1 lecture 3: march 6, 2007 topic: 1. frequency-sampling methods (part i)

1

Lecture 3: March 6, 2007

Topic:1. Frequency-Sampling Methods (Part I)

2

Lecture 3: March 6, 2007

Topic:1. Frequency-Sampling Methods (Part I)

• basic principle and tasks,

• non-uniform frequency-sampling method,

• uniform frequency-sampling method A.,

• uniform frequency-sampling method A.: optimization approach included.

3

4.2. Frequency-Sampling Methods

To approximate any continuous frequency response, one could sample frequency response at N points along the frequency interval and evaluate the continuous frequency response as an interpolation of the sampled frequency response. The approximation error would then be exactly zero at the sampling frequencies and be finite between them. The smoother the frequency response being approximated, the smaller the error of interpolation between the sample points. For the solution of this approximation – interpolation task (i.e. the evaluation of a impulse response of FIR filter!) a number of methods will be given in the next parts of our presentation.

0,



4

Desired real-valued frequency response: jdrH e

Frequency-Sampling Method: Basic Principle

Approximation error:

k k

k

j jdr r

e

H e H e

k

Samples of mjdrH e

Approximation of ideal frequency response: j

rH e

5

Interpolation

( )kjH e

Frequency-Sampling Method: Basic Tasks

A. Sampling process:

( )jH e

k

Frequency selection:

k

Set the phase

Sample of frequency response is defined!

( )kjdrH e

Set the value of sample (real-valued)

B. Approximation by interpolation: 1( )jH e ( )MjH e 2( )jH e 1( )MjH e

h n

Impulse response is defined!

End of design!

6

The four cases of linear phase FIR filters. The real-valued frequency response. Summary.

1. M-even, Symmetry-S (M/2-coefficients):

12

0

1( ) 2 ( )cos

2

M

rk

MH h k k

1

2

0

(0) 2 ( )

M

rk

H h k

7

2. M-odd, Symmetry-S ((M+1)/2-coefficients): 3

2

0

1 1( ) 2 ( )cos

2 2

M

rk

M MH h h k k

3

2

0

1(0) 2 ( )

2

M

rk

MH h h k

12

0

1( ) 2 ( )sin

2

M

rk

MH h k k

(0) 0rH

3. M-even, Symmetry-A (M/2-coefficients):

8

4. M-odd, Symmetry-A ((M-1)/2-coefficients):

3

2

0

1( ) 2 ( )sin

2

M

rk

MH h k k

(0) 0rH

9

The four cases of linear phase FIR filters. Phase response. Summary.

1( ) 0

2( )1

( ) 02

r

r

Mfor H

Mfor H

1( ) 0

2 2( )1 3

( ) 02 2

r

r

Mfor H

Mfor H

3-4. M-odd:

1-2. M-even:

10

4.2.1. Non-Uniform Frequency-Sampling Method

k( )kj

drH e

Each of the equations in the previous tables constitutes a set of linear equations for determining the coefficients of a linear phase FIR filter. Consequently, if we specify response at either (M+1)/2 or (M-1)/2 or M/2 points in we can solve the corresponding set of linear equations for the coefficients h(n). Generally, the values of and

can be chosen arbitrarily. Therefore, the method is called as non-uniform frequency sampling method.

11


k( )kj

drH e

Each of the equations in the previous tables constitutes a set of linear equations for determining the coefficients of a linear phase FIR filter. Consequently, if we specify response at either (M+1)/2 or (M-1)/2 or M/2 points in we can solve the corresponding set of linear equations for the coefficients h(n). Generally, the values of and


12


k( )kj

drH e

Each of the equations in the previous tables constitutes a set of linear equations for determining the coefficients of a linear phase FIR filter. Consequently, if we specify response at either (M+1)/2 or (M-1)/2 or M/2 points in , we can solve the corresponding set of linear equations for the coefficients h(n). Generally, the values of and


13

Example:

Build up a set of linear equations for the design of the 6-th order linear phase FIR filter (M=6), which has a symmetrical unit sample response and the real-valued frequency response that satisfy the conditions:

1 1

2 2

3 3

0 (0) 1

0.83 3

0.12 2

r r

r r

r r

H H

H H

H H

14

Solution:

12

0

61

2

0

2

0

1( ) 2 ( )cos

2

12 ( )cos

2

6 12 ( )cos

2

M

rk

k

k

MH h k k

Mh k k

h k k

If M=6 and unit sample response is symmetrical, then the following expressions for are validated:( )j

rH e

15

5 5( ) 2 (0)cos 2 (1)cos 1

2 2

52 (2)cos 2

2

rH h h

h

5 3( ) 2 (0)cos 2 (1)cos 2 (2)cos

2 2 2k k k

r kH h h h

16

1 1

2 2

3 3

0 (0) 1

0.83 3

0.12 2

r r

r r

r r

H H

H H

H H

17

1 10 (0) 1r rH H

5.0 3.0 1.01 2 (0)cos 2 (1)cos 2 (2)cos

2 2 2h h h

2 2 0.83 3r rH H

1 2 (0) 2 (1) 2 (2)h h h (1)

5 30.8 2 (0)cos 2 (1)cos 2 (2)cos

6 6 6h h h

(2)

1 1 11

5 3( ) 2 (0)cos 2 (1)cos 2 (2)cos

2 2 2rH h h h

18

3 3 0.12 2r rH H

5 30.1 2 (0)cos 2 (1)cos 2 (2)cos

8 8 4h h h

(3)

