Ece 513 Chap 1 Slides

7/27/2019 Ece 513 Chap 1 Slides

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FUNDAMENTAL DSP CONCEPTS

C. Williams & W. Alexander

North Carolina State University, Raleigh, NC (USA)

ECE 513, Fall 2013

C. Williams & W. Alexander (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2013 1 / 155


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Outline

1 Introduction2 Digital Signal Processing Applications3 Why Digital Signal Processing4 Representation of DiscreteTime Signals5 Operations on Sequences6 Normalized Frequency Representation7 The ZTransform

8 Region of Convergence9 Frequency Representation of DiscreteTime Systems10 Difference Equation Representation11 The Frequency Response12 PoleZero Plots

13 System Response14 Frequency Shifting15 Inverse ZTransform for Systems with Complex Poles16 Inverse ZTransform for Systems with Multiple Poles17 Cascade Implementation of Digital Filters18 Stabilization of an Unstable FilterC. Williams & W. Alexander (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2013 2 / 155


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Introduction

Signals play an important role in many activities in our daily lives.Examples include:

SpeechMusicBiomedical signalsVideoDigital Television



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Introduction

A deterministic signalis a function of an independent variable such as time, distance,position, temperature, and pressure. Itcan be uniquely determined by

a well-dened process such as a mathematical expression of one or

more independent variables,or by table look up.

For example,

s (t ) = 3sin (2 .1 t + 0 .3198 ) u (t ) (1)

is a deterministic signal with independent variable t .



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Introduction

A speech signal can not be described functionally by amathematical expression.However, a recorded segment of speech can be represented to ahigh degree of accuracy as the sum of several sinusoids ofdifferent amplitudes and frequencies such as [1]

s (t ) =N

k = 1

Ak (t ) sin [2 F k (t )t + k (t )] (2)

A signal that is determined in a random way and can not bepredicted ahead of time is a random signal.Statistical approaches are often used to analyze random signals.



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DSP Applications

Digital signal processing is heavily used in information technology.Information technology includes such diverse subjects as

digital signal processing,image processing,multimedia applications,computational engineering,visualization of data,database management,teleconferencing,remote operation of robots,

autonomous vehicles,computer networks, etc.



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Why DSP?

Many applications involving continuoustime signals use digitalsignal processing.This often involves

1 sampling the continuoustime signal at regular intervals,2 quantizing the samples to obtain a digital sequence,3 processing the digital system using a computer or a digital system,4 converting the output digital sequence to a continuoustime

system.



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Digital signal processing of a continuoustime signal using thisapproach may be preferable over processing the signal directly inthe continuoustime domain for many of the following reasons:

1 Programmability

2 Accuracy3 Data storage without degradation4 Multiplexing of signals5 Easy to process low frequency signals6 Cost is generally low due to volume production7 It is easy to use encryption to provide security with digital signals.



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Why DSP?

Disadvantages associated with digital signal processing:1 A digital signal processing system is often more complicated than a

corresponding analog signal processing system.2

The upper frequency for digital systems is determined by thesampling frequency.3 Digital systems are constructed using active circuits that consume

power. Analog systems, that consume less power, can be designedusing only passive circuits.



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Representation of DiscreteTime Signals I

A discretetime signal x (n ) is a function of an independentvariable that can only take on discrete values.We can represent a discretetime signal using:

1 A table as given in Table 1,

n 0 1 2 3 4 5 6x(n) -3 -1 -2 5 0 4 -1

Table: A tabular representation of a discretetime function.

2 A functional representation such as

x (n ) =0 for n < 0(0 .2)n for 0 n 100 for n > 10

(3)


f l


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Representation of DiscreteTime Signals II

3 A sequence representation such as

x (n ) = 3, 1, 1 , 5,

0 , 4 , 1 (4)


O i S I


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Operations on Sequences I

We often desire to operate on an input sequence x (n ) using a set

of prescribed rules to obtain an output sequence.The basic operations for this course include

1 Modulation - y (n ) = h (n )x (n ) (point by point multiplication of twosequences) as shown in Figure 1.

x(n)

Figure: Modulation.


O ti S II


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Operations on Sequences II

2 Addition - y (n ) = h (n ) + x (n ) (point by point addition of twosequences) as shown in Figure 2.

x(n)

Figure: Addition.


O ti S III


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Operations on Sequences III

3 Multiplication - y (n ) = ax (n ) (multiplication of each point by aconstant) as shown in Figure 3.

x(n)

Figure: Multiplication.

4 Delay - y (n ) = x (n 1) as shown in Figure 4.

Z-1

x(n) x(n-1)

Figure: Delay.


Operations on Sequences IV


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Operations on Sequences IV

Division is an important operation for many practical DSPoperations.

Division is a more complicated operation than multiplication oraddition, etc.

Therefore, division for DSP operations is typically implemented by1 Replacing division by a constant with multiplication by the

reciprocal of the constant,2 Dividing by a power of two by shifting twos complement numbers to

the right or to the left as appropriate.


Normalized Frequency Representation


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When we sample a continuoustime signal:1 We obtain a sequence that can be represented by replacing the

independent variable t by nT in the case where t is the

independent variable.2 This results in a normalized representation of the frequencycontent.

