1、DESIGN OF a 5th ORDER BUTTERWORTH LPFDESIGN OF a 5th ORDER BUTTERWORTH LOW-PASS FILTER USING SALLEN & KEY CIRCUITBackground Theory:Filters are classified according to the functions that they are to perform, in terms of ranges of frequencies. We will be dealing with the low-pass filter, which has the
2、 property that low-frequency excitation signal components down to and including direct current, are transmitted, while high-frequency components, up to and including infinite ones are blocked. The range of low frequencies, which are passed, is called the pass band or the bandwidth of the filter. It
3、extends from =0 to = c rad/sec (fc in Hz). The highest frequency to be transmitted is c, which is also called the cutoff frequency. Frequencies above cutoff are prevented from passing through the filter and they constitute the filter stopband.The ideal response of a low-pass filter is shown above. H
4、owever, a physical circuit cannot realize this response. The actual response will be in general as shown below.It can be seen that a small error is allowable in the pass band, while the transition from the pass band to the stopband is not abrupt.The sharpness of the transition from stop band to pass
5、 band can be controlled to some degree during the design of a low-pass filter.The ideal low-pass filter response can be approximated by a rational function approximation scheme such as the Butterworth response.The Butterworth ResponseNormalizing H0=1 and Then finding the roots of D(s) Example:For n=
6、5All the poles are: -1.0000 -0.8090 + 0.5878i -0.8090 - 0.5878i -0.3090 + 0.9511i -0.3090 - 0.9511i 0.3090 + 0.9511i 0.3090 - 0.9511i 1.0000 0.8090 + 0.5878i 0.8090-0.5878POLE LOCATIONSThe poles are distributed over the circle of radius 1 (). Never a pole in the imaginary axis.Finding H(s) from H(s)
7、 H(-s):H(s) is assigned all RHS poles H(-s) is assigned all LHS polesFollowing this procedure, the Butterworth LPF H(s) (H0=1, wc=1rad/sec) can be found for various filters of order n.We can use MATLAB to get this denominator polynomial (Butterworth polynomial)In MATLAB (code):all_poles=roots(-1)n,
8、zeros(1,2*n-1),1)poles=all_poles(find(real(all_poles)0)Den=poly(poles)Circuit Design: We want to design of a fifth order Butterworth low-pass filter with a cutoff frequency of 10KHz. During the design we make use of magnitude and frequency scaling and also of the uniform choice of as a characterizin
9、g frequency will appear in all design steps, except for the last where the de-normalized (actual) values will be found.Circuit Implementation:Implementation of the circuit is done using the Sallen & Key Topology.this is of the general formIf k=1,taking To realize a 5th order BLPF one Sallen & Key st
10、age with a single op-amp is required for every complex-conjugate pole pair. Since n=5 (odd), an additional negative pole is required and we use an RC/voltage follower. Also we made the choice of K=1, which requires that the inverting op-amp circuit be replaced by a voltage follower as shown below.To
11、 find actual values:Make all resistors=Frequency scaling=Multiplying each capacitor by Performance Measures:Cutoff frequency=10KHzFrequency (KHz)Vin (mV)Vout (mV)1.502500498.52.009500493.55.782500493.79.0015004879.6500387.510.04500245.711.01500225.712.04500187.513.02500115.614.0250090.6215.0650075.0
12、016.0150062.5017.0150047.50Ideal response:Actual response: From the recorded values after measurements.Measured dB gain values vs. log frequency valuesCircuit Diagram:Final CircuitParts:PartQuantityLM324NLow Power Quad Operational Amplifier11 K resistor50.016 F capacitor10.019 F capacitor10.013 F ca
13、pacitor10.051 F capacitor10.0049 F capacitor19 V battery2References:Deliyannis T., Yichuang Sun, and J. K. Fidler 1999, Continuous-time Active Filter Design, CRC Press, New York.Van Valkenburg M. E., Analog Filter Design, 1982, Oxford University Press, New York.Chen Wai-Kai, Passive & Active Filter Design, Chapter 2, pp. 50-92Huelsman L. P. and P. E. Allen, Introduction to the Theory and Design of Active Filters
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