19

The set of desired equations:

1 2 (0) 2 (1) 2 (2)h h h

5 30.8 2 (0)cos 2 (1)cos 2 (2)cos

6 6 6h h h

5 30.1 2 (0)cos 2 (1)cos 2 (2)cos

8 8 4h h h

(1)

(2)

(3)

20

By solving of the above given set of linear equations we can obtain:

(0), (1), (2)h h h

Because of symmetrical impulse response:

(0) (5), (1) (4), (2) (3)h h h h h h

Therefore, by using these results, the complete impulse response can be obtained:

(0), (1), (2), (3), (4), (5)h h h h h h

21

4.2.2. Uniform Frequency-Sampling Method A

Although the values of and in the corresponding equations can be chosen arbitrarily, it is usually desirable to select equally spaced points in the frequency, in the rage . Following this idea the frequencies can be selected in accordance with the next rules:

k ( )kjdrH e

0

M-even, Symmetry-S: 2

k

k

M

0,1, , 12

Mk

M-odd, Symmetry-S: 1

0,1, ,2

Mk

2

k

k

M

22

Problem of (0)drH for symmetry-A:

Because in this case independent of our choice of h(n), it is obvious that the frequency

(0) 0drH

cannot be used in the specification. Thus we may define the frequencies in accordance with the following rules.

0

i.e the situations when

(0) ( ) 0dr drH H

0

The proposed choice of equally spaced frequencies allows to avoid completely the zeros at

23

M-even, Symmetry-A: 2

k

k

M

1,2, ,2

Mk

Method A.

M-odd, Symmetry-A: 2

k

k

M

1

1,2, ,2

Mk

Method B.

M-even, Symmetry-A:

12

2k

k

M

0,1, , 1

2

Mk

M-odd, Symmetry-A:

12

2k

k

M

1

0,1, ,2

Mk

24

4.2.2.1. Frequency Response Sample Specification

In the case of a linear phase FIR filter design by frequency sampling methods, the values of

The simplest approach:

have to be specified, too.

( )r kH

( ) 1r kH k pass-band

( ) 0r kH k stop-band

( )r kH 2

M

is not specified in the transient-band, i.e. the transient-band width is

25

In this form of specification, is changed abruptly from the pass-band, where , to the stop-band, where . Instead of having an abrupt change, if we specify an intermediate value of in the transition band, the resulting frequency response has significantly smaller side lobes in the stop-band. The following example illustrates this point.

( )rH ( )rH ( ) 1rH

( ) 0rH ( )rH

26

Example:

Determine the coefficients h(n) of a linear phase FIR filter of length M=15 which has a symmetrical unit sample response and the frequency response that satisfies the condition:

1 0,1,2,32

0 4,5,6,715r

for kkH

for k

a)

1 0,1,2,32

0.4 415

0 5,6,7r

for kk

H for k

for k

b)

2 /15Transition band width:

4 /15Transition band width:

27

The magnitude response of the resulting filter are illustrated in the next figures. In the case of the task of a), we observe that the filter has an overshoot at the edge of the pass-band, just prior to the transition region. It also has relatively large side lobes in the stop-band. In the case of the task b) it can be noted that the side lobes are significantly lower than in the previous design.

28

The magnitude response of the resulting filter are illustrated in the next figures. In the case of the task of a), we observe that the filter has an overshoot at the edge of the pass-band, just prior to the transition region. It also has relatively large side lobes in the stop-band. In the case of the task b) it can be noted that the side lobes are significantly lower than in the previous design.

29

Example: magnitude responses (linear scale)

Task of a): transition region width= 2 / M

15M

Task of b): transition region width=4 / M

k=0 k=1 k=2 k=3

k=5

k=6 k=7

a) k=4

b) k=4

30

Task of a): transition region width=2 / M

Task of b): transition region width=4 / M

1020*log jH e

15M

[dB]

Example: magnitude responses (dB scale)

31

4.2.2.2. Optimization Approach

The previous example clearly illustrate the advantages in allowing a small transition in the design of filters. The only disadvantage is that the width of the transition region is increased. Specifically, the transition region was about twice as wide as in the above example. The reduction in the stop-band side-lobes can be obtained by allowing for one or more (usually one or two) additional frequency specifications in the transition bands.

32



33



34

Solution A: The suitable values of the sample specifications in the transition band can be found experimentally (e.g. by using MATLAB software).

Solution B: The optimum values of the sample specifications in the transition band are tabulated and available in some references.

35

Solution C: A set of equations must be written and solved that mathematically expresses the desired optimization. Let us assume that the frequency samples in the transition band are labeled T1 and T2 for convenience. These samples are ones to be optimized. In order to find optimum values T1 and T2, a set of constraint equations must be written for frequencies in the pass-bad and stop-band. Typical sets of such constraints might be:

36

( )e : approximation error


Optimization task (constraint equations) C1:

( ) j jde H e H e pass-band

0 : fixed tolerance

jdH e

: desired (ideal) frequency response

jH e : the „real“ frequency response function

stop-band by selection of T1 and T2

min max j jdH e H e

37

a known weighting function on the frequency response approximation error


min max j j jdW e H e H e

by selection of T1 and T2

Solution of the C1 and C2 problem:linear programming methods

Optimization task (constraint equation) C2:

weighed error of approximation

1 lecture 3: march 6, 2007 topic: 1. frequency-sampling methods (part i)

Documents