DSP Digital Signal Processing MODULE III PART2

8/14/2019 DSP Digital Signal Processing MODULE III PART2

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N ~ log10 [(10- 0.1Kl - 1)/(10- 0.1~ - 1)]where,

A = ill (n.. - ill)/[ - ili + ill n..]B =il2 (ilu - ill)/[il~ - ill ilu]

5.1.5 ChebyshevFilters.There are two types of chebyshev fllters. Type-I filters are all-pole ftlters thathas equiripple behaviour in the passband and a monotonICrunctIon In the stopband.The second type (Type-II) filters contain both oles and zeros and has a monotonic

behaviour in the rpassband and an equiripple behaviour in the stop an. e zeros'Of thIs type of tJlters lie on the imaginary axis In the s-plane.~ The type-I -normalizedcltebyshev lowpass filter"s magnitud~quared frequencyresponse is given by,

I HN (j il) 12=1/(1 H2 T~ (ilwhere ( is the filter error parameter defIned as,. (2= 100.1Ai'- 1; ~ - Attenuation in passbandand TJ\ (il) is the Nth order chebyshev polynomial. The chebyshev polynomials canbe defined as,

(5.11)

T (il) = {cos (K cos- 1 il) for! il I ~ 1.

N cosh (N cosh-1 il) for i il I> 1It can easily be seen from Eqn (5.12) that,

IHN (j il)

12 is ~ at il =0 for even1+

Nand 1 at il =0 for odd N. These two cases of magnitude squared frequency responseof the type-I chebyshev filter are given in Fig 5.3.

(5.12)

11-1" (iQ)I' 11-1" 00)1'

1 U. 0Q, ,0Odd-N Even-N

Fig. 5.3 Magnitude squared frequency response for the normalized type-I,- chebyshev filter of odd.N and even-No


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'?From Fig 5.3 and Eqn (5.11), we can observe the following properties.

1. !HNGQ)12 oscillates between 1 and 1/(1 + ~h within the passband, the soI. 2 .called equiripple, and has a value of 1/(1 + ( ) at Q = 1, the so called cutofffrequency.

2. I HN (j Q) 12 is monotonic outside the passband. The stopband begins at~ with I HN (j Q) 12=-t-A.The normalized transfer function HN (S) of the chebyshev filter can be obtained

from Eqn (5.11). That is, we can find the poles of HN (S). HN ( - S) and select theleft half plane poles to form HN (S). The poles of HN (S) . HN ( - S) are given by rootsof the denominator of Eqn (5.11). We can show that the roots of,

1 + (2 T~ (S/j) = O.are located on an ellipse in the S-plane and defined as 8K =ak + j Qk, where,

.00 (l l .00 - 1 1 \1 . (2K - 1 'IaK=sl N'SI - sm ~ 17t() ,~1 )(1 . - l' 1 J ( 2K - 1 )K =cosh N' sm h E cosl 2N 7t(5.13)

for k= I, 2,..., 2N.We now form the normalized transfer function H!\ (8) by using the LHP poles ofHN (8). HN ( - 8), as, r--- - -. HoHN(S)=~

TI (S - Pi)i= 1(5.14)

where Pi's are SK's on LHP and we choose the constant Ho as,N

10- 0.05~ . TI ( - Pi) for even - Ni = 1Ho= NTI (-Pi)i = 1

(5.15)for odd - N

so as to achieve zero minimum passband attenuation, that is, which makes Ho equalto 1 for odd-N and 1/~ for even - N.


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omputation of N:It is essential to get a closed form formula for the order N, in terms of givenilter specifications. Let us derive an equation for N in the case of normalized lowpassFor normalized case, the cutoff frequency no =1 rad/sec. Let us choose the

assband edge frequency .01 as equal to the cutoff frequency.

