Digital Signal Processing

DIGITAL SIGNAL PROCESSING

Prof. Dr. Ljubiša Stanković

University of Montenegro, Montenegro

Digital signal processingDigital signal processing

CHAPTERS:

1. Discrete-time signals and systems

2. Discrete Fourier transform

3. z-transform

4. Spectral estimation

5. Time-frequency analysis

6. Multidimensional signal processing

Discrete-Time signals and systems

Discrete signals

Discrete signal x[n] can be represented as a

sequence of real or complex numbers, where with

n is denoted n – th number.

Some important and very often used discrete

signals (sequence):

Unite impulse

Unite impulse is called also delta function or

simply impulse, and defined as:

Discrete-Time signals and systemsUnite step

A discrete time unite step function is defined to be:

Real exponential functions

A real discrete-time exponential function is given by:

Discrete-Time signals and systemsSinusoidal and complex exponential signals

Periodicity

A signal x[n] is said to be periodic with N if x[n] = x[n+N]. A complex exponential sequence is periodic only if 2π/ω0 is integer number or if it is rational number p/q.

0jw ne

Discrete-Time signals and systemsExamples

Solutions:a)

for k = 1 N = 3.


b)

for k =7 we have N = 6.

c)

From above equation we can conclude that sequence is not periodic.


It is important to note that an arbitrary sequence

can be written in the form:

Some important definition for a discrete signals

1. Magnitude of signal:

2. Energy of a signal:


Linear shift-invariant discrete systems

Discrete systems can be described by transform

which maps output sequence x[n] into output

equence y[n]:

where T is operator which denote transform of a

system.


Linear system

System is linear if:

Examples:Consider following systems and check their linearity:


Solutions:

Thus, the system is not linear.

Thus, the system is linear.


Shift invariance

A system is shift invariant if the characteristics of

system are not function of time, i.e. if a input signal

x(n) produce output signal is y(n), than input

signal x(n- N) will produce output signal y(n − N)

Examples

Consider systems:


Solutions:

Obviously, the system is shift-invariant.

The system is not shift-invariant.

Discrete-Time signals and systemsDescription of a linear shift-invariant systems through

a response of a systems on the unite impulse:

If a systems are shift-invariant than:

This is convolution sum and it can be written as:

It is very easy to show:


Causality and stability

System is causal if :

System is stabile if for we have

for any value n.

Proof:

From above equation follows:


Difference equations

The general form of linear difference equations

with constant coefficients is presented by:

For this equation there is exist family of solutions,

so it is necessary to set initial conditions for

uniquely solution. If we assume that system is

causal i.e. if x(n) = 0 for n < N than y(n) = 0 for n <N, we can write:


Example:

Consider difference equation in the form:

We will find impulse response if we take

and assume that system is causal


This is an example of a system with infinite impulse

response-IIR system.

If we take Aj = 0 for j > 0 than in our example we

have:

This system has finite impulse response-FIR system.


Fourier transform of a discrete signals

Consider complex exponential signal:

For this signal we can write:

With relation:

is defined Fourier transform of a discrete signal h(n).


Thus:

In general case:

Uniform convergence:

Having in mind:

We can conclude that Fourier transform of discrete signals is periodic with 2π.Since is periodic we can consider it as a Fourier’s series.


A inverse transform can be obtained:

We have:

Examples:

By using definition compute Fourier transform for:


Solutions:

1.

2.

In this case the amplitude characteristic will be:


Properties of the Fourier transform of discrete

signals

1. Linearity

2. Shifting in the time

3. Modulation


4. Convolution

By substitution m = n − k we obtain:


5. Product

If we interchange order of integration and

summation, we have:

Finally we have:


Sampling of continuous-time signals

According to Shannon theorem continuous signal

can be recovered from its discrete version if

discrete signal has been sampled by periodic

sampling T which satisfy condition:

T =1/(2fm)

where fm is maximal frequency of a signal.

Consider an analog signal xa(t) that has the

Fourier representation:


Assume that signal xa(t) has limited bandwidth,

i.e.:

Consider now a periodic form Xp(jωa) of the Xa(jωa)

that has a period 2 ω0. The Fourier transform

Xa(jωa) can be recovered from Xp(jωa) if ω0 ≥ ωa.

