sevastopoulos Richard Markell

INTRODUCTION

Filter design, be it active, passive, or switched capacitor, is traditionally a mathematically intensive pursuit. There are many architectures and design methods to choose from. Two methods of high order bandpass filter design are discussed herein. These methods allow the filter designer to simplify the mathematical design process
®
and allow LTC’s switched capacitor filters (LTC
1059, LTC1060, LTC1061,LTC1064) to be utilized as high quality bandpass filters.
The first method consists of the traditional cascading of non-identical 2nd order bandpass sections to form the familiar Butterworth and Chebyshev bandpass filters. The second method consists of cascading identical 2nd order bandpass sections. This approach, although “non­textbook,” enables the hardware to be simple and the mathematics to be straightforward. Both methods will be described here.
AN27A is the first of a series of application notes from LTC concerning our universal filter family. Additional notes in the series will discuss notch, lowpass and highpass filters implemented with the universal switched capacitor filter. An addition to this note will extend the treatment of bandpass filters to the elliptic or Cauer forms.
This note will first present a finished design example and proceed to present the design methodology, which relies on tabular simplification of traditional filter design techniques.

DESIGNING BANDPASS FILTERS

Table 1 was developed to enable anyone to design But­terworth bandpass filters. We will discuss the tables in more detail later in this paper, but let’s first design a filter.
EXAMPLE 1—DESIGN
A 4th order 2kHz Butterworth bandpass filter with a –3dB bandwidth equal to 200Hz is required as shown in Figure 1.
Noting that (f
/BW) = 10/1 we can go directly to Table1
oBP
for our normalized center frequencies. From Table 1 under 4th order Butterworth bandpass filters, we go to
/BW) = 10.
(f
oBP
0
–3
–10
–20
–30
GAIN (dB)
–40
–50
–60
0.5k 0.7k 1k 2k 3k 4k 5k
Figure 1. 4th Order Butterworth BP Filter, f
FREQUENCY (Hz)
200Hz
AN27A F01
oPB
10k
= 2kHz
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AN27A-1
Application Note 27A
We find fo1 = 0.965 and fo2 = 1.036 (both normalized to
= 1). To find our desired actual center frequencies,
f
oBP
we must multiply by f and f
The Qs are Q
= 2.072kHz.
o2
1
= Q2 = 14.2 which is read directly from
= 2kHz to obtain fo1 = 1.930kHz
oBP
Table 1. Also available from the table is K, which is the product of each individual bandpass gain H
. To put it
oBP
another way, the value of K is the gain required to make the gain, H, of the overall filter equal to 1 at f
. Our filter
oBP
parameters are highlighted in the following table:
f
oBP
2kHz 1.93kHz 2.072kHz Q1 = Q2 = 14.2 2.03
f
o1
f
o2
Qs K