TN (0) = cosh (Ncosh- 10) for .01~0 ~ ".he normalized loss function, A (j .0) in dB is given by,. 2 2 1/2

A.9 .0) =20 log [1 +E TN (.0)]22=10 log [1 + E TN (O)J

t the stopband edge (critical) frequency, 02, the loss f~ction is,A (j .02)=As ' the stopband attenuation. "-

2 2 ] -As =10 log [1 + E TN (02) .= 10 log { 1 + (10.1 A"-1) [cosh (N cosh - 102)]2 }

herefore, from the above relationship, we get an expression for 02 as,

hat is,

02 = cosh ( ~ cosh- 1 [(10.1A"- 1)/(10.1,\, - 1)]1/2 )hen the ratio of 02 to 01 can be rearranged to obtain,

N = cosh-1 [(10.1 A, - 1)/(10.1 A,.- 1)]112cosh-1 (02/01)N is not an integer in the above equation, we choose the next larger integer. So

e reVl'1'ite the equajion as follows:cosh-1 [(10.1 A, -1)/(10.1 '\, -1)]112N2:

cosh- 1 (02/01) (5.16)

The values of sinh-1 (x) and cosh-1 (x) can be evaluated using the followingentities. -sinh-1 (x) =In (x + -.Jx2 + 1)cosh - 1 (x) =in (x + -.Jx2- 1 )


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To design a lowpass filter with unnoimalized transfer function, first we have toidentify the normalized one by using the given specification using Eqns (5.14) and(5.16) and then do the lowpass-to-Iowpass transformation.To design other types of chebyshev filters we can proceed as in the Butterworthfllter case. The required formulas to find the order N of chebyshev polynomial toform the transfer function of the prototype filter are listed below. I

Highpass filters:h- 1 0.1A 1AN


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he filter unstable. Therefore the filter design problem for IIR filters always involvespproximation of both magnitude and phase response specifications.

When an TIR filter is determined strictly in terms of a magnitude approximation,here the phase is completely disregarded, it is convenient to consider designs inerms of the magnitude squared function, which is defined as,

I H (ei~ 12 = I H (Z)" H (r1) !z=i'he poles and zeros of a magnitude squared function are distributed withirror-image symmetry with respect to the unit circle in the Z-plane. The poles ofZ are uniquely determined from the magnitude squared fu.nction ~ those I .side the unit circle for a stable IIR fi ter. e zeros may lie anywhere in the

:-plane, but,"li" we choose zeros ot magnitude s uared function lying InsIde or on thenit circle in e -p ane, en e resu ting filter will be a minimum ase filterthat is, lesser pnase dIstortIon filter. e pro em of designing IIR filters is nothingut fmdm~ tne niter coenlClent Djs and ai'S of Eqn (5.17) such that some aspects of

filter's response approximates a desired response. -A method for designing IIR digital filters is _direct closed form digital designn either the frequency or the time domain. Beginning with the desired response ofhe filter, one can often decide where to place poles and zeros at appropriate positionsn the Z-plane to approximate this response directly. Among the techniques that fallnto the category of direct digital design are magnitude squared function design andime domain design.

Another method by which IIR filters are often designed is by using Optimationrocedures to place poles and zeros at appropriate positions in the Z-plane topproXImate in some sense the desired response. This design procedure does notenerally yield closed form expressions for pole and zero positions (that is, filteroefficients) as a function of the desired response. This optimation procedure is usedo determine the filter coefficients that minimize some error criterion, subject to theppropriate design equations. Using this iterative procedure either the errorventually reaches a minimum value or a specified maximum number of iterationsperformed and the procedure terminates. The minimum mean squared error designone of the important methods based on optimization procedure.

The most popular technique for designing IIR filters is to digitize an analogilter that satisfies the design specifications. The technique of designing anppropriate analog filter and digitizing the resulting transfer function to give aigital fIlter is most useful for designing standard filters such as lowpass, highpass,andpass and bandstop filters where a considerable amount knowledge on such analogilter is available in the literature. This approach is highly preferred, because:

1. The art of analog filter design is highly advanced and since useful resultscan be achieved, it is advantageous to utilize the design procedures alreadydeveloped for analog filters.


