GCT535- Sound Technology for Multimedia Filters Graduate School of - - PowerPoint PPT Presentation

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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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SLIDE 1

GCT535- Sound Technology for Multimedia Filters

Graduate School of Culture Technology KAIST Juhan Nam

1

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SLIDE 2

Filters

§ Control the frequency response of input signals

– Lowpass, Highpass, Bandpass, Bandreject (notch), Allpass, Equalizers – Shape spectrum of input signals

2

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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SLIDE 3

Filters in Audio Systems

§ Tone control

– Synthesizers

  • VCF in Analog synth

– Guitar Effect

  • EQ, Wah Wah

– DJ mixer, audio Mixers

  • Filter, EQ

– Audio Players

  • EQ

3

Synthesizer (MiniMoog) Audio Mixer Wah-Wah Pedal

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SLIDE 4

Filters in Audio Systems

§ Communication

– Speech Coding

  • Vocoder: analysis/synthesis of voice
  • Formant modeling: linear prediction coding

– Audio Coding

  • Filterbank in MP3

– Band-limiting

  • Control bandwidth: 8K, 16K
  • Anti-aliasing filters

4

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

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SLIDE 5

Filters in Audio Systems

§ Audio analysis

– Short-time Fourier Transform (as a filterbank) – Constant-Q transform – Mel-scaled filterbank – Cochlear filterbank in human ears

5 Oval window Round window

Cochlear Filterbank

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SLIDE 6

Description of Filter Characteristic

§ Bode Plot

– Log scale in both magnitude and frequency axes

6

10

2

10

3

10

4

−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Lowpass Filters freqeuncy(log10) Gain(dB) 10

2

10

3

10

4

−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

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SLIDE 7

Designing Digital Filters

§ Approach 1

– Find filter coefficients that satisfy “filter specification” – Low-order filters

  • Gain at DC and Nyquist, cut-off frequency, Q

– High-order filters

  • Passband/Stopband margin and flatness, width of transition band
  • Window-based method
  • Parks-McClellan: Iterative search
  • Linear programming or convex optimization
  • mainly for band-limiting or filterbank

– Fitting filters to measured frequency response

  • Linear prediction coding (LPC)

§ Approach 2

– 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

7

filter specification We handle these two approaches, which are mainly used in tone control

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SLIDE 8

One-pole One-zero Filters

§ Three degrees of freedom

– Fixed: gain at DC (0 Hz) and Nyquist frequency

  • Lowpass: 𝐼 1 = 1, 𝐼 −1 = 0
  • Highpass: 𝐼 1 = 0, 𝐼 −1 = 1

– User parameter: cut-off frequency

8

𝐼 𝑨 = 𝐶(𝑨) 𝐵(𝑨) = 𝑐, + 𝑐. / 𝑨0. 1 + 𝑏. / 𝑨0.

  • 3dB

Cut-off frequency = 1kHz Cut-off frequency = 1kHz

  • 3dB

Lowpass Filter Highpass Filter

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SLIDE 9

One-pole One-zero Filters

§ Transfer Function

9

𝐼 𝑨 = 1 − 𝑏 2 / 1 + 𝑨0. 1 − 𝑏 / 𝑨0. 𝐼 𝑨 = 1 + 𝑏 2 / 1 − 𝑨0. 1 − 𝑏 / 𝑨0.

Z-plane Lowpass Filter Highpass Filter

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SLIDE 10

One-pole One-zero Filters

§ Associating the cut-off frequency with the coefficient

– Frequency when the gain is -3dB

10

𝐼 𝜕

4 =

1 − 𝑏 2 / 1 + 𝑓067 1 − 𝑏 / 𝑓067

4

= 1 2 𝑏 = 1 − tan (𝜕 2) 1 + tan (𝜕 2)

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SLIDE 11

One-pole One-zero Shelving Filter

§ Equalizers that control bass or treble gain

– User parameter: cut-off frequency and band gain à crossover frequency and gain – Fixed: gain at DC or Nyquist

