digital signal - ksu · to reduce statistical noise for small number of samples ... li tan, digital...

Digital Signal

1CEN543, Dr. Ghulam Muhammad

King Saud University

Continuous time

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Discrete time

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Analog Signal

Discrete-time Signal

Digital Signal

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A digital signal processing scheme

DSP (Digital Signal Processing)

To avoid aliasing for sampling

Analog to Digital Converter

Digital to Analog Converter To avoid aliasing

for sampling

Computer / microprocessor / micro controller/ etc.

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Some Applications of DSP

• Noise removal from speech.Noisy Speech

Clean Speech

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Some Applications of DSP

• Signal spectral analysis.

Time domain Frequency domain

Single tone: 1000 Hz

Double tone: 1000 Hz and 3000 Hz

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Some Applications of DSP

• Noise removal from image.

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Some Applications of DSP

• Image enhancement.

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Summary Applications of DSP

Digital speech and audio:

• Speech recognition• Speaker recognition• Speech synthesis• Speech enhancement • Speech coding

Digital Image Processing:

• Image enhancement • Image recognition• Medical imaging• Image forensics• Image coding

Multimedia:• Internet audio, video, phones• Image / video compression• Text-to-voice & voice-to-text• Movie indexing

……..

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For a given sampling interval T, which is defined as the time span between two sample points, the sampling rate is given by

samples per second (Hz).

For example, if a sampling period is T = 125 microseconds, the sampling rate isdetermined as fs =1/125 s or 8,000 samples per second (Hz).

Tf s

1

Sampling

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Sampling - Theorem 2 / 23

7 / 23 22/ 23

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Sampling - Theorem

The sampling theorem guarantees that an analog signal can be in theoryperfectly recovered as long as the sampling rate is at least twice as large asthe highest-frequency component of the analog signal to be sampled.

The condition is:

For example, to sample a speech signal containing frequencies up to 4 kHz, theminimum sampling rate is chosen to be at least 8 kHz, or 8,000 samples persecond.

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Sampling - Theorem

Sampling interval T= 0.01 sSampling rate fs= 100 HzSinusoid freq. = 4 cycles / 0.1

= 40 Hz

Sampling condition is satisfied, so reconstruction from digital to analog is possible.

Do this by yourself!

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Sampling Process

In frequency domain:

Xs(f): Sampled spectrumX(f): Original spectrumX(fnfs): Replica spectrum

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Sampling Process

Original spectrum

Original spectrum plus its replicas

Original spectrum plus its replicas

Original spectrum plus its replicas

Reconstruction not possible

Minimum requirement for Reconstruction

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Shannon Sampling Theorem

For a uniformly sampled DSP system, an analog signal can be perfectlyrecovered as long as the sampling rate is at least twice as large as thehighest-frequency component of the analog signal to be sampled.

The minimum sampling rate is called the Nyquist rate Half of the sampling frequency is called the folding frequency.

ttttx 100cos300sin1050cos3)(

Problem: Find the Nyquist rate for the following signal.

The maximum frequency present is 150 Hz = fmax.

Therefore Nyquist rate = 2× fmax = 300 Hz.

Solution:

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Problem:Example 1

Solution:

Using Euler’s identity,

Hence, the Fourier series coefficients are:

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Example 1 – contd.

a.

b. After the analog signal is sampled at the rate of 8,000 Hz, thesampled signal spectrum and its replicas centered at thefrequencies nfs, each with the scaled amplitude being 2.5/T

Replicas, no additional information.

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Signal Reconstruction

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Signal Reconstruction

Perfect reconstruction is not possible, even if we use ideal low pass filter.

Aliasing

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Example 2Problem:

Solution:Using the Euler’s identity:

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Example 2 – contd.

a.

b.The Shannon sampling theory condition is satisfied.

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Example 3Problem:

Solution:

a.b.

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Example 4Problem: Consider the analog signal ttttx 12000cos106000sin52000cos3)(

(a) What is the Nyquist rate for this signal?(b) What is the discrete-time signal after sampling it at Fs = 5000 samples/s?(c) What is the analog signal y(t) that we reconstruct from the samples if we use

ideal interpolation?

Solution: KHzFKHzFKHzF 6 ,3 ,1 321 (a)Max.