Since Xp(jωa) is periodic, it can be expanded in the

Taylor series:


where is T =π/ ω0. The Fourier’s coefficient can be

obtained from:

If we compare last equation with equations that give xa(t), we can conclude:

Finally, we have that samples of signal xa(t) and

Xp(jωa) form a Fourier pair.

Now, we have:



Examples:

1. If output signal is:

and impulse response of a system is:

Compute signal on the output of the system.

Solution:

On the output of the system amplitude and phase

of the input signal will be changed.


Namely:


2. If:

find signal h(n).

Solution:


3. Find the Fourier transform of the signal

h(t) = etu(t) and draw its amplitude characteristic.

Write a discrete form of the signal and draw its

amplitude characteristics for the cases T = 1 and

T =0. 2.

Solution:

Thus, we have

By discretization we obtain:

and its Fourier transform:


4. Compute the sum:

Consider the sequence with the Fourier transform:

From above equation follows:

Discrete Fourier Transform

Definition of the discrete Fourier transform

We have seen that the Fourier transform of a

discrete signal is continual and periodic function

with period 2π. If we want to use this transform in

digital signal processing, we need its discrete

version, i.e. we have to sample it in frequency

domain.

Consider the Fourier transform of a signal x(n). Assume that X(ejω) is sampled with rate Δω = 2π/N,

where N is number of samples along the period.


Since samples of signal and its Fourier transform

are the transformation’s pair, we have that

sampling in the frequency domain cause periodical

signal in the time domain and vice versa.

Thus, for the discrete Fourier transform we have

the periodic signal xp(n) obtained from x(n).

For the sampling rate in frequency domain

Δ ω (Δ ωa =Δ ω /T)

we have periodic series in the time domain with the period tp = NpT:

From the above equation we have:Np = N

It means the following: If we want that the

periodic signal contains the original signal x(n),

discretization must be done with the same (or

greater) number of samples as a duration of the signal x(n).


Definition of the discrete Fourier transform:

where it is assumed:

Xp(k) is the discrete Fourier transform.


If we use notation:

the discrete Fourier transform can be written in theform:

Or matrix form:

We see that the computation of the Discrete Fourier Transform require approximately N2

multiplications and additions.


Inverse discrete Fourier transform

Inverse form of the discrete Fourier transform can

be obtained by multiplication of DFT definition by

WN−km. Thus we have:

Taking:

Discrete Fourier TransformFrom the above equations follows the definition of

the inverse Fourier transform:

If the duration of the signal x(n) is smaller than

N we have:


i.e.

where w(n) denotes the window function defined

by:


Examples

1. Find the discrete Fourier transform of the sequence:

Solution:From the equation for x(n) we see that the duration of x(n) is Np = 5. Thus we have touse N ≥ Np =5. Taking N = 5 we have:

The function Xp(k) is periodic with N = 5 as well.


2. If we have the periodic sequence Xp(n) with

period Np = N, and Xp(k) is its DTF, find the DFT for

the same sequence, taking Np = 2N.

Solution

for Np =2N we have:

the above equation can be written in the form:


because x(n) = x(n + N) and W2NkN =(−1)k.

Now we can write relation between Xp(k) and

Xp ′ (k):

Xp ′ (k) = 2Xp(k) for k even, and Xp ′ (k) = 0 for k odd.


Relationship between frequency and k − th number in the discrete Fourier transform

If the signal x(n) is obtained by sampling of the

analog signal xa(t ), than the frequency of the

discrete signal ω can be represented through the

analog frequency ωa, by:

By discretization of the Fourier transform we have:

From the previous two equations we have:


Note that this equation holds only for k ≤ N/2 − 1.

Frequencies between N/2 − 1 and N are mapped

negative frequencies:


Zero padding

Number of samples of the discrete Fourier

transform N in frequency domain depends on the

number of samples of a signal in the time domain.

If we want to get more samples within the basic

period of the Fourier transform (interpolation), by

using the discrete Fourier transform, then it is necessary to take more samples as a signal period.


