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Finite Impulse Response (FIR)

Digital Filters (II)

Ideal Impulse Response Design Examples

Yogananda Isukapalli

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• FIR Filter Design Problem

y[n]

1 z-1 z-1 z-1 z-1

h0 h1 h2 h3 hN-2 hN-1

x[n]

1 1 1 1 1

Consider a general (infinite impulse response) definition:

å¥

-¥=

-=n

nznhzH ][)(

Given H(z) or H(ejw), find filter coefficients{b0, b1, b2, ….. bN-1} which are equal to {h0, h1, h2, ….hN-1} in the case of FIR filters.

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Where C is a counterclockwise closed contour in the region of convergence of H(z) and encircling the origin of the z-plane

å¥

-¥=

-=n

jnwjw enheH ][)(

From complex variable theory, the inverse transform is:

ò -= C dznzzHj

nh 1)(21][p

• Evaluating H(z) on the unit circle ( z = ejw ) :

dweeHnh jnwjw)(21][ ò

-

=p

pp where dz = jejw dw

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• Design of an ideal low pass FIR digital filter

-2p -p -wc 0 wc p 2p w

H(ejw)K

Find ideal low pass impulse response {h[n]}

ò-

=p

ppdweeHnh jnwjw

LP )(21][

ò-

=c

c

w

w

jnwdwKep21

Hence

)sin(][ cLP nwnKnhp

= n = 0, ±1, ±2, …. ±¥

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Let K = 1, wc = p/4, n = 0, ±1, …, ±10

The impulse response coefficients are

n = ±4, h[n] = 0= ±5, = -0.043= ±6, = -0.053= ±7, = -0.032

n = 0, h[n] = 0.25= ±1, = 0.225= ±2, = 0.159= ±3, = 0.075

n = ±8, h[n] = 0= ±9, = 0.025= ±10, = 0.032

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Non CausalFIR ImpulseResponse

We can make it causal if we shift hLP[n] by 10 units to the right:

))10sin(()10(

][ cLP wnnKnh --

=p n = 0, 1, 2, …. 20

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Causal FIR(N = 21) ImpulseResponse

Notice the symmetry: h[n] = h[N-1-n] whichsatisfies the linear phase condition.

• Filter Transfer Function

å=

-=20

0][)(

n

jnwLP

jwLP enheH since z = ejw

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Implies å=

-=20

0][)(

n

nLPLP znhzH

Filter Implementation for 21 coefficientsMagnitude of filter Frequency Responsefor 11, 21 and 31coefficient lengths

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Low Pass filterFrequency Response in dBfor 11 and 21 coefficients

Low pass filter phaseresponse for 21-coefficientdesign

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The above figure implies

)()( )( p-= wjLP

jwHP eHeH

nLPHP nhnh )1]([][ -=

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Now:

å¥

-¥=

--=n

wjnLP

jwHP enheH )(][)( p å

¥

-¥=

- -=n

njnwLP enh )1(][

å¥

-¥=

--=n

jnwnLP enh )1]([

][nhHP

Example: Analog high pass filter specification

0 10 kHz f

Ideal high pass

Given: fs = 50kHz, it follows:w = WT= 2pf/fs = 0.4p rad

Digital specification: 0.4p = p - wa

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\ wc (LPF) = p - wa = p - 0.4p = 0.6p

First design ideal LPF with wc= 0.6pAnswer : )6.0sin(1][ n

nnhLP p

p=

n = 0, ±1, ±2, …. ±¥Follows now:

nLPHP nhnh )1]([][ -=

))(6.0sin()(

)1(][ InIn

nhIn

HP --

-=

-

pp

The I indicates the length of the filter

Hence the above filter is a causal FIR high pass digital filter

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14

15

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Bandpass FIR filter design:

-p -wu -wo-wl 0 wl wo wu p w

H(ejw)K

Shift to right: H(ejw) ejwo

Shift to left: H(ejw) e-jwo

Follows:][]cos2[][ nhnwnh LPoBP =

n = 0, ±1, ±2, …. ±Iwhere

)(2 lowpasswww clu =-

2lu

owww +

=

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Example:

-15 -10 0 10 15 f in kHz

H(ejW)1 fs = 50kHz

Digital specifications:pp 4.0

)10)(5()10(24

4

==lw

pp 6.0)10)(5()5.1)(10(2

4

4

==uw

p5.0=ow

p1.02

=-

= luc

www

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First design LPF:

)1.0sin(1][ nn

nhLP pp

= n = 0, ±1, ±2, ….

