New Proposed Algorithms for nth Order

New Proposed Algorithms for nth Order Butterworth Active Filter Computer-Aided Design Haider Fakher Radhi Al-Saidy

New Proposed Algorithms for nth Order Butterworth Active Filter Computer-Aided Design Haider Fakher Radhi Al-Saidy

Information and Communication Department, Al-Khwarizmi Engineering College Baghdad University,Jaderia,Baghdad, Iraq. Email: [email protected] doi: 10.4156/ijact.vol2.issue3.7

Abstract This paper describes new proposed algorithms for constructing the transfer function of nth order Butterworth LPF using the idea of the cascade combination of active filters. Computer-aided design is used with new algorithms to achieve the procedure of constructing the transfer function and give the practical results of the nth order Butterworth active LPF. C++ program is used for designing procedure by linear programming. The proposed algorithms are very fast, flexible, and exact. The program allows users to design different frequency, components, and order of Butterworth active LPF with high flexibility.

Keywords: Programming algorithms, Active filter design, Frequency-dependent component. 1. Introduction This paper introduces new proposed algorithms for nth order Butterworth LPF using the idea of the cascade combination of active filters. This work is to simplify active filter design by writing a program which can be used even without any background about filter design and to get high flexibility of components selection for getting the most economical design. Analog electronic filters are present in just about every piece of electronic equipment. There are the obvious types of equipment, such as radios, televisions, and stereo systems. Test equipment such as spectrum analyzers and signal generators also need filters [1]. Detection of a wanted signal may be impossible if unwanted signals and noise are not removed sufficiently by filtering. Electronic filters allow some signals to pass, but stop others. To be more precise, filters allow some signal frequencies applied at their input terminals to pass through to their output terminals with little or no reduction in signal level [2]. Winder [1] has written simple programs for filter design like “Super FILTER” and “Filter Master”. He has written “Active_F” as an active filter design program [1]. Chen [2] has provided a lot of information in filters with significant additions in the areas of computer-aided design of active filters. Yue Wu [3] has introduced the design of active filters for RF receivers. He has also introduced the design of active bandpass filters which are based on CMOS active inductors [4].

2. Active Filters A filter is a device which passes signals of certain frequencies and rejects or attenuates those of other frequencies. Passive filters are constructed with inductors, capacitors, and resistors, but for certain frequency ranges inductors, because of their size and practical performance limitations, are undesirable. Consequently there has been, for some years, a trend toward replacing inductors by active devices which simulate the effect of inductors [2,5]. This trend has accelerated with the advances in IC integration which have made the active devices available at prices competitive with, and in many cases cheaper than, those of inductors. The active device we use to construct active filters is the operational amplifier [6].

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International journal of Advancements in Computing Technology Volume 2, Number 3, August, 2010

3. General Circuits and Equations A low-pass filter is a device which passes signals of low frequencies and attenuates those of high frequencies. There are many numbers of ways of designing LPFs using active devices. In Fig.(1), the op-amp, together with resistors R3 and R4 constitutes a voltage –controlled voltage source (VCVS) and the overall network is second order LPF which has the following transfer function [7]:

V2 ( s ) K = 2 V1 ( s ) s + as + b

(1)

Figure 1. Second order active LPF In Fig.(2), the network is first order LPF which has the following transfer function[8]:

V2 ( s ) 2B = V1 ( s ) s + B

(2)

(where R3=R4)

Figure 2. First order active LPF An analysis of Fig.(1) shows that it achieves Eq.(1):

K=

2 R1 R2C1C2

(3)

1 R1C 2

(4)

1 R1 R2 C1C 2

(5)

(where R3=R4)

a= (where R3=R4)

b=

An analysis of Fig.(2) shows that it achieves Eq.(2) with:

B=

1 RC

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(6)

New Proposed Algorithms for nth Order Butterworth Active Filter Computer-Aided Design Haider Fakher Radhi Al-Saidy Higher order filters may be obtained by cascading two or more second-order filters. To implement odd nth order, we can add a first order stage.

