E85.2607: Lecture 1 -- Introductionronw/adst-spring2010/lectures/lecture1.pdf · E85.2607: Lecture 1 { Introduction 2010-01-21 22 / 24. Fun with Matlab Signal generation linspace,

E85.2607: Lecture 1 – Introduction

1 Administrivia

2 DSP review

3 Fun with Matlab

E85.2607: Lecture 1 – Introduction 2010-01-21 1 / 24

Course overview

Advanced Digital Signal Theory

Design, analysis, and implementation of audio effects and synthesizers

EQ, reverb, chorus, phase vocoder, sinusoidal modeling, FM synthesis, ...

Emphasis on practical implementation, building complete systems.

Course web page: http://www.ee.columbia.edu/~ronw/adst

E85.2607: Lecture 1 – Introduction 2010-01-21 2 / 24

http://www.ee.columbia.edu/~ronw/adst

whoami

PhD, Electrical Engineering, Columbia University

Research interests: Source separation, speech recognition, musicinformation retrieval

http://www.ee.columbia.edu/~ronw

E85.2607: Lecture 1 – Introduction 2010-01-21 3 / 24

http://www.ee.columbia.edu/~ronw

Sound familiar?

Periodic and aperiodic signals

Discrete Fourier Transform

Convolution, filtering

Linear time-invariant systems

Impulse response

Frequency response

z-transform

E85.2607: Lecture 1 – Introduction 2010-01-21 4 / 24

Digital signals

Continuous time Discrete time

Dis

cret

eva

lue

Con

tinuo

usva

lue

Discrete-time signal sequence of samples

e.g. x [n ] = [0,−2.3,−1.3, 20, 4.2, ... ]Digital signal discrete-time signal that has been quantized

Discrete on both axes: samples can only take on a limited set ofvalues (quantization levels)Quantization introduces noise

E85.2607: Lecture 1 – Introduction 2010-01-21 5 / 24

Sampling

DAFXADC DACx(n) y(n) y(t)x(t)

f =1__Ts

f =1__Ts

0.5

0 500-0.05

0

0.05x(t)

t in !sec "0 10 20

-0.05

0

0.05

n "

x(n)

0 10 20-0.05

0

0.05

n "

y(n)

0 500-0.05

0

0.05

t in !sec "

y(t)

T

t = n1

fs

t time in seconds (real number)n time in samples (integer)

fs sampling rate(e.g. 44100 samplessecond

)E85.2607: Lecture 1 – Introduction 2010-01-21 6 / 24

Important signals: impulse

−4 −2 0 2 4 6 8 10n

0.0

0.2

0.4

0.6

0.8

1.0

δ[n] =

{1, n = 0,0, n 6= 0

Think of all discrete time signals as a sequence of scaled and time-shiftedimpulses.

x [n ] = [0,−2.3,−1.3, 20, 4.2, ... ]= 0 δ[n ]− 2.3 δ[n − 1]− 1.3 δ[n − 2] + 20 δ[n − 3] + 4.2 δ[n − 4] + ...

E85.2607: Lecture 1 – Introduction 2010-01-21 7 / 24

Important signals: sinusoid

0 5 10 15n

−1.0

−0.50.0

0.5

1.0x[n] = sin(2πfn)

sin(2πfn + φ), frequency = f cyclessample , phase = φ

Period: N = 1f samples

But what will it sound like?

Convert from samples: f cyclessample × 1fssamplessecond

How many samples in one period of a 440 Hz tone sampled at 44.1 kHz?

E85.2607: Lecture 1 – Introduction 2010-01-21 8 / 24

Important signals: sinusoid

0 5 10 15n

−1.0

−0.50.0

0.5

1.0x[n] = sin(2πfn)

sin(2πfn + φ), frequency = f cyclessample , phase = φ

Period: N = 1f samples

But what will it sound like?

Convert from samples: f cyclessample × 1fssamplessecond

How many samples in one period of a 440 Hz tone sampled at 44.1 kHz?

