Design Article
Desert Island Design: Bridging the (filter) gap without software
Dave Ritter, Principal Applications Engineer, Intersil Corporation
11/4/2008 5:19 PM EST
The Setup
Can anyone design anything without touching a keyboard and invoking a software tool? This time we're taking on filters (see Reference below for the previous article on designing bandgap references). Scan the web and you'll find hundreds of links to filter design shareware, java scripts and tutorials. But what if you didn't even have a reference book, how far could you go?
The Premise
Today we have a dilemma at sea. We desperately need to find a sunken marker for a cable that needs repair. If we don't find the marker, we'll have to bring out a crew of divers at huge expense to search manually. Our primary search tool is a sonar set that displays on a VGA video monitor. It works, but the new radar set (which the authorities made us install) causes the video screen to go crazy with a haze of noise and interference.
Our experience tells us that a high frequency signal from the radar is getting into our video. It turns out that the radar sends out pulses at 160 MHz rate, so that must be the culprit. We need a lowpass filter that will pass the 70 MHz video but reduce the 160 MHz interference by, say, 100:1.
Thinking It
What can you recall about lowpass filters? How many poles do we need? The attenuation of a 6 db/octave filter (one pole) is proportional to frequency, while a 12 db/octave filter (two poles) is proportional to frequency squared, 18 db/octave is frequency cubed, and so on. So the attenuation rate of a filter is proportional to frequency raised to the power of the number of poles.
The ratio of 160 MHz (our noise problem) to 70 MHz (our video bandwidth) is 2.3. What power of 2.3 is greater than 100? With a little pencil and paper we find out that 6 poles is more than enough: a six-pole, 70-MHz filter will reduce a 160-MHz signal by more than 100 times.
What kind of filter do we need? It needs to be flat as possible in the passband (below 70-MHz) and rolloff quickly above that.
Does the name 'Butterworth' ring a bell? Butterworth filters are maximally flat and their poles are very conveniently placed on a circle on the s-plane.
The mere mention of the s-plane causes many young engineers to convulse. Relax, the s-plane can be your friend. The x-axis is the real axis and the y-axis is the imaginary axis. Poles directly on the x-axis do not 'ring' in the time domain. Poles below the 45° diagonal have no peaking on a Bode plot (but can have slight overshoot and ringing in the time domain). Poles directly on the y-axis signify an oscillator (infinite peaking on a Bode plot, infinite ringing in the time domain). All poles that have a y-component always appear in pairs. The closer the pairs are to the x-axis, the less peaking they cause in a frequency plot, and the less overshoot and ringing in the time domain.
Deep Theory
Figure 1 shows the pole locations for a 6th-order Butterworth filter. They are placed evenly around a half circle in the left half plane, which means that adjacent poles are separated by 30°. The first pole pair is 15° from the real axis, the second pole pair is at 45°, and the third pole pair is at 75°.
Figure 1: Our low-pass filter in the s-domain
(Click on image to enlarge)
Now think hard: we need a transfer function with pole locations that translate into R's, L's and C's. The math can get obscure, but there's an easy way. Every even-ordered linear system can be factored into a series of second order systems. Basically, we can break this 6-pole system into 3 separate systems, each with one pair of poles.
A two-pole system has a characteristic frequency and a Q. The characteristic frequencies are all the same because the poles lie on a circle. The formula is:
The Q's are trickier. Fortunately we have placed a cheat sheet in the top of our tool box with important equations. And there it is:
What does Q mean? The term comes from the early radio days when the quality (thus "Q") of a bandpass filter was related to its bandwidth. In modern theory, the Q determines the degree of resonance in a second order system: the ratio of reactance to resistance, or the ratio of stored energy to dissipated energy. In a signal processing system Q shows the degree of peaking in a 2nd order low-pass filter stage.
Figure 2 shows the relative peaking of the three stages in our Butterworth filter.
Figure 2: Individual stages of a 6th-order Butterworth filter
(Click on image to enlarge)
The low Q stage (blue pair) has a gradual rolloff, about 6 db at 70 MHz. The medium Q stage (green pair) is sharper and is exactly 3 dB down at 70 MHz. The high Q stage (red pair) peaks up 6 dB, just enough to compensate for the low Q stage. The composite filter is 3 dB down at 70 MHz but it is sharper and flatter than any of the individual stages, and it rolls off 3 times faster.