  • Lowpass: 𝐼 −1 = 0, Highpass: 𝐼 1 = 1

11

Bass Shelving Filter

Crossover Freq.= 100Hz Gain = 12dB 9dB Gain = -12dB Crossover Freq. = 1kHz 9dB

  • 9dB
  • 9dB

Gain = 12dB Gain = -12dB

Treble Shelving Filter

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SLIDE 12

One-pole One-zero Shelving Filter

§ Transfer Function

– Bass Shelving Filter – Treble Shelving Filter

12

𝐼 𝑨 = 1 + 𝑕 / 𝑀𝑄 𝑨 𝐼 𝑨 = 1 1 + 𝑕@ / 𝑀𝑄 𝑨

𝑕 = 10(ABCD

4, ) − 1

(Gain ≥ 0dB) (Gain < 0dB)

𝐼 𝑨 = 1 + 𝑕 / 𝐼𝑄 𝑨 𝐼 𝑨 = 1 1 + 𝑕@ / 𝐼𝑄 𝑨

(Gain ≥ 0dB) (Gain < 0dB) 𝑕@ = 10(0ABCD

4, ) − 1

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SLIDE 13

Alternative: One-pole One-zero Shelving Filter

§ Set the gain at crossover frequency to half the maximum gain

– Bass: 𝐼 1 = 𝑕, 𝐼 𝑓67K = 𝑕

  • , 𝐼 −1 = 1

– Treble: 𝐼 1 = 1, 𝐼 𝑓67K = 𝑕

  • , 𝐼 −1 = 𝑕

13

Crossover Freq.= 1kHz Crossover Freq.= 1kHz

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SLIDE 14

Alternative: One-pole One-zero Shelving Filter

§ Associating the crossover frequency with the coefficients

14

𝐼 𝑓67K

4 =

𝑐, + 𝑐. / 𝑓067K 𝑏, − 𝑏. / 𝑓067K

4

= 𝑕 𝐼 1

4 =

𝑐, + 𝑐. 𝑏, − 𝑏.

4

= 𝑕4 𝐼 −1

4 =

𝑐, − 𝑐. 𝑏, + 𝑏.

4

= 1 𝐼 𝑓67K

4 =

𝑐, + 𝑐. / 𝑓067K 𝑏, − 𝑏. / 𝑓067K

4

= 𝑕 𝐼 1

4 =

𝑐, + 𝑐. 𝑏, − 𝑏.

4

= 1 𝐼 −1

4 =

𝑐, − 𝑐. 𝑏, + 𝑏.

4

= 𝑕4

Bass Treble 𝑐, = 𝐻 tan 𝜕, 2 + 𝐻

  • 𝑐. = 𝐻 tan 𝜕,

2 − 𝐻

  • 𝑏, = tan 𝜕,

2 + 𝐻

  • 𝑏. = tan 𝜕,

2 − 𝐻

  • 𝑐, =

𝐻

  • tan 𝜕,

2 + 𝐻 𝑐. = 𝐻

  • tan 𝜕,

2 − 𝐻 𝑏, = 𝐻

  • tan 𝜕,

2 + 1 𝑏. = 𝐻

  • tan 𝜕,

2 − 1

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SLIDE 15

Alternative: One-pole One-zero Shelving Filter

15

The cascade of bass and treble shelving filters renders flat response.

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SLIDE 16

Biquad (two-pole two-zero) Filters

§ Five degrees of freedom

– Fixed: gain at DC (0 Hz) and Nyquist frequency – User parameters

  • Cut-off frequency
  • Resonance
  • Bandwidth (sharpness of peak)

§ Mapping user parameters to filter coefficients

– Reson filter can be used but, actually, it is quite complicated – The art of analog filter design is highly advanced and so we take advantage of it

16

𝐼NO 𝑨 = 𝐶(𝑨) 𝐵(𝑨) = 𝑐, + 𝑐. / 𝑨0. + 𝑐4 / 𝑨04 1 + 𝑏. / 𝑨0. +𝑏4/ 𝑨04

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SLIDE 17

Laplace Transform

§ Laplace Transform: continuous-time version of z-transform § Fourier Transform: continuous-time version of discrete-time Fourier Transform

