Chapter 4 Recursive Filter Design

Pusan National UniversityMeasurement & Control Lab.

Digital Signal Processing 1/46

Chapter 4 Recursive Filter Design

Recursive Filter ( IIR Filter ) should be stable .

※ Filter Design Procedure

① Stability Check

② Review for Satisfying Filter Design Specification?

1 0

n m

k i k i i k i

i i

y A y B x

0

1

( )( )

( )1

mi

i

i

ni

i

i

B zY z

H zX z

A z



The transformations for converting from analog filter H(s) to digital filterH(z). We have to consider the following terms:

① What transformation should we use?

=> It should be “useful” to convert from a continuous filter

function to a discrete filter function.

② If the continuous filter is stable, will the transformation ensure that

the digital filter is stable? All its poles should be within a unit circle.

③ Does the continuous filter’s meeting the specifications mean that the

digital filter will also meet them?

If not, how do we modify the procedure do ensure that it does?

=> We use bilinear transform



4.2 Digital Filter Design by Means of the z-Transform

( Impulse Invariant Method )

1. Impulse Invariant Method: If H(s) is in factored form, we obtain H(z),

the transfer function for the digital filter, by using expression(3.38) .

2. Impulse Invariant

Definition

A continuous filter G(s) and digital filter H(z) are said to be impulse-

invariant if the pulse response of H(z) is the same as the sampled impulse

response of G(s).

※ If H(z) is the z-transform of the samples g(kT) of a continuous g(t), then

H(z) and G(s) are impulse-invariant.

*

( )

1

1 1( ( ) ( ) ( ) ( ) )

2 11

( ) ( ) ( )1

Impulse Invariant Digital Filter with the continuous filter

c j

T T s pc j

Tp

H s h t t H p dpj e

H z residues of H p at poles of H pe z

L



The values of the samples of the impulse response are obtained by

setting t equal to kT

(4.4)

The z-transform of g(kT) is

The unit-pulse response is obtained as

which is the same as (4.4).⇒ Impulse invariant

Proof of Impulse Invariant Definition

proof) Expand G(s) in partial fractions

1 1

( ) 1, , ( )

i

n ns ti

i

i ii

RG s where i n g t R e

s s

1

( ) i

ns kT

i

i

g kT R e

11 1

( )1 i i

n ni i

Ts Tsi i

R R zH z

e z z e

1

1

1 1

( ) ( ) ( )

( )i i

iTsi

k

n nTs s kTki

iTsi iz e

h k residues of H z z at poles of some inside ROC for H z

R zz e z R e

z e



3. The frequency variable 𝝎 for the digital filter bears a linear relationship

to that for the continuous filter, within the operating range (0 to 𝝎

𝒔

𝟐) of the

digital filter.

⇒∴ Critical frequencies such as cutoff and band width frequencies for the

digital filter can be used directly in section of the continuous filter.

4. Taking the z-transform of a continuous filter function is a satisfactory

procedure when applied to all pole filters such as Butterworth, Chebyshev, ,

and Bessel, provided that the designing filters are bandlimited. In a

bandlimited filter, the magnitude response of the continuous filter is

negligibly small at frequencies exceeding half the sampling frequency, in

order to reduce the aliasing effect.

Thus we must have

( ) 02

sH j for



※With the highpass filter, because the passband of 𝑯(𝒔 ) is unlimited,

the periodic extension indicates that the periodic lobes of the passband

tend to “fill up” the stopband as a result of aliasing in the operating

range −𝝎𝒔/𝟐 𝒕𝒐 𝝎𝒔/𝟐 . Consequently, the digital filter is useless as a

highpass filter.

⇒ There is a way to get around this difficulty by modifying the

procedure.



※ Filter Design Procedure

① Select a suitable continuous, lowpass prototype to satisfy the sharpness

of-cutoff specification.

② Make a lowpass-to-bandpass frequency transformation for the

specified passband.

③ Apply transformation to the resulting continuous filter function and

obtain H(z)

④ Use the transfer function obtained to express the digital filter in

difference equation form for implementation as a computer algorithm.