It can be easy obtained by adding zero values at

end of the signal sequence.

Number of zero values depends from our will, i.e.,

on how many samples we want to have in the

discrete Fourier transform.

This procedure can be understood as an

interpolation of the discrete Fourier transform.


Some properties of the discrete Fourier transform -Convolution of periodic signals

Consider following properties of the discrete

Fourier transform:

1. If :

Then:

We have the sequence:


By substitution n − m = l follows:

It can be shown that:


Convolution of periodic versions of signals

Consider signals xp1(n) and xp2(n) which are periodic versions of signals x1(n) and x2(n).

In order to derive period for this convolution consider the following example.

Illustrative example

In this example we will consider signals: x1(n) = u(n) − u(n − 5), and x2(n) = u(n) − u(n − 5).

Discrete Fourier TransformSince signals have period N = 5, then for the

DFT calculation, we can form periodicals versions

with N ≥ 5. In this example assume N = 7.


From the Figure it is easy to obtain

xp3(n) = xp1(n) ∗ xp2(n):

From the Figures we see that results are

different for the original signals and their periodic

versions.

The reason is in overlapping of fictive periods.

Discrete Fourier TransformIf we want to avoid this effect we have to

introduce enough number of zero values.

Namely, if either sequence has duration N then

the period has to be 2N-1, or in general if duration

of the first one is N and duration of the second one

is M, then the period has to be M N−1.


When the duration of input signal is

significantly different from the duration of

sequence of impulse response (duration of the

impulse response is significantly shorter), we can

decompose the input sequence into few

subsequence, i.e.



Thus output is obtained as:

Here we must be careful since before every

convolution we must add L − 1 zero values (it is

assumed that L duration of a subsequence).

Discrete Fourier TransformFast Fourier transform – FFT

Algorithm called the Fast Fourier transform or FFT

algorithm plays very important role in digital signal

processing.

This algorithm is an interesting research topic.

That is the reason why exist various forms of this

algorithm.

By using the FFT time needed for computation

of the discrete Fourier transform can significantly

be reduced comparing time for computation of the

discrete Fourier transform by definition.


In our consideration we will present an

approach that belongs to the group of algorithms

called Decimations-in-Frequency.

The aim of this decimation is to decompose

Xp(k) Into subsequences, then further,

subsequences into subsequences etc.

For this algorithm it is necessary that the

number of samples is of the form:


Now decompose sequence Xp(k) into two

sequences:


From the previous equation we have:

Having in mind:

and that summations in both terms are from 0 to

N/2 − 1, we can write:


If we separate the previous equation for k = 2r and

k = 2r + 1, we get:

Where:

and


Since:

we have:

We see that the resulting transform is in the form of two transforms with N/2 terms.


Thus, one discrete Fourier transform with N terms is decomposed into two discrete Fourier transform with N/2 terms.

We have concluded that for the calculation of the DFT with N elements, by definition, we need approximately N2 operations. For two DFTs of N/2 elements we need 2(N/2)2 = N2/2 calculations.


This procedure can be continued in n steps, we

have the elements as a simple multiplication and

summation.

Consider an example with N = 23.

Illustration

Finally, we can conclude that:

N log2N

is number of necessary multiplications and

summations, as well.


It is interesting to find ratio N2/(N log2N) which illustrates efficiency of the FFT algorithm in comparison with calculation of the discrete Fourier transform by definition.

For example, for N = 512, if we need 1 minute to compute the discrete Fourier transform on a computer, by using the FFT it will be calculated for only 1 second.


Examples

1. Find the discrete Fourier transform of the following sequences:

a) x(0) = −1, x(1) = 1, x(2) = −1

b) x(n) = an (u(n) − u(n − N))

Solutions:

a) Taking N = 3 we can write:


b) Taking period N we have:

Taking for example a =1, follows:


2. Find relationship between Xp(k) and Xp(N − k) in

the case of real sequences xp(n).

Solution

Since:

We can conclude that in the case of real

sequence xp(n) we have:


3. If g(n) and f(n) are real sequences, show that

their discrete Fourier transforms (G(k) and

F(k)) can be obtained from the discrete Fourier

transform Y(k) of the sequence y(n) = g(n)+ jf(n).

