Chapter 14 Finite Impulse Response (FIR) Filtersgalia.fc.uaslp.mx/~rmariela/RTDSP/ch62.pdf · Chapter 1, Slide 1 Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004 Chapter

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004Chapter 1, Slide 1

Chapter 14Chapter 14

Finite Impulse Response (FIR) FiltersFinite Impulse Response (FIR) Filters

Learning ObjectivesLearning Objectives

�� Introduction to the theory behind FIR Introduction to the theory behind FIR filters:filters:

�� Properties (including aliasing).Properties (including aliasing).

�� Coefficient calculation.Coefficient calculation.

�� Structure selection.Structure selection.


�� Implementation in Matlab, C, assembly Implementation in Matlab, C, assembly and linear assembly.and linear assembly.

IntroductionIntroduction

�� Amongst all the obvious advantages that Amongst all the obvious advantages that digital filters offer, the FIR filter can digital filters offer, the FIR filter can guarantee linear phase characteristics.guarantee linear phase characteristics.

�� (either analogue or IIR filters can achieve (either analogue or IIR filters can achieve this.this.

There are many commercially available There are many commercially available


�� There are many commercially available There are many commercially available software packages for filter design. software packages for filter design. However, without basic theoretical However, without basic theoretical knowledge of the FIR filter, it will be knowledge of the FIR filter, it will be difficult to use them.difficult to use them.

Properties of an FIR FilterProperties of an FIR Filter

�� Filter coefficients:Filter coefficients:

[ ] [ ]∑−

=

−⋅=1

0

�

k

k knxbny

x[n]x[n] representsrepresents thethe filterfilter input,input,


x[n]x[n] representsrepresents thethe filterfilter input,input,

bbk k represents the filter coefficients,represents the filter coefficients,

y[n]y[n] representsrepresents thethe filterfilter output,output,

�� isis thethe numbernumber ofof filterfilter coefficientscoefficients

(order(order ofof thethe filter)filter)..



[ ] [ ]∑−

=

−⋅=1

0

�

k

k knxbny

z-1 z-1 z-1 z-1x(n)

FIR equationFIR equation


z-1

+

z-1 z-1

+ +

z-1

y(n)

x(n)

x xxxb0 b1 b2 bN-1

Filter structureFilter structure



[ ] [ ]∑−

=

−⋅=1

0

�

k

k knxbny

�� If the signal x[n] is replaced by an impulse If the signal x[n] is replaced by an impulse



δδ[n] then:[n] then:

[ ] [ ]∑−

=

−=1

0

�

k

k knbny δ

[ ] [ ] [ ] [ ]�bbby k −++−+= δδδ L100 10



[ ] [ ]∑−

=

−⋅=1

0

�

k

k knxbny





[ ] [ ]∑−

=

−=1

0

�

k

k knbny δ

[ ] [ ] [ ] [ ]�nbnbnbny k −++−+= δδδ L110



[ ] [ ]∑−

=

−⋅=1

0

�

k

k knxbny





[ ] [ ]∑−

=

−=1

0

�

k

k knbny δ

[ ]

≠=

=−kn for 0

knfor 1knδ



[ ] [ ]∑−

=

−⋅=1

0

�

k

k knxbny

�� FinallyFinally::


�� FinallyFinally::

[ ][ ]

[ ]khb

hb

hb

k =

=

=

M

1

0

1

0



[ ] [ ]∑−

=

−⋅=1

0

�

k

k knxbny

WithWith:: [ ]khb =


WithWith:: [ ]khbk =

�� The coefficients of a filter are the same as the The coefficients of a filter are the same as the

impulse response samples of the filter.impulse response samples of the filter.

Frequency Response of an FIR FilterFrequency Response of an FIR Filter

�� By taking the zBy taking the z--transform of h[n], H(z):transform of h[n], H(z):

�� Replacing z by eReplacing z by ejjωω in order to find the in order to find the

( ) [ ]∑−

=

−=1

0

�

n

nznhzH


�� Replacing z by eReplacing z by e in order to find the in order to find the

frequency response leads to:frequency response leads to:

( ) ( ) [ ]∑−

=

−=

==1

0

�

n

jnj

ezenheHzH j

ωωω


�� Since eSince e--j2j2ππππππππkk = 1 then:= 1 then:

�� Therefore:Therefore:

( ) [ ] ( ) [ ]∑∑−

=

−−

=

+−=

==+

1

0

1

0

22

�

n

jn�

n

jn

ezenhenhzH ωπω

πω


( ) ( )ωπω jkj eHeH =+ 2

�� FIR filters have a periodic frequency FIR filters have a periodic frequency

response and the period is 2response and the period is 2ππ..


