ECE 6560 Multirate Signal Processing Chapter 8bazuinb/ECE6560/Chap_08.pdfECE 6560 Multirate Signal...

ECE 6560Multirate Signal Processing

Chapter 8

Dr. Bradley J. BazuinWestern Michigan University

College of Engineering and Applied SciencesDepartment of Electrical and Computer Engineering

1903 W. Michigan Ave.Kalamazoo MI, 49008-5329

ECE 6560 Notes and figures are based on or taken from materials in the course textbook: fredric j. harris, Multirate Signal Processing for Communication Systems, Prentice Hall PTR, 2004. ISBN 0-13-146511-2.

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Chapter 8: Half-band Filters 8.1 Half-band Low Pass Filters 2028.2 Half-band High Pass Filters 2048.3 Window Design of Half-band Filters 2058.4 Remez Alorithm Design of Half-band Filters 207

8.4.1 Half-band Remez Algorithm Design Trick 2088.5 Hilbert Transform Band-pass Filter 210

8.5.1 Applying the Hilbert Transform Filter 2128.6 Interpolating with Low Pass Half-band Filters 2148.7 Dyadic Pass Half-band Filters 217


3

Rectangular Low Pass Filter

12

rectf

ffH

tffth 11 2sinc2

tf

tffth

1

11 2

2sin2

tf

tf

fth

222

222sin

21

1

1

11

11

21

222:

ft

tfzeroFirst

zerost

zerost


4

Half-Band Low Pass Filter

sfffH 2rect

tffth ss

2sinc

2

tf

tffth

s

s

s

142

42sin

42

tf

tffth

s

s

s

2

2sin

2

szerost

zerosts

ft

tfzeroFirst

22

:

1

1

41sffLet


5

Half-Band Low Pass Filters

• Spectral Characteristics and Filter Coefficients

ss fbyscaleand

fntLet 1

tf

tffth

s

s

s

2

2sin

2

2

sinc21

2

2sin

21 n

n

n

nh

Even integers, except 0 or all coefficeints!


6

Half-Band Coefficients

• Filter length “2N+1”– But there are N zero coefficients for N even or

N-1 zero coefficients for N odd– Note: always select N as odd

(i.e. if N=2M+1, then 2N+1=>4M+3)

Chap8_1.m - odd

Chap8_1c.m - even

4M+3 and 4M+5 use the same

non-zero coefficients!


7

Computation Load

• There is one multiply for every unique coefficient– Conventional filter

• 4M-taps 4M Multiplies, 4M-1 Adds– Half-Band filter

• 4M+3 taps 2M+3 Multiplies, 2M+2 Adds– Half-Band filter, if h(0) doesn’t require a multiply,

• 4M+3 taps 2M+2 Multiplies, 2M+2 Adds– Half-Band filter, if h(0) doesn’t require a multiply and symmetric,

• 4M+3 taps M+1 Multiplies, M+M+1 Adds

• Potential savings: approximately ¼ the multiplies and ½ the adds!

Half-Band Length Consideration

• Since the half-band filter is originally defined by the sincfunction, it must be “windowed” to a finite length.– Apply a rectangular filter of length 4M+3 or 4M+5(?!)

• The finite length filter will always appear as a “perfect half-band” convolved by a frequency domain function related to the window used (rect or other).

• Expect ripples in the passband and stopband!– Improved if windowed by other than rect function.