Solution

From the signal y(n) = g(n) + jf(n) we can write:

The discrete Fourier transform of y(n) is:


Conjugate complex value of the previous

equation gives:

From the above equation the discrete Fourier

transform of the signal y ∗(n) follows:


4. The relationship between k − th number in the

discrete Fourier transform and analog value of the

frequency is given by:

If we want to avoid shifting of the discrete Fourier

transform for k = N/2 − 1 we can multiply input

signal x(n) by (−1).

Proof:

The discrete Fourier transform of the (−1)x(n)

is:


for k ≤ N/2 − 1 we have:

Thus we have:

For the case k > N/2 − 1 follows:

Thus we have

Illustration additionally can show results of this transformation.

Z Z –– TransformTransform

The z-transform can be understood as a

generalization of the Fourier transform.

Applications of this transform are mainly for

description and realization of systems.

Definition of the z-transform

Z-transform of the signal x(n) is defined as:

where z is complex.


X(z) is defined for z where previous sum converges.

The region of convergence of the z-transform is

defined by two annular ring with r1 and r2 which

contain the poles of the function X(z).

The values r1 and r2 depend from the behavior

of the signal x(n) in the cases when n tends plus

infinity and minus infinity.


Example 1

Find the z-transform of the signal x(n)=u(n)

Solution:

According to definition we have:

We know that previous sum converge for

|z−1 | <1 i.e. |z| > 1.


Thus the region of convergence is exterior (to the

pole location z = 1) of the unit circle |z| = 1. The

poles are denoted by ”x ”, while the zeros by ”o ”.


Example 2

Find the z-transform of the signal x(n) = −u(−n − 1).

Solution:

According to definition we have:

where the region of convergence is defined by

|z| < 1.


From the previous two examples we can conclude

that either have the same X(z).

Thus we can conclude that by using z-transform a

signal is not uniquely determined. However if we

have also the region of convergence uniquely will

be satisfied.

Consider now, four important sequence and find

their z-transform.

1. Causal series x(n) = 0 for n < 0

The z-transform of this signal is:


We see that z → ∞ belong to the region of

convergence.

Thus we can conclude that region of convergence

will be annular ring exterior to the pole location

with the longest distance R from origin, so we

have: R < |z| < ∞.


2. Non causal series x(n) = 0 for n > 0.

We se that sum converge for z = 0. The region

of convergence is the disk centered at the origin

and interior to the pole location R. Where R is the

pole the nearest to the origin, 0 ≤ |z| < R.


3. Sum of the causal and anticausal series

For this case we have:

From the previous considerations we have

concluded that first series converge for:

0 ≤ |z| < R1

and second one for:

R2 < |z| < ∞


The resultant region of convergence is:

R2 < |z| < R1

This is the annular ring. If R2 > R1 than the region of convergence is ⊘.


Example Find the z-transform and the region of convergence for the series:

X(n) = an u(n) − bn u(−n − 1).

Solution

The first sum converge for |z| >a, while the second one for |z| < b. thus the region ofconvergence is:

a < |z| < b


4. Finite length sequences x(n) = 0 for n ≤ n1 and

n ≥ n2

We conclude that sum converge for any z except 0

and/or ∞ what depends from conditions are

n1 and n2 positive or negative numbers.

Z Z –– TransformTransformInverse z-transform

The inverse z-transform is defined by:

If we multiply right and left side of the previous

equations by: zk−1 and if we perform integration

along restricted closed path C which resides

within the region of convergence, we obtain:


Since integral on the right side of the equation is

different from zero only for k = n, we

have:

This is the general form for determination of the

inverse z-transform. The previous integral can

be calculated by using the theorem of residuum:

The residuum of the function F(z) in the pole

z = z0, that is pole of order k, can be calculated

with:


The inverse z-transform will be calculated on the

base of expansion of X(z) in the series with

respect z−1. In that case we write X(z) in the form:

Than by comparing the previous equation and

definition of the inverse z-transform we see that

X(n) = Xn.