3 We can replace the continuoustime radial frequency by thediscretetime frequency = T .




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Example (1.1)

Consider the signalx (t ) = Ae j t u (t ) (5)

After sampling, we have

x (n ) = Ae j nT = Ae j n (6)

Thus, we see that the normalized discretetime frequency isobtained by multiplying the continuoustime frequency by the

sampling interval T . = T (7)



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Example (1.1)

We know from the Sampling Theorem that the largest possiblesampling interval to avoid aliasing is

T max =

N (8)

Thus, the normalized discretetime Nyquist frequency is always

N = N T max =N N

= (9)

We will be primarily concerned with the normalized discretetimefrequencies (10)

in this course since we will be studying discretetime signals.



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The Z Transform for a discretetime sequence x (n ) is dened as

the power series [1]

X (z ) =

n = x (n )z n (11)

In most of our examples, the sequence of interest begins at somedesignated sample which we arbitrarily assign the index 0.In this case, the sequence does not exist prior to n = 0 and

X (z ) =

n = 0x (n )z

n (12)

This is called the right sided ZTransform .


Region of Convergence


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The ZTransform of a discretetime sequence only exists forthose values of z for which the innite power series converges.The region of convergence (ROC) for the Z Transform is the set ofall values of z for which this innite power series converges.

An innite power series may have more than one ROC and thediscretetime sequence associated with each of these ROCs maybe different.Thus, Z Transform for a specic discretetime sequence must alsoinclude the ROC.We will use an example to illustrate this concept [1].




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Example (1.2)

Consider the sequence

x (n ) = a n , n 0

b n

, n 1(13)

Using the denition, the Z Transform for this sequence is given by

X (z ) =1

n = ( b n )z n +

n = 0

a n z n (14)




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g g

Example (1.2)

We can obtain a closed form expression for X (z ) since it is ageometric series. Let

S 1 =1

n =

( b n )z n (15)

and

S 2 =

n = 0

a n z n (16)

Then

S 1 =

m = 1

b m z m =

m = 1(b 1z )m (17)




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g g

Example (1.2)

A geometric series in the form

Y =N

n = 0

c n (18)

can be written in closed form as

Y =1 c N + 1

1 c (19)

where c may be complex as well as real.



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Example (1.2)

Ifb 1z < 1 (|z | < |b |) (22)

thenlim

M (b 1z )M + 1 = 0 (23)

andS 1 =

11 (b 1z )

1 (24)

Simplifying this result, we obtain

S 1 =z

z b (25)




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Example (1.2)

We can obtain a closed form expression for S 2 in a similar way.

S 2 = limN

1 (az 1 )N + 1

1 (az 1)(26)

Ifaz 1 < 1 (|a | < |z |) (27)

thenlim

N

(az 1)N + 1 = 0 (28)

andS 2 =

11 (az 1 )

(29)



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Example (1.2)

On the other hand, if |b | < |a |, we have chosen the wrong way topartition X (z ). The two sequences S

1and S

2have no common

region of convergence.Figure 5 shows the region of convergence (in gray) for S 1 withb = 3.



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Example (1.2)

Figure 6 shows the region of convergence (in gray) for a = 0 .5.

Figure: Region of convergence for S 2 with a = 0.5.



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Region of convergence


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We can obtain the discretetime sequence corresponding to a ZTransform by evaluating it within the ROC.We will use an example to illustrate this concept.




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Example (1.3)

This example involves nding a bounded sequence with thefollowing ZTransform

X (z ) =0 .81 z

z + 0 .62+

0 .5815 z z + 1 .6129

(34)

Note that X (z ) has been expanded into partial fractions.Let

X 1(z ) =0 .81 z

z + 0 .62(35)

andX 2(z ) =

0 .5815 z z + 1 .6129

(36)

such thatX (z ) = X 1(z ) + X 2(z ) (37)




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Example (1.3)

We will nd the appropriate sequence for each term and then addthe two sequences.

X 1(z ) =0 .81

1 ( 0 .62 z 1)(38)

If 0 .62 z 1 < 1 (39)

so thatlim

N ( 0 .62 z 1)N + 1 = 0 (40)

then

X 1(z ) = 0 .81

n = 0

( 0 .62 )n z n (41)



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Example (1.3)

It follows that

x (n ) = 0 .5815 ( 1 .6129 )n for n 10 .81 ( 0 .62 )n for 0 n (45)

Figure 8 gives a stem plot of x (n ) for n = 50 to n = 50.




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Example (1.3)

50 40 30 20 10 0 10 20 30 40 500.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

Stem plot of x(n) for n = 50 to n = 50

sample number

m a g n

i t u d e

Figure: Stem plot of x(n) for n = - 50 to n = 50.


Frequency Representation


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We will consider the frequency representation of a discretetimesequence.We rst consider an aperiodic sequence and we will later considerthe frequency representation of periodic and nite durationsequences.The two-sided ZTransform is given by

X (z ) =

n = x (n )z n (46)

The two-sided ZTransform can represent an aperiodic sequenceover the range of n .The region of convergence must include the unit circle in order forthe sequence x (n ) to be nite for all values of n.




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The unit circle is dened where z = e j

e j = 1 (47)

We can evaluate X (z ) on the unit circle as follows.

