1

Fundamentals of digital signalprocessing

2

Sound modeling

soundsymbols

aimsanalysissynthesisprocessing

classification:signal modelssource modelsabstract models

3

0 500-0.05

0

0.05x(t)

t in µs ec →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 µs ec →

y(t)

Digital signals

analogsamplingprocessingreconstruction

sampling interval Tsampling frequency fs= 1/T

4

Digital signals: time representations

8000 samples100 samplesline with dots

vertical quantizationintegere.g. -32768 .. 32767normalizede.g. -1 .. (1-Q)Q = quantization step

0 1000 2000 3000 4000 5000 6000 7000 8000-0.5

0

0.5

x(n)

0 100 200 300 400 500 600 700 800 900 1000-0.5

0

0.5

x(n)

0 10 20 30 40 50 60 70 80 90 100-0.05

0

0.05

x(n)

n →

5

Spectrum: analog vs. digital signal

sampling leads to a replication of the baseband spectrum

6

Spectrum: analog vs. digital signal

Sampling leads to a replication of the analog signal spectrumReconstruction of the analog signal:

low pass filtering the digital signal

7

Discrete Fourier Transform

Magnitude

Phase

8

0 0.5 1 1.5 2 2.5 3 3.5

x 104

-40

-20

0

20

|X(f)

| in

dB →

f in Hz →

Magnitude s pectrum |X(f)| in dB

Discrete Fourier Transform (example)

FFT with 16 pointscosine (16 points)

magnitude (16 points)normalization:0 dB for sinusoid ±1

magnitude (frequency points)

step fs/N

magnitude dB vs. Hz

0 2 4 6 8 10 12 14 16

-1

0

1

a)

n →

Cos ine s ignal x(n)

0 2 4 6 8 10 12 14 16

0

0.5

1

b)

k →

Magnitude s pectrum |X(k)|

0 0.5 1 1.5 2 2.5 3 3.5

x 104

0

0.5

1

c)f in Hz →

Magnitude s pectrum |X(f)|

Nkfs /

9

Inverse Discrete Fourier Transform (DFT)

if X(k) = X*(N-k)then IDFT gives N discrete-time real values x(n)

10

Frequency resolution

Zero padding: to increasefrequency resolution

0 2 4 6

-1

0

1

2

x(n)

→

8 s amples

0 2 4 6

0

2

4

6

8

10

|X(k

)| →

8-point FFT

0 5 10 15

-1

0

1

2

n →x(

n) →

8 s amples + zero-padding

0 5 10 15

0

2

4

6

8

10

k →

|X(k

)| →

16-point FFT

11

Window functions

to reduce leakage:weight audio samples bya window

Hamming windowwH(n) = 0.54 – 0.46 cos(2πn/N)

Blackman windowwB(n) = 0.42 – 0.5cos(2πn/N) + 0.08(4πn/N)

12

Window

Reduction of the leakage effect by window functions:

(a) the original signal, (b) the Blackman window function

of length N =8, (c) product x(n)w(n) with 0≤n ≤ N-i, (d) zero-padding applied to z(n)

w(n) up to length N = 16

The corresponding spectra are shown on the right side.

13

Spectrograms

14

Waterfall representation

0 1000 2000 3000 4000 5000 6000 7000 8000-1

-0.5

0

0.5

1S ignal x(n)

x(n)

→

n →

0 5 10 15 20

02000

40006000

-100

-50

0

Waterfall Repres entation of S hort-time FFTs

f in Hz →n →

Mag

nitu

de in

dB

→

15

Digital systems

16

Definitions

Unit impulse

Impulse reponse h(n) = output to a unit impulseh(n) describes the digital sistem

Discrete convolution:y(n)=x(n)*h(n)

17

Algorithms and signal graphs

Delaye.g. y(n) = x(n-2)

Weighting factore.g. y(n) = a x(n)

Additione.g.y(n) = a1 x(n) + a2 x(n)

18

Simple digital system

weighted sum over several input samples

19

Transforms

Frequency domain desciption of the digital system

Z transform

Discrete time Fourier transform

Transfer function H(z):Z transform of h(n)

Frequency response:Discrete time Fourier transform of h(n)

20

Causal and stable systems

Causality: a discrete-time system is causalif the output signal y(n) = 0 for n <0 for a given input signalu(n) = 0 for n <0.This means that the system cannot react to an input before the input is applied to the system

Stability: a digital system is stable if

stability implies that transfer function H(z) and frequency response are related by

21

IIR systems

= system with infinte impulse responsee.g. second order IIR system

Difference equation

Transfer function

22

IIR systems

= system with infinte impulse response h(n)

Difference equation

Z transform of diff. eq.

Transfer function

23

FIR systems

= system with finite impulse response h(n)e.g. second order FIR system

Difference equation

Z transform of diff. eq.

Transfer function

24

Fir example

computation of frequency response

impulse responsemagnitude resp.pole/zero plotphase resp. 0 2 4

-0.1

0

0.1

0.2

0.3

(a) Impuls e Res pons e h(n)

n →

-1 0 1 2

-1

0

1

Re(z)

Im(z

)

(c ) P ole/Zero plot

4

0 10 20 30 400

0.2

0.4

0.6

0.8

f in kHz →

|H(f)

| →

(b) Magnitude Res pons e |H(f)|

0 10 20 30 40-2

-1.5

-1

-0.5

0

f in kHz →

∠ H

(f)/ π→

(d) P has e Res pons e ∠ H(f)