Example 1. Find the inverse z-transform for:

Solution:Expanding the previous equations into series for |z| > 1/4 we have:

Thus we can conclude:

X(n) = (1/4)nu(n)


In the case when the region of convergence is

|z| < 1/4 coefficients of series must be less

than 1, so X(z) has to be transformed in the form:

Thus,


Example 2.

Find x(n) if

Solution

Consider first:

thus we have:

x(n) = an u(n)


If we find differential of X1(z), we obtain:

Now we have:

x(n) = an-1 u(n)


Table of the z-transform


Properties of the z-transform

Derivations of the properties of the z-transform are

analogy with the properties of the Fourier

transform.

1. Linearity

If we have y(n) = ax(n) + bh(n) than

Y(z) = aX(z) + bY(z).

2. Shifting in the time domain

For the signal x(n − n0), we have:


Example

Consider the difference equation

x(n − 1) − 2x(n − 2) = y(n)+ y(n + 1)

and represent its in z domain.

Solution:


3. Multiplication by complex exponential sequence

4. Convolution

If we have y(n) = x(n) ∗ h(n) than follows:


Example

By using z-transform find convolution of the

signals:

x(n) = u(n) and h(n) = (1/3)n u(n)

Solution

By using property 3, we obtain:


Now, we have:

The region of convergence is |z| >1. y(n) will

be obtain by using inverse z-transform. First we

will write previous equation in the following form:

where B = 3/2 and C = −1/2.

Thus we can write:


Relationship between z-transform, Fourier

transform and discrete Fourier transform

If we compare definition of the Fourier transform

of discrete signals and z-transform definition we

see that Fourier transform is equal to the z-

transform for |z| = 1. Thus the values of the z-

transform on the |z| = 1 in z domain are the values

of the Fourier transform of the sequence.

By expressing the complex variable z in polar form

as z = re j ω , we obtain:


taking r = 1 follows:

In general case the z-transform on the circuits

defined by r is equal to the Fourier transform of

the sequence x(n) multiplied by r−n.This is

reason why the z-transform exist in the some cases

when the Fourier transform does not exist.

Z Z –– TransformTransformOne example that confirm the previous

statement is x(n) = u(n).

The Fourier transform of this sequence does not

converge, but the z-transform converge for r > 1.

From the previous considerations we know that

the values of discrete Fourier transform are the

samples of the Fourier transform of discrete

signals.

This means that the values of

discrete Fourier transform are equal to the samples

of the z-transform for |z| = 1.


Example

Find the z-transform of the sequence:

x(n) = u(n) − u(n − 4)

and in the case N = 8 find the Fourier transform

and the discrete Fourier transform by using result

obtained for z-transform.

Solution:


System function

Consider a system where is:

y(n) = x(n) ∗ h(n)

Having in mind properties of the z-transformfollows:

Y(z) = X(z)H(z)


The z-transform of the impulse response is referred

to as the system function.

The system function evaluated on the unit circle

(|z| = 1) is the frequency impulse response of the

system.

From the previous considerations we know that

stable system must satisfied condition:

Z Z –– TransformTransformConsider now the z-transform of the h(n):

From the previous equation follows that in the

case of stabile systems unit circle |z| = 1

must belong to the region of convergence of the

function H(z).

For causal system the region convergence must be

exterior of a circle passing through the pole of

H(z) that is farthest from the origin.


Example

Check the causality of the system:

Solution

We see that the region of convergence is |z| > 1/2.

Thus the system is causal.


Consider now the system described by a linear

Constant – coefficient difference equation, i.e.

the system that satisfy the general N − th order

difference equation:

Applying the z-transform to each side of

previous equation, we have:


where the property of the z-transform:

is used.

Now, we can write:

In the case where Aj = 0 for j > 0, the system

with finite impulse response is obtained (FIR) and

in that case we have:


Example

Find the impulse response of the causal system

described by:

and check its stability.

Solution:


Poles of this function are: z−1 = 1 and z−1 = 1/4.

Since the system is causal the region of

convergence is |z| > 4. This means that the

system is not stabile (unit circle does not belong to

the region of convergence).