1 0

n m

k i k i i k i

i i

y A y B x

( )n nH s

( )H s

( )H z



Ex 4.1) Use the impulse-invariant method to design a bandpass digital filter

that is monotonic in passband and stopband , has a passband from 500 to

550Hz, and is as low in order as practical. Sampling rates of 2kHz and

10kHz should be examined.

Solve)

Because of the simplicity requirement for band pass filter, a second order

filter will suffice. Because a 2nd -order bandpass filter is required, the

design procedure may begin with a first-order, continuous, lowpass

prototype of the Butterworth type.

Applying the bandpass transformation to 𝑯𝒏(𝒔𝒏) gives

where,

2 2

0 0 0

0

( )

n

s ss

Bs B s

1( )

1n n

n

H ss

22 20 2 20

0 00

1( )

( ) / 1 ( )2 4

Bs BsH s

s B Bs BsB s

s

2

0,u l u lB



12

1 12 2

2

2 2

2 2 2 2 2 2 2 2 2

( ) (where, , ( ) ),( ) 2 4

Magnitude response is

( )( ) (2 ) ( ) ( )

Bs B BLet H s a b

s a b

B BM

a b a b a b a

1

1

1

1 2 2 2 2

1( ) ( )

1 ( ) ( )

1( )

1 ( ) ( )

1 ( sin cos )( )

1 2 cos 2 (cos )

( sin cos

,

Tp

p a jb

Tp

p a jb

aT

aT aT

aT

BpH z p a jb

e z p a jb p a jb

Bpp a jb

e z p a jb p a jb

aB bT bT e z

Bz z qb

e bTz e z z r bT z r

awhere r e q bT b

b) aTT e

1

1( ) ( )

1

Tp

H z residues of H p at poles p a jbe z

To obtain a digital filter that is impulse-invariant with 𝑯(𝒔), apply transformation

to 𝑯(𝒔)



2

2 2

( )

1 2 cos

1 2 cos( ) 1 2 cos( )

M T

q q TB

r r b T r r b T

If let T = 0.0005 ( 2KHz ) or T = 0.0001 ( 10KHz ) ,

Magnitude diagram of M(w) is shown as

From the geometry of Fig. 4.2, the

magnitude response is



From 𝑯(𝒛) we can get the following difference equation

The block diagram for the canonic implementation of this filter is

shown in the following figure.

1 1 2 2 0 1 1

2

1 2

0 1

2 cos ,

, ( sin cos )

k k k k k

aT aT

aT

y A y A y B x B x

A e bT A e

aB B B B bT bT e

b

2 2

( )( )

2 (cos )

Bz z qH z

z r bT z r

( sin cos )aT aTar e q bT bT e

b



Higher-order lowpass and bandpass filters may be designed in a similar

manner. They should be implemented as a series or parallel arrangement

of first - and second - order filters. For example, take the z-transform of

the third-order Butterworth filter.

ex) After first expanding 𝑯 𝒔 in partial fractions as follows:

Taking z-transform of 𝑯(𝒔) function, the z-transfer function is derived as

This gives a digital filter consisting of a first-order and a second-order

filter in parallel.

3

2 31

2 2 2 2( )

( )( )

c

c c c c c c

R s RRH s

s s s s s s

1

1 1 2

2 2

1 1( )

1 1

3 3 3where, 2 cos , cos sin / 3

2 2 2

c c

c c

c T T

T T

c c c

zH z

e z z e z

T T Te e



Because of aliasing, we cannot design a bandstop or highpass digital

filter by taking the z-transform without first bandlimiting the function

𝑯(𝒔)to be transformed. This involves multiplying 𝑯(𝒔) by a so-called

guardband filter 𝑮 𝒔 , which is a lowpass filter arranged such that

( ) ( ) 02

sG j H j for

Ex) A bandstop filter is required with stop band cutoff frequencies 𝒇𝒍 and

𝒇𝒖, and with 𝒇𝒉 being the highest frequency in the signal to be processed.

First design a suitable continuous-filter 𝑯(𝒔) of order 𝒏𝟏 that has the

required stopband with the specified sharpness of cutoff.