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2. Many useful analog design methods have relatively simple closed form designformulas. Therefore, digital filter design methods based on such analog designformulas are rather simple to compute.

3. In many applications it is of interest to use a digital filter to simulate theperformance of an analog linear time-invariant filt::!r.There are four most widely used procedures for mapping the analog transfer

function (digitizing) into digital transfer function. They are,1. Method of mapping of differentials.2. Matcl1ed z-transform technique.3. Impulse invariant transformation.4. Bilinear transformation.

Mapping of differentials is the simplest way of digitizing the continuous system.The method is to replace differentials in the differential equations of the continuoussyi;tem with finite differences in order to obtain a difference equation thatapproximates the giveu differencial equation. This simplest replacement that can bemade is to use a forward or backward difference to replace a first' differential. Theattractive feature of replacing differentials by simple. difference is that rationaltransfer functions in S become rational transfer functions in Z by using simplesubstitutions like 8 =(2 - l)/T, 8 = (1 :- Z- l)/T, etc, where T is the sampling period.This mapping has the desirable property, that a stable analog filter is transformedinto a stable digital f1lter. However, the possible location of the poles of the digitalfilter are confined to relatively small frequencies and as a consequence, the mappmgis restricted to the design of lowpass filters, and bandpass filters having relativelysmall resonant frequencies. That is, this mapping method does not adaquatelypreserve the filter characteristics.

Another method of converting an analog fllter into an equivalent digital filteris to map the poles and zeros of H (8) directly into poles and zeros in the Z-plane.This mapping has the property that an 8-plane pole (zero) at 8 =- a maps to aZ-plane pole (zero) at Z = e- aT where T is the sampling period. Thus each factor ofthe form (8 + a) in H (S) is mapped into the factor (1 - e- aT Z- \ This mapping iscalled the matched Z-transformation To preserve the frequency characteristic ofthe analog filter, the sampling interval in the matched Z-transformation must beproperly selected to yield the pole and zero locations at the equivalent positions inthe Z-piane. Although the matched Z-transformation is easy to apply, there are manycase when it is not a suitable mapping. For example, if the analog system has zeros.with ce~fL.fi:.e.qu.e.ncies greater than half of the sanlPling- freqmmcy, the matchedZ-transformation is unsuitable. In general, use of impulse invariant or bilineartransformation is to be preferred over the matched Z-transformation.


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H(eiT)=* L Haun+jlQ,.)1=-- --- (5.18)~here ~ =-2T7tis the ra..cIian_samPli?[ frequency_of ~he-~gital s!~tem. This mappin~f~om the S-plane to the Z-plane is shown in Fig ~-- - - - -- --- .

jn S-plane Z-plane------------------------.37tIT7tIT

-7tIT

}E[C-------.Fig. 5.4. The mapping from the S-plane to the Z-plane correspondingto impulse invarian~ transformation.Each horizontal strip of the S-plane of width 2'; is mapped into the entire4

Z-plane. The left half the strip maps into the interior of the unit circle. the rightalf of the strip maps into the e),,1:erior of the unit circle, and the imaginary axis inthe strip maps onto the unit circle. Adjacent strips in the S-plane are thus aliasedr folded over into each other in the Z-plane. That is, the impulse invariance methodoes not correspond to a simple algebraic mapping of S-plane to the Z-plane. Fromhe Fig 5.4, it is clear that the frequency response of an analog filter and thequivalent digital filter obtained by impulse invariant transformation are to bedentical, the analog filter must be band limited to the range ~7t::;n s .Otherwise,he mapping produces aliasing errors (interferences).


