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

GCT535- Sound Technology for Multimedia Digital Systems Graduate School of Culture Technology KAIST Juhan Nam 1

Systems Input Output System § 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

Digital System Digital Input Output System 𝑦 𝑜 𝑧 𝑜 § 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

Amplifying Sounds 𝑧 𝑜 = 𝑏 & 𝑦 𝑜 Acoustic Electrical Digital 4

Basic Operations in Digital Systems Digital Input Output System 𝑦 𝑜 𝑧 𝑜 § Multiplication: 𝑧 𝑜 = 𝑐 ( & 𝑦 𝑜 § Delaying: 𝑧 𝑜 = 𝑦 𝑜 − 1 § Addition: 𝑧 𝑜 = 𝑦 𝑜 + 𝑦 𝑜 − 1 5

Linear Time-Invariant (LTI) System Digital Input Output System 𝑦 𝑜 𝑧 𝑜 § 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

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

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

The Simplest Lowpass Filter § Difference equation 𝑧 𝑜 = 𝑦 𝑜 + 𝑦(𝑜 − 1) § Signal flow graph + 𝑦 𝑜 𝑧 𝑜 𝑨 4. “Delay Operator” 9

The Simplest Lowpass Filter: Sine-Wave Analysis § Measure the amplitude and phase changes given a sinusoidal signal input 10

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

The Simplest Lowpass Filter: Frequency Response § Mathematical approach – Use complex sinusoid as input: 𝑦 𝑜 = 𝑓 678 – Then, the output is: 𝑧 𝑜 = 𝑦 𝑜 + 𝑦 𝑜 − 1 = 𝑓 678 + 𝑓 67(84.) = 1 + 𝑓 467 & 𝑓 678 = 1 + 𝑓 467 & 𝑦(𝑜) – Frequency response: 𝐼 𝜕 = 1 + 𝑓 467 = 𝑓 6 ; < + 𝑓 46 ; < 𝑓 46 ; / )𝑓 46 ; 7 < = 2cos ( < 7 – Amplitude response: 𝐼(𝜕) = 2 cos / 7 – Phase response: ∠𝐼 𝜕 = − / 12

The Simplest Highpass Filter § Difference equation: 𝑧 𝑜 = 𝑦 𝑜 − 𝑦(𝑜 − 1) § Signal flow graph + 𝑦 𝑜 𝑧 𝑜 𝑨 4. −1 Frequency response 13

Finite Impulse Response (FIR) System § Difference equation 𝑧 𝑜 = 𝑐 ( & 𝑦 𝑜 + 𝑐 . & 𝑦 𝑜 − 1 + 𝑐 / & 𝑦 𝑜 − 2 + ⋯ + 𝑐 D & 𝑦 𝑜 − 𝑁 § Signal flow graph 𝑧 𝑜 + 𝑦 𝑜 𝑐 ( 𝑨 4. 𝑦 𝑜 − 1 𝑐 . 𝑨 4. 𝑦 𝑜 − 2 . . . 𝑐 / 𝑨 4. 𝑦 𝑜 − 𝑁 𝑐 D 14

Impulse Response Digital Input Output System 𝑧 𝑜 = ℎ(𝑜) 𝑦 𝑜 = 𝜀 𝑜 ℎ 𝑜 § 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 / … , 𝑐 D ] (for FIR system) § Characterizes the digital system as a sequence of numbers – A system is represented just like audio samples! 15

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

Convolution § The output of LTI digital filters is represented by convolution operation between 𝑦 𝑜 and ℎ 𝑜 Q Q 𝑧 𝑜 = 𝑦 𝑜 ∗ ℎ 𝑜 = O 𝑦 𝑗 & ℎ 𝑜 − 𝑗 = O ℎ(𝑗) & 𝑦(𝑜 − 𝑗) RS4Q RS4Q This is more practical expression when the input is an audio streaming § Examples – The simplest lowpass filter • 𝑧 𝑜 = 1, 1 ∗ 𝑦 𝑜 = 1 & 𝑦(𝑜) + 1 & 𝑦(𝑜 − 1) = 𝑦 𝑜 + 𝑦(𝑜 − 1) 17