In order to determinate h(n), write H(z) in the form:


Examples:

1. Find the z-transform of the sequence

x(n) =δ(n − 5).

Solution:

2. If X(z) is the z-transform of x(n), find the z-transform of:


By substitution n + k = m, we have:

Y(z) = X(z)X(1/z)

3. Find the impulse response of the system with z-transform:

Solution:

Having in mind expansion in the series:

we can write:


Thus,

4. Find the causal sequence x(n), if its z-transform has the form:

Solution:

Write X(z) in the form:


Thus we have:

x(n) =[ 1.25 − 0.25(0.2)n]u(n)


5. For the system shown in Figure, find system

function, check stability, and determine response

on the signal

x(n) = δ(n) − 2 δ(n − 1).

Solution:

From the Figure we have:


The system function is:

The pole of this system is z = 2, this fact means if

the system is causal it is not stabile.

If x(n) = δ(n) − 2 δ(n − 1) than:

ESTIMATION THEORYIntroduction to random signals

At the beginning, we will give some important

definitions:

Mean of the process is defined as:

The operator of mean value E is linear, i.e.:

If the random variables are independent or

uncorrelated then:

ESTIMATION THEORYA sufficient condition for independence is:

In this case the random variables are statistically

independent.

Mean square value of x(n) is:

ESTIMATION THEORYCorrelations and covariances

The autocorrelation is defined as:

or,

where ∗ denotes complex conjugation.

The cross-correlation of two random processes x(n)

and y(n) is defined as:

ESTIMATION THEORYThe autocovariance is defined as:

If n = m the variance is obtained:

In the case of stationary process, the variance is

independent of time and denoted as

In the case of random processes that are

stationary in the wide sense we have:

ESTIMATION THEORYIn this case autocorrelation depends only on the

time difference m − n, thus:

Also we have:

ESTIMATION THEORYWhite noise

The signal that has the autocorrelation in the form:

is called white noise.

The name comes from the fact that the Fourier

transform of this is constant

It means that power density spectrum of this

function is constant, what is the property of the

white light.

In the case of real noise w we have:

ESTIMATION THEORYPower density spectrum

Consider the z-transform of the autocorrelation

rzz(n) (in the case of stationary signals):

Define now Sxx(ω) as values of the z-transform

on the unit circle:

Note that Sxx(ω) is a real-valued function, since

ESTIMATION THEORYFrom the above equation wecan write:

Having in mind the definition of rxx(n), we have:

Thus, the expected signal power is equal to the integral of Sxx(ω) . This is the reason why Sxx(ω) is called power spectral density. Later, it will be shown that the signal energy within the frequency region [ω1, ω2] is equal to the integral of Sxx(ω) from ω1 to ω2.

ESTIMATION THEORYLinear systems and random signals

For a linear system we know that:

If the signal x(n) is stationary, i.e. E{x(n − k)}= MIx,

we can write:

ESTIMATION THEORY

The auto-correlation of the output signal is

defined as:

In the case of stationary signal when

rxx(n − i,m − k)= rxx(n − m k − i), we have:

We can conclude,if the signal x(n) is stationary

in the wide sense, then the signal at the output of

the linear system is stationary in the wide sense,as

well.

ESTIMATION THEORYFind, now, the z-transform of the ryy(n,m).

We have:

By substitution l=p − k + i, we obtain:

If h(n) is real:

ESTIMATION THEORYPower spectral density is:

Therefore if |H(e j)|2 is an ideal band-pass filter for the interval [ω1, ω2] then the expectedpower of the output signal is:

since Sxx(ω1) could be considered as a constant within [ω1, ω2] for small ω2 − ω1 . This provesthat Sxx(ω1) is the spectral power density.

ESTIMATION THEORYOptimal filtering

Consider the signal in the form:

x(n) = s(n) + w(n)

where s(n) is desired signal and w(n) is the noise.

If we assume that the signal and noise are

uncorrelated we can write:

ESTIMATION THEORY

For the case when the power spectral density of the signal and noise are not overlapped, wecan easily obtain denoised signal..

Namely, passing the signal through the bandpass Filter which passes only the frequency components of Sss(ω), denoised signal is obtained.