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Let us consider a analog system with transfer function in its partial fractionexpansion, in order to show the method of mapping in impulse invarianttransformation.N AK

Ha (8) = L 8 - 8k=l KThe corresponding impulse response is given by,

(5.19)

Nha (t) = L AK' eSKt u (t)k=l

where u (t) is a continuous-time unit-step function.The unit sample response of the digital filter is then obtained by replacing t by

nT, where T is the sampling period. That is,N

h (n) = ha (nT)= L AK eSKnT u (n).K=lN ( \n= L AK eSK T ) u (n).K=l

The transfer function of the digital filter is given by the Z-transform of h (n) as,N AK

H (Z) = L 1 - eSK T ZK=l (5.20)In comparing Eqns (5.':'9) and (5.20) we observed that a pole at S =SK in the 8-planetransforms to a pole at eSKT in the Z-plane and the coefficients in the partial fractionexpansions of Ha (8) and H (Z) are equal. Therefore, if the analog filter is stable, thenthe resulting digital filter is also stable. However, for complex poles and multi-orderpoles cases, this mapping approach should be modified.

Except for aliasing (due to absense of band limitation) the relationship betweenanalog and digital frequencies is linear, that is, the shape of the frequency responseis preserved. It should be noted that the impulse invariance transformation isappropriate for band limited filters. For example, highpass or band-stop filters wouldrequire additional band limiting (guard) filter to avoid severe aliasing distortion.

Let us illustrate the way of converting an analog filter's transfer function intoa digital transfer function through two examples.


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xample 5.3: Convert the analog filter transfer function,1H (8) = - -

to H (Z) using impulse invariance transformation. Assume an sampling frequencyf 5 sps.olution: The partial fraction expansion of given H (s) yields,

1 --1-H (8) = 8 + 1 - 8 + 2he impulse response can be obtained, by taking inverse Laplace, which results in,

h (t) =e- t - e- 2t ; t ~ O.y replacing t by nT, we obtain,

h(nT)=e-nT-e-2nT; n~O.y applying Z-transform, we have,

1 --1:H (Z) = - T - 1 - 1 - 2TZ- 11-e Z -e1 - e- 2TZ-l - 1 + e- Tz- 1

1 - e- T Z- 1 - e- 2T z- 1 + e- 3T Z- 2- (e-T -e-2T)Z-1-1-(e-T+e-ZT)Z-I+e-3TZ-2

or T =i see, the transfer function of the corresponding digital filter is given by,H (Z) = 0.148r 11 - 1.489Z- 1 + 0.549Z - 2

In general, there are four steps involved in the design of digital filter from itsalog counterpart through impulse invariant method. They are,1. Use the given specifications and fmd the order N and formulate H (8).2. Find the il1Verse Laplace transform of H (8) ~ h (t).

Replace t by nT ~ h (nT).:J.4. Find the Z-transform of h (nT) ~ H (Z).


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xample 5.4: Convert the analog filter with system function H (8) = 8 + 0'21(8 + 0.1) +nto a digital IIR filter by means of the impulse invariant method.Solution:

The impulse response of the given analog filter can be obtained using inverseaplace table. (see appendix - F). Thus, we have,h (t) = e- O.lt cos 3t; t ;::0

['3t - i3t 1- - O.lt e1 + e. i- e 2 J

=1.e(-0.1+ j3) t + 1. e(- 0.1 - j3) t2 2hen, replacing t by nT, provides,

h lnT ) =1. e(- 0.1 + j3) nT + 1. el- 0.1 - j3) nT .2 2 ' n;::Opplying Z-transform leads to the required digital transfer function,

1 1.- 22 + .H (Z) =1- e(- 0.1 + j3) T Z- 1 1- e(- 0.1- .13)T Z- 11. (1- e(- 0.1-j3i T Z-l) + 1. (1- e(- 0.1+j3) TZ-l)2 2

(1 - e(- 0.1+j3)TZ-1) (1- e(- 0.1 - j3) T Z-l)- 1 - (e- 0.1cos 3T) Z- 1 .- 1- (2e- 0.1 cos 3T) Z-l + e- 0.2TZ- 2

hus we obtained the digital IIR filter transfer function in a realizable form..2.3. Bilinear Transformation