8

Chap8_1.m

Chap8_1c.m


9

Half-Band High Pass Filter

• Spectral Characteristics and Filter Coefficients– Complex mix by fs/2

nihnh LPHP exp

2

2sin

121cos

2

2sin

21

n

n

nn

n

nh nHP

Chap8_2.m

All odd n coefficients are negative


10

Mirror Property of Half-Band Filters

• When the high-pass and low-pass versions of the filter are summed, the coefficients for n ≠ 0 cancel!

nn

n

n

n

nhnh HPLP

cos

2

2sin

21

2

2sin

21

nnnn

nhnh HPLP

cos2

sin2

sin1

2

sin12

sin1 nnn

nhnh nHPLP

002

sin2

sin1:

00sinsin1:

0121

21:

nfornnn

nhnhoddn

nforkkn

nhnhevenn

nfornhnhevenn

HPLP

HPLP

HPLP

NnNfornhnhTherefore HPLP ,0,


11

For a Causal Half-Band

• For causality in the sample domain, the entire filter is time shifted by a factor of N (or 2M+1)

• The resulting filter summation is

NnforNnnhnh HPLP 20,

Chap8_3.m


12

Window Design of Half-Band Filter

• Windowing provides additional side-lobe attenuation– Kaiser windowing– FirPM windowing

nwn

n

nhLP

2

2sin

21

;8.5,1*2*.2:sinc*5.0

NkaiserNNnhLP

;1*2_*.2:sinc*5.0

NwindowremezNNnhLP

Chap8_4.mPassband narrowsRipple decreases

Stopband improves


13

Design Example 8.1

Design a half-band, kaiser windowed filter• Sample Rate 20 kHz• Transition BW 4 kHz (20% of Fs)• Out-of-Band Attenuation 60 dB

The parameter β of the Kaiser window is estimated from Figure 3.8 of chapter 3 to be = 5.8

The estimated filter length is obtained from (3.11) is N=19, but 21 was used. (Note: zero coefficients!)

Chap8_5.m


14

Design Example Results

h=0.5*sinc(-5:0.5:5).*kaiser(21,5.8);


15

Design Example 8.1 Repeated

Design a half-band, remez windowed filter• Sample Rate 20 kHz• Transition BW 4 kHz• Out-of-Band Attenuation 60 dB

h2=remez(20,[0 5-2 5+2 10]/10,[1 1 0 0],[1 1]);

Note: Zeros should be forced.

• Matlab has computation round-off; therefore, the zeros may be very small values after filter generation.

• Correct them by forcing the appropriate coefficients to zero.


16

Design Example Results

h2=remez(20,[0 5-2 5+2 10]/10,[1 1 0 0],[1 1]);


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Half-Band Remez Design Trick

• Can only the non-zero coefficients be generated? Yes.

h3a=remez(9, [0 5-2 5 5]/5, [1 1 0 0]);

h3 = zeros(1, 21);

h3(2:2:21)=0.5*h3a;

h3(10)=0.5;

Chap8_6.m

Interpolate the no-zeros filter by 2 and add the “center tap”.

The sum of two filters!


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Base Filter Responses

h3a=remez(9, [0 5-2 5 5]/5, [1 1 0 0]);


19

Coefficient Comparison

• Full remez half-band filter• Design trick coefficients

Thoughts on the Trick

• The initial “trick” filter is interpolated by 2 with no filtering.– A spectral replica must exist at the high-frequencies– The only way to remove the HF is to have perfect cancellation

• The sum of a delay element.– The unit coefficients provide phase delay.


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Chap8_6.m see the final 2 plots

When Windowing, Why use 21 instead of 19?

• The windows are different!

• For a 21 tap filter, w(1) and w(21) are multiplied by zero– We still must convolve in the frequency domain,

but the two windows are actually different.

• Selecting the “best window” means using 21 taps, not 19 taps.

• Note that the firpm filter is identical for 19 and 21. (4k-1 [or 4k+3] versus 4k+1 [or 4k+5])


21


22

Hilbert Transform Band-Pass Filter

• Could we keep just the positive frequency segment of the spectrum?– Real signal are conjugate symmetric, therefore the output would

have to be complex!

• Forming the Hilbert Transform using a half-band filter

2exp nihnh LPHT 2exp

2

2sin

21

nin

n

nhHP

Chap8_7.m

Why we might want this?

