FILTERS: ACTIVE & PASSIVE. Introduction

Transcription

1 FILTERS: ACTIVE & PASSIVE Introduction Filters pervade electronic design, as there is always a need to shape the frequency response of signals propagating through the system of interest. To achieve the correct shaping, one considers the Fourier transform of the filter, and designs it so that the magnitude of this transform has the desired shape. In analog design, there are standard filters one encounters, namely the low-pass, high-pass, bandpass, notch and (not as often) all-pass filters. To be sure, there are numerous other filters, but the others such as FIR and IIR filters occur in digital or mixed-signal systems that employ digital signal processing. For this lab, only the analog filters are considered. A distinction can be made between active and passive filters. While passive filters are nice in that they may allow signals of any amplitude to be processed linearly, active filters (which restrict the allowable input signal, due to power supply limitation) afford the advantage of allowing one to cascade filters, without each filter loading the preceding filter, and hence altering the response of that filter. Such isolation is the major advantage of active filters vs. passive filters.

2 Theory Passive Filters In your first circuits course, you were introduced to the first-order low-pass filter, described by the following transfer function: H () 0 H() 0 H() 0 H() 0 s jω πf + ( πf p ) H πf j tan p ω e j ω pω pω pω () H(0) is the DC gain, s is the Laplace variable, ω is the radial frequency, p ω is the radial pole frequency, f is the frequency in Hertz and p f is the pole frequency in Hertz. The complex Laplace variable has been replaced by the purely imaginary Fourier variable jω since the focus of this lab is on steady-state behavior, not the transients involved in approaching steady-state operation. The resultant Fourier description is complex, with an attendant magnitude and phase. For this first-order filter, the magnitude and phase are sketched in Fig. : H(jω) (db) Phase of H(jω) (degrees) H(0) 0-0 db/decade f/p f f/p f 0. 0 Fig. (a) First-order response magnitude (linear-log axes), (b) phase (linear-log axes). To implement this filter with passive components, one requires one reactive component (i.e., an inductor or capacitor) and one resistor. The standard capacitor verson is shown in Fig. for reference, and the inductor equivalent is left as a prelab exercise.

3 V in R V out C Fig. First-order low-pass filter using a capacitor. As was pointed out in the introduction to this lab, a major drawback of passive filters is that simply cascading them alters the response one desires; this is because the input and output impedances of each filter are not infinite and zero, respectively, thus the filters are not ideal voltage-to-voltage devices. As an example, consider a cascade of two passive filters, one designed with a pole at khz, the other at 5 khz, as shown in Fig. 3: R kω R kω V in V out πr C khz C 59nF C 3nF πr C 5kHz Fig. 3 Cascade of two first-order filters. Solving the appropriate Kirchhoff equations leads to the following result: H H + + π ( R C + R C + R C ) s + ( R R C C ) s + ( 807Hz) π ( 6.7kHz) s s () As this simple example demonstrates, loading is not negligible in passive filters: the first pole is low by 9.3%, the second high by 3.4%. Active filters alleviate this problem by providing the requisite interstage buffering so that each stage may be designed independent of the others, and later hitched together without

4 performance degradation. Because active filters are able to supply dynamic power, they make it possible to implement low-pass and high-pass filters with gains greater than 0-dB. Active Filters Active filters employ active elements (e.g., op-amps, transistors, etc.) in addition to passive elements. As an elementary example, recall the finite DC-gain integrator circuit from lab #3 if you did the prelab correctly, you should have determined that this circuit is actually a low-pass filter! Furthermore, since the op-amp has very low-output impedance (further diminshed by the shunt-shunt feedback employed, but let s not confuse the issue with more advanced topics!), appending a second integrator does not alter appreciably the locations of either of the poles. A particularly useful active filter that can implement both low-pass and high-pass responses is the Sallen- Key filter. The beauty of this filter is that it is scalable, i.e., if you know the parameter values that give you one cutoff frequency, you simply scale all component values by the same factor to give you any other cutoff you desire. To determine the scaling factor, it pays to review the transfer function of a second-order filter: as s ω 0 + bs + c H (3) s + + Q ω 0 A low-pass filter requires ab0, a high-pass filter bc0, a bandpass filter ac0 and a notch-filter b0. Only the denominator determines the form of the time-domain response; the numerator merely alters the coefficients of each term in the time-domain response equation. Since the denominator is so critical, let s examine it more carefully. Solving for the poles leads to: p, ω 0 ± (4) Q ( ) Q The important point here is that for Q > 0.5, the poles are complex. Whereas purely real poles cause the filter response to roll off monotonically with frequency, complex poles with Q > cause the magnitude to rise in the vicinity of the imaginary part before ultimately rolling off at 40 db/decade (i.e., - 0dB/decade per pole). Decreasing Q causes the imaginary portion to dominate, and in the limit of infinite Q, the poles become purely imaginary (i.e., they lie on the imaginary axis), leading to an impulse at radial frequency ω 0. Now, back to the Sallen-Key filter if the component values can be expressed in terms of Q, it should be clear that one may achieve either real or complex poles by suitable adjustment.