X () = X (z )| z = e j =

n = x (n )e j n (48)

We dene Equation 48 as the Fourier Transform of the aperiodic,discretetime sequence x (n ).




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It is easy to show that X () is periodic since

e j ( + 2k ) = e j e j 2k = e j ; k (49)

where k is an integer and

e 2 jk = 1 k (50)

Thus, X ( j ) is periodic with period 2 or

X () = X ( + 2k ) ; k (51)

We can obtain the inverse transform of X () to obtain thesequence x (n ) if the region of convergence for X (z ) includes theunit circle.




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Let us multiply X () by e j m and integrate it over one period toobtain

X ()e j m d =

n = x (n )e j n e j m d (52)

If X () exist (implies that the region of convergence for X (z )includes the unit circle), then we can change the order ofsummation and integration.

X ()e j

m d =

n = x (n )

e j

(m n )d (53)



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We can obtain the frequency representation of a sequence byevaluating its Z-Transform on the unit circle as shown in theprevious section.We will present an example to illustrate the frequencyrepresentation of a discretetime sequence.


Frequency Representation Example


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Example (1.4)

Consider the discretetime sequence

x (n ) = 2 .0(0 .75 )n u (n ) (59)

The corresponding ZTransform is given by

X (z ) = 2 .0

n = 0(0 .75 )n z n =

2 .01 0 .75 z 1

; |z | > 0 .75 (60)

Since the region of convergence includes the unit circle, X () isdened and

X () =2 .0

1 0 .75 e j (61)



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Example (1.4)

0 0.5 1 1.5 2 2.5 3 3.51

2

3

4

5

6

7

8

Frequency Plot

Frequency(radians)

M a g n

i t u

d e



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Example (1.4)

These plots can be generated by using the following Matlab code.We can load the system transfer function coefcients usingb = [2.0];

a = [1 -0.75];We can then compute the frequency response usingw = p i * (0.0:0.01:1.0);h = freqz(b,a,w);hmag = abs(h);hphase = angle(h);



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Example (1.4)

These plots provide the magnitude and phase of the systemresponse for the range 0 .It is important to note that since we have assumed that x (n ) is real

X ( ) = X ()

|X ( )| = |X ()|arg (X ()) = arg (X ()) (64)

We can observe this by computing X () for the full range of .Note that X () is periodic so we can easily obtain X () fornegative values by subtracting multiples of 2 from the givenfrequencies.




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Example (1.4)

We can use Matlab to compute X () for .The full range plots can be obtained with the following Matlabscript.dw = pi/100;w 3 = d w * (-100:100);h3 = freqz(b,a,w3);hmag3 = abs(h3);hphase3 = angle(h3);

The corresponding magnitude response is given in Figure 11 andthe corresponding phase response is given in Figure 12.



E l (1 4)


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Example (1.4)

4 3 2 1 0 1 2 3 41

2

3

4

5

6

7

8Frequency Plot (Full Range)

Frequency(radians)

M a g n i

t u d e

Figure: Magnitude plot of X () for the full range for Example 1.4.



E l (1 4)


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Example (1.4)

4 3 2 1 0 1 2 3 41

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1Phase Plot ( Full Range)

Frequency (radians/sec)

P h a s e

A n g l e

( r a

d i a n s

)

Figure: Phase plot of X () the full range for Example 1.4.




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Example (1.4)

Alternately, the full range from 0 2 can be obtained usingthe following Matlab code.[h4,w4] = freqz(b,a,201, whole);hmag4 = abs(h4);hphase4 = angle(h4);

The corresponding magnitude response is given in Figure 13 andthe corresponding phase response is given in Figure 14.



Example (1 4)


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Example (1.4)

0 1 2 3 4 5 6 71

2

3

4

5

6

7

8Magnitude Plot

Frequency (radians)

M a g n i

t u d e

Figure: Magnitude plot of X () for the full range for Example 1.4.



Example (1 4)


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Example (1.4)

0 1 2 3 4 5 6 71

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1Phase Plot

Frequency (radians)

M a g n

i t u

d e

Figure: Phase plot of X () the full range for Example 1.4.


Difference Equation Representation


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The general form for the system transfer function of a linear,shift-invariant discretetime system is given by

H (z ) =

L

k = 0

b (k )z k

1 .0 +L

k = 1

a (k )z k (65)

Figure 15 gives a block diagram of the discretetime system.




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

Figure: Block diagram for a discretetime system.




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We can also write the system transfer function in terms of the ratioof the ZTransforms of the output and the input.

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

(66)

It follows that

Y (z ) 1 .0 +L

k = 1

a (k )z k = X (z )L

k = 0

b (k )z k (67)



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Similarly, if y (n k ) = 0 n < k , we can write

z k Y (z ) =

n = 0

y (n k )z n (71)

We can use this result to modify Equation 67 to obtain

n = 0[y (n ) + a (1)y (n 1) + + a (L)y (n L)] z n (72)

=

n = 0 [b (0)x (n ) + b (1)x (n 1) + + b (L)x (n L)] z n



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It follows that the linear shiftinvariant discretetime system can


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ybe represented in either of the following standard forms.