�� Frequency Frequency response:response:

( ) ( )ωπω jkj eHeH =+2

FIRFIR y[n]y[n]x[n]x[n]


FFss/2/2FFss/2/2

FreqFreq

FreqFreq

x[n]

x[n]

y[n]

y[n]


�� Solution: Use an antiSolution: Use an anti--aliasing filter.aliasing filter.

FIRFIR y[n]y[n]x[n]x[n]

ADCADC

Analogue Analogue

AntiAnti--AliasingAliasing

x(t)x(t)


FFss/2/2FFss/2/2FreqFreqFreqFreq

x(t)

x(t)

y[n]

y[n]

Phase Linearity of an FIR FilterPhase Linearity of an FIR Filter

�� A causal FIR filter whose impulse A causal FIR filter whose impulse response is symmetrical is guaranteed to response is symmetrical is guaranteed to have a linear phase response.have a linear phase response.


0n

h(n)

1 n n+1 2n+12n

N = 2n + 2

0n

h(n)

1 n n+1 2n2n-1n-1

N = 2n + 1

Even symmetryEven symmetry Odd symmetryOdd symmetry

Phase Linearity of an FIR FilterPhase Linearity of an FIR Filter

�� Application of 90Application of 90°° linear phase shift:linear phase shift:

Signal Signal

separationseparation

9090oo

delaydelay

delaydelay

++

++

II

ReverseReverse

IHIH


separationseparation

9090oo

delaydelay

++

--

QQ

ForwardForward

QHQH

tBtAI rf ωω sincos +=

tBtAQ rf ωω cossin += tBtA

tBtAIH

rf

rf

ωω

πωπω

cossin

2sin

2cos

+−=

++

+= tBQIH rωcos2=+

tBIQH fωsin2=−

Design ProcedureDesign Procedure

�� To fully design and implement a filter five To fully design and implement a filter five steps are required:steps are required:

(1)(1) Filter specification.Filter specification.

(2)(2) Coefficient calculation.Coefficient calculation.

(3)(3) Structure selection.Structure selection.


(4)(4) Simulation (optional).Simulation (optional).

(5)(5) Implementation.Implementation.

Filter Specification Filter Specification -- Step 1Step 1

(a)

1

f(norm)fc : cut-off frequency

pass-band stop-band

pass-band stop-bandtransition band

fs/2

|H(f)| |H(f)|

|H(f)|


pass-band stop-bandtransition band

1

sδ

pass-bandripple

stop-bandripple

fpb : pass-band frequency

fsb : stop-band frequencyf(norm)

(b)

p1 δ+

s∆

p∆0

-3

p1 δ−

fs/2

fc : cut-off frequency

|H(f)|(dB)

|H(f)|(linear)

Coefficient Calculation Coefficient Calculation -- Step 2Step 2

�� There are several different methods There are several different methods available, the most popular are:available, the most popular are:

�� Window method.Window method.

�� Frequency sampling.Frequency sampling.

�� ParksParks--McClellan.McClellan.


�� We will just consider the window method.We will just consider the window method.

Realisation Structure Selection Realisation Structure Selection -- Step 3Step 3

( ) ∑−

=

−=1

0

�

k

k

k zbzH

( ) ( ) ( )zXzHzY ⋅= ( ) ( ) ( ) ( )1....1 110 +−++−+= − �nxbnxbnxbny �

�� Direct form structure for an FIR filter:Direct form structure for an FIR filter:


z-1 z-1 z-1

+ + +

b0

b2

bN-1

y(n)

x(n)

b1


( ) ∑−

=

−=1

0

�

k

k

k zbzH

�� Linear phase structures:Linear phase structures:



�� N even:N even:

�� N Odd:N Odd:

( ) ( )∑−

=

−−− +=1

2

0

1

�

k

k�k

k zzbzH

( ) ( )∑

−

=

−−

−−−− ++=

2

1

0

2

1

2

1

1

�

k

�

�

k�k

k zbzzbzH


(a) N even.(a) N even.

