GCT535- Sound Technology for Multimedia Filters
Graduate School of Culture Technology KAIST Juhan Nam
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GCT535- Sound Technology for Multimedia Filters Graduate School of - - PowerPoint PPT Presentation
GCT535- Sound Technology for Multimedia Filters Graduate School of Culture Technology KAIST Juhan Nam 1 Filters Control the frequency response of input signals Lowpass, Highpass, Bandpass, Bandreject (notch), Allpass, Equalizers
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f/Hz |H(f )| LP fc f/Hz |H(f )| HP fc f/Hz |H(f )| BP fcl fch f/Hz |H(f )| BR fcl fch f/Hz |H(f )| All pass
[DAFx book]
+12 +6 −6 −12 f fc fc fc fb fb fc
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Synthesizer (MiniMoog) Audio Mixer Wah-Wah Pedal
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Channel Vocoder
2 4 6 8 f/kHz → −100 −80 −60 −40 −20
LPC modeling of Formant
x(n) BP 1 x2
BP1(n)
x2
BP2(n)
x2
BPk(n)
( )2 ( )2 ( )2 LP xRMS1(n) xRMS2(n) xRMSk(n) BP 2 LP BP k LP
5 Oval window Round window
Cochlear Filterbank
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−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Lowpass Filters freqeuncy(log10) Gain(dB) 10
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−30 −20 −10 10 20 30 Q =0.5 Q =1 Q =2 Q =4 Lowpass Filters freqeuncy(log10) Gain(dB)
Cut-off Frequency Q or Resonance Roll-Off (e.g. -12dB/oct) Passband Stopband
– Find filter coefficients that satisfy “filter specification” – Low-order filters
– High-order filters
– Fitting filters to measured frequency response
– Digitizing analog filters which is already well-developed – Biquad filters: “prototype” analog filters (tone control) – High-order IIR filters: Butterworth filter, Chebyshev filter, Elliptical filter
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filter specification We handle these two approaches, which are mainly used in tone control
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Cut-off frequency = 1kHz Cut-off frequency = 1kHz
Lowpass Filter Highpass Filter
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Z-plane Lowpass Filter Highpass Filter
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4 =
4
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Bass Shelving Filter
Crossover Freq.= 100Hz Gain = 12dB 9dB Gain = -12dB Crossover Freq. = 1kHz 9dB
Gain = 12dB Gain = -12dB
Treble Shelving Filter
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= 10(ABCD
4, ) − 1
(Gain ≥ 0dB) (Gain < 0dB)
(Gain ≥ 0dB) (Gain < 0dB) @ = 10(0ABCD
4, ) − 1
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Crossover Freq.= 1kHz Crossover Freq.= 1kHz
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4 =
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4 =
4
4 =
4
4 =
4
4 =
4
4 =
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Bass Treble 𝑐, = 𝐻 tan 𝜕, 2 + 𝐻
2 − 𝐻
2 + 𝐻
2 − 𝐻
𝐻
2 + 𝐻 𝑐. = 𝐻
2 − 𝐻 𝑏, = 𝐻
2 + 1 𝑏. = 𝐻
2 − 1
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The cascade of bass and treble shelving filters renders flat response.