A simple conformal mapping from the S-plane to the Z-plane which eliminatese aliasing problem mentioned above and preserves the desired simple algebraicrm is the bilinear transformation defined by,

2 (1 - Z- 1)S ~ T (1 + Z- 1) (5.21)

he nature of this mapping is shown in Fig 5.5. As seen in this figure, the entirn axis in the S-plane is mapped onto the unit circle. The Eqn (5.21) can be rewrittenr Z in terms of S as,


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z =(2/T)+ 8(2 T)-8j?=

(5.22)hen S =j n, we find, \;f\ ~

Z == (2/T) + j n(2/T) - j nhere, we have, I Z I= 1. Also from Eqn (5.22), we find that for real part of Segative, the IZ I< 1 and for real part of S positive, the IZ I> 1. This implies, theeft of the S-plane is mapped inside the unit circle and the right half of the S-planes mapped outside the unit circle in the Z-plane. Therefore, the bilinearransformation yields stable digital filters from stable analog filters. Also, the bilinearransformation avoid the problem ()f aliasing encountered with the use of impulsenvariance, because it maps the entire j Q axis in the S-plane onto the unit circle ~he Z-plane.

S-planeZ-plane

Fig. 5.5 The mapping of the S-plane into theZ-plane using the bilinear transformation.The transfer function of the digital filter H (Z) is obtained from the bilinearansformation by making the algebraic substitution of Eqn (5.21).

hat is.

H(Z)= H(S) Is=! (l_Z-I)T (l+Z-I)owever, there is a highly non-linear relationship between analog frequency n andigital frequency w. Evaluation Eqn (5.21) for Z ==e?'"and S = j n giving,

(s,tV


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hich can be written as,. 2 (l-e-j(J))JQ=T (1 + e-j(J)) ~ ( t~~. 2 [ei(J)/2e-j(J)/2] ::: 2-J Q = T [ei(J)/2 e-jw/2] T

. c)}'1J.' ~""~ 'l-,Ve.-~j 5;\,> CJ/~.;l ~ '112-

j Q = "j. tan ( ~ )I n=tan( ~n-he inverse relationship is given by,

OJ= 2 tan-l(n T I2 )he nonlinear relationship between Q and OJis shown in Fig 5.6 for the case T =2or small values of OJ,the mapping is almost linear. But for most of the frequencycale, the mapping is highly non-linear (Frequency compression effect). Thisposes a strong restriction on bilinear transformation. It implies that the amplitude

(5~24

(5.25)

OJOJ=2 tan" (nT/2)

n


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he analog filter transfer function is transformed to the digital domain using Eqn5.23) and the resulting digital filter will meet the desired specifications.8 ro

s Q = 2fT tan (ro'2)"""""''''''''''''''''.

rQ)~

Qp: ;ils Q

IH(jQ)1

Qp Q

Fig. 5.7. Techniques for compensating the nonlinear frequencywarping of the bilinear transformation.

Although the bilinear transformation can be used to transform the stable analogilter into the stable digital filter, the distortion in the frequency axis will manifesttself in terms of distortion in the phase characteristic associated with the filter.herefore, a linear phase analog filter cannot be mapped into a linear phase digitalilter through the bilinear transformation.

Let us illustrates the procedure for bilinear transformation to design a digitalIR filter from analog transfer function through the following examples.

xample 5.5 Design a single pole lowpass digital IIR filter with - 3 dB bandwidthf 0.2 11,by use of bilinear transformation.olution: The first-order (single pole), normalized analog transfer function is giveny, (Butterworth case),

I"H (8) = 8 + 1

he cutoff frequency given is 0.2 1trad. It is prewarped using.


