university of manchester department of computer science cs3291 digital signal processing...

Nov'05 CS3291: Section 4 1

University of Manchester

Department of Computer Science

CS3291

Digital Signal Processing '05-'06

Section 4:

‘A design technique for FIR digital filters’


4.1.Introduction:

FIR digital filter of order M implemented by programming the signal flow graph shown below. Its difference equation is:

y[n] = a0x[n] + a1x[n-1] + a2x[n-2] + ... + aMx[n-M]

z-1 z-1 z-1 z-1 x[n]

y[n]

a0 a1

...

aM-1 aM


• Its impulse-response is {..., 0, ..., a0, a1, a2,..., aM, 0, ...} • Its frequency-response is: M H(e j ) = a n e - j n n=0 •Consider problem of choosing a0, a1,..., aM such that H( ej ) is close to some target frequency-response. •Use the inverse DTFT:

deeHnh njj )(

2

1][


• Assume we require low-pass filter whose gain-response approximates ideal 'brick-wall' gain-response:.

0

1

Gi( )

-

/3 - /3

If we take phase-response () = 0 for all , the required frequency-response is:

<</3 : 0

3/ : 1)()( )(jj eGeH


By the inverse DTFT,

0n :

0=n :

)3/sin()/1(

3/1

12

1 ][

3/

3/

nn

denh nj

= (1/3)sinc(n/3) for all n.

0:1

0:)sin(

)(sinc wherex

xx

xx


x

1 2-1-2-3 3

1 sinc(x)


• A digital filter with this impulse-response would have exactly the required low-pass frequency-response.

• But {h[n]} has non-zero samples extending from n = - to ,

• Not a finite impulse-response.

• Also not causal.

1/3

h[n]

n

Ideal impulse response

for low-pass filter

cut-off pi/3


To produce a realisable impulse-response:

(1) Truncate {h[n]} to a FIR by setting h[n] to zero for all values of n outside the range -M/2 n M/2 ( assume M is the order of the required FIR filter and is even ).

(2) Delay resulting sequence by M/2 samples to ensure that the first non-zero sample occurs at n = 0.


n

M=10

h[n] = (1/3)sinc(n/3)

Starting with ideal impulse response:


n

n

h[n]

h[n]

M=10Truncate to M/2

Delay by M/2 samples


• Resulting causal impulse response realised by setting an = h[n] for n=0,1,2,...,M.

• Taking M=4, for example, the finite impulse response obtained for the /3 cut-off low-pass specification is : {..,0,..,0, 0.14, 0.28, 0.33 , 0.28 , 0.14 , 0 ,..,0,..}


• Gain & phase responses given in Figure 4.4.

z-1 z-1 z-1 z-1 x[n]

y[n]

0.28 0.28 0.14 0.330.14

• Resulting FIR filter is as shown below:

( Note:4th order FIR filter has 4 delays & 5 multipliers ).


G(

dB

0

-10

-20

-30

/3

-6 dB

Fig 4.4


• Truncating {h[n]} to M/2 & delaying by M/2 samples produces gain & phase responses different from those originally specified.

• Gain-response: cut-off rate not sharp, two "ripples" appear in stop-band, peak of the first ripple at about -21dB.

• Phase-response: not zero for all as orig specified, ( ) = - ( M/2 ) for | | /3;

linear phase in pass-band slope arctan(M/2) with M = 4 . phase-delay M/2 = 2 samples.


Question.Why does delaying {h[n]} by M/2 produce this effect? Answer: If DTFT of {h[n]} is H(ej) = G()ej(), the DTFT of the delayed impulse-response:

/2) M-)(j(

)(j/2Mj-2/

2/

)2/(

e )G(

)eG(e = )(=

][ =

M/2-nm settingby ][ = ]2/[

jjM

n

jmjM

m

jMm

n

jn

eHe

emhe

emheMnh

Delaying {h[n]} by M/2 samples multiplies H(ej) by e-Mj/2 which increases () by -M / 2. Increases phase-delay -() / by M/2 sampling intervals.


• Can we improve low-pass filter by increasing order to ten?

• Taking 11 terms of { (1 / n) sin (n / 3) } we get, after delaying by 5 samples: {...0,-0.055,-.069, 0,.138,.276,.333,.276,.138,0,-.069,-.055,0,...}.

• If we analyse, by computer, a tenth order FIR filter with this impulse-response, we obtain the gain response shown in fig 4.5.


for n=1:11; a(n)= 0.3333*sinc((n-6)/3); end; H=freqz(a,1,1000); plot( [0:999]/1000, 20*log10(abs(H) ); axis( 0,1,-60,10] ); grid on; xlabel( ‘rel_freq / pi ); ylabel( ‘Gain (dB)’ );

freqz( [-0.055, -0.069, 0, 0.138, 0.276, 0.333, 0.276, 0.138, 0, -0.069, -0.055] );

Alternative graph may be produced by the single statement:

The graph was produced by the following MATLAB statements:


A note on the phase-response graphs produced by ‘freqz’

• Plots () against rather than phase lag -().