Then design a lowpass filter G(s) with cutoff frequency 𝒇𝒉 and with

attenuation 𝒂 dB, at frequency 𝒇𝒓, where

, :r s u sf f f f sampling frequency



The order of guardband filter is given by

( ※ Refer to Butterworth filter design)

The attenuation 𝒂 must be sufficient to ensure negligible aliasing into the

periodic response at 𝒇𝒓. The magnitude of 𝑮(𝒔)𝑯(𝒔) after multiplying the

guardband filter is shown in

The product 𝑮(𝒔)𝑯(𝒔) yields a digital filter of order 𝒏 = 𝒏𝟏 + 𝒏𝟐.

Note that the if the sampling frequency is increased, the cutoff requirement

–and hence the order of the guardband filter –may be reduced.

(∵ Large 𝒇𝒓 ⇒ Large 𝒇𝒔 )

2

20log r

h

an

f

f



4.3 Designing Digital Filters By Numerical Approximation to an Integration

4.3.1 Introduction

We have seen that in designing a digital filter by means of the z-transform,

we use the mapping 𝒛 = 𝒆𝒔𝑻 to go from points in the 𝒔-plane to points in the

𝒛-plane. ⇒We have to be cautious for keeping sampling time to avoid

aliasing problem.

As mentioned earlier, there are several different ways of going from the 𝒔-

plane to the 𝒛-plane.

★Numerical Approximation

Let the continuous system be

where,

( )( ) ,

( )

Y s aH s y ay ax

X s s a

0

( ) ( ) ( )t

dyay ax

dt

y t a y x d

t kT



( 1)

0( 1) ( ) ( )

k T

y k a y x d

0

( 1)

0 ( 1)

( ) ( ) ( )

( ) ( ) ( ) ( )

( 1)

kT

k T kT

k T

y k a y x d

a y x d a y x d

y k y

⇒ This is obtained as a recursive form if Δy is calculated.



※We will consider three different ways to approximate Δy:

1. Forward rectangular approximation

2. Backward rectangular approximation

3. Trapezoidal approximation

We shall see that each of these numerical approximations indicates how to

convert a continuous filter to a digital filter by means of an appropriate

replacement for 𝒔.

Note that there is no actual sampling involved in going from a continuous

filter function 𝑯(𝒔) to a digital filter function 𝑯(𝒛) via any of the following

transformations.

However, the notation T is retained to denote the time step of the digital filter

𝑯(𝒛).The sampling frequency 𝝎𝒔 is the frequency that corresponds to time step 𝑻.

It is defined as 2s

T



When we compare this with

, we can get a digital filter from continuous filter by replacing in the

. That means that s is replaced as .

( )

aH s

s a

1 1

1

1

( ) 1 ( )

( )( )

1( ) 1 1

Y z aT z Y z aTz X z

Y z aTz aH z

zX z aT za

T

4.3.2 Forward Rectangular Approximation

1 1y ay k ax k T

1zs

T

( ) 1 ( ) ( 1) 1 1

1 1 1

y k y k y k y k aTy k aTx k

aT y k aTx k

1

zs

T

( )a

H ss a



Let us check this transformation to see whether it gives a stable digital

filter from stable continuous filter. The mapping is implies as follows: 1

1

1 ( (1 )

1 ,

0 1 , 0 1 , 0 1

zs z Ts

Tz u jv T j T jT

u T v T

u u u

However it also contains a much larger area of unstable points. It follows

that a stable continuous filter could produce an unstable digital filter by the

forward rectangular approximation.

ex) 𝑯(𝒛) has a pole 𝒛 = 𝟏 − 𝒂𝑻. Hence if 𝒂𝑻 > 𝟐, or 𝑻 > 𝟐/𝒂, the pole lies

outside the unit circle, and H(z) is unstable.



