Active Filters, EQs & Crossovers

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Transcript Active Filters, EQs & Crossovers

Active Filters, EQs &
Crossovers
Dennis Bohn
Rane Corporation
Bohn 6-03
Rane Corporation
It’s All About the Mathematics
Electronic filters are all about the mathematics.
You cannot escape the math.
We will study the math;
… you will love the math.
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Rane Corporation
Simplified Laplace Transforms
• Represents complex (frequency dependent)
impedance, i.e., magnitude & phase
• Uses the Laplace Operator, s, where
s = complex frequency variable = jω = j2πf
– Resistor Impedance = R (freq. independent)
– Capacitor Reactance = 1/sC
– Inductor Reactance = sL
• Allows writing a circuit’s transfer function by
summing circuit currents using Kirchoff’s Law
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Transfer Functions (TF)
• Transfer functions mathematically describe
the frequency domain behavior of filters.
• TF = ratio of Laplace Transforms of a circuit’s
input and output voltages:
Vin(s)
Filter
Vout(s)
T(s) = Vout(s) / Vin(s)
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Filter Transfer Functions
• General filter transfer function is the ratio of
two polynomials:
a 1 s  b1 s  c1
2
T(s) 
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a 2 s  b2 s  c 2
2
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TF Poles & Zeros
a 1 s  b1 s  c1
2
T(s) 
a 2 s  b2 s  c 2
2
• “Zeros” = values that make numerator equal
zero, i.e., the roots of the numerator.
– Makes amplitude response rolloff 6
dB/oct.
– Shifts phase +90°/zero (+45° @ fc)
• “Poles” = values that make denominator equal
zero, i.e., the roots of the denominator.
– Makes amplitude response rise 6 dB/oct.
– Shifts phase –90°/zero (–45° @ fc)
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Rane Corporation
Audio Filter Order
• The order or degree (equivalent terms) is the
highest power of s in the transfer function.
• For analog circuits usually equals the number of
capacitors (or inductors) in the circuit.
• 2nd-order most common.
• For common audio filters the order equals the
rolloff rate divided by 6dB/oct, e.g. 24 dB/oct
rolloff = 4th order (24 6 = 4)
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Rane Corporation
Audio Filter Order (cont.)
Rule: 6 dB/oct & 90° per order
Examples:
1st-order = 6 dB/oct; θ = 90° ( 45° @ fc)
2nd-order = 12 dB/oct; θ = 180° ( 90° @ fc)
3rd-order = 18 dB/oct; θ = 270° (135° @ fc)
4th-order = 24 dB/oct; θ = 360° (180° @ fc)
… etc.
Bohn 6-03
Rane Corporation
Why 6 dB/octave Slope?
The impedance of a capacitor is half with twice the
frequency, i.e., XC = 1/sC = 1/2fC
The impedance of an inductor is twice when
frequency doubles, i.e., XL = sL = 2fL
Twice or Half Impedance = 6 dB change
Twice or Half Frequency = One Octave change
Bohn 6-03
Rane Corporation
Why Phase Shift?
• Phase shift is the flip side of time
• It takes time to build up a charge on a
capacitor -- that’s why you cannot change the
voltage on a capacitor instantaneously.
• It takes time to build up a magnetic field (flux)
in an inductor -- that’s why you cannot
change the current through an inductor
instantaneously.
• All this time = phase shift
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Rane Corporation
Why 2nd-Order?
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Maximum phase shift is 180 degrees
Guarantees circuit is unconditionally stable
No oscillation problems under any conditions
Get higher order circuits by cascading 2nd-order
sections … or
• Design 4th-order section to mathematically
emulate two cascaded 2nd-order (Rane’s L-R)
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Rane Corporation
Normalized Transfer Function
Low-Pass (LP) =
(2 poles)
Amplitude
1
s  s 1
2
2 poles = -12 dB/oct
Frequency
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Normalized Transfer Function
• Bandpass (BP)
(1 zero, 2 poles)
=
s
s  s 1
2
1 pole = -6 dB/oct
1 pole = -6 dB/oct
Amplitude
1 zero = +6 dB/oct
Frequency
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Normalized Transfer Function
High-Pass (HP)
s
=
(2 zeros, 2 poles)
2
s  s 1
2
2 poles = -12 dB/oct
Amplitude
2 zeros = +12 dB/oct
Frequency
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Rane Corporation
Coefficients Determine Performance
LP =
K
s  As  B
2
K
=
s 
2
ωo
Q
sω
2
o
• Butterworth: maximally flat passband
s2 + 1.414s + 1
• Chebyshev: steeper rolloff w/magnitude ripples
s2 + 1.43s + 1.51
• Bessel: best step response, but gentle rolloff
s2 + 3s + 3
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Response Comparison
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Q Effects
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Butterworth Q = 0.707
Bessel Q = 0.5
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Group Delay Comparison
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Step Responses
Butterworth
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Bessel
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Active or Passive?
