[PPT] - GCT535- Sound Technology for Multimedia Digital Systems Graduate PowerPoint Presentation

SLIDE 1

GCT535- Sound Technology for Multimedia Digital Systems

Graduate School of Culture Technology KAIST Juhan Nam

1

SLIDE 2

Systems

§ Audio systems modify the acoustic characteristics of the input sound § Examples

– Amplifiers, mixers, equalizers – Microphone, speakers – Audio plug-ins in DAW – Musical Instruments, human voice system, rooms (in a broad sense)

2

Input Output System

SLIDE 3

Digital System

§ We are interested in digital systems:

– Take the input signal 𝑦 𝑜 as a sequence of numbers – Perform mathematic operations upon the input signal

Addition, multiplication and delay

– Returns the output signal 𝑧 𝑜 as another sequence of numbers

3

Input Output Digital System 𝑦 𝑜 𝑧 𝑜

SLIDE 4

Amplifying Sounds

4

Acoustic Electrical Digital

𝑧 𝑜 = 𝑏 & 𝑦 𝑜

SLIDE 5

Basic Operations in Digital Systems

§ Multiplication: 𝑧 𝑜 = 𝑐( & 𝑦 𝑜 § Delaying: 𝑧 𝑜 = 𝑦 𝑜 − 1 § Addition: 𝑧 𝑜 = 𝑦 𝑜 + 𝑦 𝑜 − 1

5

Input Output Digital System 𝑦 𝑜 𝑧 𝑜

SLIDE 6

Linear Time-Invariant (LTI) System

§ Linearity

– Homogeneity: if 𝑦 𝑜 → 𝑧 𝑜 , then a & 𝑦 𝑜 → a & 𝑧 𝑜 – Superposition: if 𝑦. 𝑜 → 𝑧. 𝑜 and 𝑦/ 𝑜 → 𝑧/ (n), then 𝑦. 𝑜 + 𝑦/ 𝑜 → 𝑧. 𝑜 + 𝑧/ 𝑜

§ Time-Invariance

– If 𝑦 𝑜 → 𝑧 𝑜 , then 𝑦 𝑜 − 𝑂 → 𝑧 𝑜 − 𝑂 for any 𝑂 – This means that the system does not change its behavior over time

6

Input Output Digital System 𝑦 𝑜 𝑧 𝑜

SLIDE 7

LTI System

§ LTI systems in frequency domain

– No new sinusoidal components are introduced – Only existing sinusoids components changes in amplitude and phase.

§ Examples of non-LTI systems

– Clipping – Distortion – Aliasing – Modulation

7

SLIDE 8

Two Ways of Defining LTI Systems

§ By the relation between input 𝑦 𝑜 and output 𝑧 𝑜

– Difference equation – Signal flow graph

§ By the impulse response of the system

– Measure it by using a unit impulse as input – Convolution operation

8

SLIDE 9

The Simplest Lowpass Filter

9

𝑧 𝑜 = 𝑦 𝑜 + 𝑦(𝑜 − 1) 𝑨4. 𝑦 𝑜 𝑧 𝑜

+

§ Difference equation § Signal flow graph

“Delay Operator”

SLIDE 10

The Simplest Lowpass Filter: Sine-Wave Analysis

§ Measure the amplitude and phase changes given a sinusoidal signal input

10

SLIDE 11

The Simplest Lowpass Filter: Frequency Response

§ Plot the amplitude and phase change over different frequency

– The frequency sweeps from 0 to the Nyquist rate

11

SLIDE 12

The Simplest Lowpass Filter: Frequency Response

§ Mathematical approach

– Use complex sinusoid as input: 𝑦 𝑜 = 𝑓678 – Then, the output is: 𝑧 𝑜 = 𝑦 𝑜 + 𝑦 𝑜 − 1 = 𝑓678 + 𝑓67(84.) = 1 + 𝑓467 & 𝑓678 = 1 + 𝑓467 & 𝑦(𝑜) – Frequency response: 𝐼 𝜕 = 1 + 𝑓467 = 𝑓6;

< + 𝑓46; < 𝑓46; < = 2cos

(

7 /)𝑓46;