4.3.3 Backward Approximation

( ) ( )

( ) ( 1)

y au k ax k T

y k y k ay k ax k T

1

1 1

1 ( ) ( )

aT y k y k aTx k

aT z Y z aTX z

1

1

( )( )

( ) 1 1 /

1 Comparing ( )

1Replacing ( ) ( )

,

zs

Tz

Y z aT aH z

X z aT z z Tz a

a zH s s

s a Tz

zs H z H s

Tz



1 1 1 1 1 1 1

1 2 1 2 2 2 1

Tswhere z

Ts Ts Ts

2 2 2 2

2 2 2

1 1 1 1 1

2 2 1 2 1

1 11

2 1 1

1 1 2

2 1

Ts T j Tz

Ts T j T

T j T T j T

T j T T j T

T T j T

T T



2 2

2 2

2 2 2 2 2 2 2 2 2

2 2 2 2 2 2

0 ( )

1 1 1 2

2 2 1

1 1 (1 ) 4 1 (1 ) 1

2 4 (1 ) 4 (1 ) 4

1 1

2 2

,if s j

T j Tz

T

T T Tz

T T

z

It shows that a large part of interior of the unit circle can be mapped only

from unstable continuous filters.



4.3.4 Trapezoidal Approximation ( Bilinear Transform )

1 1 ( ( ) )2

1 12

Ty a y k x k a y k x k

Ta y k y k a x k x k

∴ The transfer function is

1 1

11 11

1

(1 ) (1 )

2 2( )2 12 1

1 11 12 12 2 1

2 1

1

z zaT aT

a aH z

aT zaT aT zz z az a

T zT z

zs Bilinear Transform

T z

( ) 1

1 ( 1) ( )1

2 2

y k y k y

y k y k x k x ky k aT aT



If we let ,

then equating the real and imaginary parts gives ,

2c

T

( 1)

( 1)

u jvs j c

u jv

2 2

2 2 2 2

2 2

2 2 2 2

2 2 2

2 2 2

( 1) ( 1) ( 1) 2

( 1) ( 1)

1 2

( 1) ( 1)

1 0

1 0

u jv u jv u v jvc c

u v u v

u v vc c

u v u v

r u v

r u v

1, ,

1

jzs c s j z re u jv

z



Because , maps onto the unit circle

in the z-plane.

2 2 2r u v 2 20 ( 1) u v

0 2 2

2 2

1

( 1)

u vc

u v

2 2 1 u v 2 1That is r

0 2 2 1, 1 u v r

Therefore, a stable continuous filter gives a stable digital filter with this

bilinear transformation



※ The relation between the frequency 𝝎 on s-plane and the frequency 𝝎𝑫 on z-

plane :

Because ,

From the bilinear transform, If we let ,

using

Thus the relationship between frequencies with respect

to the continuous time domain and the discrete time domain

is given by

⇒ It is evident that frequency distortion,

or warping is the price paid for having

no aliasing.

sTz e Dj T j T

De e

, exp( )Ds j z j T

/ 2 / 2

/ 2 / 2

exp( ) 1tan

exp( ) 1 2

D D

D D

j T j T

D D

j T j T

D

j T e e Tj c c jc

j T e e

tan2

DTc

1

1

zs c

z



Alternatively, the transformation could be carried out in two steps from s-

plane to p-plane via,

And from p-plane to s-plane via

tan2

pTs c

pTz e

,2

,2

s

s

j p

j p



※

The magnitude response 𝑯(𝒋𝝎) of a suitable continuous lowpass filter

function is transformed into the magnitude response 𝑯(𝒋𝝎𝑫𝑻) of a digital

filter by the bilinear transform.

( ) ( )DH j H j T



At near around 𝝎𝒔/𝟐, the digital filter would be in error because of the

frequency warping. ⇒ Therefore, we have to pre-warp the critical

frequencies, 𝝎𝑫 , of the continuous filter.

Note that the magnitude response of the digital filter terminates at 𝝎𝒔/𝟐 .

Furthermore, the passband and stopband do not undergo much distortion,

because the response is reasonably flat in those ranges.

Thus, although this method of designing digital filters is suitable for filters

that have piecewise-constant magnitude responses (such as LPF, BPF, HPF,

and BSF), it would not be suitable for designing, say, a differentiator, whrein

the magnitude response is continuously changing.

However, where the response is rapidly changing, as in the transition region,

there is considerable distortion of the response, compared with that of the

continuous filter.

Because of the frequency distortion, the phase response of a digital filter that

results from the bilinear transform of, say, a continuous filter with

approximately linear phase (such as Bessel filter) is decidedly nonlinear.



※ The bilinear transform of a continuous filter function

Substitute

When the critical frequency 𝝎𝑫 of digital filter is given, the filter

specification should be transformed to the specification of a continuous

filter. And then design H(s). ⇒ Design H(z).