• Segments at phase intervals of 2 or 360o

-()()

/

2

4

2

1

freqz


Gain response of 10th order lowpass FIR filter with C = /3


Gain response of 20th order lowpass FIR filter with C = /3


• Cut-off rate for 10th order FIR filter sharper than for 4th order.

• More stop-band ripples.

• Gain at peak of first ripple after cut-off remains at -21 dB.

• This effect is due to Gibb's phenomenon.

• Linear phase in passband with phase delay of 5 samples.

• Going to 20th order produces even faster cut-off rates,

& more stop-band ripples,

but main stop-band ripple remains at about -20dB.

• To improve matters we need to discuss " windowing ".


4.2. Windowing:

To design FIR filters we multiplied {h[n]} as calculated by inverse DTFT by a rectangular window sequence {r[n]} where

2/n :

M/2n M/2- :

0

1 = ][

Mnr

r[n]

n


• Causes sudden transition to zero at window edges.• It is these transitions that produce the stop-band ripples. • Levels of ripples reduced if {r[n]} replaced by non-rectangular window sequence { w[n] }. • Produces a more gradual transition at the window edges. • Simple non-rectangular window sequence is Hann window • It is a "raised cosine "

M/2 > n :

M/2n -M/2:

0

))2/1/(cos(1(5.0][

Mn

nw

w[n]

n

M/2 -M/2


• Other types of window exist ( e.g. Hamming, Kaiser ). • Multiplying {h[n]} by {w[n]} instead of {r[n]} gradually tapers impulse-response towards zero at window edges. • Consider again low-pass filter example with rel cut-off /3. • Ideal impulse-response was found to be:

{h[n]} = { …..... , 0.14, 0.28, 0.33, 0.28, 0.14, ………}

• When M = 4, the Hann window {w[n]} = {..,0,..,0, 0.25, 0.75, 1, 0.75, 0.25, 0,..,0,..}

• Multiplying term by term & delaying by M/2 = 2 samples we get: {..,0,..,0, 0 .04, 0.21, 0.33, 0.21, 0.04, 0,..,0,..}


Resulting "Hann-windowed" FIR filter of order 4 is as shown below with a0=0.04, a1=0.21, a2=0.33, a3 = 0.21, a4 = 0.04.

Its gain-response is approximately as shown in Figure 4.9.

z-1 z-1 z-1 z-1 x[n]

y[n]

a0 a1 a3 a4 a2


G(dB

0

-10

-20

-30

-6dB

2

Fig. 4.9


Figure 4.9: 4th order FIR filter (Hann)


4.3 Effect of windowing on freq-response of FIR digital filter

• Effect is to gradually reduce amplitude of ideal impulse-response towards zero at edges of window rather than to abruptly truncate.

• Effect on gain-response of FIR filter obtained is:

i) to greatly reduce stop-band ripples ( good ). ii) to reduce the cut-off rate ( bad ).

• Phase-response is not affected in the pass-band.

• We can improve the cut-off rate by going to higher orders. • Graphs below are for 10th & 20th order ( Hann windowed ):


Tenth order FIR filter with C = /3 ( Hann window )


MATLAB program to design & graph 10th order FIR lowpass filter with Hann window

w=hanning(11); for n=1:11; a(n) = 0.3333*sinc((n-6)/3)*w(n); end; h = freqz(a,1,1000); plot([0:999]/1000,20*log10(abs(h)),'k'); axis([0,1,-50,0]); grid on; xlabel('Rel_freq / pi'); ylabel('Gain(dB)');


20th order FIR filter with C = /3 (Hann window)


4.4. Highpass, band-pass & band-stop linear phase FIR filters

• Can be designed almost as easily as low-pass.• Remember to define required gain-response GI() from - to +• Make GI(-) = GI(). • Band-pass filter with pass-band from fS/8 to fS/4 has following gain response ideally:-

Gain

1


• Applying the inverse DTFT (not forgetting negative ):

|| 2/ : 0

2/ || 4/ : 1

4/ || : 0

)( jeH

• Taking () = 0 for all initially as before, we obtain:

dededeeHnh njnjnjj 2/

4/

4/

2/ 1

2

1 1

2

1 )(

2

1 ][

• Can evaluate this, but there is a nicer way


H(ej) = H1(ej) H2(ej)

- /4/2- -/2 -/4

11

deHdeHnh njnj

)(e

2

1 )(e

2

1 ][ j

2j

1

= (1/2)sinc(n/2) - (1/4)sinc(n/4) for - < n <


Exercise: By a similar method, or otherwise, show that the impulse-response for an ideal zero phase 'brick-wall' high-pass filter with cut-off frequency /6 radians per sample (i.e. one twelfth of the sampling frequency) is: h[n] = sinc(n) - (1/6)sinc(n/6) for - < n <

0 : )6/(sin)6/1(

0 : 6/11

nnc

n

Exercise: Derive the impulse-response for an ideal zero phase 'brick-wall' band-stop digital filter with cut-off frequencies L and U radians per sample.