Follows: ][)5.0cos(2[][ nhnnh LPBP p= n = 0, ±1, ±2, ….

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20

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Bandpass filter frequency response

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Homework # 6

1. Design a length-5 FIR bandpass filter with an antisymmetric impulse response h[n], i.e. h[n]= -h[4-n], 0£n £4, satisfying the following magnitude response values : Determine the exact expression for the frequency response of the filter designed.

2. An FIR filter of length 5 is defined by a symmetric impulse response i.e. h[n]= h[4-n], 0£n £4,. Let the input to this filter be a sum of 3 cosine sequences of angular frequencies: 0.2 rad/samples, 0.5 rad/samples, and 0.8 rad/samples, respectively. Determine the impulse response coefficients so that the filter passes only the midfrequency component of the input.

3. The frequency response H(ejω) of a length-4 FIR filter with a real and antisymmetric impulse response has the following specific values: H(ejπ)=8, and H(ejπ/2)= -2+j2. Determine H(z).

.1|)(| and 5.0|)(| 24 ==pp jj

eHeH

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HLP(ejw)

-2p-wc -p -wc wc p 2p-wc w

Frequency transformations of ideal filters.

The ideal low filter frequency response is shown below. hLP[n]represents the non-causal impulse response of HLP(ejw)

0-2p 2p

HLP(ej(w-p))

-p-wc -p -p+wc p-wc p p+wc w0-2p 2p

Shifting the low pass filter frequency by p, we get

High pass Filter

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The above figure is the frequency response of an ideal high pass filter with a cutoff frequency of p-wc (HHP(ejw) ).Hence,

)()( )( p-= wjLP

jwHP eHeH

From the shifting property of the Fourier transform, we get

njLPHP enhnh p][][ =

][)1(][ nhnh LPn

HP -=

Hence, to design an ideal high pass filter with cutoff frequencyof wa, first design a ideal low pass filter with a cutoff frequencyof (p-wa). Then use the above transformation to obtain the impulseresponse of the ideal high pass filter.

)sin(1][ cLP nwn

nhp

=where

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KHBP(ejw)

-p -wu -wo wl wl wo wu p

The frequency response for an ideal band pass filter is shown below. The center frequency is wo.

Looking at the frequency responses of a band pass filter and a lowpass filter, we can observe that a band pass filter is obtained byshifting the low pass filter to the left and to the right by wo andadding the two shifted responses.

HLP(ejw)

-p -wc wc p w

Low pass frequency response

Band pass Filter

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HLP(ej(w-wo))+HLP(ej(w+wo))

-p -wc-wo -wo wc-wo -wc+wo wo wc+wo p

Shifting the low pass filter by wo to the left and to the right and adding , we get

Comparing the ideal band pass filter and the above figure, we canobserve that

ocl www +-=

ocu www +=

2lu

cwww -

=

2lu

o

www

+=

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The frequency response of the band pass filter HBP(ejw) is

)()()( )()( oo wwjLP

wwjLP

jwBP eHeHeH +- +=

From the shifting property of the Fourier transform , we have

njwLP

njwLPBP

oo enhenhnh -+= ][][][

)}cos(2]{[][

}]{[][

oLPBP

njwnjwLPBP

nwnhnh

eenhnh oo

=

+= -

To design a band pass filter, first design a low pass filter with cutoff frequency wc given by

2lu

cwww -

=

And use the above transformation to get the impulse response ofthe band pass filter .

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Band stop Filter

HBS(ejw)

-wu -wo -wl wl wo wu w0-p p

The frequency response of an ideal band stop filter (HBS(ejw))is

Looking at the frequency responses of band stop and bandpass filters, we observe that the band stop filter is obtainedby subtracting the band pass from 1.