1

1

zs c

z

1 11

1

1 1

1( ) ( )

1( ) ( ) ( )

1( ) ( )

1

m m

i i

i iz

n ns cz

i i

i i

zs c

zH s H z H s

zs c

z



※ The procedure for using the bilinear transform to design a digital

filter to meet certain specifications is as follows:

1. Convert the critical frequencies 𝝎𝑫𝒊 specified for the digital filter to

the corresponding frequencies in the s-domain by relationship as

follows:

2. Design a continuous filter 𝑯(𝒔) that has the desired properties of the

digital filter at the frequencies 𝝎𝒊. The continuous filter does not

have to be implemented.

3. Replace s by in 𝑯(𝒔) to obtain 𝑯(𝒛), which is the

required digital filter.

tan2

Dii

Tc

1

1

zs c

z



Ex 4.2)

Using the bilinear transform, design a highpass filter, monotonic in

passband and stopband, with cutoff frequency of 1000 Hz and down 10

dB at 350 Hz. The sampling frequency is 5000 Hz . The gain should

be unity in the passband.

Solve)

We must pre-warp the critical frequencies as follows:

11 1 1

22 2 2

22 700 ( / ) tan 2235 /

22

2 2000 ( / ) tan 7265 /2

DD

DD

Tf rad s rad s

TT

f rad s rad sT

41/ 2 10 sec sT f



The frequency 𝝎𝒓 for the normalized approximation to be continuous

lowpass prototype is

∴ n=1⇒ The first-order Butterworth approximation

2

1

1 2

1 2

10 10

72653.2504

2235

( 3.2504 , 1 3.2504)

100.9767

20log 20log 3.2504

r

l l l

n n n r

r

ss

an

1( )

1n n

n

H ss



① Continuous highpass filter with cutoff frequency highpass filter 𝝎𝒍 via

the transformation

Now we transform 𝑯(𝒔) into a digital filter by replacing 𝒔 with

1Substitute ( )

1

ln

l l

ss H s

s s

s

4

24

1

1

2 1( 1) ( 1)1( )

2 1 ( 1) ( 1) ( ) ( )

11

2 2where 10 , 7265

2 101 1

( ) 0.5792 0.57920.1584 1 0.1584

l l ll

ll

l

l

zc z c zT zH z

z c z z c z c

T zc z

ccz

c

cT

z zH z

z z

2 1

1

zs

T z



∴ The corresponding difference equation is

A pole-zero plot and a rough magnitude response obtained graphically

from the pole-zero plot are shown as the following figure. The maximum

gain at the operating range of the filter ends,

is attained as

( ) 0.1584 ( 1) 0.5792 ( ) ( 1)y k y k x k x k

12

sT z

1

1 1( ) 0.5792 1

1 0.1584zH z



② In case of passband

③ In case of stop band

2

0 0 00

0

, , n n l n l

s ss B

Bs B s

00

0

n

Bs

s

s



0.1584

0.3249

0.7265

1.3764

, 1, ,4( / )Di i rad s tan2

Dii

T 1, , 4i

2

1

0.32492.0511

0.1584

4

3

1.37641.8946

0.7265

200

800

1200

400

Ex 4.3) Using the bilinear transform, design a bandstop digital filter, with

stopband 100 to 600Hz. The magnitude response should be monotonic in the

stopband and should have a ripple of about 1.1 dB in the passband. The

response should be down at least 20dB at 200 Hz and 400 Hz. Use a 2000 Hz

sampling rate.

Solution)

The critical frequencies for the digital filter and the corresponding

Frequencies for the continuous filter are given as following table.

41/ 5 10 secsT f



The normalized lowpass filter frequencies 𝝎𝒏, and the continuous

bandstop filter frequencies 𝝎𝒊 are related by

4 1

2 2 2

0 0 1 4

1.218,

0.218

i

ni

i

BB

2 2

3 4

1.218 0.32493.54

0.3249 0.218

2.85 , 1 /

n

n n rad s

For the normalized filter, .