<

– Amplitude response: 𝐼(𝜕) = 2 cos

7 /

– Phase response: ∠𝐼 𝜕 = −

7 /

12

SLIDE 13

The Simplest Highpass Filter

§ Difference equation: 𝑧 𝑜 = 𝑦 𝑜 − 𝑦(𝑜 − 1) § Signal flow graph

13

𝑨4. 𝑦 𝑜 𝑧 𝑜

+

Frequency response −1

SLIDE 14

Finite Impulse Response (FIR) System

§ Difference equation § Signal flow graph

14

𝑧 𝑜 = 𝑐( & 𝑦 𝑜 + 𝑐. & 𝑦 𝑜 − 1 + 𝑐/ & 𝑦 𝑜 − 2 + ⋯ + 𝑐D & 𝑦 𝑜 − 𝑁

𝑨4. 𝑦 𝑜

+

𝑨4. 𝑨4. . . . 𝑐. 𝑐/ 𝑐D 𝑐( 𝑦 𝑜 − 1 𝑦 𝑜 − 2 𝑦 𝑜 − 𝑁 𝑧 𝑜

SLIDE 15

Impulse Response

§ The system output when the input is a unit impulse

– 𝑦 𝑜 = 𝜀 𝑜 = 1, 0, 0, 0, … → 𝑧 𝑜 = ℎ(𝑜)= [𝑐(, 𝑐., 𝑐/ … , 𝑐D] (for FIR system)

§ Characterizes the digital system as a sequence of numbers

– A system is represented just like audio samples!

15

𝑦 𝑜 = 𝜀 𝑜 𝑧 𝑜 = ℎ(𝑜) ℎ 𝑜 Input Output Digital System

SLIDE 16

Examples: Impulse Response

§ The simplest lowpass filter

– h 𝑜 = 1, 1

§ The simplest highpass filter

– h 𝑜 = 1, −1

§ Moving-average filter (order=5)

– h 𝑜 =

. M , . M , . M , . M , . M

16

SLIDE 17

Convolution

§ The output of LTI digital filters is represented by convolution operation between 𝑦 𝑜 and ℎ 𝑜 § Examples

– The simplest lowpass filter

𝑧 𝑜 = 1, 1 ∗ 𝑦 𝑜 = 1 & 𝑦(𝑜) + 1 & 𝑦(𝑜 − 1) = 𝑦 𝑜 + 𝑦(𝑜 − 1)

17

𝑧 𝑜 = 𝑦 𝑜 ∗ ℎ 𝑜 = O 𝑦 𝑗 & ℎ 𝑜 − 𝑗

Q RS4Q

= O ℎ(𝑗) & 𝑦(𝑜 − 𝑗)

Q RS4Q This is more practical expression when the input is an audio streaming

SLIDE 18

Proof: Convolution

§ Method 1

– The input can be represented as the sum of weighted and delayed impulses units

𝑦 𝑜 = 𝑦(, 𝑦., 𝑦/ … , 𝑦D = 𝑦( & 𝜀 𝑜 + 𝑦. & 𝜀 𝑜 − 1 + 𝑦/ & 𝜀 𝑜 − 2 + ⋯ + 𝑦D & 𝜀 𝑜 − 𝑁

– By the linearity and time-invariance

𝑧 𝑜 = 𝑦( & ℎ 𝑜 + 𝑦. & ℎ 𝑜 − 1 + 𝑦/ & ℎ 𝑜 − 2 + ⋯ + 𝑦D & ℎ 𝑜 − 𝑁 = ∑

𝑦(𝑗) & ℎ(𝑜 − 𝑗)

D RS(

§ Method 2

– The impulse response can be represented as a set of weighted impulses

ℎ 𝑜 = 𝑐(, 𝑐., 𝑐/ … , 𝑐D = 𝑐( & 𝜀 𝑜 + 𝑐. & 𝜀 𝑜 − 1 + 𝑐/ & 𝜀 𝑜 − 2 + ⋯ + 𝑐D & 𝜀 𝑜 − 𝑁