Proof: Convolution § Method 1 – The input can be represented as the sum of weighted and delayed impulses units • 𝑦 𝑜 = 𝑦 ( , 𝑦 . , 𝑦 / … , 𝑦 D = 𝑦 ( & 𝜀 𝑜 + 𝑦 . & 𝜀 𝑜 − 1 + 𝑦 / & 𝜀 𝑜 − 2 + ⋯ + 𝑦 D & 𝜀 𝑜 − 𝑁 – By the linearity and time-invariance D • 𝑧 𝑜 = 𝑦 ( & ℎ 𝑜 + 𝑦 . & ℎ 𝑜 − 1 + 𝑦 / & ℎ 𝑜 − 2 + ⋯ + 𝑦 D & ℎ 𝑜 − 𝑁 = ∑ 𝑦(𝑗) & ℎ(𝑜 − 𝑗) RS( § Method 2 – The impulse response can be represented as a set of weighted impulses • ℎ 𝑜 = 𝑐 ( , 𝑐 . , 𝑐 / … , 𝑐 D = 𝑐 ( & 𝜀 𝑜 + 𝑐 . & 𝜀 𝑜 − 1 + 𝑐 / & 𝜀 𝑜 − 2 + ⋯ + 𝑐 D & 𝜀 𝑜 − 𝑁 – By the linearity, the distributive property and 𝑦 𝑜 ∗ 𝜀 𝑜 − 𝑙 = 𝑦(𝑜 − 𝑙) D • 𝑧 𝑜 = 𝑐 ( & 𝑦 𝑜 + 𝑐 . & 𝑦 𝑜 − 1 + 𝑐 / & 𝑦 𝑜 − 2 + ⋯ + 𝑐 D & 𝑦 𝑜 − 𝑁 = ∑ ℎ(𝑗) & 𝑦(𝑜 − 𝑗) RS( 𝑦 𝑜 ∗ ℎ . 𝑜 + ℎ / 𝑜 = 𝑦 𝑜 ∗ ℎ . 𝑜 + 𝑦 𝑜 ∗ ℎ / 𝑜 18

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

Example: Convolution § Given 𝑦 𝑜 = 𝑦 ( , 𝑦 . , 𝑦 / , … , 𝑦 M and ℎ 𝑜 = ℎ ( , ℎ . , ℎ / D § 𝑧 𝑜 = ∑ ℎ(𝑗) & 𝑦(𝑜 − 𝑗) RS( – 𝑧 0 = ℎ(0) & 𝑦(0) Transient region – 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 Fully overlapped region – 𝑧 4 = ℎ 0 & 𝑦 4 + ℎ 1 & 𝑦 3 + ℎ 2 & 𝑦 2 (steady-state) – 𝑧 5 = ℎ 0 & 𝑦 5 + ℎ 1 & 𝑦 4 + ℎ 2 & 𝑦 3 – 𝑧 6 = ℎ 1 & 𝑦 5 + ℎ 2 & 𝑦 4 Transient region – 𝑧 7 = ℎ 2 & 𝑦 5 § The size of transient region is equal to the number of delay operators (i.e. memory size ) 20

Demo: Convolution 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 21

Examples: FIR System § Convolution Reverb Lobby Church Room Impulse Response 22

A Simple Feedback Lowpass Filter § Difference equation 𝑧 𝑜 = 𝑦 𝑜 + 𝑏 & 𝑧(𝑜 − 1) § Signal flow graph + 𝑦 𝑜 𝑧 𝑜 𝑨 4. 𝑏 – When 𝑏 is slightly less than 1, it is called “Leaky Integrator” 23

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 = 𝑏 8 § 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

A Simple Feedback Lowpass Filter: Frequency Response § More dramatic change than the simplest lowpass filter (FIR) – Phase response is not linear 𝑧 𝑜 = 𝑦 𝑜 + 0.9 & 𝑧(𝑜 − 1) 25

Reson Filter § Difference equation 𝑧 𝑜 = 𝑦 𝑜 + 2𝑠 & cos𝜄 & 𝑧 𝑜 − 1 − 𝑠 / & 𝑧 𝑜 − 2 § Signal flow graph + 𝑧 𝑜 𝑦 𝑜 + 𝑨 4. 2𝑠 & cos𝜄 + 𝑨 4. −𝑠 / 26

Reson Filter: Frequency Response § Generate resonance at a particular frequency – Control the peak height by 𝑠 and the peak frequency by 𝜄 For stability: 𝑠 < 1 27

Infinite Impulse Response (IIR) System § Difference equation 𝑧 𝑜 = 𝑐 ( & 𝑦 𝑜 − 𝑏 . & 𝑧 𝑜 − 1 − 𝑏 / & 𝑧 𝑜 − 2 − ⋯ − 𝑏 a & 𝑧 𝑜 − 𝑂 § Signal flow graph + 𝑧 𝑜 𝑦 𝑜 𝑐 ( 𝑨 4. 𝑧 𝑜 − 1 - 𝑏 . 𝑨 4. 𝑧 𝑜 − 2 . . . - 𝑏 / 𝑨 4. 𝑧 𝑜 − 𝑂 - 𝑏 a 28

Examples: IIR System § Feedback Delay + Delay Line 𝑦 𝑜 feedback Wet Dry y 𝑜 + 29