We must design Chebyshev type I filter because a ripple in the passband is

specified

2 3

1 4

min , 2.85n nr

n n

12

102

120log ( 1.1 )

1R dB

1.11 1

10 102 2(10 1) (10 1) 0.5369R

1 2

1.218 0.15841

0.1584 0.218n

2 2

0

n

Bss

s



And also

∴ n=3

For a third-order Chebyshev type I filter

10 10( 20) 20log 6( 1) 20 log rL n n

10

10

( 6 20log )2.1

(6 20log )r

Ln

2 1 2 1sinh cos cosh sin , , , 1, , 1

2 2 2 2 2

1, , 0, 1,sinh co 2, ,

2

0

s co

.4766 , 0

sh sin

.2383 0.9593

kk k

jn

k k n n nj n even k

n ns

nn odd k

s j

n

1

2

1 1 1 1 1sinh ln( 1) 0.46013

sinh 0.4766, cosh 1.10772 2

n n

e e e e

10 10 10

1

10 10

20log ( ) 20log 20log ( )

20log 20log 2

r n r

n n

r

L H j C

L

1( )

( )r

n r

H jC

2.85r



Use the frequency transformation

To obtain the sixth-order continuous bandstop filter

2 2

0

n

Bss

s

1

2 2 2

0

( )2

n

n n n

KH s

s s s as a b

2 2

0

2 2 3

2 0

2 2 4 3 2

0 3 2 1 0

0

212 4 1 0 1 42 2

0

22

3 2 02 2 2 2

240

1 0 02 2

( )( ) ( )

, ,( )

2, , 2

( ) ( )

2,

( )

n

Bsn ss

K sH s H s

Bss s c s c s c s c

s

KK B

s a b

aB Bc c

a b a b

aBc c

a b

단



Applying the bilinear transform (c=1 )

1

1

zs

z

2

2 0

2

01

2 4 3 21 1 0 3 2 1 0

2 3

0 2

2

0 3 2 1 0

0

2

02

0 0 3 1 0

1 0 32 2 3 20 0

0 0

2(1 )1

(1 )( ) ( )

( 2 )( )

(1 ),

(1 )(1 )

(1 )(1 ) (2 2 )

, , 2(1

(1 ) (1 )

zs

z

zK z

H z H sz d z d z e z e z e z e

KK

Bc c c c

s

B

s c c cd d e

B B c c

s s

단

1 0

2 0 3 1 0 3 2 1 0

2 1 2

3 2 1 0 3 2 1 0 3 2 1 0

2 3

2 2

)

(3 3 ) (2 2 ) (1 )2 , 2 ,

(1 ) (1 ) (1 )

0.1285( 1.2841 1)( )

( 0.8376)( 0.4232)( 1.7950 0.8872)( 0.5229 0.6893)

c c

c c c c c c c c ce e e

c c c c c c c c c c c c

z zH z

z z z z z z

2 2 3

2 0

2 2 4 3 2

0 3 2 1 0

0

( )( )

K sH s

Bss s c s c s c s c

s



At

the filter has 6 zero.

Magnitude goes to zero

at 1747.40 rad/s = 278.11 Hz, and

should be stopband

exp( 1747.40 )z j T

0( )H j T

0( ) 0H j T

z transform using MatLab

Ex)

num=[b1 b2 b3 b4 b5 b6];

den=[a1 a2 a3 a4 a5 a6];

[numz1,denz1]=

c2dm(num,den,0.0005,'tustin')

2 2 3

2 0

2 2 4 3 2

0 3 2 1 0

0

( )( )

K sH s

Bss s c s c s c s c

s



Elliptic Filter>> numz=0.3182*[1 -2.4665 3.4253 -2.4665 1]

numz =

0.3182 -0.7848 1.0899 -0.7848 0.3182

>> denz=[1 -1.3166 0.6226 -0.4650 0.3368];

>> dsys=tf(numz,denz, 0.0005)

dsys =

0.3182 z^4 - 0.7848 z^3 + 1.09 z^2 - 0.7848 z + 0.3182

------------------------------------------------------

z^4 - 1.317 z^3 + 0.6226 z^2 - 0.465 z + 0.3368

Sample time: 0.0005 seconds

Discrete-time transfer function.