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2 Wene ="Ttan :2

ssuming T = 1 see, we h'ave" 2 0.21t 0 ~ 5~


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INFINITE IMPULSE RESPONSE FILTER"3. Apply bilinear transformation to H (8) and find H (Z). That is,"

H (Z) = H (8) I 8 - i (l - Z-1)T (l+Z"15.6. Design a digital Butterworth lowpass filter using the bilinearansformation method to satisfy the following specifications:

a) - 3.01 dB cutoff frequency at 0.5 1tradb) at least 15 dB attenuation at 0.75 1trad

2 001 0.5 1t01 =-tan -=2 tan-= 2.T 2 2

~~Lp.\-O"'"t-0'

tep 1: Prewarping 001and OJ:!using T =1 sec.

02 =2 tan 0.75 1t=4.8282.2he order of normalized Butterworth lowpass filter can be obtained using,

Jog10 [ (103.01/10 - 1)/(1015/10 - 1) ]N>.- 2 log [2/4.8282]~ 1.9412.

e choose the next larger integer as N=2 and formulate the transfer function.1

H (8) =82 + "28 + 1tep 2: The prewared transfer function is obtained using.

H(8) = H (8) 18= 8/n14


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tep 3: Applying the bilinear transformation to get H (Z) yields,"H (Z)= H (8) Is=1 (1 - Z-l)T (1+Z-l)

4= 2[

2 (1 - Z-l)]

+ 2 { 2[

2 (1- Z-l)]+4

(1 + Z-1) (1 + Z- 1) .1+2Z-1+Z-2

3.144 + 0.586 Z- 2- 0.318 + 0.636 Z-1 + 0.318 Z- 2- 1+ 0.186Z-z

his is the required digital filter transfen.function.In the similar way, other types of digital IIR filters (highpass, bandpass, aandstop) are designed using required frequency transformation formulas. It is

ote that the factor ~ gets cancelled in the above procedure through steps 2 andhat is, in,

H (8) = H (8) Is=(2/T) t~ (0)/2)"H (Z) = H (8) Is=1 (1 - Z-l)T (1+Z-1). That is why we assumed T =1 when it was not given. Therefore, for simplici

e can avoid this factor ~ in steps 2 and 3.xample 5.7: Design a first-order digital Butterworth highpass filter whichquivalent to an analog filter with cutoff frequency 1 KHz at a sampling rate04 sps. Use Bilinear transformation.

he first-order normalized transfer function is given by,1

H (8) = 8 + 1sing lowpass to highpass transformation, we have,


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HI (S)= H(S) IS=I/SS

-S+lrewarped cutoff frequency is given by,

nc = tan ( 2 7t (100~) (10- 4) )

=tan ( ;~ J= 0.325ote: Digital frequency OJ= n T,wheren = 2 7tf).herefore, the prewarped transfer function is given by,

H(S)= H (S) I1 IS=S/O.325=[ S:1 1 =S/O.325

S- S + 0.325

he required digital transfer function is obtained through bilinear transformation a

H (Z) = H(S) Is= 1 - Z-l1 + Z-l1- Z- 11+Z-1

- 1-Z-1 1 + 0.3251 +Z-1-Z-1

- (1 - Z- 1)+ 0.325 (1 + Z- 1)1-Z-1

- 1.325 - 0.675 Z- 1- 0.755- 0.755Z-1- 1 - 0.509Z-1

his transfer function is in a suitable form for realization.


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.2.4. Digital - to - Digital TransformationIn Section 5.3.1, we have seen analog-to-analog transformation to obtain th

on-normalized lowpass, highpass, bandpass and bandstop filters from the normalizedwpass prototype filter transfer function. Similarly, a set of digital transformationan be obtained that take a digital prototype lowpass filter and turn it into a digitaighpass, bandpass, bandstop or another digital iowpass filter. These transformationsre given below.owpass to Lowpass

Z-l Z-l-a1- aZ-1

a = sin [(OOp- OOn)/2]sin [(OOp+ OOn)/2]