E85.2607: Lecture 1 – Introduction 2010-01-21 8 / 24

Discrete Fourier Transform

Decompose any periodic signal into sum of re-scaled sinusoids

Inverse DFT

x [n ] =1

N

N−1∑n=0

X [k ] e j2πnk/N k = 0, . . . ,N − 1

Forward DFT

X [k ] =N−1∑n=0

x [n ] e−j2πnk/N

Note that X [k ] are complex: X [k ] = XR [k ] + jXI [k ]

E85.2607: Lecture 1 – Introduction 2010-01-21 9 / 24

Magnitude and Phase spectra

X [k ] = |X [k ] | e j∠X [k ]

Magnitude: amount of energy ateach frequency

|X [k ]| =√

X 2R [k ] + X2I [k ]

Phase: delay at each frequency

∠X [k ] = arctan XI [k ]XR [k ]

−1.0−0.5

0.0

0.5

1.0

x[n] = cos(2πfn)

0

2

4

6

8

10|X[k]|

0 2 4 6 8 10 12 14 16−4−3−2−1

01234

∠X[k]

E85.2607: Lecture 1 – Introduction 2010-01-21 10 / 24

DFT symmetry and aliasing

24000 28000 32000 36000 40000-80

-60

-40

-20

0

X(f

)in

dB

f in Hz !

0 4000 8000 12000 16000 20000 44000 48000 52000 56000 60000

24000 28000 32000 36000 40000-80

-60

-40

-20

0

X(f

)in

dB

Spectrum of analog signal

!

0 4000 8000 12000 16000 20000 44000 48000 52000 56000 60000

f in Hz

Spectrum of digital signalSampling frequency f =40000 kHzS

The spectrum of a discrete time signal is periodic with period fs

The spectrum of a real valued signal is symmetric around fs/2

Any energy at frequencies greater than fs/2 will wrap around

E85.2607: Lecture 1 – Introduction 2010-01-21 11 / 24

Short-time Fourier Transform

What if frequency content varies withtime?

Break signal up into short (optionallyoverlapping) segments

Multiply by window function

Take DFT of each segment

E85.2607: Lecture 1 – Introduction 2010-01-21 12 / 24

STFT example

0 1 2 3 4 5 6 7 8Time (sec)

0

2

4

6

8

10

Freq

uenc

y(k

Hz)

−30−24−18−12−60

6

12

18

24

30

Pow

er(d

B)

E85.2607: Lecture 1 – Introduction 2010-01-21 13 / 24

http://www.ee.columbia.edu/~ronw/adst/lectures/matlab/wavs/lib-excerpt.wav

Linear time-invariant systems

Process input signal using delays, multiplications, additions

Describe using a difference equation, e.g.

y [n ] = b0 x [n ] + b1 x [n − 1]

Can also have feedback, e.g.

y [n ] = b0 x [n ] + b1 x [n − 1] + a1 y [n − 1] + a2 y [n − 2]

E85.2607: Lecture 1 – Introduction 2010-01-21 14 / 24

LTI systems – Block diagram

E85.2607: Lecture 1 – Introduction 2010-01-21 15 / 24

LTI systems – Properties

E85.2607: Lecture 1 – Introduction 2010-01-21 16 / 24

Impulse response and convolution

h(n)

y(n) = h(n)

-1 0 1 2 3 4n

x(n) = (n)! y(n) = h(n)

x(n) = (n)!

-1 0 1 2 3n

1

Characterize LTI system in time-domain by its impulse response h[n ]An LTI system is just another signal

Given input signal x [n ], output is convolution with h[n ]

Convolution

y [n ] = x [n ] ∗ h[n ] =∞∑

k=−∞x [k ] h[n − k ]

But how do we compute the impulse response from a differenceequation?

E85.2607: Lecture 1 – Introduction 2010-01-21 17 / 24

z-transform

z-transform

X (z) =∞∑

n=−∞x [n ] z−n

Maps discrete-time signal to a continuous function of a complexvariable

Incredibly useful for analyzing LTI systems

Turns difference equations into polynomials:

x [n −m ] z←→ z−M X (z)

Convolution becomes multiplication:

x [n ] ∗ h[n ] z←→ X (z) H(z)

If z = e jΩ, get discrete-time Fourier transform (DTFT)

E85.2607: Lecture 1 – Introduction 2010-01-21 18 / 24

Transfer function

Transfer function H(z) is the z-transform of the impulse response

Can read off z-transform from difference equation

y [n ] = b0 x [n ] + b1 x [n − 1] + a1 y [n − 1] + a2 y [n − 2]

z←→ Y (z) = (bo + b1 z−1) X (z) + (a1 z−1 + a2 z−2) Y (z)

H(z) =Y (z)

X (z)=

bo + b1 z−1

1− a1 z−1 − a2 z−2

But why? δ[n ]z←→ 1

Compute impulse response analytically by finding inverse z-transform ofH(z) (lots of algebra)