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X ,
X [\,
X [\, Laplace Transform Z-Transform
X , Fourier Transform Discrete-Time Fourier Transform
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(JOS Filter book)
c 𝑡 − 𝐷𝑤 0 = 𝑊 b(𝑡)
c(𝑡)
a(𝑡) =
c(𝑡)
b 𝑡 + 𝑊 c(𝑡) =
(Assuming the initial voltage is zero) 𝑗 𝑢 = 𝑤b(𝑢) 𝑆
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(JOS Filter book)
l 𝑡 = 𝑡𝑀𝐽l(𝑡)
k 𝑡 = 𝐽l(𝑡)
b 𝑡 = 𝑆𝐽b(𝑡)
c(𝑡)
a(𝑡) =
c(𝑡)
b 𝑡 + 𝑊 c(𝑡) =
c(𝑡)
l 𝑡
k 𝑡
c(𝑡)
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v = 1 = 0𝑒𝐶
v 67wv = v 6vwv = . 6.w. = 4
v 67 = v 7
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s-plane 𝑡 = 𝑘𝜕
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s-plane 𝑡 = 𝑘𝜕
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67 7K
67K 7
7K €
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s-plane 𝑡 = 𝑘𝜕
−𝜕,( 1 2𝑅 ± 𝑘 1 − 1 2𝑅
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poles Pole locations for varying Q
7K 7 )4
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s-plane 𝑡 = 𝑘𝜕
−𝜕,( 1 2𝑅 ± 𝑘 1 − 1 2𝑅
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poles Pole locations for varying Q
Peak is not exactly on 𝜕,
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Resonant Lowpass Resonant Highpass Bandpass
Notch
Allpass
Peaking-EQ
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𝑦(𝑜 − 1) 𝑦(𝑜) 𝑦 𝑢 → R 𝑒𝑢
𝑧 𝑢 = R 𝑦 𝑢 𝑒𝑢
𝑡 𝑧 𝑜 = 𝑧 𝑜 − 1 + 𝑦 𝑜 + 𝑦(𝑜 − 1) 2 / 𝑈 Laplace Transform z-Transform 𝑍(𝑨) 𝑌(𝑨) = 𝑈 2 / 1 + 𝑨0. 1 − 𝑨0. 𝑍 𝑨 1 − 𝑨0. = 1 + 𝑨0. 2 𝑌(𝑨) / 𝑈 𝑍(𝑡) 𝑌(𝑡) = 1 𝑡 𝑍(𝑡) 𝑌(𝑡) ↔ 𝑍(𝑨) 𝑌(𝑨)
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π π (Ω0,ω0) Ω ω
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10
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−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Lowpass Filters freqeuncy(log10) Gain(dB) 10
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−30 −20 −10 10 20 30 Q =0.5 Q =1 Q =2 Q =4 Lowpass Filters freqeuncy(log10) Gain(dB)
𝐼 𝑨 = (1 − cos 𝜄 2 ) 1 +2𝑨0. +𝑨04 (1 + 𝛽) −2cos 𝜄𝑨0. +(1 − 𝛽)𝑨04 𝛽 = sin 𝜄 2𝑅 𝜄 = 2𝜌 𝑔
c
𝑔
U
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10
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10
3
10
4
−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Highpass Filters freqeuncy(log10) Gain(dB) 10
2
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−30 −20 −10 10 20 30 Q =0.5 Q =1 Q =2 Q =4 Highpass Filters freqeuncy(log10) Gain(dB)
𝛽 = sin 𝜄 2𝑅 𝜄 = 2𝜌 𝑔
c
𝑔
U
𝐼 𝑨 = (1 + cos 𝜄 2 ) 1 −2𝑨0. +𝑨04 (1 + 𝛽) −2cos 𝜄𝑨0. +(1 − 𝛽)𝑨04
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10
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−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Bandpass Filters freqeuncy(log10) Gain(dB) 10
2
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−30 −20 −10 10 20 30 Q =0.5 Q =1 Q =2 Q =4 Bandpass Filters freqeuncy(log10) Gain(dB)
𝐼 𝑨 = (sin 𝜄 2𝑅 ) 1 −𝑨04 (1 + 𝛽) −2cos 𝜄𝑨0. +(1 − 𝛽)𝑨04 𝛽 = sin 𝜄 2𝑅 𝜄 = 2𝜌 𝑔
c
𝑔
U
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10
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−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Notch Filters freqeuncy(log10) Gain(dB) 10
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−30 −20 −10 10 20 30 Q =0.5 Q =1 Q =2 Q =4 Notch Filters freqeuncy(log10) Gain(dB)
𝐼 𝑨 = 1 −2cos 𝜄𝑨0. +𝑨04 (1 + 𝛽) −2cos 𝜄𝑨0. +(1 − 𝛽)𝑨04 𝛽 = sin 𝜄 2𝑅 𝜄 = 2𝜌 𝑔
c
𝑔
U
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Q=1 𝐼 𝑨 = (1 + 𝛽 / 𝐵) −2cos 𝜄𝑨0. +(1 − 𝛽 / 𝐵)𝑨04 (1 + 𝛽/𝐵) −2cos 𝜄𝑨0. +(1 − 𝛽/𝐵)𝑨04 𝛽 = sin 𝜄 2𝑅 𝜄 = 2𝜌 𝑔
c
𝑔
U
𝐵 = 10(ABCD(Ž•)
)
Q=4
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