• Technique not restricted to “conventional” gain-responses. • It is not difficult to design a linear phase filter whose gain-response approximates that below:

Gain

1

0.5


4.5. Summary of design technique

To design an FIR digital filter of even order M, with gain response G I () and linear phase by the windowing method,1) Set H(ej) = G I () the required gain-response. This assumes () = 0.2) IDTFT to produce the ideal impulse-response {h[n]}.3) Window to M/2 using chosen window.4) Delay windowed impulse-response by M/2 samples.5) Realise by setting multipliers of FIR filter.


• Instead of obtaining H( ej ) = G I ( ), we get e-jM/2G()

• G() is distorted version of GI() due to windowing.

• Phase-response is () = -M/2 which is linear phase . • -() / = M/2 for all .

• Filter coeffs symmetric about M/2.

e.g. {…2, -3, 5, 7, 5, -3, 2, …} M =6 (even)

{…, 1, 3, 5, 5, 3, 1, …} M=5 (odd)

This is because the filters are linear phase.


= e-5j/2 (e5j/2 +3e3j/2 +5ej/2 +5e-j/2 +3e-3j/2 +e-5j/2 ) = e-5j/2 (2cos(2.5) + 6cos (1.5) + 10cos(/2) )

= G()ej() with () = -5/2.

• Hence () / = -5/2 = constant, so H(ej) is linear phase.

• FIR digital filters whose impulse-responses are symmetric are linear phase. • Let {h[n]} = {…, 1, 3, 5, 5, 3, 1, …}

jjjjj

n

njj eeeeeenheH 5432 35531][)(


• Windowing or "Fourier series approxn technique ".

• It is possible to design FIR filters which are not linear phase


4.6 Further applications of windowing design technique

• Technique even more powerful than has been yet indicated• Not restricted to linear phase filters. • Consider some further examples of its use.

4.6.1 Fractional sampling interval delay filter:

4.6.2. Differentiator:

4.5.3. Hilbert transformer:

See notes for details


4.6.1. Fractional sampling interval delay filter: Gain-response is one for all , but whose phase-response is () = - 0.5 . Therefore:

e -j / 2 : 0 H(e j ) =

e j / 2 : - 0


4.6.2. Differentiator:

Would produce cos(t) from sin (t). FIR filter can act as a differentiator if it outputs {(/T) cos(n)} when the input is {sin(n)} for any in the range - to . Required gain-response is therefore G I () = /T.Instead of specifying a phase response of zero, specify a required phase lead of /2. Required frequency response is:

(/T) e j / 2 : 0 j /T : 0 H(e j ) = =

(/T e - j / 2 : - 0 -j /T : - 0


4.5.3. Hilbert transformer:

G() = 1 and () = -/2 for 0 < < . “Quadrature phase” rather than linear phase. A linear phase component is added to the 90o phase shift when we delay the windowed impulse response for causality. The required frequency response is:

e - j / 2 : 0 -j : 0 H(e j ) = 0 : = 0 = 0 : = 0

e j / 2 : - 0 j : - 0


Designing & implementing FIR digital filters using SP Toolbox

• Design of linear phase FIR digital filters by windowing technique carried out by command 'FIR1'. • Applied to segment of sampled sound by command 'filter'.

• Design & implement 128th order FIR band-pass digital filter with passband 300 Hz to 3.4 k Hz • Apply it to wav file of mono music sampled at 11.025 kHz.

• Notes: (1) FIR cut-off frequencies specified relative to fS/2 .(2) By default FIR1 uses Hamming window; (other windows such as Hann can be specified).(3) Filter scaled so centre of pass-band has magnitude one.


clear all;[x, fs, nbits] = wavread('caprice.wav');%To design:wlow = 2 * pi * 300 / fs ; % radians/sample rel to fswup = 2 * pi * 3400 / fs ; % radians/sample rel to fsa = fir1(128, [wlow wup] / pi ) ;freqz ( a , 1) ; % plot gain & phase% To implement:L=length(x);y = filter(a, 1, x ); % Remember the delay of 64 samples.wavwrite(x,fs,nbits,'capnew.wav');


Remez Exchange Algorithm method:

• Better than windowing technique, but more complicated. • Available in MATLAB. • Design 40th order FIR lowpass filter whose gain is unity (0 dB) in range 0 to 0.3 radians/sample & zero in range 0.4 to .