)(1)( jwBP

jwBS eHeH -=

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åå¥

-¥=

-¥

-¥=

- -=n

jnwBP

n

jnwBS enhenh ][1][

If we work out the first few terms, we get

wjBP

jwBPBP

jwBP

wjBP

wjBS

jwBSBS

jwBS

wjBS

ehehhehehehehheheh

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]2[]1[]0[1]1[]2[......

]2[]1[]0[]1[]2[......--

--

++-+-+-+=

+++-+-+

Hence, the impulse response coefficients for a band stop filter areobtained as

]0[21]0[1]0[ LPBPBS hhh -=-=

0 }cos(2]{[][][ ¹-=-= nnwnhnhnh oLPBPBS

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Low pass

)sin(][ cLP nwnKnhp

= In ±±±= ,...,2,1,0

H(ejw)K

-p -wc wc pHigh pass

],[)1(][ nhnh LPn

HP -= In ±±±= ,...,2,1,0

K

-p -(p -wc) (p- wc) p

H(ejw)

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2,2

],[)]cos(2[][

0lu

clu

LPoBP

wwwwww

nhnwnh+

==-

= In ±±±= ,...,2,1,0

K

-p -wu -wo wl wl wo wu p

H(ejw)

Band pass

2,2

],[][],0[]0[

0lu

clu

BPBS

BPBS

wwwwww

nhnhhKh

+==-

-=-=

In ±±±= ,...,2,1

K

-p -wu -wo wl wl wo wu p

H(ejw)

Band stop

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Examples1. An FIR filter is defined by a symmetric impulse response, i.e. h[0] = h[2]. Input to the filter

is a sum of two cosine sequences of angular frequencies 0.2 rad/s and 0.5 rad/s

Determine the impulse response coefficients so that it passes only the high frequency component of the input

Solution:

Since h[0] = h[2]

9.5631 h[1] and 4.8788- h[2] h[0]

equations, two theseSolving

1 h[1] .5)2h[0]cos(0 )H(e

0 h[1] .2)2h[0]cos(0 )H(e

1)H(e 5.0

and 0)H(e 2.0

,filter, he through tpasscan component frequency high only Since

) h[1] )cos( 2h[0] (e

e h[1] )e(ee h[0]

]1[)1](0[)(

j0.5

j0.2

j0

j0

0j-

j-j-jj-

2

0

0

===

=+=

=+=

=Þ=

=Þ=

+=

++=

++= --

w

w

w

wwww

www

w

w

w

when

eheheH jjj

i.e.

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2. Design a length – 4 FIR bandpass filter with an anti symmetric impulse responsei.e. h[n] = h[-n-4], for 0�n�4 satisfying the following magnitude responseValues: 5.0|)(| and 1 |)(| 2/4/ == pp jj eHeH

0.7452 h and 0.0381- h

get weequations, Solving

0.5 /4)sin(2h /4)3sin(2)H(e

1/8)sin(2/8)3sin(2)H(e

5.0|)H(e| and 1|)H(e| have weSince

))2/sin(2)2/3sin(2(

))(()(

)1()1(

)(

10

10/2j

10/4j

/2j/4j

102/3

2/2/2/1

2/32/32/30

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0

30

211

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221

==

=+=

=+=

==

+=

-+-=

-+-=

--+=

+++=

-

-----

---

---

---

pp

pp

p

p

pp

w

h

andhh

whwhje

eeeeheeeh

eeheh

ehehehh

ehehehheH

wj

jwjwjwjwwjwjwj

jwjwwj

wjwjjwo

wjwjjwo

j

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3. The frequency response of a length-4 FIR filter has values:

0|)(| and 37|)(| 2|)(| 2/0 =-== pp jjj eHjeHeHDetermine H(z)

Solution

Using the symmetry property of DTFT of a real sequence, we observe that

37|)(*||)(| 2/2/3 jeHeH jj +== pp

Thus the 4 point DFT of the sequence is given by:

]37, 0 ,37, 2[)( jjkH +-=

The inverse DFT of the H(k) gives h(n) as:

h(n) = (4 , 2 , -3 , -1) (using ifft in MATLAB)

Therefore H(z) is:

32-1 32z 4 )( -- --+= zzzH