Simple Digital Filters - Catholic University of Americasip.cua.edu/res/docs/courses/ee515/chapter04/ch4-5.pdf · Simple Digital Filters • Later in the course we shall review various

1Copyright © 2001, S. K. Mitra

Simple Digital FiltersSimple Digital Filters

• Later in the course we shall review variousmethods of designing frequency-selectivefilters satisfying prescribed specifications

• We now describe several low-order FIR andIIR digital filters with reasonable selectivefrequency responses that often aresatisfactory in a number of applications


Simple FIR Digital FiltersSimple FIR Digital Filters

• FIR digital filters considered here haveinteger-valued impulse response coefficients

• These filters are employed in a number ofpractical applications, primarily because oftheir simplicity, which makes them amenableto inexpensive hardware implementations


Simple FIR Digital FiltersSimple FIR Digital FiltersLowpass FIR Digital Filters• The simplest lowpass FIR digital filter is the

2-point moving-average filter given by

• The above transfer function has a zero at and a pole at z = 0

• Note that here the pole vector has a unitymagnitude for all values of ω

zzzzH2

11 121

0+=+= − )()(

1−=z



• On the other hand, as ω increases from 0 toπ, the magnitude of the zero vectordecreases from a value of 2, the diameter ofthe unit circle, to 0

• Hence, the magnitude response isa monotonically decreasing function of ωfrom ω = 0 to ω = π

|)(| 0ωjeH



• The maximum value of the magnitudefunction is 1 at ω = 0, and the minimumvalue is 0 at ω = π, i.e.,

• The frequency response of the above filteris given by

01 00

0 == |)(|,|)(| πjj eHeH

)2/cos()( 2/0 ω= ω−ω jj eeH



• The magnitude responsecan be seen to be a monotonicallydecreasing function of ω

)2/cos(|)(| 0 ω=ωjeH

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

First-order FIR lowpass filter


Simple FIR Digital FiltersSimple FIR Digital Filters• The frequency at which

is of practical interest since here the gainin dB is given by

since the dc gain

cω=ω

)(2

1)( 000

jj eHeH c =ω

)( cωG

dB32log20)(log20 100

10 −≅−= jeH)( cωG )(log20 10

cjeH ω=

0200 010

== )(log)( jeHG



• Thus, the gain G(ω) at isapproximately 3 dB less than the gain at ω= 0

• As a result, is called the 3-dB cutofffrequency

• To determine the value of we set

which yields

cω=ω

cω

cω

2/π=ωc2122

0 )2/(cos|)(| =ω=ωc

j ceH



• The 3-dB cutoff frequency can beconsidered as the passband edge frequency

• As a result, for the filter the passbandwidth is approximately π/2

• The stopband is from π/2 to π• Note: has a zero at or ω = π,

which is in the stopband of the filter

cω

)(zH0

)(zH0 1−=z


Simple FIR Digital FiltersSimple FIR Digital Filters• A cascade of the simple FIR filter

results in an improved lowpass frequencyresponse as illustrated below for a cascadeof 3 sections

)()( 121

0 1 −+= zzH

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

First-order FIR lowpass filter cascade



• The 3-dB cutoff frequency of a cascade ofM sections is given by

• For M = 3, the above yields• Thus, the cascade of first-order sections

yields a sharper magnitude response but atthe expense of a decrease in the width of thepassband

)2(cos2 2/11 Mc

−−=ωπ=ω 302.0c


Simple FIR Digital FiltersSimple FIR Digital Filters• A better approximation to the ideal lowpass

filter is given by a higher-order moving-average filter

• Signals with rapid fluctuations in samplevalues are generally associated with high-frequency components

• These high-frequency components areessentially removed by an moving-averagefilter resulting in a smoother outputwaveform



Highpass FIR Digital Filters• The simplest highpass FIR filter is obtained

from the simplest lowpass FIR filter byreplacing z with

• This results in)()( 1

21

1 1 −−= zzH

z−


Simple FIR Digital FiltersSimple FIR Digital Filters• Corresponding frequency response is given

by

whose magnitude response is plotted below)2/sin()( 2/

1 ω= ω−ω jj ejeH

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

First-order FIR highpass filter



• The monotonically increasing behavior ofthe magnitude function can again bedemonstrated by examining the pole-zeropattern of the transfer function

• The highpass transfer function has azero at z = 1 or ω = 0 which is in thestopband of the filter

)(zH1

)(zH1



• Improved highpass magnitude response canagain be obtained by cascading severalsections of the first-order highpass filter

• Alternately, a higher-order highpass filter ofthe form

is obtained by replacing z with in thetransfer function of a moving average filter

z−

nMn

nM

zzH −−=∑ −= 1

01

1 1)()(



• An application of the FIR highpass filters isin moving-target-indicator (MTI) radars

• In these radars, interfering signals, calledclutters, are generated from fixed objects inthe path of the radar beam

• The clutter, generated mainly from groundechoes and weather returns, has frequencycomponents near zero frequency (dc)