H (z ) =

L

k = 0

b (k )z k

1 .0 +L

k = 1

a (k )z k

y (n ) =L

k = 0

b (k )x (n k ) L

k = 1

a (k )y (n k ) (75)

Note the relationship between the coefcients b (k ) and a (k ) in thesystem transfer function and the corresponding coefcients in thedifference equation.This relationship makes it easy to write the difference equation fora given system transfer function by inspection or vice versa.


Frequency Response


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The general form for the transfer function for a causal,discretetime is

H (z ) =Y

(z

)X (z ) =

L

k = 0

b (k )z k

1 .0 +L

k = 1

a (k )z k (76)

Note that H (z ) is represented using negative powers of z.


Frequency Response


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We can also write H (z ) in the form

H (z ) =Y (z )

X (z )=

L

k = 0

b (k )z Lk

z L + L

k = 1

a (k )z Lk (77)

We can obtain this form by multiplying the numerator anddenominator of the previous form by z L.



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Frequency Response


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It is easy to show this by considering the difference equation for adiscretetime system

y (n ) +L

k = 1

a (k )y (n k ) =L

k = 0

b (k )x (n k ) (81)

If we assume thaty (n ) = H ()e j n (82)

then, we have

y (n k ) = H ()e j (n k ) = y (n )e j k (83)


Frequency Response

Substituting these values in the difference equation, we obtain


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H ()e j n

1 .0 +

L

k = 1 a (k )e j k

= e j n

L

k = 0 b (k )e j k

(84)

Solving for H (), we obtain

H () =

L

k = 0b (k )e

j k

1 .0 +L

k = 1

a (k )e j k (85)

This result is equivalent to evaluating the system transfer function,H (z ), on the unit circle where

z = e j (86)


Frequency Response


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It follows that

H () = H (z )|z = e j =

L

k = 0

b (k )e j k

1 .0 +L

k = 1

a (k )e j k (87)

We will provide examples of the representation of the frequencyresponse for both FIR and IIR discretetime systems.



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FIR Filter Frequency Response Example


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Example (1.5)

Thus, we letx (n ) = e j n (90)

Then

x (n 1) = e j (n 1) = e j x (n )x (n 2) = e j (n 2) = e j 2 x (n )x (n 3) = e j (n 3) = e j 3 x (n )

x (n 4) = e j (n

4)

= e j 4

x (n ) (91)



Example (1.5)


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It follows that

H () = 0 .0110 + 0 .1637 e j + 0 .6947 e j 2

+ 0 .1637 e j 3 0 .0110 e j 4 (92)

H () = 0 .0110 e 2 j e 2 j + e 2 j

+ 0 .1637 e 2 j e j + e j

+ 0 .6947 e 2 j (93)

H () = e 2 j {0 .6947 + 0 .3274 cos () 0 .0220 cos (2)} (94)

We can determine the magnitude and phase of H () at someparticular value of by substituting for in H ().



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Example (1.5)


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Figure 16 gives the magnitude spectrum of H ().

0 0.5 1 1.5 2 2.5 3 3.50.3

0.4

0.5

0.6

0.7

0.8

0.9

1Magnitude Spectrum for FIR Example

Frequency (Radians/sec)

M a g n

i t u

d e

Figure: Magnitude Spectrum for FIR lter.



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IIR Filter Frequency Response


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Example (1.6)

Read in the lter coefcients.b = [0.6283 -1.2565 0.6283];a = [1.0000 -1.1804 0.4816];

Compute the frequency response using the Matlab function freqz .w = p i * (0:0.01:1.0);h = freqz(b, a, w);hmag = abs(h);hphase = angle(h);



Example (1.6)


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Plot the results.figure(1);plot(w, hmag)title(Frequency Response Plot)xlabel(Normalized Frequency (radians))

ylabel(Magnitude)print -dps iirfreq.psfigure(2)plot(w, hphase)title(Phase Response Plot)xlabel(Normalized Frequency (radians))ylabel(Magnitude)print -dps iirphase.ps



Example (1.6)


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Figure 18 gives the resulting magnitude response.

0 0.5 1 1.5 2 2.5 3 3.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Frequency Response Plot

Normalized Frequency (radians)

M a g n

i t u

d e

Figure: Magnitude plot of H () for Example 1.6.



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Example (1.6)

Note that the magnitude of the frequency response is zero at

= 0.The plot shows the phase equal to zero at this value of as well.


PoleZero Plots


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The location of the poles and zeros in the complex planedetermine the frequency response of the system.We can use polezero plots to help estimate the frequencyresponse of discretetime systems.For example, it is usually desirable to have the magnitude of a lowpass lter approach or equal to zero as the frequency approachesthe Nyquist frequency ( for normalized frequencies).

C Williams & W Alexander (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513 Fall 2013 82 / 155

PoleZero Plots


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Therefore, we expect to nd the zeros to be at or clustered aroundz = 1 for low pass lters.The location of the poles will then determine the value of the cutofffrequency which is typically dened as the point where themagnitude gain is equal to 22 or the power gain is equal to 0.5.For example, Figure 20 gives the polezero plot for a secondorder low pass lter with cutoff frequency of c = 0 .6 .


PoleZero Plots

1

PoleZero Plot for Low Pass Filter


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1 0.5 0 0.5 1

1

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

2

Real Part

I m a g

i n a r y

P a r t

Figure: PoleZero Plot of H(z) for low pass lter with cutoff frequencyc = 0 .6 .