However, if the noise is white (existing in the whole frequency range) by using previous method it is possible to obtain only partial denoised signal.

In the general case the problem is in determination of d(n) = s(n + m) in the most accurateway.

ESTIMATION THEORY

If m = 0 we have the case of optimal filtering.

In some cases, it is necessary to predict the values

of the signal in the future, then m > 0.

However, in some cases we need to determine

some previous value of the signal. In this case we

have optimal smoothing, and m < 0.

Processing by using IIR system

Consider an IIR system defined by:

ESTIMATION THEORY

The mean square error is:

From this we have:

Define, now, correlation functions:

From the above equations we have:

If the signal and noise are uncorrelated, we have:

ESTIMATION THEORY

Fourier domain form of the optimal filter, when

d(n) ≡ s(n), is:

ESTIMATION THEORY

Power spectrum estimation

The mean value of the n − th sample of the

sequence x(n) can be estimated by:

where xi (n) is the n − th sample in the i − th measurement.

The special class of the random processes are

ergodic processes. In this case probability

averages are equal to time average, i.e.

ESTIMATION THEORY

The process is ergodic if we can estimate its

statistically properties on the base of only one

random signal.

In previous equation μx is random value

because we have finite number of samples 2N + 1.

In the case N → ∞ the value of μx will be

sufficiently accurate.

Estimate, now, autocorrelation function (which

is the mean value of the product

x(n + m)x ∗(n)):

ESTIMATION THEORY

In the case of stationary process we have:

If we know x(n) only within the interval

−N ≤ n ≤ N, then x(n + m) will be known only for

n ≤ N − m for positive number m, and

n ≥ −N |m| for negative number m.

If we want to avoid calculations for positive and

negative value of m, we will introduce symmetric

product:

ESTIMATION THEORY

In this sum we have 2N + 1 − |m| terms, but we average it with 2N + 1. This is the reason whywe have systematic error that can be avoided by:

Thus is the biased estimate of the autocorrelation

ESTIMATION THEORY

Definition and variance of the Periodogram

Define the Fourier transform of the biased

autocorrelation function:

Since

it can be shown that:

ESTIMATION THEORYThe spectrum estimate IN (ω) is called the

periodogram. The expected value of the

periodogram is:

Taking

ESTIMATION THEORYThus, the periodogram is a biased estimate of

the power spectrum Sxx (ω). The previous

equation can be written, using convolution, in the

form:

where WB is the Fourier transform of the so called

Bartlett window ( for

|m| < 2N − 1) given by:

ESTIMATION THEORYVariance of the Periodogram

Express the periodogram in the form:

The covariance at frequencies ω1 and ω2 of IN (ω) is:

ESTIMATION THEORYIn the case of white Gaussian process we have:

Thus,

ESTIMATION THEORYTherefore, we have:

The variance is:

ESTIMATION THEORYSmoothed spectrum estimators

If the sequence x(n), 0 ≤ n ≤ N − 1, is divided

into K segments of M samples, the periodogram

will be:

If we assume that periodograms are independent

of one another, we have

ESTIMATION THEORY

By assumption that K periodograms are

statistically independent, then Bxx(ω) is the

mean of the set of K independent observations of

the periodogram IM(ω) :

From previous equation it is clear that as K becomes large, the variance approaches zero, so

this smoothed estimate is a consistent estimate.

ESTIMATION THEORYEFFECTS OF FINITE REGISTER LENGTH

Signal values in digital signal processing are stored in a binary format, using registers witha finite length. This can cause the error.

Namely, if we have number with b bits multiplied by another one with b bits, the result will be data with 2b bits. If the length of register is less than2b we will have truncation error. This error is:

where Q[x] and x are numbers after and before thetruncation.

ESTIMATION THEORYIf we consider, now, effects of quantizations of

analog signal, we know:

Every samples must be represent by finite length number, so we will have truncation orrounding to the nearest quantization level and it will cause quantization error. This error can beexpressed by noise e(n), than we have:

where x(n) is exact value and e(n) quantization error.