17

𝑌 𝑡 = R 𝑦(𝑢)𝑓0UV𝑒𝑢

X ,

𝑌 𝑨 = Y 𝑦 𝑜 / 𝑨0[

X [\,

𝐼 𝑓67 = Y ℎ 𝑜 / 𝑓067[

X [\, Laplace Transform Z-Transform

𝑌 𝑔 = R 𝑦(𝑢)𝑓067V𝑒𝑢

X , Fourier Transform Discrete-Time Fourier Transform

𝑡 = 𝑘𝜕 𝑨 = 𝑓67

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SLIDE 18

Example: Laplace Transform

§ RC lowpass filter

– Kirchhoff’s law: – For capacitor: – For resistor: – By taking Laplace transform: – Transfer function:

18

𝑤a 𝑢 = 𝑤b 𝑢 + 𝑤c(𝑢)

(JOS Filter book)

𝑒𝑦 𝑒𝑢 ⇔ 𝑡𝑌 𝑡 − 𝑦(0) 𝑒𝑅(𝑢) 𝑒𝑢 = 𝑗 𝑢 = 𝐷 𝑒𝑤c(𝑢) 𝑒𝑢 𝐽 𝑡 = 𝐷𝑡𝑊

c 𝑡 − 𝐷𝑤 0 = 𝑊 b(𝑡)

𝑆 𝐼 𝑡 = 𝑊

c(𝑡)

𝑊

a(𝑡) =

𝑊

c(𝑡)

𝑊

b 𝑡 + 𝑊 c(𝑡) =

1 1 + 𝑡𝑆𝐷 = 1 𝑆𝐷 ( 1 𝑡 + 1 𝑆𝐷 )

(Assuming the initial voltage is zero) 𝑗 𝑢 = 𝑤b(𝑢) 𝑆

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SLIDE 19

Example: Laplace Transform

§ RLC lowpass filter

– Kirchhoff’s law: – For inductor: – For capacitor: – For resistor: – Transfer function:

19

(JOS Filter book)

𝑤a 𝑢 = 𝑤b 𝑢 + 𝑤c(𝑢), 𝑤k 𝑢 = 𝑤l 𝑢 𝑤l 𝑢 = 𝑀 𝑒𝑗l(𝑢) 𝑒𝑢 ⇔ 𝑊

l 𝑡 = 𝑡𝑀𝐽l(𝑡)

𝑗b 𝑢 = 𝑗l 𝑢 + 𝑗k(𝑢) 𝑗c 𝑢 = 𝐷 𝑒𝑤c(𝑢) 𝑒𝑢 ⇔ 𝑊

k 𝑡 = 𝐽l(𝑡)

𝑡𝐷 𝑗b 𝑢 = 𝑤b(𝑢) 𝑆 ⇔ 𝑊

b 𝑡 = 𝑆𝐽b(𝑡)

𝐼 𝑡 = 𝑊

c(𝑡)

𝑊

a(𝑡) =

𝑊

c(𝑡)

𝑊

b 𝑡 + 𝑊 c(𝑡) =

𝑊

c(𝑡)

𝑆 𝑊

l 𝑡

𝑡𝑀 + 𝑡𝐷𝑊

k 𝑡

+ 𝑊

c(𝑡)

= 1 𝑆 1 𝑡𝑀 + 𝑡𝐷 + 1 = 1 𝑆𝐷 𝑡 𝑡4 + 1 𝑆𝐷 𝑡 + 1 𝑀𝐷

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SLIDE 20

LTI System in continuous-time domain

§ In general, Laplace transform of LTI systems is represented with:

– Compared to z-transform, 𝑨0. is replaced with 𝑡

§ Poles and Zeros

– roots of denominator and numerator: 𝑞O (poles), 𝑟O (zeros) – Plot the frequency response of the system using poles and zeros

20

𝑍(𝑡) 𝑌(𝑡) = 𝐼 𝑡 = 𝑐, + 𝑐. / 𝑡 + 𝑐4 / 𝑡4 + … + 𝑐q / 𝑡q 1 + 𝑏. / 𝑡 + 𝑏4 / 𝑡4 + … + 𝑏r / 𝑡r 𝐼 𝑡 = 𝑡 − 𝑟. 𝑡 − 𝑟4 𝑡 − 𝑟s … (𝑡 − 𝑟q) 𝑡 − 𝑞. 𝑡 − 𝑞4 𝑡 − 𝑞s … (𝑡 − 𝑞r)