Nyquist rate = 2×6K Hz = 12K Hz.

(b)

nn

nnn

nnn

nnn

F

nxnTxnx

s

5

22sin5

5

12cos13

5

12cos10

5

22sin5

5

12cos3

5

112cos10

5

212sin5

5

12cos3

5

62cos10

5

32sin5

5

12cos3

)()(

(c)

ttty 4000sin52000cos13)(

Alias

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Decomposition and Synthesis

Any signal can be decomposed into additive components, and the component signals can be added (synthesis) to produce the original signal.

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Decomposition-1

Impulse decomposition:

N samples signal is decomposed into N component signals each containing N samples.

Zero

A component signal contains one point from the original signal, with the remainder of the values being zero.

Step decomposition:

k-th component signal, xk[n], contains zeros for points through 0 to k-1, while the remaining points have a value equal to x[k] – x[k-1].

CEN543, Dr. Ghulam Muhammad King Saud University

Impulse decomposition Step decomposition

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Even/Odd decomposition:

2

][][][

2

][][][

nNxnxnx

nNxnxnx

o

e

Mirror image around x[N/2]. x[0] and x[N/2] are equal to original value. Others like: x[N/2+1] = x[N/2-1].

Even symmetry:

Odd symmetry:

Negative mirror image around x[N/2]. x[0] and x[N/2] are equal to zero.

Others like: x[N/2+1] = -x[N/2-1].

Odd positions are zero!

Even positions are zero!

Decomposition-2

You can check the result by using the following formula:

][][][ nxnxnx oe

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Fourier Decomposition

N point signal

(N/2)+1 cosine signals, each with N points

(N/2)+1 sine signals, each with N points

Zero!

Zero!

So, total N significant signals!

CEN543, Dr. Ghulam Muhammad King Saud University

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Example 5

Problem:

Answer:

A signal is defined as x[n]={2, 3, 5, -2, -3, -5, -2, 2}; Find the result of even / odd decomposition. Index starts from 0.

Even symmetry: xe = {2, 2.5, 1.5, -3.5, -3, -3.5, 1.5, 2.5}

Odd symmetry: xo = {0, 0.5, 3.5, 1.5, 0, -1.5, -3.5, -0.5}

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Sinusoid Drawing

44 ,4

2sin)(

tttx

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Mean and Standard Deviation

1

0

1 N

iix

NMean:

1

0

2)(1

1 N

iix

NStandard Deviation:

Mean = DC Value

S.D. = How the signal fluctuates around Mean (AC)

To reduce statistical noise for small number of samples

CEN543, Dr. Ghulam Muhammad King Saud University

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Histogram

More sample produces smoother histogram

1

0

M

ii

HN

M = Number of points in the histogram

i

M

i

M

ii

HiN

iHN

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0

1

0

)(1

1

1

Mean and S.D. using histogram:

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Histogram, pmf, pdf

Look at the values along y-axis

Discrete

Continuous

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Waveforms and pdfs

Look at the pdf of random noise.

It is Gaussian!

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Histogram Bins

The range 0-4 is divided by 600 in (b) and by 8 in (c).

Poor resolution in vertical axis: (b).

Poor resolution in horizontal axis: (c).

Using more samples makes better resolution.

Look at the vertical axis

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Normal Distribution (Gaussian)

2

)( xexy Basic shape: 2

2

2

)(

2

1)(

x

exPEquation for normal distribution:

: Mean

: Standard Deviation

The normalization term is to make the area under curve = 1

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Digital Noise Generation

Uniform distribution: one RND function (0~1)

Two times RND functions and add

Twelve times RND functions and add

Central Limit Theorem:

A sum of random numbers becomes normally distributed as more and more of the random numbers are added together.

Gaussian

For each sample:

S.D. desired : mean, desired:

)6(12

1~0

RND

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CEN543, Dr. Ghulam Muhammad King Saud University

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Figure Acknowledgement

Most of the figures are taken from the following books:

Li Tan, Digital Signal Processing, Fundamentals and Applications, Elsevier, 2008.

Steven W. Smith, Digital Signal Processing: A Practical Guide for Engineers and Scientists, Newnes, Elsevier, 2003.