PoleZero Plots


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The zeros are both located at z = 1 and the poles are located at

p (1) = 0 .1848 + 0 .4021 j p (2) = p (1) = 0 .1848 0 .4021 j (97)

Figure 21 gives the magnitude and phase response plots for thislter.


PoleZero Plots

1.5Magnitude of Frequency Response for H(z)


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4 3 2 1 0 1 2 3 40

0.5

1


M a g n

i t u

d e

4 3 2 1 0 1 2 3 44

2

0

2

4Phase of Frequency Response for H(z)


P h a s e

( r a

d i a n s

)

Figure: Magnitude and Phase plots of H(z) for low pass lter with cutofffrequency c = 0 .6 .


PoleZero Plots


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We also expect the gain of the low pass lter to be equal to nearly

equal to one at = 0 or where z = 1.In this case, the second order low pass lter, with two zeros atz = 1, can be generally represented as

H (z ) =A(z + 1)2

z 2 2 e {p (1)}z + |p (1)|2 (98)

We can set the gain to one at z = 1 to obtain the value of A. Itfollows that

A =1 2 e {p (1)} + |p (1)|2

4 (99)


PoleZero Plots


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On the other hand, the gain for a high pass lter is desired to be ator near zero at = 0.Therefore, we expect the zeros of a high pass lter to be at orclustered around z = 1.The locations of the poles then determine the cutoff frequency.Figure 22 gives the polezero plot for a second order high passlter with a cutoff frequency of c = 0 .6 .



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PoleZero Plots

0.8

1Magnitude of Frequency Response for H(z)

e


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4 3 2 1 0 1 2 3 40

0.2

0.4

0.6


M a g n

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4 3 2 1 0 1 2 3 44

2

0

2



P h a s e

( r a

d i a n s

)

Figure: Magnitude and Phase plots of H(z) for high pass lter with cutofffrequency c = 0 .6 .



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PoleZero Plots


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The gain for a band stop lter is desired to be at or near zero forthe stop frequency which is normally given by s = 0 .5(1 + 2)where 1 and 2 are the low and high cutoff frequenciesrespectively.

The locations of the poles then determine the cutoff frequencies.Figure 24 gives a polezero plot for a fourth order band stop lterwith cutoff frequencies of 1 = 0 .6 and 2 = 0 .75 .


PoleZero Plots

0.8

1

2

PoleZero Plot for Band Stop Filter


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1 0.5 0 0.5 1

1

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

2

Real Part

I m a g

i n a r y

P a r t

Figure: PoleZero Plot of H(z) for band stop lter with cutoff frequencies at1 = 0 .6 and 2 = 0 .75 .


PoleZero Plots

There are two zeros located at

r (1) = r (3) = 0 5373 + 0 8434 j (102)


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r (1) = r (3) = 0 .5373 + 0 .8434 j (102)

and two zeros located at

r (2) = r (4) = 0 .5373 0 .8434 j (103)

The poles are located at

p (1) = 0 .5810 + 0 .6372 j p (2) = 0 .5810 0 .6372 j p (3) = 0 .3187 + 0 .7678 j p (4) = 0 .3187 0 .7678 j (104)

Figure 25 gives the magnitude and phase response plots for thislter.

C Willi & W Al d (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513 F ll 2013 95 / 155

PoleZero Plots

0.8


d e


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4 3 2 1 0 1 2 3 40

0.2

0.4

0.6


M a g n

i t u

d

4 3 2 1 0 1 2 3 410

5

0

5



P h a s e

( r a

d i a n s )

Figure: Magnitude and Phase plots of H(z) for band stop lter with cutofffrequency 1 = 0 .6 and 2 = 0.75 .


PoleZero Plots


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The gain for a band pass lter is desired to be at or near zero forboth = 0 and = .Therefore, we expect to nd zeros of the transfer function at orclustered near both z = 1 and at z = 1.

The locations of the poles then determine the cutoff pass bandfrequencies.Figure 26 gives a polezero plot for a fourth order band pass lterwith cutoff frequencies of 1 = 0 .6 and 2 = 0 .75 .


PoleZero Plots

0.8

1

PoleZero Plot for Band Pass Filter


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1 0.5 0 0.5 1

1

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

2 2

Real Part

I m a g

i n a r y

P a r t

Figure: PoleZero Plot of H(z) for band pass lter with cutoff frequencies at1 = 0 .6 and 2 = 0 .75 .


PoleZero Plots


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There are two zeros at z = 1 and two zeros at z = 1.Since the critical frequencies for this example are the same asthose for the band stop lter, the poles are located in the sameplace as there are for the band stop lter above.Figure 27 gives the frequency magnitude and phase plots for thislter.


PoleZero Plots

0 6

0.8


u d e


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4 3 2 1 0 1 2 3 40

0.2

0.4

0.6


M a g n

i t u

4 3 2 1 0 1 2 3 410

5

0

5



P h a s e

( r a

d i a n s )

Figure: Magnitude and Phase plots of H(z) for high pass lter with cutofffrequency 1 = 0 .6 and 2 = 0.75 .