HARDWARE IMPLEMENTATION

Universal switched capacitor filters are simple to imple­ment. A bandpass filter can be built from the traditional state-variable filter topology. Figure 2 shows this topology for both switched capacitor and active operational amplifier implementations. Our example requires four resistors for each 2nd order section. So eight resistors are required to build our filter.
We start with two 2nd order sections (1 LTC1060, 2/3 LTC1061 or 1/2 LTC1064), Figure 3.
We associate resistors as belonging to 2nd order sections, so R1x belongs to the x section. Thus R12, R22, R33 and
R42 all belong to the second of two 2nd order sections in our example.
Our requirements are shown in the following table:
SECTION 1 SECTION 2
= 1.93kHz
f
o1
Q1 = 14.2 H
= 1
oBP1
Note that H choosing H
oBP1
oBP2
× H
oBP2
= 2.03.
= K and this is the reason for
For this example we choose the fo=
= 2.072kHz
f
o2
Q2 = 14.2
H
= 2.03
oBP2
f
CLK
50
R2 R4
mode,
so we will tie the 50/100/Hold pin on the SCF chip to V+, generally (5V to 7V). We choose 100kHz as our clock and calculate resistor values. Choosing the nearest 1% resistor values we can implement the filter using Figure3’s topol­ogy and the resistor values listed below.
R11 = 147k R12 = 71.5k R21 = 10k R22 = 10.7k R31 = 147k R32 = 147k R41 = 10.7k R42 = 10k
Our design is now complete. We have only to generate a TTL or CMOS compatible clock at 100kHz, which we feed to the clock pin of the switched capacitor filter, and we should be “on the air.”
STATE VARIABLE SCF ACTIVE (OP-AMP) STATE VARIABLE
R4
R3
R2 R2
HP
R1
V
IN
+
AGND
f
R2
R3
fo=
CLK
100(50)
|
R4
Q =
R2
|
H
oHP
R2
R4
S BP LP
+
1/2 LTC1060 1/3 LTC1061 1/4 LTC1064
= –R2 /R1 H
MODE 3
= –R3 /R1 H
oBP
oLP
= –R4 /R1
R1
V
IN
R5
+
1
fO=
2πRC
Figure 2. Switched Capacitor vs Active RC State Variable Topology
AN27A-2
R4
C
C
R
R
+
R6
3/4 LTC1014
R2 R4
+
AN27A F02
an28f
Application Note 27A
R41
R31
R21
R11
V
IN
+
fo=
f
CLK
50
R3
R2
; Q=
R2
R4
Figure 3. Two 2nd Order Sections Cascaded to Form 4th Order BP Filiter
HP
+
1/2 LTC1060
R2
H
oBP
R4
S1
A
= –R3 /R1
BP
LP
DESIGNING BANDPASS FILTERS—THEORY BEHIND THE DESIGN
Traditionally, bandpass filters have been designed by labori­ous calculations requiring some time to complete. At the present time programs for various personal or laboratory computers are often used. In either case, no small amount of time and/or money is involved to evaluate, and later test, a filter design.
R42
R32
R12
R22
+
HP
+
1/2 LTC1060
S1
A
BP
LP
BP OUTPUT
AN27A F03
with the required characteristics (generally the Qs are too high). We wish to explore here the use of cascaded identical 2nd order sections for building high Q bandpass filters.
For a 2nd order bandpass filter
2
1–G
Q =
G
×
1– f /f
f/f
o
( )
o
2
(1)
Many designers have inquired as to the feasibility of cas­cading 2nd order bandpass sections of relatively low Q to obtain more selective, higher Q, filters. This approach is ideally suited to the LTC family of switched capacitor filters (LTC1059, LTC1060, LTC1061 and LTC1064). The clock to center frequency ratio accuracy of a typical “Mode1” design with non “A” parts is better than 1% in a design that simply requires three resistors of 1% tolerance or better. Also, no expensive high precision film capacitors are required as in the active op amp state variable design.
We present here an approach for designing bandpass filters using the LTC1059, LTC1060, LTC1061 or the LTC1064 which many designers have “on the air” in days instead of weeks.

CASCADING IDENTICAL 2ND ORDER BANDPASS SECTIONS

When we want to detect single frequency tones and si­multaneously reject signals in close proximity, simple 2nd order bandpass filters often do the job. However, there are cases where a 2nd order section cannot be implemented
Where Q is the required filter quality factor
f is the frequency where the filter should have gain, G, expressed in Volts/V.
is at the filter center frequency. Unity gain is assumed
f
o
.
at f
o
EXAMPLE 2—DESIGN
We wish to design a 2nd order BP filter to pass 150Hz and to attenuate 60Hz by 50dB. The required Q may be calculated from Equation (1):
2
So,Q =
1– 3.162× 10
( )
3.162× 10
–3
–3
60 /150
×
1– 60/ 150
( )
= 150.7
2
This very high Q dictates a –3dB bandwidth of 1Hz.
Although the universal switched capacitor filters can realize such high Qs, their guaranteed center frequency accuracy of ±0.3%, although impressive, is not enough to pass the 150Hz signal without gain error. According to the
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AN27A-3
Application Note 27A
f
–R3
R3
previous equation, the gain at 150Hz will be 1 ±26%; the rejection, however, at 60Hz will remain at –50dB. The gain inaccuracy can be corrected by tuning resistor R4 when mode 3, Figure 2, is used. Also, if only detection of the signal is sought, the gain inaccuracy could be acceptable.
This high Q problem can be solved by cascading two identi­cal 2nd order bandpass sections. To achieve a gain, G, at frequency f the required Q of each 2nd order section is:
1–G
Q =
G
×
1– f / f
f / f
o
( )
o
2
(2)
The gain at each bandpass section is assumed unity.
In order to obtain 50dB attenuation at 60Hz, and still pass 150Hz, we will use two identical 2nd order sections.
We can calculate the required Q for each of two 2nd order sections from Equation (2):
So,Q =
1–3.162× 10
3.162× 10
–3
–3
×
60 /150
1– 60/ 150
( )
= 8.5!!
2
With two identical 2nd order sections each with a poten­tial error in center frequency, f
, of ±0.3% the gain error
o
at 150Hz is 1 ±0.26%. If lower cost (non “A” versions of LTC1060 and LTC1064) 2nd order bandpass sections are used with an fo tolerance of ±0.8%, the gain error at 150Hz is 1 ±1.8%! The benefits of lower Q sections are therefore obvious.