Working the Numbers
A little diagram on our cheat sheet shows that the θ in the Q formula is the angle from the real axis (x-axis) to the pole. So:
Q = ½ cos 15°,
Q = ½ cos 45°, and
Q = ½ cos 75°.
Everybody knows that cos 45° = 0.707 which makes:
so that's one down and two to go.
Our cheat sheet tells us that sin θ ≈ θ when θ is small, and since we're on a desert island we'll assume that 15° is small enough. Unfortunately θ has to be in radians. A full circle is:
which makes 15°
= (15/360) × 2π
= 2π/24 ≈ 0.25 = ¼.
We also remember that:
so that
We need to figure out the √(15/16). 15/16 is close enough to 1 for our purposes, so we conclude that:
That makes Q = 0.5.
Whew, on to Q! We need the cos (75°). Again our cheat sheet says that:
We already decided that 15° was small enough that:
and Q ≈ 2.
We have all the Q's and the ω0, now on to the hardware.
Building It
We can build a 2nd order filter section from an L, a C and an R (Figure 3). Fortunately there are a few 0.5 μH inductors in the parts bin, so: L = 0.5 μH.
Figure 3: Desert island 70-MHz Butterworth filter
(Click on image to enlarge)
ω0 is adjusted by choosing the right cap, and Q is adjusted by choosing the right resistor. We keep the three stages from interacting by putting buffers between them, using our transistor pack. Our cheat sheet reminds us that:
What about the Q's and the R's? There's a little confusion because our cheat sheet says that:
Which is it? It depends on where the R is placed, and whether the circuit resonates in series or parallel. The easiest way to figure out the right formula is to let the R get really big. When R is large, our circuit becomes more resonant (less damped, more peaking) so the Q is high. We use the 2nd version.
Now we calculate
Then we calculate:
R = Q /4.5×x10-3 = 0.5/4.5×103 = 111 Ω
R = Q /4.5×10-3 = 0.707/4.5×10-3 = 157 Ω
R = Q /4.5×10-3 = 2/4.5×10-3 = 444 Ω.
We can make these from 5% resistor values:
111 ≈ 100 + 10
157 ≈ 100 + 56
444 ≈ 390 + 56
(we might put a 500 Ω or 1 kΩ potentiometer here).
That's the whole circuit!
Testing It
If we really were out at sea we would just try our filter and hope it cleaned up our video problem. But later, back in the lab we could run some curves to see how well we did. Figure 4 shows the response of an ideal 6th-order Butterworth and our approximation. The error is about 0.5 dB right near the cutoff, so we did pretty well.
Figure 4: Response plots of ideal and desert-island filters: 1× and 10× detail
(Click on image to enlarge)
The high Q section is usually the most trouble, so we might just replace the 444 Ω with a potentiometer and adjust for optimum. Cranking the 444 Ω up to a little over 500 Ω increases the Q to make up for some losses due to the emitter resistance and inductor losses.
Back to the Real World
We did very well on our desert island design, and back in the lab we tweaked our filter to look like the textbook plots. Filters are important signal processing blocks. Real designs often need good filters to remove noise and out of band artifacts in a signal just like our sonar set.
A more typical case is the reconstruction of video waveforms from a digital stream. Most video today is created in the digital realm and converted to analog for the CV or YPrPb inputs of your TV set. The output of the final DAC in you set top box or DVD player has clock noise and spectral artifacts from the digital conversion. Parts such as the Intersil ISL59151 are designed to remove this noise from both HD and standard-resolution video signals; it just happens to be a 6-pole, 70-MHz Butterworth filter.
Reference
Dave Ritter, "Desert Island Design: Bridging the (band) gap without software," Planet Analog, click here.
About the author
Dave Ritter grew up outside of Philadelphia in a house that was constantly being embellished with various antennas and random wiring. By the age of 12, his parents refused to enter the basement anymore, for fear of lethal electric shock. He attended Drexel University back when programming required intimate knowledge of keypunch machines. His checkered career wandered through NASA where he developed video-effects machines and real-time disk drives. Finally seeing the light, he entered the semiconductor industry in the early 90's.
Dave has about 20 patents, some of which are actually useful. He has found a home at Intersil Corporation as a principal applications engineer. Eternally youthful and bright of spirit, Dave feels privileged to commit his ideas to paper for the entertainment and education of his soon-to-be-massive readership.