System Response


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We can determine the response of a discretetime system for aninput sequence by obtaining the inverse Z Transform of theproduct of the Z Transforms of the input sequence X (z ) and thesystem transfer function H (z ).We can use partial fraction expansion to nd the inverse ZTransform in a convenient way.Consider the output Z Transform given by

Y (z ) = H (z )X (z ) (105)


System Response

We can expand Y (z )z to obtain an expansion of the form

Y (z ) M c k ( )


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( )

z =

k = 1

ck

z p k (106)

where there are M poles, c k represents the expansion coefcientsand p k represents the corresponding poles.We can then write

Y (z ) =M

k = 1

c k z z p k

=M

k = 1

c k 1 p k z 1

(107)

If all of the p k have magnitudes less than 1 ( |p k | < 1 k ), then

y (n ) =M

k = 1

c k (p k )n u (n ) (108)


System Response Example

l ( )


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Example (1.7)

The system response for a discretetime system is given by

H (z ) =0 .1593 z 2 + 0 .3187 z + 0 .1593

(z + 0 .472 )(z 0 .567 )(109)

We want to nd the output for this system for 0 n when theinput is

x (n ) = 2 .5(0 .76 )n u (n ) (110)



Example (1.7)


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Example (1.7)

We rst nd X(z) and Y(z),

X (z ) =2 .5z

z 0 .76(111)

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

=2 .5z (0 .1593 z 2 + 0 .3187 z + 0 .1593 )

(z 0 .76 )(z + 0 .472 )(z 0 .567 )(112)



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Example (1 7)


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Example (1.7)

We can nd A,B,C as follows

A =Y (z )

z (z 0 .76 )|z = 0 . 76 = 5 .1889

B = X (z )z

(z + 0 .472 )|z = 0 . 472 = 0 .0866

C =X (z )

z (z 0 .567 )|z = 0 . 567 = 4 .877 (114)



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Example (1.7)


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We used the geometric series to obtain the inverse transform

X (z ) =

n = 0

x (n )z n (117)

If |p | < 1, thenX (z ) =

cz z p

=c

1 .0 pz 1(118)

and

x (n ) = c (p )n u (n ) (119)


Frequency Shifting

Frequency shifting can be used to perform a frequencytransformation of a discretetime system.We will explore this analytically in this section.


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The frequency response of a linear, shiftinvariant system can beobtained from its ZTransform.

H (z ) =

n = 0

h (n )z n (120)

where h (n ) is the impulse response of the system.

H () = H (z )|z = e j =

n = 0

h (n )e j n (121)

Assume that we multiply h (n ) by a complex exponential e j 1 toobtain

h 1(n ) = h (n )e j 1n (122)


Frequency Shifting

The resulting frequency response is given by

H 1() =

h (n )e j 1 n e j n (123)


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1( )n = 0

( ) ( )

H 1() =

n = 0

h (n )e j ( 1 )n

= H ( 1) (124)

We see that multiplying the impulse response h (n ) by the complexexponential e j 1 shifts the frequency response by 1 and thefrequency response at = 0 is shifted to the location =

1.

If we multiply h (n ) by the complex exponential e j 1 to obtain

h 2(n ) = h (n )e j 1 n (125)



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Inverse ZTransform for Systems with Complex Poles


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We will consider the case where the system has complex poles inthis section.If the coefcients of both the numerator and denominatorpolynomials of H (z ) are real, then

1 the poles and zeros of H (z ) are real, or2 they occur in complex conjugate pairs.

We will consider the case where all of the poles are real except fortwo complex poles that are complex conjugates.



We can therefore use partial fraction expansion to expand thetransfer function H(z ) as


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transfer function H (z ) as

H (z ) =N 1

m = 1

F m (z ) +N 2

k = 1

C k z z p k

(127)

where all of the C k and p k are real and

F m (z ) =D 1m z

z p 1m +

D 2m z z p 2m

(128)

The constants D 1m and D 2m are complex conjugates and thepoles p 1m and p 2m are complex conjugates.



We can compute the inverse ZTransform of F m (z ) to obtain


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f m (n ) = [ D 1m (p 1m )n + D 2m (p 2m )n ] u (n ) (129)

We can write p 1m and p 2m in complex exponential form as

p 1m = e + j

p 2m = e j (130)

Then, we can write

f m (n ) = D 1m

e ( + j )n + D 2m

e ( j )n (131)



However,

e j n = cos (n) + jsin (n)


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e cos (n ) + jsin (n )e j n = cos (n ) jsin (n ) (132)

Thus,

f m (n ) = ( D 1m + D 2m ) e n

cos (n )u (n )+ j (D 1m D 2m ) e n sin (n )u (n ) (133)

We observe that

D 1m + D 2m = 2 e (D 1m )D 1m D 2m = 2 j m (D 1m ) (134)




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Thus

f m (n ) = 2 e (D 1m ) e n cos (n )u (n )

2 m (D 1m ) e n sin (n )u (n ) (135)

We can further simplify f m (n ) by using the trigonometric equation

cos (n + ) = cos () cos (n ) sin () sin (n ) (136)



We can then write f n in the form


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We can then write f m (

n )

in the form

f m (n ) = Ae n cos (n + ) (137)

where

= tan 1 m (D 1m )e (D 1m )

