Simple Digital Filters

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1

Simple Digital Filters Simple Digital Filters

• Later in the course we shall review various methods of designing frequency-selective filters satisfying prescribed specifications

• We now describe several low-order FIR and IIR digital filters with reasonable selective frequency responses that often are satisfactory in a number of applications

2

Simple FIR Digital Filters Simple FIR Digital Filters

• FIR digital filters considered here have integer-valued impulse response coefficients

• These filters are employed in a number of practical applications, primarily because of their simplicity, which makes them amenable to inexpensive hardware implementations

3

Simple FIR Digital Filters Simple FIR Digital Filters

Lowpass 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 atz= 0

• Note that here the pole vectorhas a unity magnitude for all values ofω

z z z

z

H 2

1 ¹ 1

2 0 1

= + +

= ( ⁻ ) )

(

−1

= z

4

Simple FIR Digital Filters Simple FIR Digital Filters

• On the other hand, asωincreases from0to π, the magnitude of the zero vector

decreases from a value of2, the diameter of the unit circle, to0

• Hence, the magnitude response is a monotonically decreasing function of ω fromω= 0toω= π

| ) (

|H₀ e^j^ω

5

Simple FIR Digital Filters Simple FIR Digital Filters

• The maximum value of the magnitude function is1 atω= 0, and the minimum value is0 atω= π, i.e.,

• The frequency response of the above filter is given by

0

1 ₀

0

0( )|= , | ( )|=

|H e^j H e^j^π

) 2 / cos(

)

( ^/²

0 e^j^ω =e⁻^j^ω ω H

6

Simple FIR Digital Filters Simple FIR Digital Filters

• The magnitude response can be seen to be a monotonically decreasing function ofω

) 2 / cos(

| ) (

|H₀ e^j^ω = ω

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

First-order FIR lowpass filter

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7

Simple FIR Digital Filters Simple FIR Digital Filters

• The frequency at which

is of practical interest since here the gain in dBis given by

since the dc gain ωc

= ω

) 2 (

) 1

( ₀ ⁰

0 j j

e H e

H ^ω^c =

) (ω_c G

dB 3 2 log 20 ) ( log

20 ₁₀ ⁰ − ₁₀ ≅−

= H e^j

) (ω_c

G ²⁰^log10 ⁽ ^c⁾

ej

H ^ω

=

0 20

0 ⁰

10 =

= log ( ) )

( H e^j

G

8

Simple FIR Digital Filters Simple FIR Digital Filters

• Thus, the gainG(ω) at is

approximately3 dB less than the gain at ω

= 0

• As a result, is called the3-dB cutoff frequency

• To determine the value of we set which yields

ωc

= ω

ωc

2 π/

=

ω_c ²

2 1 2

0( )| cos ( /2)

|H e^j^ω^c = ω_c =

9

Simple FIR Digital Filters Simple FIR Digital Filters

• The 3-dBcutoff frequency can be considered as the passband edge frequency

• As a result, for the filter the passband width is approximatelyπ/2

• The stopband is fromπ/2 toπ

• Note: has a zero at orω= π, which is in the stopband of the filter

ωc

) (z H₀

) (z

H₀ z=−1

10

Simple FIR Digital Filters Simple FIR Digital Filters

• A cascade of the simple FIR filter results in an improved lowpass frequency response as illustrated below for a cascade of3 sections

) ( )

( ₂¹ ¹

0 z = 1+z⁻ H

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

First-order FIR lowpass filter cascade

Simple FIR Digital Filters Simple FIR Digital Filters

• The 3-dB cutoff frequencyof a cascade of Msections is given by

• For M= 3, the above yields

• Thus, the cascade of first-order sections yields a sharper magnitude response but at the expense of a decrease in the width of the passband

) 2 ( cos

2 ¹ ¹^/²^M

c= − −

ω

π

= ω_c 0.302

Simple FIR Digital Filters Simple 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 sample values are generally associated with high- frequency components

• These high-frequency components are essentially removed by an moving-average filter resulting in a smoother output

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Simple FIR Digital Filters Simple FIR Digital Filters

Highpass FIR Digital Filters

• The simplest highpass FIR filter is obtained from the simplest lowpass FIR filter by replacingzwith