– By the linearity, the distributive property and 𝑦 𝑜 ∗ 𝜀 𝑜 − 𝑙 = 𝑦(𝑜 − 𝑙)

𝑧 𝑜 = 𝑐( & 𝑦 𝑜 + 𝑐. & 𝑦 𝑜 − 1 + 𝑐/ & 𝑦 𝑜 − 2 + ⋯ + 𝑐D & 𝑦 𝑜 − 𝑁 = ∑

ℎ(𝑗) & 𝑦(𝑜 − 𝑗)

D RS(

18

𝑦 𝑜 ∗ ℎ. 𝑜 + ℎ/ 𝑜 = 𝑦 𝑜 ∗ ℎ. 𝑜 + 𝑦 𝑜 ∗ ℎ/ 𝑜

SLIDE 19

Properties of Convolution

§ Commutative: 𝑦 𝑜 ∗ ℎ. 𝑜 ∗ ℎ/ 𝑜 = 𝑦 𝑜 ∗ ℎ/ 𝑜 ∗ ℎ. 𝑜 § Associative: (𝑦 𝑜 ∗ ℎ. 𝑜 ) ∗ ℎ/ 𝑜 = 𝑦 𝑜 ∗ (ℎ. 𝑜 ∗ ℎ/ 𝑜 ) § Distributive: 𝑦 𝑜 ∗ ℎ. 𝑜 + ℎ/ 𝑜 = 𝑦 𝑜 ∗ ℎ. 𝑜 + 𝑦 𝑜 ∗ ℎ/ 𝑜

19

SLIDE 20

Example: Convolution

§ Given 𝑦 𝑜 = 𝑦(, 𝑦., 𝑦/, … , 𝑦M and ℎ 𝑜 = ℎ(, ℎ., ℎ/ § 𝑧 𝑜 = ∑ ℎ(𝑗) & 𝑦(𝑜 − 𝑗)

D RS(

– 𝑧 0 = ℎ(0) & 𝑦(0) – 𝑧 1 = ℎ 0 & 𝑦 1 + ℎ(1) & 𝑦(0) – 𝑧 2 = ℎ 0 & 𝑦 2 + ℎ 1 & 𝑦 1 + ℎ 2 & 𝑦 0 – 𝑧 3 = ℎ 0 & 𝑦 3 + ℎ 1 & 𝑦 2 + ℎ 2 & 𝑦 1 – 𝑧 4 = ℎ 0 & 𝑦 4 + ℎ 1 & 𝑦 3 + ℎ 2 & 𝑦 2 – 𝑧 5 = ℎ 0 & 𝑦 5 + ℎ 1 & 𝑦 4 + ℎ 2 & 𝑦 3 – 𝑧 6 = ℎ 1 & 𝑦 5 + ℎ 2 & 𝑦 4 – 𝑧 7 = ℎ 2 & 𝑦 5

§ The size of transient region is equal to the number of delay operators (i.e. memory size )

20

Transient region Fully overlapped region (steady-state) Transient region

SLIDE 21

Demo: Convolution

21

If the length of 𝑦 𝑜 is M and the length of ℎ 𝑜 is N, then the length of 𝑧 𝑜 is M+N-1

SLIDE 22

Examples: FIR System

§ Convolution Reverb

22

Room Impulse Response Lobby Church

SLIDE 23

§ Difference equation § Signal flow graph

– When 𝑏 is slightly less than 1, it is called “Leaky Integrator”

A Simple Feedback Lowpass Filter

23

𝑧 𝑜 = 𝑦 𝑜 + 𝑏 & 𝑧(𝑜 − 1) 𝑨4. 𝑦 𝑜 𝑧 𝑜

+

𝑏

SLIDE 24

A Simple Feedback Lowpass Filter: Impulse Response

§ Impulse response: exponential decays

– 𝑧 0 = 𝑦 0 = 1 – 𝑧 1 = 𝑦 1 + 𝑏 & 𝑧 0 = 𝑏 – 𝑧 2 = 𝑦 2 + 𝑏 & 𝑧 1 = 𝑏/ – … – 𝑧 𝑜 = 𝑦 𝑜 + 𝑏 & 𝑧 𝑜 − 1 = 𝑏8

§ Stability!