(b) N odd.(b) N odd.+

b0+

+

+

+b

1

+b

2

+b

N/2-1

y(n)

(a)

z-1

z-1

z-1

z-1

z-1


+

b0+

+

+

+b1

+b2

+b(N-3)/2

y(n)

x(n)

b(N-1)/2

+

(b)

z-1

z-1

z-1

z-1

z-1

z-1


( ) ∑−

=

−=1

0

�

k

k

k zbzH

�� Cascade structures:Cascade structures:



( ) ( )

( )

( )∏

∑

=

−−

−−−−−

−−−

−−−

=

−

++=

++++=

++++==

M

k

kk

��

�

�

�

k

k

k

zbzbb

zb

bz

b

bz

b

bb

zbzbzbbzbzH

1

2

2,

1

1,0

1

0

12

0

21

0

10

1

1

2

2

1

10

1

0

1

...1

...


( ) ∑−

=

−=1

0

�

k

k

k zbzH

�� Cascade structures:Cascade structures:



z -1

+b

1,1

x(n)

z -1

+

b1,2

z -1

+b

2,1

z -1

+

b2,2

z -1

+b

M,1

z -1

+

bM,2

y(n)b0

Simulation, Step 4Simulation, Step 4

0 20 40 60 80 100 120 140-0.1

0

0.1

0.2

0.3

0.4

Coefficient number, n

h(n)

Truncated Impulse Response


0 0.5 1 1.5 2

x 104

-6000

-4000

-2000

0

Frequency (Hz)

Pha

se (

degr

ees)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

-60

-40

-20

0

Frequency (Hz)

Mag

nitu

de (

dB)

Implementation Implementation -- Step 5Step 5short fir_filter (short input)

{

int i;

short output;

int acc=0;

int prod;

R_in[0] = input; /* Update most recent sample */

acc = 0; /* Zero accumulator */

for (i=0; i<128; i++) /* 128 taps */


{

prod = (h[i]*R_in[i]); /* Perform Q.15 multiplication */

acc = acc + prod; /* Update 32-bit accumulator, catering */

} /* for temporary overflow */

output = (short) (acc>>15); /* Cast output to 16-bits */

for (i=127; i>0; i--) /* Shift delay samples */

R_in[i]=R_in[i-1];

return output;

}

Filter coefficientsFilter coefficients-0.0002 0.0002 0.0004 0.0002 -0.0002 -0.0005 -0.0003 0.0003

0.0007 0.0004 -0.0004 -0.0009 -0.0005 0.0005 0.0012 0.0007

-0.0007 -0.0016 -0.0009 0.0009 0.0021 0.0012 -0.0013 -0.0027

-0.0015 0.0016 0.0035 0.0019 -0.0020 -0.0044 -0.0024 0.0026

0.0055 0.0030 -0.0032 -0.0069 -0.0037 0.0040 0.0085 0.0045

-0.0049 -0.0105 -0.0056 0.0061 0.0131 0.0070 -0.0076 -0.0165

-0.0089 0.0097 0.0212 0.0117 -0.0128 -0.0285 -0.0159 0.0180

0.0411 0.0239 -0.0284 -0.0699 -0.0452 0.0634 0.2119 0.3183

0.3183 0.2119 0.0634 -0.0452 -0.0699 -0.0284 0.0239 0.0411

0.0180 -0.0159 -0.0285 -0.0128 0.0117 0.0212 0.0097 -0.0089

-0.0165 -0.0076 0.0070 0.0131 0.0061 -0.0056 -0.0105 -0.0049

0.0045 0.0085 0.0040 -0.0037 -0.0069 -0.0032 0.0030 0.0055

0.0026 -0.0024 -0.0044 -0.0020 0.0019 0.0035 0.0016 -0.0015

-0.0027 -0.0013 0.0012 0.0021 0.0009 -0.0009 -0.0016 -0.0007


-0.0027 -0.0013 0.0012 0.0021 0.0009 -0.0009 -0.0016 -0.0007

0.0007 0.0012 0.0005 -0.0005 -0.0009 -0.0004 0.0004 0.0007

0.0003 -0.0003 -0.0005 -0.0002 0.0002 0.0004 0.0002 -0.0002

Implementation Implementation -- Step 5Step 5

short R_in[128]; /* Input samples R_in[0] most recent, R_in[127] oldest */

/* coefficient calculated with sampling frequency of 48kHz and cutoff at 8kHz*/

short h[]= /* Impulse response of FIR filter */

{ -7, 7, 14, 7, -8, -17, -9 , 10,

22, 12, -13, -29, -16, 18, 39 , 22,

-24, -52, -29, 31, 69, 38, -41 , -89,

-49, 53, 114, 62, -67, -144, -78 , 84,


181, 97, -104, -225, -121, 130, 278 , 149,

-160, -345, -185, 199, 429, 231, -250 , -541,

-293, 319, 696, 382, -421, -933, -522 , 589,

1347, 783, -932, -2292, -1480, 2079, 6945 , 10429,

10429, 6945, 2079, -1480, -2292, -932, 783 , 1347,

589, -522, -933, -421, 382, 696, 319 , -293,

-541, -250, 231, 429, 199, -185, -345 , -160,

149, 278, 130, -121, -225, -104, 97 , 181,

84, -78, -144, -67, 62, 114, 53 , -49,

-89, -41, 38, 69, 31, -29, -52 , -24,

22, 39, 18, -16, -29, -13, 12 , 22,

10, -9, -17, -8, 7, 14, 7 , -7};

void main()