A = 2 m (D 1m )2 + e (D 1m )2 (138)


Complex Poles Inverse ZTransform Example

Example (1.8)

We wish to nd the impulse response of the discretetime system


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with its transfer function given by

H (z ) =0 .5276 z 3 + 1 .5829 z 2 + 1 .5829 z + 0 .5276

z 3 + 1 .76 z 2 + 1 .1829 z + 0 .2781(139)

If x (n ) is an impulse, then

X (z ) =

n = (n )z n = 1 (140)

Thus,Y (z ) = H (z )X (z ) = H (z ) (141)



Example (1.8)


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For convenience in obtaining the geometric series form, weexpand

H (z )z =

0 .5276 z 3 + 1 .5829 z 2 + 1 .5829 z + 0 .5276z (z 3 + 1 .76 z 2 + 1 .1829 z + 0 .2781 )

=D 1

z p 1+

D 2z p 2

+C 1

z p 3+

C 2z

(142)



Example (1.8)

It follows that


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p 1 = 0 .6252 + 0 .3935 j p 2 = 0 .6252 0 .3935 j = p 1p 3 = 0 .5097p 4 = 0 (143)

and

D 1 = 0 .3221 + 0 .1501 j D 2 = 0 .3221 0 .1501 j = D 1C 1 = 0 .7254C 2 = 1 .8972 (144)



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Example (1.8)


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We have

p 1 = e + j = e (cos () + jsin ())

tan () =m (p 1)e (p 1)

=0 .3935

0 .6252

= tan 1 0 .3935 0 .6252

= 2 .5799

e cos () = 0 .6252 e = 0 .6252cos ()

= 0 .7387

= log e (0 .7387 ) = 0 .3029 (146)



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Example (1.8)


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We can simplify f 1(n ) further as

f 1 (n ) = Ae 0 . 3029 n cos (2 .5799 n + )u (n ) (148)

where

A = 2 (0 .1501 )2 + ( 0 .3221 )2 = 0 .7107 = tan 1

0 .1501 0 .3221

= 2 .7055 (149)



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Inverse ZTransform for Systems with Multiple Poles

If the Z Transform has one or more multiple order poles, then thepartial fraction expansion must contain additional terms.For example, if the Z Transform has a pole of order 3 such that

( ) ( )


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Y (z )z

= B (z )M 1k = 1 (z p k ) (z p M )3

(152)

Then, the inverse Z Transform will have the form

Y (z )z

=M

1

k = 1

C k z p k

+ C 1M z p m

+ C 2M (z p m )2

+ C 3M (z p m )3

(153)

The general equation for nding the residues for a system with Lpoles located at p m is given by [2]

C k =1

(L k )!( p m )Lk d Lk

dz (z p m )L

Y (z )z

(154)

We now consider an example where Y (z ) has multiple poles.C. Williams & W. Alexander (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2013 126 / 155

Multiple Poles Inverse ZTransform Example

Example (1.9)

C id h di i i h f f i i b


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Consider the discretetime system with transfer function given by

H (z ) =0 .1584 z 3 + 0 .4752 z 2 + 0 .4752 z + 0 .1584

z 3 0 .2z 2 0 .2275 z + 0 .0612(155)

This can be rewritten as

H (z ) =0 .1584 (z + 1)3

(z 0 .35 )2 (z + 0 .5)(156)

We want to nd the impulse response of this system.



Example (1.9)

h l f ( ) l d


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The poles of H (z ) are located at

p (1) = 0 .5p (2) = p (3) = 0 .35 (157)

We perform partial fraction expansion of H (z )z to obtain

H (z )z

=C 1

z + 0 .5+

C 2z 0 .35

+C 3

(z 0 .35 )2+

C 4z

(158)



Example (1.9)


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C 1 =(z + 0 .5) H (z )

z z = 0 . 5=

0.1584 ( 0.5 + 1)3

( 0 .5 0.35 )2 ( 0.5)= 0 .05481

C 3 = (z 0 .35 )2

H (z )z z = 0 . 35= 0 .1584 (0.35 + 1)

3

0 .35 (0 .35 + 0 .5) = 1 .31

C 4 =z H (z )

z z = 0=

0 .1584 (0 + 1)3

(0 + 0 .5)(0 0 .35 )2= 2 .5861 (159)



Example (1.9)

We can nd C 2 as

d ( 0 35 )2 H(z )


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C 2 =d (z 0 .35 )2 H (z )z

dz z = 0 . 35

= 2 .3729 (160)

Thus,

H (z ) = 0 .05481 z

z + 0 .5

2 .3729 z z 0 .35

+1 .31 z

(z 0 .35 )2+ 2 .5861 (161)

Computing the inverse ZTransform, we obtain

h (n ) = [ 0 .05481 ( 0 .5)n 2 .3729 (0 .35 )n ] u (n )

+1 .310 .35

n (0 .35 )n u (n ) + 2 .5861 (n ) (162)


Cascade Implementation of Digital Filters

A given system transfer function for a discretetime system

( )

L

b (k )


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H (z ) =B (z )A(z )

= k = 0b (k )

1 .0 +L

k = 1

a (k )

(163)

can be represented as a product of rst order or higher ordersystem transfer functions such that

H (z ) =M

m = 1

H m (z ) (164)

The a (k ) = 0 .0 for a FIR lter.