– If 𝑏 < 1, the filter output converges (stable) – If 𝑏 = 1, the filter output oscillates (critical) – If 𝑏 > 1, the filter output diverges (unstable)

24

SLIDE 25

A Simple Feedback Lowpass Filter: Frequency Response

§ More dramatic change than the simplest lowpass filter (FIR)

– Phase response is not linear

25

𝑧 𝑜 = 𝑦 𝑜 + 0.9 & 𝑧(𝑜 − 1)

SLIDE 26

Reson Filter

26

§ Difference equation § Signal flow graph 𝑧 𝑜 = 𝑦 𝑜 + 2𝑠 & cos𝜄 & 𝑧 𝑜 − 1 − 𝑠/ & 𝑧 𝑜 − 2 𝑨4. 𝑦 𝑜 𝑧 𝑜

+

2𝑠 & cos𝜄

+

𝑨4.

−𝑠/

+

SLIDE 27

Reson Filter: Frequency Response

§ Generate resonance at a particular frequency

– Control the peak height by 𝑠 and the peak frequency by 𝜄

27

For stability: 𝑠 < 1

SLIDE 28

Infinite Impulse Response (IIR) System

§ Difference equation § Signal flow graph

28

𝑧 𝑜 = 𝑐( & 𝑦 𝑜 − 𝑏. & 𝑧 𝑜 − 1 − 𝑏/ & 𝑧 𝑜 − 2 − ⋯ − 𝑏a & 𝑧 𝑜 − 𝑂

𝑦 𝑜

+

𝑐( 𝑨4. 𝑧 𝑜 𝑨4. 𝑨4. . . .

𝑏.
𝑏/
𝑏a

𝑧 𝑜 − 1 𝑧 𝑜 − 2 𝑧 𝑜 − 𝑂

SLIDE 29

Examples: IIR System

§ Feedback Delay

29

+

𝑦 𝑜

feedback

y 𝑜

Dry

+

Wet

Delay Line

SLIDE 30

General Form of LTI Systems

§ Difference equation

– 𝑧 𝑜 = 𝑐( & 𝑦 𝑜 + 𝑐. & 𝑦 𝑜 − 1 + 𝑐/ & 𝑦 𝑜 − 2 + … + 𝑐D & 𝑦 𝑜 − 𝑁 −𝑏. & 𝑧 𝑜 − 1 − 𝑏/ & 𝑧 𝑜 − 2 − ⋯ − 𝑏a & 𝑧 𝑜 − 𝑂

§ Signal flow graph

30

𝑨4. 𝑦 𝑜

+

𝑨4. 𝑨4. . . . 𝑐. 𝑐/ 𝑐D 𝑐( 𝑦 𝑜 − 1 𝑦 𝑜 − 2 𝑦 𝑜 − 𝑁 𝑨4. 𝑧 𝑜 𝑨4. 𝑨4. . . .

𝑏.
𝑏/
𝑏a

𝑧 𝑜 − 1 𝑧 𝑜 − 2 𝑧 𝑜 − 𝑂

SLIDE 31

Frequency Response

31

Input Output Filter x(n) = ejωn y(n) = G(ω)ej(ω+Θ(ω))

20 40 60 80 100 120 140 160 180 200 −1 −0.5 0.5 1 Time [Samples] Amplitude Input Frequency = 100Hz input

utput

20 40 60 80 100 120 140 160 180 200 −1 −0.5 0.5 1 Time [Samples] Amplitude Input Frequency = 1000Hz input

utput

SLIDE 32

Frequency Response

§ Sine-wave Analysis

– 𝑦 𝑜 = 𝑓678 à 𝑦 𝑜 − 𝑛 = 𝑓67(84c) = 𝑓467c𝑦 𝑜 for any 𝑛 – Let’s assume that 𝑧 𝑜 = 𝐻 𝜕 𝑓6(78ef 7 ) à 𝑧 𝑜 − 𝑛 = 𝑓467c𝑧 𝑜 for any 𝑛