• The 41 coefficients will be found in array ‘a’. • Produces 'equi-ripple' gain-responses where peaks of stop-band ripples are equal rather than decreasing with increasing frequency. • Highest peak in stop-band lower than for FIR filter of same order designed by windowing technique to have same cut-off rate. • There are 'equi-ripple' pass-band ripples.


a = remez (40, [0, 0.3, 0.4,1],[1, 1, 0, 0] );h = freqz (a,1,1000);plot([0:999]/1000,20*log10(abs(h)),'k');axis([0,1,-50,0]);grid on;xlabel('Rel_freq / pi');ylabel('Gain(dB)');


Gain of 40th order FIR lowpass filter designed by “ Remez ”


4.9. Some implementation issues4.9.1. Fixed point implementation of FIR digital filters• FIR digital filters often implemented in mobile equipment. • Low power fixed point DSP processors are norm.• Typically with a basic 16-bit word-length. • Must be programmed using only integer arithmetic.• Take 4th order FIR filter with impulse response: {….. 0.04, 0.21, 0.33, 0.21, 0.04, …...}. • Rounding each coeff to nearest integer clearly a mistake. • Multiply all coeffs by a large constant then round: . A0 = 4, A1 = 21, A2 = 33, A3 = 21 , A4 = 4.

We must divide the output by same constant, in this case 100. Instead of 100, we choose a power of two for the constant. Dividing by a power of two (e.g. 1024) is very simple.


• The larger the constant, the more accurate the coefficients. • Careful not to choose too large a constant• If integers produced get too large, we risk overflow • Difficult balancing act between inaccuracy & overflow. • FIR filters easy to program in fixed point arithmetic. • Never become unstable as there is no feedback. • In some cases, overflows can be allowed to occur • Can risk overflow more readily with FIR digital filters than with IIR digital filters, and thus have greater coefficient accuracy.

• Scaling by 1024, is adopting a 'Q-format' of ten.• Programmer assumes a binary point to exist ten bit positions from the right within the 16-bit word.


MATLAB implementation of 4th order low-pass filter using integer arithmetic only:

A = [4 21 33 21 4 ] ; x = [0 0 0 0 0 ] ; while 1

x(1) = input( 'X = '); Y = A(1)*x(1); for k = 5 : -1: 2

Y = Y + A(k)*x(k);x(k) = x(k-1);

end; Y = round( Y / 100) ;

disp([' Y = ' num2str(Y)]); end;


4.9.2. Alternative signal flow graphs

•Signal flow-graph below has same impulse-response as fig 4.1.

•Same input x[n] applied to all multipliers.

•Could have advantages & other alternatives exist.

z-1 z-1 z-1+ + +

AMA0AM-2AM-1

x[n]

y[n]


PROBLEMS:1 Design a 10th order FIR low-pass digital filter with cut-off at fS/4 with and without a Hann window. Use MATLAB to compare the gain responses obtained.2 Design a 10th order FIR bandpass digital filter with lower and upper cut-off frequencies at /4 and /2 respectively.3 Write a MATLAB program for one of these filters using integer arithmetic only.4 Design a 4th order FIR high-pass filter with cut-off at /3.5 Do all FIR filters have exactly linear phase responses?6 Show that if { h[n] } is real, H( e-j ) = H*( ej ) and hence that:

dneHnh j ))(cos(|)(| 1

][0

7 Show that if H( ej ) is linear phase with a phase delay N samples, with N an integer, its impulse-response is symmetric about n=N.


8. Design 6th order FIR filter which is linear phase & delays input by half a sampling interval (plus usual 3 for causality).9. Design 6th order linear phase FIR filter whose gain-response approximates that shown in fig 4.13.10. Show that if input to ideal Hilbert transformer is {cos(n)} output will be {sin(n)} for all . 11.Design 8th order FIR approx to all-pass Hilbert transformer . Sketch a graph of its phase-response.12. Show that ( ) = -k corresponds to a delay of k samples.13 Give H(z), {h[n]} and a program for Fig 4.15 & mention any other advantages it may have fig 4.1. 14 Rearrange figs 4.1 and 4.15 for greater efficiency when M is even & the filter is linear phase.15 Explain why IIR digital filters cannot have exactly linear phase responses, whereas FIR filters can.

university of manchester department of computer science cs3291 digital signal processing...

Documents