• There exists no sound quality attributable to
active or passive circuits per se.
• TF determines the overshoot, ringing and
phase shift regardless of implementation.
• A transfer function is a transfer function is a
transfer function … no matter how it is
implemented -- all produce the same
fundamental results as long as the circuit
stays linear: same magnitude response,
same phase response, same time response;
however there are secondary differences.
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Rane Corporation
Active vs. Passive
Active
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Gain & adjustable
No loading effects
Parameters adjustable
Smaller Cs
No inductors
Smaller, lighter & cheaper
No magnetic coupling
High Q circuits easy
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Passive
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Less noise
No power supply
More reliable
Less EMI susceptible
Better at RF frequency
No oscillations
No on/off transients
No hard clipping
Handles large V & I
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Creating An Equalizer
In
Out
Input Signal
1
BP
BP Filter
fc
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Rane Corporation
Boost = Original + Bandpass
Boost (Lift)
+
In
BP
1 + BP
Out
1
fc
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Rane Corporation
Cut = Reciprocal
In
+
Out
Cut (Dip)
BP
1
1
1+BP
fc
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Rane Corporation
Why 1/3-Octave Centers?
• 1/3-Octave (21/3 oct = x1.26) approximately
represents the smallest region humans reliably
detect change.
• Relates to Critical Bands: a range of frequencies
where interaction occurs; an auditory filter.
• About 1/3-octave wide above 500Hz (latest info
says more like ~1/6-oct); 100 Hz below 500 Hz
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Rane Corporation
Creating A Crossover:
Use LP & HP To Split Signal
HP1
Input
HP2
LP2
LP1
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High Out
Mid Out
Low Out
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1st-Order & Butterworth Crossovers
1st-order
plus 2nd
through 4thorder
Butterworth
vector
diagrams
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Linkwitz-Riley Crossover
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Two Cascaded Butterworth Filters
Outputs Down 6 dB at Crossover Frequency
Both Outputs Always in Phase
No Peaking or Lobing Error at Crossover
Frequency
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Rane Corporation
Creating A LR Crossover
Cascaded Butterworth
Input
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BW-HP
BW-HP
High Out
BW-LP
BW-LP
Low Out
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Linkwitz-Riley Crossovers
LR-4
LR-2
LR-8
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Rane Corporation
Ray Miller (Rane)
Bessel Crossover
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Rane Corporation
Successfully Crossing-Over
• Must know the exact amplitude and phase
characteristics of the loudspeakers.
• Driver response strongly interacts with active
crossover response.
• True response = loudspeaker + crossover
• DSP multiprocessors à la Drag Net allow
custom tailoring the total response.
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Rane Corporation
Accelerated-Slope Tone Controls
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Rane Corporation
Stop Kidding Yourself
(Rick Chinn Request)
Why low-cut and high-cut filters are a must for
sound system bandwidth control;
or,
Why cutting the end sliders on your EQ doesn’t
do diddly-squat.
Bohn 6-03
Rane Corporation
Analog vs. Digital Filters
Analog
• Speed 10-100x faster
• Dynamic Range
– Amplitude: 140 dB
e.g., 12 Vrms & 1 V
noise
– Frequency: 8 decades
e.g., 0.01 Hz to 1 MHz
• Cheap, small, low power
• Precision limited by noise &
component tolerances
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Digital
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Very complex filters
Full adjustability
Precision vs. cost
Arbitrary magnitude
Total linear phase
EMI & magnetic noise
immunity
• Stability (temp & time)
• Repeatability
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Digital Filters and DSP
Allow circuit designers to do new things.
We can go back and solve old problems ...
like the truth-in-slider-position bugaboo of
graphic equalizers:
– Proportional-Q was good
– Constant-Q was better
– Perfect-Q is best
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Rane Corporation
Truth in Slider Position
Proportional-Q
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Truth in Slider Position
Constant-Q
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Rane Corporation
Truth in Slider Position
Perfect-Q
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PERFECT-Q™ & DEQ 60
Rick Jeffs
Sr. Design Engineer
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Rane Corporation
DEQ 60 Graphic 1/3-Oct EQ
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Rane Corporation
DEQ 60 Features
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Rane Corporation
DEQ 60 Performance
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