§ Putting this into the different equation

32

𝑧 𝑜 = 𝑐( + 𝑐. & 𝑓467 + 𝑐/ & 𝑓46/7 + … + 𝑐D & 𝑓467D 1 + 𝑏. & 𝑓467 + 𝑏/ & 𝑓46/7 + … + 𝑏a & 𝑓467a 𝑦(𝑜) 𝑧 𝑜 = 𝑐( & 𝑦 𝑜 + 𝑐. & 𝑓467 & 𝑦 𝑜 + 𝑐/ & 𝑓46/7 & 𝑦 𝑜 + … + 𝑐D & 𝑓467D & 𝑦 𝑜 −𝑏. & 𝑓467 & 𝑧 𝑜 − 𝑏/ & 𝑓46/7 & 𝑧 𝑜 − ⋯ − 𝑏a & 𝑓467a & 𝑧 𝑜

𝐼(𝜕) = 𝑐( + 𝑐. & 𝑓467 + 𝑐/ & 𝑓46/7 + … + 𝑐D & 𝑓467D 1 + 𝑏. & 𝑓467 + 𝑏/ & 𝑓46/7 + … + 𝑏a & 𝑓467a

𝐼(𝜕) : frequency response 𝐻 𝜕 = 𝐼(𝜕) : amplitude response 𝜄 𝜕 = ∠𝐼(𝜕) : phase response

SLIDE 33

Examples: Frequency Response

§ The simplest highpass filter: 𝑧 𝑜 = 𝑦 𝑜 − 𝑦(𝑜 − 1)

– 𝐼 𝜕 = 1 − 𝑓467 = 2 &

.4ghi; /

= 2𝑘 & 𝑓4i;

<

g

i; < 4ghi; <

/6

= 2sin (

7 /) & 𝑓4i;hm

<

– 𝐻 𝜕 = 2sin (

7 /), 𝜄 𝜕 = − 674n /

– In this simple case, we can obtain the amplitude/phase response analytically

§ How about this? 𝑧 𝑜 = 𝑦 𝑜 + 0.9 & 𝑧(𝑜 − 1)

– 𝐼(𝜕) =

. .4(.o&ghi;

à 𝐻 𝜕 , 𝜄 𝜕 ?

§ In general, it is very difficult to obtain analytic expressions of the amplitude and phase response

33

SLIDE 34

Z-Transform

§ 𝑎-transform

– Define z to be a variable in complex plane: we call it z-plane – When z = ejω (on unit circle), the frequency response is a particular case of the following form – We call this 𝒜-transform or transfer function of the filter – “𝑨4.” corresponds to one sample delay: delay operator or delay element – Digital systems are often expressed as 𝑨-transform: polynomial of 𝑨4.

FIR: 𝐼 𝑨 = 1 + 𝑨4. , 𝐼 𝑨 = 1 − 𝑨4.
IIR: 𝐼 𝑨 =

. .4(.o&rhs , 𝐼 𝑨 = . .4(.o&rhse(.t.&rhs

34

𝐼 𝑨 = 𝐶(𝑨) 𝐵(𝑨) = 𝑐( + 𝑐. & 𝑨4. + 𝑐/ & 𝑨4/ + … + 𝑐D & 𝑨4D 1 + 𝑏. & 𝑨4. + 𝑏/ & 𝑨4/ + … + 𝑏a & 𝑨4a

SLIDE 35

Z-Transform

§ More formally, z-transform is defined as: § With delay by 𝑁 § If this is applied to the difference equation of general form of LTI systems

35

𝑌 𝑨 = O 𝑦 𝑜 & 𝑨48

Q 8S4Q

𝑌 𝑨 𝑨4D = O 𝑦 𝑜 − 𝑁 & 𝑨48

Q 8S4Q

𝑍 𝑨 = 𝑐( & 𝑌 𝑨 + 𝑐. & 𝑨4.𝑌 𝑨 + 𝑐/ & 𝑨4/𝑌 𝑨 + … + 𝑐D & 𝑨4D𝑌 𝑨 −𝑏. & 𝑨4.𝑍 𝑨 − 𝑏/ & 𝑨4/𝑍 𝑨 − ⋯ − 𝑏a & 𝑨4a𝑍 𝑨 𝑍(𝑨) 𝑌(𝑨) = 𝑐( + 𝑐. & 𝑨4. + 𝑐/ & 𝑨4/ + … + 𝑐D & 𝑨4D 1 + 𝑏. & 𝑨4. + 𝑏/ & 𝑨4/ + … + 𝑏a & 𝑨4a = 𝐼 𝑨

