Second Order Filter Functions.doc

Second Order Filter FunctionsThe standard form of the transfer function for second order filters isBecause the denominator is a second-degree polynomial, all such filters have two poles. Since the poles are the zeros of the denominator polynomial, we can find them by applying the quadratic formula. The location, and nature, of the poles depends only on the parameters and :If , then the two poles of are given byNote that when that the radicand is positive so that the radical is real. Also note that the radical is smaller than so that both poles lie on the negative real axis of the complex s-plane.If , then the two poles of are coincident and are given byIf , then the radicand above is negative so that the radical is pure imaginary. Thus the two poles of are given byororFor , therefore, the poles form a complex conjugate pair and lie in the left half of the complex s-plane.Notice that as , both poles approach the axis. (Poles on the axis correspond to constant amplitude oscillations of the kind we will encounter when we consider oscillators.)Although the denominator determines the poles of the filter transfer function, the type of filter is determined by the numerator polynomial. We consider several cases.1. Low Pass Filter: n1 = n2 = 0If we let to obtain we see that for frequencies thatFor frequencies ,Thus, the transfer function with n1 = n2 = 0 is a low pass filter with a break frequency of . Because of the factor in the denominator, the gain in dB, , falls off at an asymptotic rate of 40 dB/decade at low frequencies.Any circuit that exhibits a transfer function of the formcan act as a low pass filter. The equal component Sallen-Key circuithas the transfer functionThus, this Sallen-Key circuit acts as a low pass filter withOne of the advantages of an equal component Sallen-Key circuit for realizing a second order low pass filter is that and can be adjusted independently by adjusting and , respectively. Well appreciate this advantage more, later.2. High Pass Filter: n0 = n1 = 0If we let to obtain we see that for frequencies that, a constantFor frequencies, Thus, the transfer function with n0 = n1 = 0 is a high pass filter with a break frequency of . Because of the factor in the numerator, the gain in dB, , increases at an asymptotic rate of 40 dB/decade at frequencies below .Any circuit that exhibits a transfer function of the formcan act as a high pass filter. The equal component Sallen-Key circuithas the transfer functionThus, this Sallen-Key circuit acts as a high pass filter withOne of the advantages of an equal component Sallen-Key circuit for realizing a second order high pass filter is that and can be adjusted independently by adjusting and , respectively. Well appreciate this advantage more, later.3. Band Pass Filter: n0 = n2 = 0It is straightforward to show that exhibits a maximum at and, for sufficiently large , the passband of the filter is quite narrow. An equal component Sallen-Key band pass circuit does not work well in practice. Later, well consider a band pass filter circuit that contains more than one operational amplifier, a requirement necessary to achieve narrow (say a few per cent of the center frequency) passbands in practice.4. Notch Filter: n1 = 0where is the notch frequency, a frequency corresponding to much reduced output. Note that has zerosand so that the response, ideally, is zero at the notch frequency and in a narrow band of frequencies about it. Far away from , is a constant.5. All Pass Filter:It is easy to see that . The all pass filter therefore affects only the phase of the input, not its amplitude.To view the poles and zeros of the various types of filters, as well as the corresponding frequency responses, double click on the following:Note especially the frequency response for a low pass filter. Regardless of your choice of the coefficients, the frequency response, differs considerably from the ideal:In the ideal low pass filter, note that the frequency response, , is constant up to the critical frequency, , at which the response drops to zero. (The angle of , the phase shift, should be zero, ideally.) Regardless of how clever your choice of and , the ideal frequency response is impossible to realize in practice. Indeed, the impulse response (the inverse Fourier transform of ) that corresponds to the ideal frequency response is easily shown to be non-causal. That is, the impulse response that corresponds to the ideal frequency response for a low pass filter requires the filter to produce an output at times before the impulse is applied.To demonstrate this problem, we take a brief detour into Fourier transform theory. Recall that a time function, , can be expressed in terms of its Fourier transform, , as a superposition of sinusoids (represented as ) as follows:where , typically a complex number, gives the amplitude and phase of the component at each frequency, . Indeed, is essentially the phasor representation of the sinusoid with frequency, . The integral sums up all of the frequency components. The constant is merely cosmetic. What may seem strange is that the integral sums over negative, as well as positive, frequencies, . Recall, however, that if we write an ordinary sinusoid in terms of complex exponential functions, that we write, from Eulers identity,so that a sinusoid represented in terms of complex exponential functions requires both positive and negative frequencies. The fact that must exist for negative as well as positive frequencies makes it slightly different from the usual phasor representations of sinusoids. Nevertheless, the Fourier transform representation and the phasor representation clearly are closely related.Recall, also, the second part of the Fourier transform pair,which permits the Fourier transform, , to be calculated for a given time function, . From this equation, we see that the complex conjugate of , which we denote as , isIn specific cases in which is real, this relation becomesThat is, if is real, it is necessarily an even function of . Note, also, specifically, that the Fourier transform of a unit impulse, , applied at time is simplyThe fact that the Fourier transform of the impulse is a constant indicates that the impulse contains equal components of all frequencies (with zero phase), from the highest to the lowest. If an impulse were applied to the input of an ideal low pass filter, therefore, the output would consist of constant components of frequencies up to and would have no components at higher frequencies. Because the ideal low pass filter introduces no phase shift and the frequency components of the impulse have zero phase angle, the spectrum, , of the output voltage, , is real and hence must be an even function. Thus, is a real constant for frequencies between and and zero outside this range. Thus, the impulse response of the ideal low pass filter is:Notice that the output of an ideal low pass filter in response to an impulse applied to its input at time begins (is non-zero) for negative times, before the impulse is even applied. In effect, an ideal low pass filter therefore would need to predict the future, ordinarily impossible, to respond as required.The closest we can come to realizing the ideal low pass filter in practice is when we do not require the output from the low pass filter to appear immediately, but are willing to tolerate a long delay before the output is available. For example, suppose we receive an analog data stream from a deep space probe and record the entire stream before we begin the filtering operation. In such a case, we can calculate the Fourier transform of the entire data stream and simply chop off all frequencies greater than the critical frequency, . In effect, the filter in this case can know what happens in the future in that it can use later values of the signal in calculating its output at a given time the values at all times having been recorded beforehand.In the usual case in which we require immediate output from the low pass filter, however, the ideal frequency response is simply impossible to realize. The ideal frequency response for high pass and band pass filters is similarly impossible to realize. The question then naturally arises as to what is the best approximation to the ideal frequency responses that we can achieve in practice. The answer varies depending on precisely what we mean by “best.” It turns out that the response of all second order low pass filters at frequencies sufficiently below the critical frequency, , falls off at 40 dB per decade. Subject to that unavoidable constraint, Butterworth filters give the flattest response at frequencies below , Bessel filters exhibit the least overshoot in response to a step input, and Chebyshev filters give the fastest fall-off for frequencies above for a specified “ripple” in the frequency response for frequencies in the pass band, that is, for frequencies less than .To realize a transfer function,for any one of the low pass filters with critical frequency, , choose and according to the following table:Filter TypeButterworth1.000Chebyshev (1 dB ripple)0.8631.045Chebyshev (2 dB ripple)0.8520.895Chebyshev (3 dB ripple)0.8410.767Bessel1.2741.732The following Excel spreadsheet gives comparative plots of these second order low pass filters:Here is one of the graphs from the spreadsheet:Notice that the Chebyshev filter with 3 dB ripple in the pass band offers the largest attenuation above , as well as the largest hump in response near . The Bessel filter achieves the least attenuation above . Notice that the response of all of these filters falls off at the same rate 40 dB per decade at frequencies higher that a few . This slope is determined, of course, by the degree of the denominator 20 dB for each power of s in the denominator.A happy note is that the “best” choices for and turn out to be the identical for both high pass and low pass filters.A steeper fall off outside the pass band can be achieved by increasing the degree of the denominator in the transfer function, . A simple way of effectively increasing the degree of the denominator is to cascade two second order filters. Cascading two filters effectively multiplies the transfer functions, thus multiplying their denominators and thereby increasing the order. Although it may not be obvious, cascading two second order Butterworth filters does not give a fourth order Butterworth filter (flattest passband). To achieve a fourth order Butterworth (or other) filter by cascading two second order filters, and for each filter should be chosen differently, according to the following table:First SectionSecond SectionType of FilterButterworth1.0001.8481.0000.765Chebyshev (1 dB)0.5021.2750.9430.281Chebyshev (2 dB)0.4661.0880.9460.224Chebyshev (3 dB)0.4430.9290.9500.179Bessel1.4361.9161.6101.241State Variable Second Order Filter SystemConsider the following circuit (originally used for solving second order differential equations during the heyday of analog computers) which turns out to be a surprisingly versatile filter circuit:For simplicity, we consider the analysis of this circuit in sections. First consider the summing operational amplifier, the one on the left:First, we write node equations at the two inputs of the operational amplifier and write the relationship between the amplifier output and its inputs:From equations (1) and (2), we findIf we multiply equation (1) by and equation (2) by 3, we find:If we subtract equation (1) from equation (2), we find:The combination of equation (3) with this result gives:If (ideal operational amplifier), we see thatorIn terms of Laplace transforms, we have, neglecting initial conditions,The remaining two operational amplifiers form two integrators of the type we considered earlier:Recall our earlier result for this