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SLIDE 21

Frequency Response

§ Frequency response 𝐼 𝜕 can be obtained from 𝐼 𝑡 when 𝑡 = 𝑘𝜕

– Analogous to frequency response 𝐼 𝜕 when 𝑨 = 𝑓67 at z-transform s-plane 𝑡 = 𝑘𝜕 z-plane 𝑨 = 𝑓67

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SLIDE 22

Example: RC lowpass filter

§ Transfer function § Amplitude response

– for 𝜕 ≪ 𝑏, 𝐼 𝑘𝜕 ≈ v

v = 1 = 0𝑒𝐶

– at 𝜕 = 𝑏, 𝐼 𝑘𝜕 =

v 67wv = v 6vwv = . 6.w. = 4

  • 4 = −3𝑒𝐶

– for 𝜕 ≫ 𝑏, 𝐼 𝑘𝜕 ≈

v 67 = v 7

§ Phase response

– for 𝜕 ≪ 𝑏, ∠𝐼 𝑘𝜕 = 0 – at 𝜕 = 𝑏, ∠𝐼 𝑘𝜕 = −45° – for 𝜕 ≫ 𝑏, ∠𝐼 𝑘𝜕 = −90°

22

𝐼 𝑡 = 𝑏 𝑡 + 𝑏 (𝑏 = 1 𝑆𝐷)

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SLIDE 23

Example: RC lowpass filter

23

𝐼 𝑡 = 100 𝑡 + 100

s-plane 𝑡 = 𝑘𝜕

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SLIDE 24

Example: highpass filter

24

𝐼 𝑡 = 𝑡 𝑡 + 100

s-plane 𝑡 = 𝑘𝜕

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SLIDE 25

Biquad Filters

§ The basic filter block that can represent all types of filters

– Filter types (e.g. LP, BP, HP, BR, AP and EQ) are determined by settings in the numerator – High-order filters can be realized by a cascade of biquad filters

  • E.g. 4-pole LP (-24 dB/oct), multi-band EQ

25

𝐼 𝑡 = 𝑐, + 𝑐.𝑡 + 𝑐4𝑡4 𝑏, + 𝑏.𝑡 +𝑏4 𝑡4

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SLIDE 26

Example: Bandpass filter

§ Transfer function

– for 𝜕 ≪ 𝜕,, 𝐼 𝑘𝜕 ≈

67 7K

– for 𝜕 ≫ 𝜕,, 𝐼 𝑘𝜕 ≈

67K 7

– at 𝜕 = 𝜕,, 𝐼 𝑘𝜕 = 𝑅 (asymptotic gains intersect at 𝜕 = 𝜕, is 0 dB) – For large Q, bandwidth between -3dB points is 𝐶𝑋 =

7K €

26

𝐼 𝑡 = 𝑡 𝜕, ( 𝑡 𝜕,)4+ 1 𝑅 ( 𝑡 𝜕,) + 1

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SLIDE 27

Example: Bandpass filter

27

s-plane 𝑡 = 𝑘𝜕

−𝜕,( 1 2𝑅 ± 𝑘 1 − 1 2𝑅

4

  • )

poles Pole locations for varying Q

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SLIDE 28

Example: Resonant Lowpass filter

§ Transfer Function

– for 𝜕 ≪ 𝜕,, 𝐼 𝑘𝜕 ≈ 1 = 0𝑒𝐶 – for 𝜕 ≫ 𝜕,, 𝐼 𝑘𝜕 ≈ −(

7K 7 )4

– at 𝜕 = 𝜕,, 𝐼 𝑘𝜕 = −𝑘𝑅

28

𝐼 𝑡 = 1 ( 𝑡 𝜕,)4+ 1 𝑅 ( 𝑡 𝜕,) + 1

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SLIDE 29

Example: Resonant Lowpass filter

29

s-plane 𝑡 = 𝑘𝜕

−𝜕,( 1 2𝑅 ± 𝑘 1 − 1 2𝑅

4

  • )

poles Pole locations for varying Q

  • 12db/oct

Peak is not exactly on 𝜕,

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SLIDE 30

Summary: Analog Filters

30

𝐼 𝑡 = 𝑡 𝜕, ( 𝑡 𝜕,)4+ 1 𝑅 ( 𝑡 𝜕,) + 1 𝐼 𝑡 = 1 ( 𝑡 𝜕,)4+ 1 𝑅 ( 𝑡 𝜕,) + 1 𝐼 𝑡 = ( 𝑡 𝜕,)4 ( 𝑡 𝜕,)4+ 1 𝑅 ( 𝑡 𝜕,) + 1