{

DSK6713_AIC23_CodecHandle hCodec;

Int16 OUT_L,OUT_R;

Uint32 IN_L,IN_R;

/* Initialize the board support library, must be called first */

DSK6713_init();

/* Start the codec */

hCodec = DSK6713_AIC23_openCodec(0, &config);

// Set codec frequency to 48KHz

DSK6713_AIC23_setFreq(hCodec, DSK6713_AIC23_FREQ_48KHZ );

while (1)

{


// Read sample from the left channel

while (!DSK6713_AIC23_read(hCodec, &IN_L));

// Read sample from the right channel

while (!DSK6713_AIC23_read(hCodec, &IN_R));

OUT_L = fir_filter(IN_L);

OUT_R = IN_R;

/* Send the FILTERED sample to the left channel */

while (!DSK6713_AIC23_write(hCodec, OUT_L));

/* Send a NON FILTERED sample to the right channel */

while (!DSK6713_AIC23_write(hCodec, OUT_R));

}

/* Close the codec */

DSK6713_AIC23_closeCodec(hCodec);

}

Floating Point ImplementationFloating Point Implementation

float R_in[128]; /* Input samples R_in[0] most recent, R_in[127] oldest. */

float h[]= /* Impulse response of FIR filter. Taken from fir_coef.txt */

{

-0.0001,0.0003,0.0004,0.0002,-0.0002,-0.0005,-0.0004,0.0001,

0.0006,0.0006,0.0001,-0.0006,-0.0010,-0.0005,0.0005,0.0013,

0.0010,-0.0002,-0.0016,-0.0017,-0.0003,0.0016,0.0025,0.0012,

-0.0013,-0.0032,-0.0025,0.0006,0.0036,0.0039,0.0007,-0.0036,

-0.0054,-0.0027,0.0029,0.0068,0.0052,-0.0012,-0.0076,-0.0081,


-0.0015,0.0074,0.0112,0.0055,-0.0059,-0.0139,-0.0108,0.0026,

0.0159,0.0173,0.0033,-0.0165,-0.0254,-0.0129,0.0145,0.0355,

0.0291,-0.0075,-0.0507,-0.0623,-0.0141,0.0897,0.2093,0.2890,

0.2890,0.2093,0.0897,-0.0141,-0.0623,-0.0507,-0.0075,0.0291,

0.0355,0.0145,-0.0129,-0.0254,-0.0165,0.0033,0.0173,0.0159,

0.0026,-0.0108,-0.0139,-0.0059,0.0055,0.0112,0.0074,-0.0015,

-0.0081,-0.0076,-0.0012,0.0052,0.0068,0.0029,-0.0027,-0.0054,

-0.0036,0.0007,0.0039,0.0036,0.0006,-0.0025,-0.0032,-0.0013,

0.0012,0.0025,0.0016,-0.0003,-0.0017,-0.0016,-0.0002,0.0010,

0.0013,0.0005,-0.0005,-0.0010,-0.0006,0.0001,0.0006,0.0006,

0.0001,-0.0004,-0.0005,-0.0002,0.0002,0.0004,0.0003,-0.0001

};

Floating Point ImplementationFloating Point Implementation

short fir_filter (short sample)

{

int i;

float acc=0;

float prod;

R_in[0] = (float) sample; /* Update most recent sample */

acc = 0; /* Zero accumulator */

for (i=0; i<128; i++) /* 128 taps */


for (i=0; i<128; i++) /* 128 taps */

{

prod = (h[i]*R_in[i]);

acc = acc + prod;

}

for (i=127; i>0; i--) /* Shift delay samples */

R_in[i]=R_in[i-1];

return (acc);

}

Documents

Chapter 14 Finite Impulse Response (FIR) Filtersgalia.fc.uaslp.mx/~rmariela/RTDSP/ch62.pdf · Chapter 1, Slide 1 Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004 Chapter