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Cascade Implementation of Digital Filters I

We can form each H k (z ) using one or more real q m or complexconjugate pairs of q m so that the coefcients of all of the H k (z ) are

also real


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also real.A standard practice for partitioning H (z ) into second ordersections is to:

Order the q m and p m in increasing (or decreasing) magnitude.Combine the complex conjugate pairs in increasing (or decreasing)magnitude to form second order system transfer functions.

H k (z ) =(z q m )(z q m )(z p m )(z p m )

=z 2 2 (q m )z + q 2m z 2 2 (p m )z + |p 2m |

(166)



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Cascade Implementation of Digital Filters

It follows that

H (z ) =K

H (z ) = AK K

H (z ) (170)


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H (z ) =k = 1

H k (z ) = Ak = 1

H k (z ) (170)

if b (0) is positive and

H (z ) = K

k = 1

H k (z ) = AK K

k = 1

H k (z ) (171)

if b (0) is negative.

We can illustrate this concept using an FIR lter as an example.



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Cascade Implementation Example

Example (1.10)

Note that the q m have been arranged in decreasing magnitudeorder.


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order.Using the procedure described above, we can form

H 1(z ) = ( z q 2)(z q 3) = z 2 2 .0790 z + 11 .8225

H 2(z ) = ( z q 7)(z q 8) = z 2

0 .1759 z + 0 .0846H 3(z ) = ( z q 1)(z q 4) = z 2 0 .3321 z 11 .7114H 4(z ) = ( z q 6)(z q 9) = z 2 + 0 .0284 z 0 .0854H 5(z ) = ( z q 5) = z + 1 .0000 (174)

A = ( 0 .0049 )15 = 0 .3454 (175)



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Stabilization of an Unstable Filter

The magnitude of a stable lter and the magnitude of an unstablelter can have the same frequency response.A t bl digit l lt ill h it l i id th it i l h


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A stable digital lter will have its poles inside the unit circle suchthat

|p k | < 1 .0 1 k L (177)

for a lter with L poles.We can stabilize an unstable lter with a pole p m outside the unitcircle by replacing it by its reciprocal and then adjusting themagnitude of gain of the lter appropriately.We will illustrate this concept using an example.


Stabilization Example I


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Example (1.11)


Stabilization Example II

The system transfer function for an unstable discretetime systemis given by

H (z ) =B (z )A(z )

(178)

where


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where

B (z ) = 0 .4360 + 1 .8677 z 1 + 3 .4742 z 2

+ 3 .4742 z 3 + 1 .8677 z 4

+ 0 .4360 z 5 (179)

and

A(z ) = 1 .0000 + 2 .4705 z 1 + 4 .0048 z 2

+ 2 .9435 z 3 + 1 .2649 z 4

0 .1281 z 5 (180)



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Stabilization Example

Example (1.11)

We can write H (z ) in the form


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H (z ) =B (z )

(z p 1 )(z p 1 )(z p 3)(z p 3)(z p 5)(181)

wherep 1 = 0 .6445 + 1 .1156 j p 3 = 0 .6324 + 0 .7262 j p 5 = 0 .0832 (182)



Example (1.11)

A stable transfer function with the same magnitude response canbe determined as


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be determined as

H 1(z ) =z 2 B (z )

(z 1 p 1)(z 1 p 1 )(z p 3)(z p 3)(z p 5 )(183)

Simplifying H 1(z ), we obtain

H 1(z ) =B (z )

|p 1 |2 (z 1p 1 )(z 1

p 1)(z p 3 )(z p 3)(z p 5)

(184)



Example (1.11)

B1(z )


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H 1(z ) =B 1(z )A1(z )

B 1(z ) = 0 .2627 + 1 .1252 z 1 + 2 .0931 z 2

+ 2 .0931 z 3

+ 1 .1252 z 4

+ 0 .2627 z 5

A1(z ) 1 .0 + 1 .9581 z 1 + 2 .3420 z 2

+ 1 .2730 z 3 + 0 .4353 z 4

0 .0465 z 5

(185)



Example (1.11)

We can use the Matlab freqz function to compute 200 values ofthe magnitude response of the original system transfer function asgiven in Equation 178.


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We can compute the magnitude response of the original H (z )using the following Matlab script.b1 = [0.4360 1.8677 3.4742 3.4742 1.8677 0.4360];

a1 = [1.0000 2.4705 4.0048 2.9435 1.2649 -0.1281];n = 0:199;w = n * pi/199;h1 = freqz(b1, a1, w);h1mag = abs(h1);

figure(1)plot(w, h1mag);print -dps stab1a.ps



Example (1 11)


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Example (1.11)

Figure 28 gives the plot for the magnitude response of H (z ).Note that this magnitude response is not in decibels.



Example (1.11)

1.2

1.4


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0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

Figure: Magnitude response plot of H(z) for Example 1.11.



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Example (1 11)


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Example (1.11)

Figure 29 gives the plot for the magnitude response of H 1(z ).Note that this magnitude response is not in decibels.



Example (1.11)

1.2

1.4


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0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

Figure: Magnitude response plot of H 1 (z ) for Example 1.11.



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