• This results in

) ( )

( ¹

2 1 1 z = 1−z⁻ H

−z

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Simple FIR Digital Filters Simple FIR Digital Filters

• Corresponding frequency response is given by

whose magnitude response is plotted below ) 2 / sin(

)

( ^/²

1e^j^ω = je⁻^j^ω ω H

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

First-order FIR highpass filter

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Simple FIR Digital Filters Simple FIR Digital Filters

• The monotonically increasing behavior of the magnitude function can again be demonstrated by examining the pole-zero pattern of the transfer function

• The highpass transfer function has a zero atz= 1 orω= 0 which is in the stopband of the filter

) (z H₁

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Simple FIR Digital Filters Simple FIR Digital Filters

• Improved highpass magnitude response can again be obtained by cascading several sections of the first-order highpass filter

• Alternately, a higher-order highpass filter of the form

is obtained by replacingzwith in the transfer function of a moving average filter

−z

n M

n n

M z

z

H ⁼

∑

₌⁻0¹⁻ ⁻ 1

1( ) ( 1)

17

Simple FIR Digital Filters Simple FIR Digital Filters

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

• In these radars, interfering signals, called clutters, are generated from fixed objects in the path of the radar beam

• The clutter, generated mainly from ground echoes and weather returns, has frequency components near zero frequency (dc)

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Simple FIR Digital Filters Simple FIR Digital Filters

• The clutter can be removed by filtering the radar return signal through atwo-pulse canceler, which is the first-order FIR highpass filter

• For a more effective removal it may be necessary to use a three-pulse canceler obtained by cascading two two-pulse cancelers

) ( )

( ₂¹ ¹

1 z = 1−z⁻ H

(4)

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Simple IIR Digital Filters Simple IIR Digital Filters

Lowpass IIR Digital Filters

• We have shown earlier that the first-order causal IIR transfer function

has a lowpass magnitude responsefor α> 0 1

0 , 1 )

( ₁ <α<

α

= − ₋ z z K H

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Simple IIR Digital Filters Simple IIR Digital Filters

• An improved lowpass magnitude response is obtained by adding a factor to the numerator of transfer function

• This forces the magnitude response to have a zero at ω= πin the stopbandof the filter

) 1 ( +z⁻¹

1 0 1 ,

) 1 ) (

( ₁

1 <α<

α

−

= + ₋⁻ z

z z K

H

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Simple IIR Digital Filters Simple IIR Digital Filters

• On the other hand, the first-order causal IIR transfer function

has a highpass magnitude responsefor α< 0

0 1 , 1 )

( ₁ − <α<

α

= − ₋ z z K H

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Simple IIR Digital Filters Simple IIR Digital Filters

• However, the modified transfer function obtained with the addition of a factor

to the numerator

exhibits a lowpass magnitude response )

1 ( +z⁻¹

0 1 1 ,

) 1 ) (

( ₁

1 − <α<

α

−

= + ₋⁻ z

z z K

H

Simple IIR Digital Filters Simple IIR Digital Filters

• The modified first-order lowpass transfer functionfor both positive and negative values of αis then given by

• As ωincreases from 0to π, the magnitude of the zero vectordecreases from a value of 2to 0

1 0 , 1

) 1 ) (

( ₁

1 <α<

α

−

= + ₋⁻ z z z K

H_LP

Simple IIR Digital Filters Simple IIR Digital Filters

• The maximum values of the magnitude function is at ω= 0and the minimum value is 0at ω= π, i.e.,

• Therefore, is a monotonically decreasing functionof ωfrom ω= 0to ω= π

) 1 /(

2K −α

0 ) ( 1 ,

) 2

( ⁰ =

α

= − _LP ^j^π

j

LP K H e

e H

| ) (

|H_LP e^j^ω

(5)

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Simple IIR Digital Filters Simple IIR Digital Filters

• For most applications, it is usual to have a dc gainof 0dB, that is to have

• To this end, we choose resulting in the first-order IIR lowpass transfer function

• The above transfer function has a zero at i.e., at ω= π which is in the stopband

