Slides i Intro Dsp

8/8/2019 Slides i Intro Dsp

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1

Fundamentals of digital signalprocessing


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Sound modeling

sound

symbols

aims

analysis

synthesis

processing

classification:

signal models

source models

abstract models


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0 500-0.05

0

0.05

x(t)

t in s 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 s ec

y(t)

Digital signals

analog

sampling

processingreconstruction

sampling interval T

sampling frequency fs= 1/T


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Digital signals: time representations

8000 samples

100 samples

line with dots

vertical quantization

integere.g. -32768 .. 32767

normalizede.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


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Spectrum: analog vs. digital signal

sampling leads to a replication of the baseband spectrum


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Spectrum: analog vs. digital signal

Sampling leads to a replication of the analog signal spectrum

Reconstruction of the analog signal:

low pass filtering the digital signal


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Discrete Fourier Transform

Magnitude

Phase


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0 0.5 1 1.5 2 2.5 3 3.5

x 104

-40

-20

0

20

|X(f)|indB

f in Hz

Magnitude s pectrum |X(f)| in dB

Discrete Fourier Transform (example)

FFT with 16 points

cosine (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/


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Inverse Discrete Fourier Transform (DFT)

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


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Frequency resolution

Zero padding: to increasefrequency resolution

0 2 4 6

-1

0

1

2

x(n)

8 sa mples

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 samples + zero-padding

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0

2

4

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8

10

k

|X(k)|

16-point FFT


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Window functions

to reduce leakage:

weight audio samples bya window

Hamming window

wH(n) = 0.54 0.46 cos(2n/N)

Blackman window

wB(n) = 0.42 0.5cos(2n/N) + 0.08(4n/N)


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Window

Reduction of the leakage effect bywindow functions:

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

of length N=8,

(c) product x(n)w(n) with 0n N-i,(d) zero-padding applied to z(n)

w(n) up to length N= 16

The corresponding spectra areshown on the right side.


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Spectrograms


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Waterfall representation

0 1000 2000 3000 4000 5000 6000 7000 8000-1

-0.5

0

0.5

1

Signal x(n)

x(n)

n

0 5 10 15 20

0

2000

4000

6000

-100

-50

0

Waterfall Representation of Short-time FFTs

f in Hz

n

Magnitudein

dB


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Digital systems


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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)


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Algorithms and signal graphs

Delay

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

Weighting factor

e.g. y(n) = a x(n)

Addition

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


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Simple digital system

weighted sum over several input samples


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Transforms

Frequency domain desciption of the digital system

Z transform

Discrete time Fouriertransform

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

Frequency response:Discrete time Fouriertransform of h(n)


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Causal and stable systems

Causality: a discrete-time system is causal

if the output signal y(n) = 0 for n


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IIR systems

= system with infinte impulse response

e.g. second order IIR system

Difference equation

Transfer function


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IIR systems

= system with infinte impulse response h(n)

Difference equation

Z transform of diff. eq.

Transfer function


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FIR systems

= system with finite impulse response h(n)

e.g. second order FIR system

Difference equation

Z transform of diff. eq.

Transfer function


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Fir example

computation of frequency response

impulse response

magnitude resp.

pole/zero plot

phase resp. 0 2 4-0.1

0

0.1

0.2

0.3

(a) Impuls e Re s ponse h(n)

n

-1 0 1 2

-1

0

1

Re(z)

Im(z)

(c) Pole/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)/

(d) Phase Response H(f)

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