E85.2607: Lecture 1 – Introduction 2010-01-21 19 / 24

Transfer function

Transfer function H(z) is the z-transform of the impulse response

Can read off z-transform from difference equation

y [n ] = b0 x [n ] + b1 x [n − 1] + a1 y [n − 1] + a2 y [n − 2]z←→ Y (z) = (bo + b1 z−1) X (z) + (a1 z−1 + a2 z−2) Y (z)

H(z) =Y (z)

X (z)=

bo + b1 z−1

1− a1 z−1 − a2 z−2

But why? δ[n ]z←→ 1

Compute impulse response analytically by finding inverse z-transform ofH(z) (lots of algebra)

E85.2607: Lecture 1 – Introduction 2010-01-21 19 / 24

Transfer function

Transfer function H(z) is the z-transform of the impulse response

Can read off z-transform from difference equation

y [n ] = b0 x [n ] + b1 x [n − 1] + a1 y [n − 1] + a2 y [n − 2]z←→ Y (z) = (bo + b1 z−1) X (z) + (a1 z−1 + a2 z−2) Y (z)

H(z) =Y (z)

X (z)=

bo + b1 z−1

1− a1 z−1 − a2 z−2

But why? δ[n ]z←→ 1

Compute impulse response analytically by finding inverse z-transform ofH(z) (lots of algebra)

E85.2607: Lecture 1 – Introduction 2010-01-21 19 / 24

Transfer function

Transfer function H(z) is the z-transform of the impulse response

Can read off z-transform from difference equation

y [n ] = b0 x [n ] + b1 x [n − 1] + a1 y [n − 1] + a2 y [n − 2]z←→ Y (z) = (bo + b1 z−1) X (z) + (a1 z−1 + a2 z−2) Y (z)

H(z) =Y (z)

X (z)=

bo + b1 z−1

1− a1 z−1 − a2 z−2

But why? δ[n ]z←→ 1

Compute impulse response analytically by finding inverse z-transform ofH(z) (lots of algebra)

E85.2607: Lecture 1 – Introduction 2010-01-21 19 / 24

Frequency response

0 5 10 15 200.00.20.40.60.81.01.2

|H(f

)|

Magnitude Response

0 5 10 15 20

Frequency (kHz)

−4−3−2−1

01234

∠H

(f)

Phase Response

Often more intuitive to analyze system in the frequency-domain

H(e jΩ) DTFT of impulse response

Slice of z-transform corresponding to the unit circleDFT (discrete freq) is just sampled DTFT (continuous freq)

E85.2607: Lecture 1 – Introduction 2010-01-21 20 / 24

Poles and zeros

−2 −1 0 1 2Real

−1.0−0.5

0.0

0.5

1.0

Imag

inar

y

Zeros: roots of numerator of H(z)

Correspond to valleys in frequency response

Poles: roots of denominator of H(z)

Correspond to peaks in frequency response

System is unstable if it has poles outside of (or on) the unit circleImpulse response goes to infinity

E85.2607: Lecture 1 – Introduction 2010-01-21 21 / 24

FIR and IIR filters

Finite Impulse Response

No feedback ⇒ all zeroes ⇒ always stable∗∗ if coefficients are finiteEasy to design by drawing frequency response by hand, then usinginverse DFT to get impulse responseOften needs a long impulse response ⇒ expensive to implement

Infinite Impulse Response

Feedback ⇒ has poles ⇒ can be unstableCan implement complex filters using fewer delays than FIRBut harder to design

E85.2607: Lecture 1 – Introduction 2010-01-21 22 / 24

Fun with Matlab

Signal generation linspace, rand, sinI/O wavread, wavwrite, soundscPlotting plot, imagescTransforms fft, ifftFiltering conv, filterAnalyzing filters freqz, zplane, iztrans

E85.2607: Lecture 1 – Introduction 2010-01-21 23 / 24

Reading

Review your DST notes

Skim Introduction to Digital Filters

Linear Time-Invariant FiltersTransfer Function AnalysisFrequency Response Analysis

DAFX, Chapter 1 (if you have it)

E85.2607: Lecture 1 – Introduction 2010-01-21 24 / 24

https://ccrma.stanford.edu/~jos/filters/Linear_Time_Invariant_Digital_Filters.htmlhttps://ccrma.stanford.edu/~jos/filters/Transfer_Function_Analysis.htmlhttps://ccrma.stanford.edu/~jos/filters/Frequency_Response_Analysis.html

AdministriviaDSP reviewFun with Matlab