5 C v in R R v out C Fig. 4 Sallen-Key low-pass filter. Analysis of the Sallen-Key low-pass filter of Fig. leads to the following design equations: C C RC ( Q) Q πf 0 (5) For a given Q, the ratio of C and C is fixed, independent of the cutoff frequency. For instance, if you are designing for a maximally flat response (i.e., no peaking in the frequency domain), Q would be 0.707, mandating C C. Furthermore, for a given f 0, the product RC is fixed as well. The beauty is that there are only two restrictions for the three component values, leaving one of the component values arbitrary. This flexibility is particularly handy when building the filter, for it allows one the freedom to vary one component value until the other two assume realistic values. For Q (a common choice for a lowpass filter, since it maximizes bandwidth under the constraint of no peaking), a frequency of Hz, and choosing R to be kω (simply because this is the most common value in any lab), C C 0.707/000π 5µF. With these component known for Hz, it is easy to design for any other cutoff frequency just divide the capacitors by the new frequency, and presto, you re done! Suppose you desire a cutoff frequency of 0 khz: just divide 5µF by 0,000 to get C C.5nF. If the computed capacitor values turn out to be unrealistic, simply scale the capacitor values by an appropriate factor, then divide the resistor by the same factor. Pretty simple! Note that this is only one version of the Sallen-Key filter, suitable for low-pass responses. In the prelab, you will consider a few variations which you will build in the lab to achieve various filter responses.

6 Butterworth & Chebychev filters In practice, one often needs low-pass filters that roll off beyond the 3-dB frequency substantially faster than the 0 db/decade achievable with a st -order filter, or even the 40 db/decade achievable with a nd -order filter. A Butterworth filter is a filter that is maximally flat (i.e., maximum bandwidth without any peaking) and can achieve arbitrarily high roll off rates beyond the cutoff frequency, the requirement being a highorder denomiator in the transfer function. For a filter of order n, the rolloff rate is 0n db/decade. A Chebychev filter is similar, except that ripple (i.e., non-monotonic characteristic) is permitted in the passband and stopband, with the advantage that this lack of smoothness leads to an initial roll off beyond the cutoff frequency steeper than that of a Butterworth filter of the same order and cutoff frequency. The point is that for a desired attentuation of x-db beyond some critical frequency, one must decide whether minimizing electrical component count is most important (Chebychev), or preservation of signal shape in the passband is most important (Butterworth). It turns out that a Butterworth filter is obtained by placing the poles of the transfer function evenly about the negative half of the circle of radius f c in the complex plane. One key result is that for an odd filter order (i.e., odd number of poles), one pole is real and lies on the real axis, with the other complex poles symmetric about the real axis; if the filter order is even, all poles are complex and are symmetric about the real axis. A Chebychev filter places the poles evenly about an ellipse in the complex plane. A sketch of the pole locations for a 5 th -order filter realized with both types of filters is shown below in Fig. 5: jω jω σ σ Fig. 5 Pole locations for (a) 5 th -order Butterworth filter, (b) 5 th -order Chebychev filter. The polynomials for an arbitrary order Butterworth filter are easy to determine, and are left as a prelab exercise. However, since ellipses are not so simple as circles, the following is a table of Chebychev denominator polynomials (normalized to the cutoff frequency) through third order:

7 Cheb Cheb Cheb 3 s (6) Cheb Plots of the magnitude and phase of the Butterwoth and Chebychev transfer functions is left as a prelab exercise.

8 Reference Reading ) Roland E. Thomas & Albert J. Rosa. The Analysis and Design of Linear Circuits, chapter 5. Prentice-Hall Inc., Englewood Cliffs, New Jersey, 994. ) L. P. Huelsman, Active and Passive Analog Filter Design: an Introduction. New York: McGraw- Hill, 993.

