Filter Notes. You may have memorized a formula for the voltage divider - if not, it is easily derived using Ohm's law, Vo Vi

Transcription

1 Filter Notes You may have memorized a formula for the voltage divider - if not, it is easily derived using Ohm's law, Vo Vi R2 R+ R2 If you recall the formula for capacitive reactance, the divider formula works fine when R2 is replaced by a capacitor. j Vo 2π fc Vi + j2π frc R j 2π fc This is the formula for the transfer function of an elementary low pass filter. Consider two frequency ranges:. Very low frequencies ( f 0) Vo Vi + 0 Flat response - no change in amplitude and no phase shift. 2. Very high frequencies (denominator dominated by second term) Vo Vi j2π frc Amplitude dropping like /f - phase shift of -90 We define a frequency midway between these two extreme ranges, f f C. The cutoff frequency is that frequency at which Vo Vi + j Vo * Vi - phase shift is -45 So 2πf C RC, and f C 2π RC

2 Filter Prep. Sketch what you expect for "Signal + Noise" for both cases. Time domain - Amplitude say from -.2 volts to +.2 volts and time from 0-00 milliseconds. Frequency domain - Amplitude arbitrary and frequency from 0Hz to 0KHz (logarithmic). 2. Characterize the noises - can you associate one case with additive jitter (static) and the other with drift?. Which is which? 3. Recall filters from ac circuits - draw circuit diagrams for high pass and low pass RC filters. Choose resistors and capacitors to realize ) low pass with fc 500Hz, and 2) high pass with fc 20Hz. What is the order of these filters? If these analog filters were used on our two cases, how much improvement in signal/noise ratio should we expect? What order digital filter shall we use in LabView? Solution: The elemental RC filters have similar forms The formula for cutoff frequency is the same, f C /(2πRC). Choose a plentiful C, say C 0. μf. So, R /(2πf C C) and I get: Low pass - fc 500Hz - R 3,85 Ohms (Use a 3.2K), and High pass - fc 20Hz - R 79,68 Ohms (Use an 80K). These are both first order filters - Response in the stop band is rolling off at -20 db/decade or -6 db/octave. For both filters the cutoff frequency is one octave away from the noise - therefore the noise would be smashed by a factor of 2. Expect the 0. volt noise amplitudes to be reduced to 0.05 volts - a 6 db improvement in S/N ratio in both cases. In LabView let's specify sixth order filters (no problem if we allow plenty of taps - filter length at least 30). We would expect noise attenuation of 20 db/ decade or 36 db/octave. With our one octave separation we would expect to smash the noise by a factor of 64. The 0. volt noise amplitudes should be reduced to about volts - a 36 db improvement in S/N ratio in both cases

3 Design of Butterworth Filters An example of pole placement using op amps

4 What in the World are Poles? Circuit designers use pole location in the complex frequency plane as an indication of expected circuit performance. Poles are the (perhaps complex) roots of the denominator of the circuit s transfer function, V o /E i. The Butterworth filter response and the Tchebyscheff filter response are among the most popular, physically realizable approximations to the ideal low pass filter. The Butterworth response is defined in terms of the location of its poles in the complex plane. We will concentrate on the Butterworth response, but learn to pronounce and spell Tchebyscheff. It s even more impressive than Kirchhoff.

5 Approximations to an ideal low pass filter

6 Circuit s Transfer Function squared, Vo / Ei 2

7 Poles of F 2 in the Complex Frequency Plane

8 Let us show that the low pass filters in Chapter are truly Butterworth. The authors of your text give us the transfer function (equation -b) for the -20 db/decade filter shown as figure -2a. A F(j ω) CL + jω RC In order to be Butterworth, the poles of F(jω) F(-jω) must divide the circle into equal arcs while avoiding the imaginary axis (2 o clock). Let s just look at the poles of F(jω) in the left half plane. The -20 db/decade filter clearly has one real pole at jω - /(RC). The circle is divided equally (two 80 arcs) while avoiding the imaginary axis. The filter is Butterworth, and its cutoff frequency is at ω c /(RC). This might be called a first order or a one pole Butterworth..

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10 The authors of your text don t give us the transfer function for the -40 db/decade filter shown as figure -4a. It is fairly easy to show that: A F(j ω) CL j 2ωRC + (j ω) R CC By the quadratic formula, we see that the roots of the denominator are at: 2 roots at ± RC 2 C C RC 2 2 Our -40 db/decade filter recipe calls for C 2 2 C ; so our two complex poles are at (-+j) / (RC 2 ) and (--j) / (RC 2 ). In polar form, the poles are at.44 / (RC 2 ) 35 and.44 / (RC 2 ) 225. The circle is divided equally (four 90 arcs). The filter is Butterworth, and its cutoff frequency is at ω c.44 / (RC 2 ). or ω c.707/ (RC ) This might be called a second order or a two pole Butterworth.

11 The -60 db/decade filter introduces the idea of cascading stages in order to introduce more poles. The -20 db/decade stage places a negative real pole at -/RC 3. The -40 db/decade stage again places a complex pair of poles according to: roots at ± RC 2 C C RC 2 2 This time a different recipe is required: C /2 C 3 and C 2 2 C 3. Now the roots are at (-+j 3) /(RC 2 ) and (--j 3) /(RC 2 ). In polar form (and in terms of C 3 ) the poles are at /(RC 3 ) 20 and /(RC 3 ) 240 The circle is divided equally (six 60 arcs). The filter is Butterworth, and its cutoff frequency is at ω c /(RC 3 ) This might be called a third order or a three pole Butterworth.

12 A Butterworth Summary The Butterworth response is popular because it is so transparent to data in the passband. The amplitude response is as flat as possible, and the phase response is very gentle as well. The Butterworth response is realized by placing poles - at equal spacing - on a circle in the complex frequency plane. The filters in Chapter are indeed Butterworth. We could generalize the procedure, and design Butterworth filters of higher order. Just place poles where they need to be. We could generalize the procedure, and design most any circuit that was defined in term of pole location.