1 0 , 1

1 2 ) 1

( ₁

1 ⎟⎟ <α<

⎠

⎞

⎜⎜

⎝

⎛ α

− + α

= − ⁻₋

z z z

H_LP

1

| ) (

|H e^j⁰ = 2 / ) 1 ( −α

= K

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Simple IIR Digital Filters Simple IIR Digital Filters

Lowpass 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., atω= π which is in the stopband

⎟⎟⎠

⎞

⎜⎜⎝

⎛ α

− + α

= − ⁻¹₋₁ 1

1 2 ) 1 (

z z z

H_LP

−1

= z

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Simple IIR Digital Filters Simple IIR Digital Filters

• has a real pole atz= α

• As ωincreases from 0to π, the magnitude of the zero vectordecreases from a value of 2to 0, whereas, for a positive value of α, the magnitude of the pole vectorincreases from a value of to

• The maximum value of the magnitude function is1atω= 0, and the minimum value is0atω= π

α

−

1 1+α

) (z H_LP

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Simple IIR Digital Filters Simple IIR Digital Filters

• i.e.,

• Therefore, is a monotonically decreasing function ofωfromω= 0 toω= π as indicated below

0

| ) (

| , 1

| ) (

|H_LP e^j⁰ = H_LP e^j^π =

| ) (

|H_LP e^j^ω

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

α = 0.8 α = 0.7 α = 0.5

10^-2 10^-1 10⁰

-20 -15 -10 -5 0

ω/π Gain, dB α = 0.8

α = 0.7 α = 0.5

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Simple IIR Digital Filters Simple IIR Digital Filters

• The squared magnitude function is given by

• The derivative of with respect toωis given by

) cos 2 1 ( 2

) cos 1 ( ) 1

| ( ) (

| 2

2 2

ω α

− α +

ω + α

= −

ω j LP e H

|2

) (

|H_LP e^j^ω

2 2 2 2

2

) cos 2 1 ( 2

sin ) 2 1 ( ) 1 (

| ) (

|

α + ω α

−

ω α + α + α

−

=− ω

ω

d e H d _LP ^j

30

Simple IIR Digital Filters Simple IIR Digital Filters

in the range verifying again the monotonically decreasing behavior of the magnitude function

• To determine the3-dB cutoff frequency we set

in the expression for the square magnitude function resulting in

0 /

| ) (

|H e^ω ²dω≤

d _LP ^j 0≤ω≤π

2

| 1 ) (

|H_LP e^j^ω^c ²=

(6)

31

Simple IIR Digital Filters Simple IIR Digital Filters

or

which when solved yields

• The above quadratic equation can be solved for αyielding two solutions

2 1 ) cos 2 1 ( 2

) cos 1 ( ) 1 (

2

2 =

ω α

− α +

ω + α

−

c c

c

c = +α − α ω

ω + α

− ) (1 cos ) 1 2 cos 1

( ² ²

1 2

cos 2

α +

= α ω_c

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Simple IIR Digital Filters Simple IIR Digital Filters

• The solution resulting in a stabletransfer function is given by

• It follows from

that is a BR functionfor|α| < 1 ) cos 2 1 ( 2

) cos 1 ( ) 1

| ( ) (

| 2

2 2

ω α

− α +

ω + α

= −

ω j LP e H

) (z H_LP

c

ωωc

= − α cos

sin 1

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Simple IIR Digital Filters Simple IIR Digital Filters

Highpass IIR Digital Filters

• A first-order causal highpassIIR 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., atω= 0 which is in the stopband

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

= + ⁻₋

1 1

1 1 2 1

z z z

H_HP

α ) α

(

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Simple IIR Digital Filters Simple 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

c

c ω

ω

−

=

α (1 sin )/cos ωc

) (z H_LP

) (z H_HP

10^-2 10^-1 10⁰

-20 -15 -10 -5 0

ω/π

Gain, dB

α = 0.8 α = 0.7 α = 0.5

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

α = 0.8 α = 0.7 α = 0.5

Simple IIR Digital Filters Simple IIR Digital Filters

• is a BR function for|α| < 1

• Example -Design a first-order highpass digital filter with a 3-dB cutoff frequency of 0.8π

• Now, and

• Therefore ) (z H_HP

587785 . 0 ) 8 . 0 sin(

)

sin(ω_c = π = 80902 . 0 ) 8 . 0 cos( π =−

5095245 .