• Create an “analytic” signal– Only have positive frequency elements

• Easier to perform some functions on complex signals instead of a real signals– AM demodulation, FM demodulation– SSB signals

• If you want to decimate a complex signal• Complex to real conversion

– What happens when you take the real part of a complex signal?– What happens to the Nyquist frequency bound … it must change!


23


24

Hilbert Transform Band-Pass Filter

2exp4

2exp

nihnFF

ihnh LPs

sLPHT

ninn

n

nhHP 2sin

2cos

2

2sin

21

.....,1,,1,,1,,12exp iiiini

• Mix the filter by a complex carrier at Fs/4– Alternating sequence for mixing: even coefficients real, odd imaginary– Half-band filter zeros: odd coefficients exist, even coef are 0 except for h(0)


25

Hilbert Transform Coefficients

• Resolving the even and odd samples in n

ninn

n

nhHT 2sin

2cos

2

2sin

21

n

inn

n

inhoddn

nnn

n

nhevenn

HT

HT

12

sin

2

2sin

21:

2cos

2

2sin

21:

• One coefficient is real, all others are complex.– Interpretation: the complex filter coefficients cause the negative

portion of the signal spectrum to be cancelled


26

Hilbert Transform Response

• Notice:– The real and (anti-symmetric) imaginary coefficients– The shifting of the half-band filter spectrum

Chap8_7.m

Chap8_8.m

Application of HB and Hilbertin MRSP

• Filter Decimation (by a factor of 2)– After application, the signal can be decimated by a factor of 2– Real signal HB filtered to Fs/4,

decimate by 2 to allow signal-of-interest to “fill the spectrum”– Complex Signal filtered to positive frequencies 0 to Fs/2,

decimate by by 2 to allow signal-of-interest to “fill the spectrum”

• Filter Interpolation (by a factor of 2)– Interpolate by 2 and apply the filter to remove spectral replicas


27

Unique Filter Coefficients

• h(0) = 0.5, all other even coefficients h(2n)=0– When decimating by 2, this results in only 1 non-zero coefficient

• h(odd) are non-zero and anti-symmetric– When decimated by 2, this sequence remains

• Two decimation filters h0(n) and h1(n)– h0(n) has one non-zero coefficient, a pure time delay

• Always purely real– h1(n) has all the other coefficients

• For HB they are all purely real• For Hilbert they are all purely imaginary


28


29

Applying the Hilbert Transform

• Filter Decimation with the half-band filter.


30

Applying the Hilbert Transform

• Filter Decimation with the half-band Hilbert filter. – Note that since the values are real and imaginary, no addition of

the H0 and H1 filter outputs is required.


31

Applying Noble Identity

• Simplifying the structure


32

Complex Nyquist Representation

• Note that the input signal is effectively being decimated by a factor of 2, but that there are now both real and imaginary outputs for every 2 samples.

• As a result, the overall data rate has not been reduced!– For real data outputs, the Nyquist rate is 2x– For Hilbert transformed outputs, the Nyquist rate is 1x,

but there are two “information containing” samples per output.


33

Using Hilbert Transformed Data

• The passband spectrum is centered about fs/4. – Significant attenuation at 0 and fs/2

• Therefore, downconvert a communication signal to be centered at fs/4.

• Perform a Hilbert Transform Half-Band Filter– Only the positive complex spectrum remains

• Process as required– AM: envelope detection I^2+Q^2– FM: mix to baseband and use narrowband derivative function– PM: mix to baseband and extract phase


34

Interpolating with LP Half-Band Filters


35

Interpolating

• A standard polyphase interpolation structure


36

Computation Load (Repeat)

• There is one multiply for every unique coefficient– Conventional filter

• 4M-taps 4M Multiplies, 4M-1 Adds– Half-Band filter

• 4M+3 taps 2M+3 Multiplies, 2M+2 Adds– Half-Band filter, if h(0) doesn’t require a multiply,

• 4M+3 taps 2M+2 Multiplies, 2M+2 Adds– Half-Band filter, if h(0) doesn’t require a multiply and symmetric,

• 4M+3 taps M+1 Multiplies, M+M+1 Adds

– Potential savings: approximately ¼ the multiplies and ½ the adds!