9 Prelab Exercises ) Design a passive first-order low-pass filter using an inductor, assuming your input and output variables are both voltages. Choose element values for a 0 khz pole frequency. Simulate the magnitude and phase responses in SPICE to confirm your design. What are the gain (in db) and phase (in degrees) at the pole frequency? To develop good habits, both frequency axes should be logarithmic, the magnitude output should be expressed in decibels on a linear scale, and the phase output should be expressed in degrees on a linear scale. ) This question deals with the idea of cascading active filters. Recall the simple op-amp integrator circuit used in earlier labs. If you complicate this circuit by placing a shunt resistor across the integrating capacitor, what is the new transfer function? What is the DC gain, 3-dB frequency, and approximate unity-gain frequency (assume f u >> f 3-dB )? Assuming that the circuit is supposed to act like an integrator at frequencies well beyond the 3-dB frequency, what is the purpose of the additional resistor? Design one of these st -order filters for a DC gain of 9.5-dB with a pole at khz, and a second filter with a DC gain of 6-dB and a pole at 5 khz. Cascade these two filters, and work out (symbolically) the transfer function of the cascade, assuming ideal op-amps. Does the filter have the desired DC gain of 5.5-dB and poles at khz and 5 khz? Why or why not? Verify your design with SPICE. What is the phase at the second pole? What is the slope of the magnitude plot (in db/decade) at high frequency (i.e., well beyond the second pole)? 3) Consider equation (3). Prove that Q is the boundary between peaked and unpeaked responses (in the frequency domain, not time domain). Do this by showing that with Q < 0.707, the response decreases monotonically, whereas with Q > 0.707, the response peaks at other than DC. (Realize that the numerator is of no consequence here while it is true that complex zeros may cause a dip in the frequency response, we re only concerned with what the poles do.) For this value of Q, the filter is called maximally flat. Design a maximally flat Sallen-Key filter with a 3- db frequency of 50 khz. Using a 74 op-amp for a gain stage, is it possible to achieve a DC gain of 0 as well? Is there a limit on the gain, given a corner frequency of 50 khz? (Remember the unity-gain frequency you determined for the 74 in a previous lab.) What is the input impedance of this filter in terms of R, C and C? What is the DC input impedance? What is the input impedance at the corner frequency? 4) Consider the following Sallen-Key filter. Determine the transfer function. What kind of filter is this (pay attention to the numerator)? Determine expressions similar to those obtained in equation (5). Again, determine the magic Q value for no (pole) peaking. Design for a cutoff frequency of 0 khz. Verify your design in SPICE by plotting magnitude and phase responses. R v in C C v out R

10 5) Consider the following circuit. Determine the transfer function. What kind of filter is this (pay attention to the numerator)? Determine expressions similar to those obtained in equation (5). Design for a center frequency of 0 khz and a Q of 5. Keeping everything else fixed, how does the peak gain vary with Q? For a given power supply level and peak gain, what restriction must you place on the input signal? Design a simple (active) attenuator that will enable your filter to handle an input signal that is twice the maximum input level when no attenuator is present. What are the penalties for using an active attenuator? Verify your design results in SPICE by plotting magnitude and phase responses. C R v in R C v out 6) Determine the denominator polynomials (normalized to the cutoff frequency) for Butterworth polynomials up to 6 th -order. Also, plot the magnitude, phase and delay of the first four Chebychev transfer functions (i.e., the reciprocals of the denominator polynomials) and the first six Butterworth polynomials. To get the most meaningful results, put all Chebychev magnitudes on one graph, all Chebychev phases on one graph, and all Chebychev delays on one graph; do the same for the Butterworth plots. (Delay is defined as the negative derivative of phase with respect to radial frequency.) Additionally, to draw a comparison, plot the 3 rd -order Butterworth and Chebychev magnitudes on one graph together, and plot both 4 th -order magnitudes together on a separate graph. Verify that the Chebychev magnitude is also less than the Butterworth beyond the cutoff frequency. Which type of filter provides more uniform delay in the passband? Why? 7) Design a Butterworth filter to achieve 0 db DC gain, a 3-dB cutoff of 0 khz and at least 50 db attenuation at 90 khz. First determine the minimum-order necessary to meet the 50-dB attenuation specification. To do this, examine the magnitude of each Butterworth transfer function at ω 9 (remember, we re working with normalized frequency). You should find that the order required is higher than, so it is prudent to cascade low-order filters, each no higher than nd -order. Recall that cascades are best done with active filters. To implement a second-order with complex poles, you may find one of the Sallen-Key filters useful. Verify your design with SPICE before building in the lab!

11 Lab Exercises ) Build each filter that you designed in the prelab, and verify all relevant parameters specified in each prelab exercise. Use care with the bandpass filter (see the prelab), and use your judgement as to whether the oscilloscope or spectrum analyzer is most appropriate for each measurement.