• The clutter can be removed by filtering theradar return signal through a two-pulsecanceler, which is the first-order FIRhighpass filter

• For a more effective removal it may benecessary to use a three-pulse cancelerobtained by cascading two two-pulsecancelers

)()( 121

1 1 −−= zzH


Simple IIR Digital FiltersSimple IIR Digital FiltersLowpass IIR Digital Filters• A first-order causal lowpass IIR digital

filter has a transfer function given by

where |α| < 1 for stability• The above transfer function has a zero at

i.e., at ω = π which is in the stopband

α−+α−= −

−

1

1

11

21)(

zzzHLP

1−=z


Simple IIR Digital FiltersSimple IIR Digital Filters• has a real pole at z = α• As ω increases from 0 to π, the magnitude

of the zero vector decreases from a value of2 to 0, whereas, for a positive value of α,the magnitude of the pole vector increasesfrom a value of to

• The maximum value of the magnitudefunction is 1 at ω = 0, and the minimumvalue is 0 at ω = π

α−1 α+1

)(zHLP


Simple IIR Digital FiltersSimple IIR Digital Filters• i.e.,• Therefore, is a monotonically

decreasing function of ω from ω = 0 to ω = πas indicated below

0|)(|,1|)(| 0 == πjLP

jLP eHeH

|)(| ωjLP eH

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

α = 0.8α = 0.7α = 0.5

10-2 10-1 100-20

-15

-10

-5

0

ω/π

Gai

n, d

B α = 0.8α = 0.7α = 0.5


Simple IIR Digital FiltersSimple IIR Digital Filters

• The squared magnitude function is given by

• The derivative of with respectto ω is given by

)cos21(2)cos1()1(|)(| 2

22

ωα−α+ω+α−=ωj

LP eH

2|)(| ωjLP eH

22

222

)cos21(2sin)21()1(|)(|

α+ωα−ωα+α+α−−=

ω

ω

deHd j

LP



in the rangeverifying again the monotonically decreasingbehavior of the magnitude function

• To determine the 3-dB cutoff frequencywe set

in the expression for the square magnitudefunction resulting in

0/|)(| 2 ≤ωω deHd jLP π≤ω≤0

21|)(| 2 =ωcj

LP eH



or

which when solved yields

• The above quadratic equation can be solvedfor α yielding two solutions

21

)cos21(2)cos1()1(

2

2=

ωα−α+ω+α−

c

c

cc ωα−α+=ω+α− cos21)cos1()1( 22

212cos

α+α=ωc


Simple IIR Digital FiltersSimple IIR Digital Filters• The solution resulting in a stable transfer

function is given by

• It follows from

that is a BR function for |α| < 1)cos21(2

)cos1()1(|)(| 2

22

ωα−α+ω+α−=ωj

LP eH

)(zHLP

)(zHLP

cc

ωω−=α cos

sin1



Highpass IIR Digital Filters• A first-order causal highpass IIR digital filter

has a transfer function given by

where |α| < 1 for stability• The above transfer function has a zero at z = 1

i.e., at ω = 0 which is in the stopband

−−+= −

−

1

1

11

21

zzzHHP

αα

)(


Simple IIR Digital FiltersSimple IIR Digital Filters• Its 3-dB cutoff frequency is given by

which is the same as that of• Magnitude and gain responses of

are shown below

cc ωω−=α cos/)sin1(cω

)(zHLP)(zHHP

10-2 10-1 100-20

-15

-10

-5

0

ω/π

Gai

n, d

B

α = 0.8α = 0.7α = 0.5

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

α = 0.8α = 0.7α = 0.5



• is a BR function for |α| < 1• Example - Design a first-order highpass

digital filter with a 3-dB cutoff frequency of0.8π

• Now, and

• Therefore

)(zHHP

587785.0)8.0sin()sin( =π=ωc80902.0)8.0cos( −=π

5095245.0cos/)sin1( −=ωω−=α cc



• Therefore,

+−= −

−

1

1

50952450112452380

zz

..

−−+= −

−

1

1

11

21

zzzHHP

αα

)(


Simple IIR Digital FiltersSimple IIR Digital FiltersBandpass IIR Digital Filters• A 2nd-order bandpass digital transfer

function is given by

• Its squared magnitude function is

α+α+β−−α−= −−

−

21

2

)1(11

21)(

zzzzHBP

2)( ωj

BP eH

]2cos2cos)1(2)1(1[2)2cos1()1(

2222

2

ωα+ωα+β−α+α+β+ω−α−=



• goes to zero at ω = 0 and ω = π• It assumes a maximum value of 1 at ,

called the center frequency of the bandpassfilter, where

• The frequencies and where becomes 1/2 are called the 3-dB cutofffrequencies

2|)(| ωjBP eH

2|)(| ωjBP eH

oω=ω

)(cos 1 β=ω −o

1cω 2cω



• The difference between the two cutofffrequencies, assuming is calledthe 3-dB bandwidth and is given by