SLIDE 36

Z-Transform

§ This, by replacing z = ejω § The frequency response is discrete-time Fourier transform of impulse response

36

𝐼 𝑓67 = O ℎ 𝑜 & 𝑓4678

Q 8S4Q

SLIDE 37

Poles and Zeros in Z-Transform

§ The polynomial of 𝑨4. in 𝐼 𝑨 can be factorized

– 𝑋e can find roots for both numerator and denominator – Zeros: roots of numerator – Poles: roots of denominator

§ We can analyze frequency response more easily using poles and zeros than numerator or denominator coefficient

37

𝐼 𝑨 = 𝐶(𝑨) 𝐵(𝑨) = 1 − 𝑟.𝑨4. 1 − 𝑟/𝑨4. 1 − 𝑟{𝑨4. … (1 − 𝑟D𝑨4.) 1 − 𝑞.𝑨4. 1 − 𝑞/𝑨4. 1 − 𝑞{𝑨4. … (1 − 𝑞a𝑨4.)

SLIDE 38

Pole-Zero Analysis: Amplitude Response

§ Amplitude Response

– Computed using distances between poles and unit circles and distances between zeros and units circles on Z-plane

38

𝐻 𝜕 = 𝐼(𝜕) = 𝐶(𝜕) 𝐵(𝜕) = 1 − 𝑟.𝑓467 1 − 𝑟/𝑓467 1 − 𝑟{𝑓467 … (1 − 𝑟D𝑓467) 1 − 𝑞.𝑓467 1 − 𝑞/𝑓467 1 − 𝑞{𝑓467 … (1 − 𝑞D𝑓467) 𝐼(𝜕) = 𝐶(𝜕) 𝐵(𝜕) = 𝑓67 − 𝑟. 𝑓67 − 𝑟/ 𝑓67 − 𝑟{ … (𝑓67 − 𝑟D) 𝑓67 − 𝑞. 𝑓67 − 𝑞/ 𝑓67 − 𝑞{ … (𝑓67 − 𝑞D) 𝐼(𝜕) = 𝐶(𝜕) 𝐵(𝜕) = (𝑓67 − 𝑟.) 𝑓67 − 𝑟/ 𝑓67 − 𝑟{ … (𝑓67 − 𝑟D) 𝑓67 − 𝑞. 𝑓67 − 𝑞/ 𝑓67 − 𝑞{ … (𝑓67 − 𝑞D)

SLIDE 39

Example: Reson Filter

§ Difference equation

– 𝑧 𝑜 = 𝑦 𝑜 + 2𝑠 & cos𝜄 & 𝑧 𝑜 − 1 − 𝑠/ & 𝑧 𝑜 − 2 − 𝑦(𝑜 − 2)

§ Transfer function – 𝐼 𝑨 =

.4rh< .4/}~•€•&rhse}<&rh<

– Poles: 𝑠(cos𝜄 + 𝑘sin𝜄),𝑠(cos𝜄 − 𝑘sin𝜄) – Zeros: 1, -1

§ Amplitude response – 𝐻 𝜕 = 𝐼(𝜕) =

‚(7) ƒ(7) = „s(7)„<(7) „…(7)„†(7)

39

Z-plane 𝑒{ 𝑒ˆ 𝑒. 𝑒/

SLIDE 40

Poles and Stability

§ If Poles are inside the unit circle

– The system is stable – 𝑠 < 1: the impulse response decays with oscillation

§ If Poles are on the unit circle

– The system is critically-stable – 𝑠 = 1: the impulse response oscillates with constant amplitude (sine generation)