4. Low-Pass Butterworth Filters A filter which approximates the ideal low-pass filter with a relatively flat pass band characteristic is Butterworth filter. Its amplitude response [5] is:

H ( jw) =

k 1 + ( w / wc ) 2 n



(7)

Where n is the order of the filter. It may be seen that the filter is improved as n increases. The Butterworth filter has the advantage of maximally flat response in the pass band. The poles of the nth order Butterworth filter can be calculated [9] by:

p k = − sin

(2k − 1)π (2k − 1)π + j cos  2n 2n

(8)

For k=1, 2,…,n The poles of nth order are complex conjugates only if the n is odd there is an additional real pole equal to s= -1. The complex conjugate poles can be calculated using the equation [2]:

p k ⋅ p k* = s 2 + (2 sin

(2k − 1)π ) s + 1 2n

(9)

For k=1, 2,…,|n/2|

5. Design Principles The nth order Butterworth active filter can be implemented using multistage. The transfer function of each is multiplied by the other. The representation of the stages by second order for each and the final stage is first order if the order n is an odd number. For example if the designed filter is 7th order then the stages will be:

Figure 3. Stages of 7th order LPF The calculation of the transfer function of each stage depends on the calculated conjugate poles from Eq. (9).

6. Realization of the Prototype Filter The designed filter for the calculated poles is prototype filter with cut-off frequency 1rad./sec. , therefore the realization steps must be performed to get the real cut- off frequency and real filter components. The impedance scaling can be made without changing the transfer function or the cut-off frequency value by multiplying by the same factor each of resistors and capacitive reactances [10] . Therefore the transformation is by multiplying each resistance by a factor (k) and dividing the capacitance by the same factor (k). In the other hand to get frequency transformation from 1 rad./sec. to required cut-off frequency (f) , we must divide each capacitance by ( ( 2π ⋅ f ) [2,11].

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7. Frequency Transformation In this paper we have only discussed the design of low pass (LP) filters. There are many other types of filters which find use in signal processing. Once the low-pass prototype filter has been designed, the other filters (High-pass, band-pass, and band-stop) can be obtained by a simple frequency transformation [6,9].

8. Design Algorithms Flowchart (1) shows the algorithms of nth order active filter which gives flexibility to choose the resistors or capacitors as well as other parameters of filter. The algorithms are optimized to give highest possible accuracy and flexibility for choosing the order and components.

9. Design Example Results Example results of the program (in C++ Language) which follows the algorithms of Flowchart (1) are as follows: • For third order with freq.=100kHz and with selected capacitance C=1nF • The first stage: R1=1591.5 Ω , R2= 1591.5 Ω • The second stage : R=1591.5 Ω • Fig.( 4) shows above third order filter

Figure 4. Third order LPF • • • • •

For sixth order with freq.=100kHz and with selected capacitance C=1nF The first stage: R1=3074.6 Ω , R2=8238 Ω The second stage: R1= 1125 Ω , R2= 2251 Ω The third stage: R1= 823.8 Ω ,R2=3074.6 Ω Fig.(5) shows above sixth order filter

Figure 5. Sixth order LPF • • • • • • •

For tenth order with freq.=100kHz and with selected capacitance C=1nF The first stage:R1=5087 Ω , R2= 497.9 Ω The second stage: R1=1753 Ω , R2=1445 Ω The third stage: R1=1125 Ω , R2= 2250 Ω The fourth stage: R1=893.1 Ω , R2= 2836 Ω The fifth stage: R1= 8057 Ω , R2 =3143.9 Ω Fig.(6) shows above tenth order filter

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New Proposed Algorithms for nth Order Butterworth Active Filter Computer-Aided Design Haider Fakher Radhi Al-Saidy

Figure 6. Tenth order Start Input n , f

Input C

Is the choice for resistance ?

Input R1

.k=1

.k=1

.kkk =(2k-1)*pi/2n .a= 2sin(kkk)

.kkk =(2k-1)*pi/2n .a= 2sin(kkk)

R1=1/(2.pi.f.a.C) R2=1/(4.pi.pi.f.f.R1.C.C)

C=1/(2.pi.f.a.R1) R2=1/(4.pi.pi.f.f.R1.C.C)

Print the results of the kth stage.

Print the results of the kth stage. .k=k+1

Is k