>> bode(dsys)

>> grid

>> hold on



4.4 Designing Digital Filters by Means of The Matched Z-transform.

The matched z-transform is based on heuristic reasoning derived partly from

experience gained with the standard z-transform.

Poles and zeros of H(s) are matched with poles and zeros of H(z)

(4.79)

H(s) is transformed according to the following rules to get the digital filter

function H(z)

1. Replace real poles and zeros 𝒑𝒊 and 𝒔𝒊 by 𝐞𝐱𝐩(𝑻𝒑𝒊) and 𝐞𝐱𝐩(𝑻𝒔𝒊)

2. Replace complex poles and zeros, say

where

1

1

( )

( ) ,

( )

m

i

i

n

i

i

K s s

H s n m

s p

, by exp( ) i ip or s a jb r j

exp( ) ,r aT bT

1

1

( )

( ) ,

( )

i

i

ms T

D

i

np T

i

K z e

H z n m

z e



3. All zeros of H(s) at 𝒔 = ∞ are mapped in H(z) the point 𝒛 = −𝟏.

𝑳 = 𝒏 −𝒎, If s →∞, z→ −1

4. Select the gain constant of H(z) to match the gain of H(s) at passband

center or at some other critical point.

Like the z-transform, the matched z-transform is restricted to bandlimited

functions(unless a guardfilter is used) because of the aliasing problem.

Rule 4 deals with the prescribed gain constant of the digital filter. It is often

required that the gain of a filter be unity (zero dB) in the passband.

, where L = n-m

1

0 1

1

( )

lim ( ) lim ( 1)

( )

i

i

ms T

L iD ns z

p T

i

z e

H s K z

z e

1

( )z

H z

1

1

(1 )

2 (1 )

i

i

mp T

iD n

s TL

i

e

K

e

1

1

( )

( 1)

( )

i

i

ms T

L i

D np T

i

z e

K z

z e

1

lim lim( 1)L L

s zs z



Ex 4.4) Use the matched z-transform to design a digital lowpass filter

from the continuous lowpass filter

The gain is unity at 𝝎 = 𝟎 and T=0.1 s

Solution)

1. pole :

2. zero :

3. By rule 3, s→∞ , there are two zeros at 𝒛 = −𝟏, Therefore

2

4( )

( 1)( 2 4)

sH s

s s s

0.11 , 1 3 0.9048

0.9048 , 3 0.1 0.1732

T

aT

s s j e e

r e bT

0.44 0.6703 KTs e e

2

2

( 1) ( 0.6703)( )

( 0.9048)( 1.78252 0.8187)

DK z zH z

z z z

2

1

2 (1 0.6703)( ) 1

(1 0.9048)(1 1.78252 0.8187)

0.0026

D

z

D

KH z

K



4.6 General Transform Techniques

Two methods for designing recursive digital filters from continuous filter

functions.



’18 Digital Signal Processing TERM PROJECT

1. First, measure a signal from a sensor(ex: thermocouple, potentiometer, strain

gauge, and etc.) in your LAB and then get the digital signal through AD

converter. Second, get the FFT spectrum using 1024 poins data and 4096 points

data and then compare the results and difference between 1024 points FFT and

4096 points FFT. Third, show that the IFFT signal is almost recovered comparing

with original signal. If you have a problem getting a signal from sensor, you can use

function generator. (Please make FFT and IFFT Program algorithm by yourself)

2. Design a digital LPF(IIR type, Chevyshev and Butterworth) for eliminating

undesirable signal from mixed signals with two more frequencies(ex: 60Hz,

300Hz). Show by simulation and experiment results that the designed filter can be

used to eliminate noise in real time.

3. Design FIR filter and IIR filter to eliminate noise which is over a desired frequency

band in your LAB experiment system. And then compare the effect of eliminating

noise. If you don't have an experiment system for getting a signal from sensor in

your LAB, you can set up arbitrary system.



‘18 Digital Signal Processing Term Project Evaluation List

Name

Complexity

of Setting

System

(50 points)

Filter Design

and Filter

Type (50

points)

Evaluation

by

Simulation

(50 points)

Analysis of

Experiment Result

and Completion

Rate (100 points)

A general Review

(100 points)

(Representation,

Term Report, etc.)

Documents

Chapter 4 Recursive Filter Design