Resonant Lowpass Resonant Highpass Bandpass

𝐼 𝑡 = ( 𝑡 𝜕,)4+1 ( 𝑡 𝜕,)4+ 1 𝑅 ( 𝑡 𝜕,) + 1

Notch

𝐼 𝑡 = ( 𝑡 𝜕,)4− 1 𝑅 ( 𝑡 𝜕,) + 1 ( 𝑡 𝜕,)4+ 1 𝑅 ( 𝑡 𝜕,) + 1

Allpass

𝐼 𝑡 = ( 𝑡 𝜕,)4+ 𝐵 𝑅 ( 𝑡 𝜕,) + 1 ( 𝑡 𝜕,)4+( 1 𝐵𝑅)( 𝑡 𝜕,) + 1

Peaking-EQ

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SLIDE 31

Bi-linear Transform

§ Derivation

31

𝑦(𝑜 − 1) 𝑦(𝑜) 𝑦 𝑢 → R 𝑒𝑢

  • → 𝑧(𝑢)

𝑧 𝑢 = R 𝑦 𝑢 𝑒𝑢

  • 𝑍(𝑡) = 𝑌(𝑡)

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)

𝑡 = 2 𝑈 / 1 − 𝑨0. 1 + 𝑨0.

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SLIDE 32

Bi-linear Transform

§ Mapping s-plane to z-plane

32

s-plane 𝑡 = 𝑘𝜕 z-plane 𝑨 = 𝑓67 𝑡 = 2 𝑈 / 1 − 𝑨0. 1 + 𝑨0. 𝑨 = 1 + 𝑈 2 𝑡 1 − 𝑈 2 𝑡

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SLIDE 33

Bi-linear Transform

§ For continuous-time frequency Ω and discrete-time frequency 𝜕

– The mapping is linear in low frequency range but it becomes warped as frequency increases

33

𝑡 = 2 𝑈 / 1 − 𝑨0. 1 + 𝑨0. 𝑡 = 𝑘Ω, 𝑨 = 𝑓67 𝑘Ω = 2 𝑈 / 1 − 𝑓067 1 + 𝑓067 Ω = 2 𝑈 / tan (𝜕 2)

π π (Ω0,ω0) Ω ω

slide-34
SLIDE 34

Digitized Resonant Low-pass Filter

§ Transfer Function

– fc : cut-off frequency, Q: resonance

34

10

2

10

3

10

4

−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Lowpass Filters freqeuncy(log10) Gain(dB) 10

2

10

3

10

4

−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

slide-35
SLIDE 35

Digitized Resonant High-pass Filter

§ Transfer Function

35

10

2

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

10

3

10

4

−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

slide-36
SLIDE 36

Digitized Band-pass filter

§ Transfer Function

36

10

2

10

3

10

4

−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Bandpass Filters freqeuncy(log10) Gain(dB) 10

2

10

3

10

4

−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

slide-37
SLIDE 37

Digitized Notch filter

§ Transfer Function

37

10

2

10

3

10

4

−30 −20 −10 10 20 30 f=400 f=1000 f=3000 f=8000 Notch Filters freqeuncy(log10) Gain(dB) 10

2

10

3

10

4

−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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SLIDE 38

Digitized Equalizer

§ Transfer Function

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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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SLIDE 39

References

§ Cookbook formulae for audio EQs based on biquad filter (R. Bristow- Johnson)

– http://www.musicdsp.org/files/Audio-EQ-Cookbook.txt

§ If you want to study more on audio equalizer, check this recently published review paper:

– ” All About Audio Equalization: Solutions and Frontiers”, Vesa Välimäki and Joshua D. Reiss, Applied Science, 2016

  • http://www.mdpi.com/2076-3417/6/5/129

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