HARDWARE IMPLEMENTATION

Mode 1 Operation of LTC1060, LTC1061, LTC1064

As previously discussed, we associate resistors with each 2nd order section, so R1x belongs to x section. Thus R12, R22 and R23 belong to the second of the two 2nd order sections, Figure 4.
Each section has the same requirements as shown:
= fo2 = 150Hz
f
o1
Q1 = Q = 8.5 H
oBP1
= H
oBP2
= 1
Note that we could get gain out of our BP filter structure by letting the product of the H
terms be >1 (within the
oBP
performance limits of the filter itself).
For our example using the LTC1060 we will use f
/100. So we input a 15kHz clock and tie the 50/100/
= f
CLK
= fo2
o1
Hold pin to mid-supplies (ground for ±5V supplies).
We can implement this filter using the two sections of an LTC1060 filter operated in mode 1. Mode 1 is the fastest operating mode of the switched capacitor filters. It provides Lowpass, Bandpass and Notch outputs.
Each 2nd order section will perform approximately as shown in Figure 5, curve (a).
Implementation in mode 1 is simple as only three resistors are required per section. Since we are cascading identical sections, the calculations are also simple.
R31
R21
R11
V
IN
AN27A-4
+
AGND AGND
CLK
fo=
H
=
oBP
100
R1
S
N
+
1/2 LTC1060 1/2 LTC1060
Q =
R2
Figure 4. LTC1060 as BP Filter Operating in Mode 1
BP
LP
R12
R32
R22
+
S
N
+
BP
AN27A F04
LP
BP OUTPUT
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Application Note 27A
( )
We can calculate the resistor values from the indicated formulas and then choose 1% values. (Note that we let our minimum value be 20k.) The required values are:
R11 = R12 = 169k R21 = R22 = 20k R31 = R32 = 169k
Our design is complete. The performance of two 2nd order sections cascaded versus one 2nd order section is shown in Figure 5, curve (b). We must, however, generate a TTL or CMOS clock at 15kHz to operate the filter.

Mode 2 Operation of LTC1060 Family

Suppose that we have no 15kHz clock source readily avail­able. We can use what is referred to as mode 2, which allows the input clock frequency to be less than 50:1 or
0
–10
–20
–30
–40
BANDPASS GAIN (dB)
–50
–60
30 50 100 150 300 500
FREQUENCY (Hz)
Figure 5. Cascading Two 2nd Order BP Sections for Higher Q Response
(a): ONE SECTION
(b): TWO SECTIONS
1k
AN27A F05
100:1 [f
= 50 or 100]. This still depends on the con-
CLK/fo
nection of the 50/100/Hold pin.
If we wish to operate our previous filter from a television crystal at 14.318MHz we could divide this frequency by 1000 to give us a clock of 14.318kHz. We could then set up our mode 2 filter as shown in Figure 6.
We can calculate the resistor values from the formulas shown and then choose 1% values. The required values are:
R11, R12 = 162k R21, R22 = 20k R31, R32 = 162k R41, R42 = 205k

CASCADING MORE THAN TWO IDENTICAL 2ND ORDER BP SECTIONS

If more than two identical bandpass sections (2nd order) are cascaded, the required Q of each section may be shown to be:
2/n
1–G
Q =
1/n
G
where Q, G, f and f
f / f
1– f / f
( )
o
2
o
×
are as previously defined and n = the
o
(3)
number of cascaded 2nd order sections.
R41
R31
R21
R11
V
IN
+
f
R2
CLK
fo=
1+
Q =
R4
100
H
=R3/ R1
oBP
N
+
1/2 LTC1060
R3
R2
1+
R2
R4
S
BP
LP
R12
R42
R32
R22
S
N
BP
LP
+
+
1/2 LTC1060
BP OUTPUT
AN27A F06
Figure 6. LTC1060 as BP Filter Operating in Mode 2
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