0 cos / ) sin 1

( − ω ω =−

=

α _c _c

Simple IIR Digital Filters Simple IIR Digital Filters

• Therefore,

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ +

= − ⁻ ₋

1 1

5095245 0

1 245238 1 0

z z . .

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

= + ⁻₋

1 1

1 1 2 1

z z z

H_HP

α ) α

(

(7)

37

Simple IIR Digital Filters Simple IIR Digital Filters

Bandpass 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 ( 1

1 2

) 1

( z z

z z H_BP

)2

( ^j^ω

BP e H

] 2 cos 2 cos ) 1 ( 2 )

1 ( 1 [ 2

) 2 cos 1 ( ) 1 (

2 2

2

ω α + ω α + β

− α + α + β +

ω

− α

= −

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Simple IIR Digital Filters Simple IIR Digital Filters

• goes to zero atω= 0 andω=π

• It assumes a maximum value of1 at , called the center frequencyof the bandpass filter, where

• The frequencies and where becomes1/2 are called the3-dB cutoff frequencies

|2

) (

|H_BP e^j^ω

|2

) (

|H_BP e^j^ω ωo

= ω

) ( cos ¹β

= ω_o ⁻

1

ωc ω_c₂

39

Simple IIR Digital Filters Simple IIR Digital Filters

• The difference between the two cutoff frequencies, assuming is called the3-dB bandwidthand is given by

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

1

2 c

c >ω ω

) (z H_BP

1 1 2−ω =cos⁻ ω

= _c _c

Bw ⎟

⎠

⎜ ⎞

⎝

⎛ α +

α 1 2

2

40

Simple IIR Digital Filters Simple IIR Digital Filters

• Plots of are shown below|H_BP(e^j^ω)|

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

β = 0.34 α = 0.8 α = 0.5 α = 0.2

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

α = 0.6 β = 0.8 β = 0.5 β = 0.2

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Simple IIR Digital Filters Simple IIR Digital Filters

• Example -Design a 2ndorder bandpass digital filter with center frequency at0.4π and a 3-dB bandwidthof0.1π

• Here and

• The solution of the above equation yields:

α= 1.376382 andα= 0.72654253

9510565 .

0 ) 1 . 0 cos(

) cos(

1 2

2= = π =

α +

α

Bw

309017 . 0 ) 4 . 0 cos(

)

cos(ω = π =

=

β _o

42

Simple IIR Digital Filters Simple IIR Digital Filters

• The corresponding transfer functions are

and

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

± 11425636

1.

j

2 1

2

37638 1 7343424 0

1 18819 1

0 ₋ ₋

−

+

−

− −

=

z z

z z H_BP

. .

. )

' (

2 1

2

72654253 0

533531 0 1 13673 1

0 ₋ ₋

−

+

−

= −

z z

z z H_BP

. .

. )

" (

)

' (z

H_BP

(8)

43

Simple IIR Digital Filters Simple IIR Digital Filters

• Thus, the poles of are outside the unit circle making the transfer function unstable

• On the other hand, the poles of are atz= and have a magnitude of0.8523746

• Hence is BIBO stable

• Later we outline a simpler stability test )

' (z

H_BP

)

" (z H_BP 8095546 0

2667655

0. ± j .

)

" (z H_BP

44

Simple IIR Digital Filters Simple IIR Digital Filters

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

)

" (z H_BP

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

0 0.2 0.4 0.6 0.8 1

-1 0 1 2 3 4 5 6 7

ω/π

Group delay, samples

45

Simple IIR Digital Filters Simple IIR Digital Filters

Bandstop IIR Digital Filters

• A 2nd-order bandstop digital filter has a transfer function given by

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

⎟⎟⎠

⎞

⎜⎜⎝

⎛

α + α + β

−

+ β

− α

= + ⁻¹₋₁ ⁻² ₋₂ )