ESTIMATION THEORYIn the case of rounding the errors is in the range:

−Δ/2 ≤ e(n) ≤ Δ/2

while in the case of truncation it is:

−Δ ≤ e(n) ≤ 0,

where Δ is quantization width Δ = 2−b.

If we want to give a model to describe the effects

of quantization we will assume:

1. The sequences of error samples {e(n)} is a sample sequence of stationary random process.

2. The error sequence is uncorrelated with the sequence of exact samples {x(n)} .

ESTIMATION THEORY

3. The error is a white-noise process.

4. The probability distribution of the error process

is uniform over the range of quantization

error.

Find signal to noise ratio in the case of rounding.

According assumption 4, we have that probability

distribution pen(e) = 1/Δ.

Thus:

ESTIMATION THEORY

Now we have:

Multidimensional discrete signals and systems

Discrete N-dimensional signal can be defined as:

where n1, n2,.....,nN are integers.


By analogy with the one-dimensional case we can

define:

1. Unite impulse:

2. Unite step:


3. Complex exponential series

Discrete multidimensional system can be defined by:

with x(n) and y(n) are defined input and output signal, respectively.


System is linear if:

If we denote multidimensional unite impulse

response with:

Where . The previous equation is

N-dimensional convolution denoted by:


Causality and stability are defined in full analogy

with the one-dimensional case.

Fourier transform of N-dimensional discrete signals

The Fourier transform of an N-dimensional

discrete signal is defined by:

The inverse Fourier transform is given by:


In the case of two-dimensional signal we have:

Example:

Find the Fourier transform of the signal:


Solution:

Also, by analogy with the one dimensional

sampling theorem, it is easy to show that:


where it has been assumed:


Multidimensional discrete Fourier transform and FFT algorithms

Consider two-dimensional discrete Fourier transform: the simplest 2D FFT algorithm arebased on the FFT algorithm for one-dimensional case. Namely:

We see that for a fixed value n1, the second sum presents one-dimensional discrete Fouriertransform which can be calculated by using some of the FFT algorithms.


Thus:

This procedure should be repeated for all n1.

Two-dimensional discrete Fourier transform

will be obtained as:

Calculations should be performed for all k2.


Ratio of number of additions and summations

for discrete Fourier transform by using

definition and FFT algorithm is given by:

In the case of M = 128 this ratio is 1170 ( if need

one second with the FFT than 19,5 minutes would

be needed by using calculation based on the

definition).


Radon transform and computer’s Tomography

Integral along line AB is:

where AB is defined by:

The previous integral can be written in the form:


The previous integral can be written in the form:

Previous integral defines projection of function

f(x, y) with respect to variable t for an arbitrary

angle θ.

Is it possible to reconstruct function f(x, y) on

the base projections?

Answer is yes.


Proof:

Consider the Fourier transform F(u, v) of the

function f(x, y):


The Fourier transform of a projection is:

Consider as a special case the value of the F(u, v),

along the line v = 0, then we have:


Thus, we have obtained that the Fourier

transform of the function f(x, y) along axis v = 0

is equal to the Fourier transform of projection for

the angle θ = 0.

This result can be generalized. It can be shown

that the Fourier transform of f(x, y) along an

arbitrary line defined by angle θ with respect to u axis is equal to the Fourier transform of the

projection defined by angle θ with respect to x axis.


The previous claim can be proved. Denote with

f(s, t) the function f(x, y) rotated in the coordinate

system. Relationship between variables

(x, y) and (s, t) is:

Since:

The Fourier transform of the projection is:


In x, y coordinate system we obtain:

Thus, the previous claim is proved.


Finally, we can conclude:

Function f(x, y) can be obtained on the following

way:

1. Find the projection for 0 ≤ θ ≤ π.

2. Determine the Fourier transforms of the

projections which give the Fourier transform of

the function f(x, y).

3. Compute the inverse Fourier transform and it if function f(x, y).


Note that the Fourier transform of function

f(x,y) will be obtained in polar raster. If we

want to use FFT algorithms it is necessary to have

the Fourier transform in rectangular raster.

One possible solution is interpolation of values

from the polar to the values on the rectangular

raster.

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