§ If Poles are outside the unit circle

– The system is unstable (e.g. howling when mics are closely placed near speakers) – 𝑠 > 1: the impulse response diverges

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SLIDE 41

Pole-Zero Analysis: Phase Response

§ Phase Response

– Computed using angles between poles and unit circles and angles between zeros and units circles on Z-plane

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𝜄 𝜕 = ∠𝐼(𝜕) = ∠𝐶(𝜕) ∠𝐵(𝜕) = ∠ 1 − 𝑟.𝑓467 1 − 𝑟/𝑓467 1 − 𝑟{𝑓467 … (1 − 𝑟D𝑓467) ∠ 1 − 𝑞.𝑓467 1 − 𝑞/𝑓467 1 − 𝑞{𝑓467 … (1 − 𝑞D𝑓467) = ∠ 1 − 𝑟.𝑓467 + ∠ 1 − 𝑟/𝑓467 + ∠ 1 − 𝑟{𝑓467 + ⋯ + ∠(1 − 𝑟D𝑓467) −∠ 1 − 𝑞.𝑓467 − ∠ 1 − 𝑞/𝑓467 − ∠ 1 − 𝑞{𝑓467 − ⋯ − ∠(1 − 𝑞D𝑓467) 𝜄 𝜕 = ∠𝐼(𝜕) = ∠𝐶 𝜕 ∠𝐵 𝜕 = ∠ 𝑓67 − 𝑟. + ∠ 𝑓67 − 𝑟/ + ∠ 𝑓67 − 𝑟{ + ⋯ + ∠(𝑓67 − 𝑟D) −∠ 𝑓67 − 𝑞. − ∠ 𝑓67 − 𝑞/ − ∠ 𝑓67 − 𝑞{ − ⋯ − ∠(𝑓67 − 𝑞D)

SLIDE 42

Example: Reson Filter

§ Difference equation

– 𝑧 𝑜 = 𝑦 𝑜 + 2𝑠 & cos𝜄 & 𝑧 𝑜 − 1 − 𝑠/ & 𝑧 𝑜 − 2 − 𝑦(𝑜 − 2)

§ Transfer function – 𝐼 𝑨 =

.4rh< .4/}~•€•&rhse}<&rh<

– Poles: 𝑠(cos𝜄 + 𝑘sin𝜄),𝑠(cos𝜄 − 𝑘sin𝜄) – Zeros: 1, -1

§ Phase response

– 𝜄 𝜕 = ∠𝐼 𝜕 =

∠‚ 7 ∠ƒ 7 = 𝜄. 𝜕 +𝜄/ 𝜕 − 𝜄{ 𝜕 − 𝜄ˆ 𝜕

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Z-plane 𝑒ˆ 𝑒. 𝜄{ 𝜄ˆ 𝜄. 𝜄/

SLIDE 43

Practical Filters

§ One-pole one-zero filters

– Leaky integrator – DC-removal filters – Bass / treble shelving filter (EQ)

§ Biquad filters

– Reson filter – Band-pass / notch filters – Equalizers

§ Any high-order filter can be factored into a cascade of one-pole one-zero filters or bi-quad filters

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𝐼‰ 𝑨 = 𝐶(𝑨) 𝐵(𝑨) = 𝑐( + 𝑐. & 𝑨4. 1 + 𝑏. & 𝑨4. 𝐼ŠR 𝑨 = 𝐶(𝑨) 𝐵(𝑨) = 𝑐( + 𝑐. & 𝑨4. + 𝑐/ & 𝑨4/ 1 + 𝑏. & 𝑨4. +𝑏/& 𝑨4/ 𝐼 𝑨 = 𝐼‰ 𝑨 ‹ 𝐼ŠR 𝑨

SLIDE 44

Digital Audio Effects

§ Filters, Equalizers

– One-pole/one-zero filter and bi-quad filters – Small number of elements and small delay lengths

§ Delay-based Effects

– Chorus, Flanger, Echo – FIR or IIR filters – Small number of elements and mid-to-long delay lengths

§ Spatial Effects

– HRTF, Reverb – FIR or IIR filters – Large number of elements and long delay lengths

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