• The transfer function is a BRfunction if |α| < 1 and |β| < 1

12 cc ω>ω

)(zHBP

α+α

212

112 cos−=ω−ω= ccwB

α+α

212



• Plots of are shown below|)(| ωjBP eH

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

β = 0.34

α = 0.8α = 0.5α = 0.2

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

α = 0.6

β = 0.8β = 0.5β = 0.2


Simple IIR Digital FiltersSimple IIR Digital Filters• Example - Design a 2nd order bandpass

digital filter with center frequency at 0.4πand a 3-dB bandwidth of 0.1π

• Hereand

• The solution of the above equation yields:α = 1.376382 and α = 0.72654253

9510565.0)1.0cos()cos(1

22 =π==

α+α

wB

309017.0)4.0cos()cos( =π=ω=β o



• The corresponding transfer functions are

and

• The poles of are at z = 0.3671712 and have a magnitude > 1

±114256361.j

21

2

3763817343424011188190 −−

−

+−−−=

zzzzHBP

...)('

21

2

726542530533531011136730 −−

−

+−−=

zzzzHBP

...)("

)(' zHBP



• Thus, the poles of are outside theunit circle making the transfer functionunstable

• On the other hand, the poles of areat z = and have amagnitude of 0.8523746

• Hence is BIBO stable• Later we outline a simpler stability test

)(' zHBP

)(" zHBP8095546026676550 .. j±

)(" zHBP



• Figures below show the plots of themagnitude function and the group delay of

)(" zHBP

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

0 0.2 0.4 0.6 0.8 1-1

0

1

2

3

4

5

6

7

ω/π

Gro

up d

elay

, sam

ples



Bandstop IIR Digital Filters• A 2nd-order bandstop digital filter has a

transfer function given by

• The transfer function is a BRfunction if |α| < 1 and |β| < 1

α+α+β−+β−α+= −−

−−

21

21

)1(121

21)(

zzzzzHBS

)(zHBS



• Its magnitude response is plotted below

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

α = 0.8α = 0.5α = 0.2

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

ω/π

Mag

nitu

de

β = 0.8β = 0.5β = 0.2



• Here, the magnitude function takes themaximum value of 1 at ω = 0 and ω = π

• It goes to 0 at , where , called thenotch frequency, is given by

• The digital transfer function is morecommonly called a notch filter

oω=ω oω

)(cos 1 β=ω −o

)(zHBS



• The frequencies and where becomes 1/2 are called the 3-dB cutofffrequencies

• The difference between the two cutofffrequencies, assuming is calledthe 3-dB notch bandwidth and is given by

2|)(| ωjBS eH1cω 2cω

12 cc ω>ω

112 cos−=ω−ω= ccwB

α+α

212



Higher-Order IIR Digital Filters• By cascading the simple digital filters

discussed so far, we can implement digitalfilters with sharper magnitude responses

• Consider a cascade of K first-order lowpasssections characterized by the transferfunction

α−+α−= −

−

1

1

11

21)(

zzzHLP



• The overall structure has a transfer functiongiven by

• The corresponding squared-magnitudefunction is given by

K

LP zzzG

α−+⋅α−= −

−

1

1

11

21)(

Kj

LP eG

ωα−α+ω+α−=ω

)cos21(2)cos1()1(|)(| 2

22


Simple IIR Digital FiltersSimple IIR Digital Filters• To determine the relation between its 3-dB

cutoff frequency and the parameter α,we set

which when solved for α, yields for a stable :

cω

21

)cos21(2)cos1()1(

2

2=

ωα−α+ω+α−

K

c

c

)(zGLP

c

ccC

CCCω+−

−ω−ω−+=αcos1

2sincos)1(1 2



where

• It should be noted that the expression for αgiven earlier reduces to

for K = 1

KKC /)( 12 −=

cc

ωω−=α cos

sin1



• Example - Design a lowpass filter with a 3-dB cutoff frequency at using asingle first-order section and a cascade of 4first-order sections, and compare their gainresponses

• For the single first-order lowpass filter wehave

π=ω 4.0c

1584.0)4.0cos(

)4.0sin(1cos

sin1=

ππ+

=ω

ω+=α

c

c



• For the cascade of 4 first-order sections, wesubstitute K = 4 and get

• Next we compute6818122 4141 ./)(/)( === −− KKC

c

ccC

CCCω+−

−ω−ω−+=αcos1

2sincos)1(1 2

)4.0cos(6818.11)6818.1()6818.1(2)4.0sin()4.0cos()6818.11(1 2

π+−−π−π−+=

2510.−=


Simple IIR Digital FiltersSimple IIR Digital Filters• The gain responses of the two filters are

shown below• As can be seen, cascading has resulted in a

sharper roll-off in the gain response

10-2 10-1 100-20

-15

-10

-5

0

ω/π

Gai

n, d

B

K=4

K=1Passband details

10-2 10-1 100

-4

-2

0

ω/π

Gai

n, d

B K=1K=4

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Simple Digital Filters - Catholic University of Americasip.cua.edu/res/docs/courses/ee515/chapter04/ch4-5.pdf · Simple Digital Filters • Later in the course we shall review various