37

Workload per Output Curves

• Filter length and attenuation factors are critical– The length of the filter shown in Figure 8.19 can be

estimated from the harris approximation presented in Chapter 3.

or from the figure


38

Filter Tap Estimation

The closer to fs/2, the more

taps are required.


39

Dyadic Half-Band Filters

• We now consider the use of a cascade of half-band filters to obtain a sample rate increase of any power of 2 such as increase by 8 or by 16.

• Suppose, for instance, we want to increase the sample rate of an input sequence by a factor 8. We have two primary options available to us. We can use an 8-path polyphase filter to accomplish this task, or we can use a cascade of three half-band filters.


40

Dyadic Half-Band Filters

• We first examine the workload for the sequence of half-band filters and then compare this workload to the M-path filter. – The sequence of half-band filters operates at successively higher

sample rates but with transfer functions that have successively wider transition bandwidths.

– There is a processing advantage to the cascade when the reduction in processing due to the wider transition bandwidth in successive filter stages compensates for operating the consecutive filter stages at successively higher sample rates.


41

Cascaded Half-Band Interpolation

• With each successive stage, the Half-Band Filter frequency responses can vary.– Notice that the desired passband becomes relatively smaller as the

sample rate increases. This allows the fractional bandwidth, alpha () to be continuously decreasing and the filter sizes getting shorter.


42

Half-Band Frequency Responses

Chap8_9.m


43

Operations Per Stage

• Notice how the original alpha has a reduced effect on the ratio as the stage number increases.

• The value of K1() including the interpolated rate factor.

Operations Per Stage Example

• Let = 45%


44

First Stage

Second Stage

Third Stage

Forth Stage

fs/f 20 3.64 2.58 2.68Ops/Input 20 7.27 10.32 21.41Ops Sum 20 27.27 37.59 59

• Let = 40%First Stage

Second Stage

Third Stage

Forth Stage

fs/f 10 3.33 2.5 2.22Ops/Input 10 6.67 10 17.78Ops Sum 10 16.67 26.67 44.45


45

Operations per Output

• Proportionality factors based on interpolation stages


46

Proportionality Factor Curves


47

M-path Polyphase Interpolation

For comparison, we can use an M-path polyphase filter to change the sample rate by a factor of M. We can recast equations (8.16) and (8.17) for the M-path filter to obtain (8.18).

As shown in (8.19) we can determine the length of each path of the M-path filter by distributing the N weights over the M-paths. If we assume that the top path, path-0, of the M-path filter contains only delays, then only (M-1) of the paths contributes to the workload and removing one of the M-paths from the workload estimate reduces the average workload. This scaled workload is shown in (8.20). Figure 8.24 presents graphical representations of (8.18) and (8.20).


48

Conventional Polyphase Curves


49

Cascaded (K2) vs. Conventional (K4)

• Cascaded becomes more efficient at higher Rate Changes!

Selecting Approach

• Polyphase Filter Implementation– Single structure or implementation– Small initial alpha

• Cascaded Half-Band Filters– Multiple Stages– Large alpha high rate change


50

In general, evaluate both methods to see which is more appropriate!

Half-band Interpolation

• Interpolate Filter

• M= 2 Interpolate Filter Polyphase


51

Half-band Decimation

• Filter Decimate

• M=2 Decimating Polyphase Filter


52

Half-band Polyphase Decimation and Interpolation

• Can you build MATLAB code that does this?– “brute force versus polyphase”– filter tap lengths change due to bandwidths


53

A Cascade of M=2 Polyphase

Filters

ECE 6560 Multirate Signal Processing Chapter 8bazuinb/ECE6560/Chap_08.pdfECE 6560 Multirate Signal...

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