Introduction to
Audio and Music Engineering
Lecture 16
Topics:
•  Tone and frequency spectra
•  Filtering and frequency content of signals
•  RC Low-pass and high-pass filters
•  RLC Band pass filters
•  Filter banks and Graphic Equalizers
3
Tone?
•  The quality or character of sound
•  The characteristic quality or timbre of a
particular instrument or voice.
•  The character or quality of a musical
sound or voice as distinct from its pitch
and intensity.
Frequency à pitch
Sound pressure level à intensity (loudness)
Attack, Spectrum, Spectral evolution à tone (timbre)
4
Tone is related to the spectrum of a waveform.
Amplitude
The spectrum is a display of the frequency content of a waveform.
Fundamental
(f0)
Overtones
Formant
(envelope)
Frequency
If the overtones are at integer multiples of the fundamental they
are called harmonics, e.g., 2f0 à 2nd harmonic, etc.
Most musical sounds have harmonic (or nearly so) overtones.
5
Back to circuits …
Applying a filter to alter tone …
Amplitude
Low pass
Frequency
Examples …
6
1st order RC low-pass and high-pass filters
VC R V ~
V C
V
C C ~
0.8 0.707 @ ω = 1 / RC
0.6 0.4 0.2 1/2
VC ⎡
⎤
1
=⎢
⎥
V ⎣ 1 + ω 2R 2C 2 ⎦
V Low Pass
1 0 0 1 R 0.8 VR
V
ωRC
3 4 5 High Pass
1 VR 2 0.707$@$ω = 1 / RC
0.6 0.4 0.2 1/2
VR ⎡ ω 2R 2C 2 ⎤
=⎢
⎥
V ⎣ 1 + ω 2R 2C 2 ⎦
0 0 1 2 ωRC
3 4 5 7
Matlab Simulation
High Pass
Begin with harmonic overtone series.
Equal amplitude sine waves
at integer multiples of f0
Low Pass
8
Band Pass Filters
Voltage divider: R & L//C
vout
vin
R
v out = v in
Vin
Z1 C
Vout
L
Z2 1.20 1.00 v out
v in
0.80 0.60 v out = v in
0.40 0.20 0.00 0 1 2 ω ω0
3 4 5 Z 1 + Z2
1
jωL
jωC
L//C =
=
1
1 − ω 2LC
jωL +
jωC
jωL ⋅
v out
Q = 1/√2 = 0.707
Z2
jωL
2
= v in 1 − ω LC
jωL
R+
1 − ω 2LC
ω
Q
ω0
2
⎡⎛
⎤
2⎞
2
ω
ω
⎢ 1−
+ 2 Q2 ⎥
⎜
⎟
2
⎢⎝
ω0 ⎠
ω0 ⎥
⎢⎣
⎥⎦
1/2
1
(LC )1/2
ω 0L
ω0 =
Q =
R
9
Graphic Equalizer
Bank of bandpass filters …
Q = 1/√2 = 0.707 à Butterworth Bandpass (2nd order)
4-band Graphic EQ
3.0#
Q = 0.707
2.5#
Response
Fletcher Munson – hearing response
combined response
2.0#
1.5#
500 1000
1.0#
2000
4000
0.5#
0.0#
0#
1#
2#
3#
f(kHz)
4#
5#
It’s OK if the filter response decreases a bit (≈ -3 dB) at high
frequencies because human hearing sensitivity increases (≈ +20 dB).
10
Octave filters (Constant “Q”)
Response
Center frequencies increase by
octaves (x2):
$fn+1 = 2fn
Fractional bandwidth is fixed:
$∆fn/fn = Q-1
1.2 500Hz 1kHz
1.0 2 kHz
0.6 0.4 0.2 0.0 1 2 Middle C
C4
4 kHz
0.8 0 C3
Q = 0.707
C5
f(kHz)
3 C6
4 5 C7
130.8 Hz 261.6 Hz 523.2 Hz 1046.4 Hz 2092.8 Hz 1.00 0.707
∆fn
0.50 0.00 0.1
0.2 0.2
f(kHz)
0.5
2 1
Logarithmic frequency axis
2
11
Filter Performance …
Flat frequency response
3.0#
Q = 0.707
1.4#
1.2#
2.0#
1.5#
500 1000
1.0#
2000
Response
Response
1.6#
Q = 0.707
2.5#
Boost 2 kHz
4000
1.0#
0.8#
0.6#
0.4#
0.5#
0.2#
0.0#
0.0#
0#
1#
2#
3#
f(kHz)
4#
0#
5#
1#
2#
3#
4#
5#
f(kHz)
Tradeoff between ripple and frequency selectivity …
Too much ripple
16.0#
14.0#
1.2#
Q=2
1.0#
10.0#
500 1000
8.0#
2000
Response
12.0#
Response
Q = 0.4
4000
6.0#
4.0#
0.8#
Poor frequency selectivity
0.6#
500 1000
0.4#
2000
4000
0.2#
2.0#
0.0#
0.0#
0#
1#
2#
f(kHz)
3#
4#
5#
0#
1#
2#
f(kHz)
3#
4#
5#
12
10-Band Graphic EQ Demonstration
using Matlab/Simulink
fc = [31.5, 63, 125, 250, 500, 1000, 2000, 4000, 8000, 16000] Hz