OOp- passband edge frequency of prototype.OOn passband edge frequency of transformed (new)

owpass to HighpassZ-1-7- Z-I+al+ar1

cos [(00n + OOp)/2]a=-cos [(OOn- OOp)/2]

wpass to BandpassZ- 2 - 2.a K Z- 1+-K - 1Z-1 I~+l 'K+l-7K ') K-= !.Z-2- ~ Z-1+ 1K+ 1 K+ 1

a = cos [(002+ Wl)/2]cos [(Wz - (01)/2]K= cot [(0)


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Lowpass to BandstopZ-2_~Z-1+1-KZ-1-. l+K l+K1- Kz- 2- ~z- 1+1l+K l+K

h cos [(~ + (01)/2]were a- cos [(002- (01)/2]

K= tan [(002 (01)/2] tan (~ )Therefore, there are generally two approaches to obtain the desired digital filterusing digitizing procedure for the given specification. The steps involved in theseapproaches are given below in block diagramatic forms.

1. I Design a II prototype II Lowpass filter~ of order N. I

Performanalog-to-analogtransformation

Digitize theresultant analogfilter

2.

Desireddigitalfilter.

Desireddigital- filter.

The frequency band transformation can be performed either in the analogdomain or in the digital domain. But depending on the types of filter we must choosethe appropriate domain for transformation. AI3we have seen that there are severealiasing problem with designing highpass and bandstop filters through the impulseinvariant method, it is better to perform mapping from an analog lowpass filter intoa digital lowpass filters and then to perform the frequency transformation in digitaldomain to avoid aliasing problem. In the case of bilinear transformation, where thereis no aliasing problem, it does not matter whether frequency transformation isperformed in the analog or in the digital domain. In fact, in this case only, the abovetwo approaches result in identical digital filters.5.2.5. Comparision of FIR and IIR Filters

AI3we have seen, there are number of design techniques to design both FIRand IIR filters. Each one has its own merits and demerit,;. Therefore, comparisionof these filters in strict sense to give a precision answer as what is best, is a difficulttask. However, a general comparision between FIR and IIR filters can be carried outin the following manner.

i Design a Digitize to Performprototype - obtain digital - digital-to-digitali Lowpass filter prototype transformation.of order N. Lovlpass filter.


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1. Linear phase FIR filter design is easy. But IIR filters are not linear phasefilters.2. Finite word length effects, like coefficients inaccuracy error, round off noise,etc are severe with IIR filters, but is not so severe with FIR filters becauseof non-recursive realizations.3. There are closed form formulas to design IIR filters. But the computationsinvolved with FIR filters are iterative and lengthy.

4. The magnitude response of IIR filters is better (sharp cutoff) than that ofFIR filters of the same order. That is, FIR filters requires more coefficientsfor sharp cutoff which needs more processing time and storage.5. Further, the inefficient direct computations involved with FIR filters can besignificantly improved using fast convolution techniques which employ theFFT.6. There are no zero-input limit cycle oscillations with non-recursive FIR filters,since these structures have no feed back.In economical point of view, the terms hardware complexity, chip area, and

mputational speed are directly related to the order of the filter required to meetgiven specifications. If we put aside phase consideration, it is generally true thatgiven magnitude specification can be met most efficiently with an IIR filter.

eview Questions:1. Give the magnitude response of lowpass Butterworth filter and show the effectof the order N on this response.Give any two properties of chebyshev filters..

3. Distinguish between the frequency response of chebyshev type I and II filters.4. Mention the most general form of the Z-transform of llR filter.5.6.

Why do we go for analog approximations to design a digital filter?What do you mean by backward difference in the mapping of differentials?

7. Why impulse invariance method is not preferred in the design of highpassand bandstop IIR filters?Using bilinear transformation, What is the image of S =~11/2in the Z-plane?.

9. Why do we need prewarping in the design procedure using bilineartransformation?

DSP Digital Signal Processing MODULE III PART2

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