1 ( 1

2 1 2 ) 1 (

z z

z z z

H_BS

) (z H_BS

46

Simple IIR Digital Filters Simple IIR Digital Filters

• Its magnitude response is plotted below

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

α = 0.8 α = 0.5 α = 0.2

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

β = 0.8 β = 0.5 β = 0.2

Simple IIR Digital Filters Simple IIR Digital Filters

• Here, the magnitude function takes the maximum value of1 atω= 0 andω= π

• It goes to0 at , where , called the notch frequency, is given by

• The digital transfer function is more commonly called anotch filter

ωo

=

ω ω_o

) ( cos¹β

= ω_o ⁻

) (z H_BS

Simple IIR Digital Filters Simple IIR Digital Filters

• The frequencies and where becomes1/2 are called the3-dB cutoff frequencies

• The difference between the two cutoff frequencies, assuming is called the3-dB notch bandwidthand is given by

|2

) (

|H_BS e^j^ω

1

ωc ω_c₂

1

2 c

c >ω ω

1 1 2−ω =cos⁻ ω

= _c _c

Bw ⎟

⎠

⎜ ⎞

⎝

⎛ α +

α 1 2

2

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49

Simple IIR Digital Filters Simple IIR Digital Filters

Higher-Order IIR Digital Filters

• By cascading the simple digital filters discussed so far, we can implement digital filters with sharper magnitude responses

• Consider a cascade ofKfirst-order lowpass sections characterized by the transfer function

⎟⎟⎠

⎞

⎜⎜⎝

⎛ α

− + α

= − ⁻₋

1 1

1 1 2 ) 1 (

z z z

H_LP

50

Simple IIR Digital Filters Simple IIR Digital Filters

• The overall structure has a transfer function given by

• The corresponding squared-magnitude function is given by

K

LP z

z z

G ⎟⎟

⎠

⎜⎜ ⎞

⎝

⎛

α

−

⋅ + α

= − ⁻₋¹₁ 1

1 2 ) 1 (

K j

LP e

G ⎥

⎦

⎢ ⎤

⎣

⎡

ω α

− α +

ω + α

= −

ω

) cos 2 1 ( 2

) cos 1 ( ) 1

| ( ) (

| ₂

2 2

51

Simple IIR Digital Filters Simple IIR Digital Filters

• To determine the relation between its 3-dB cutoff frequency and the parameterα, we set

which when solved forα, yields for a stable :

ωc

2 1 ) cos 2 1 ( 2

) cos 1 ( ) 1 (

2

2 ⎥ =

⎦

⎢ ⎤

⎣

⎡

ω α

− α +

ω + α

− ^K

c c

) (z G_LP

c c c

C

C C C

ω +

−

− ω

−

= +

α 1 cos

2 sin cos ) 1 (

1 ²

52

Simple IIR Digital Filters Simple IIR Digital Filters

where

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

forK= 1

K

C=2⁽K⁻¹⁾^/

c

ωωc

= − α cos

sin 1

53

Simple IIR Digital Filters Simple IIR Digital Filters

• Example -Design a lowpass filter with a 3- dBcutoff frequency at using a single first-order section and a cascade of4 first-order sections, and compare their gain responses

• For the single first-order lowpass filter we have

π

= ω_c 0.4

1584 . ) 0 4 . 0 cos(

) 4 . 0 sin(

1 cos

sin

1 =

π π

= + ω

ω

= + α

c c

54

Simple IIR Digital Filters Simple IIR Digital Filters

• For the cascade of4 first-order sections, we substitute K= 4and get

• Next we compute

6818 1 2

2⁽ ¹⁾^/ = ⁽⁴ ¹⁾^/⁴= .

= ^K⁻ ^K ⁻ C

c c c

C

C C C

ω +

−

− ω

−

= +

α 1 cos

2 sin cos ) 1 (

1 ²

) 4 . 0 cos(

6818 . 1 1

) 6818 . 1 ( ) 6818 . 1 ( 2 ) 4 . 0 sin(

) 4 . 0 cos(

) 6818 . 1 1 (

1 ²

π +

−

− π

−

= +

0.251

−

=

(10)

55

Simple IIR Digital Filters Simple 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

Gain, dB

K=4

K=1 Passband details

10^-2 10^-1 10⁰

-4 -2 0

ω/π

Gain, dB K=1 K=4