<< Chapter < Page Chapter >> Page >

From [link] , it is seen that the passband ripple is measured by δ 1 , the stopband ripple by δ 2 , and the normalized transition band by ω s . The previous section showed that

ω s = 1 / k

which means that the width of the transition band determines k . It should be remembered that this development has assumed a passbandedge normalized to unity. For the unnormalized case, the passband edge is ω p and the stopband edge becomes

ω s = ω p k

The stopband performance is described in terms of the ripple δ 2 normalized to a maximum passband response of unity, or in terms of the attenuation b in the stopband expressed in positive dB assuming a maximumpassband response of zero dB. The stopband ripple and attenuation are determined from [link] and [link] to be

δ 2 2 = 10 - b / 10 = 1 1 + ϵ 2 / k 1 2

This can be rearranged to give k 1 in terms of the stopband ripple or attenuation.

k 1 2 = ϵ 2 1 / δ 2 2 - 1 = ϵ 2 10 b / 10 - 1

The order N of the filter depends on k and k 1 , as shown in [link] . Equations [link] , [link] , and [link] determine the relation of the frequency-response specifications and theelliptic-function parameters. The location of the transfer function poles and zeros must then be determined.

Because of the required relationships of [link] and the fact that the order N must be an integer, the passband ripple, stopband ripple, and transition band cannot be independently set. Severalstraightforward procedures can be used that will always meet two of the specifications and exceed the third.

The first design step is generally the determination of the order N from the desired passband ripple δ 1 , the stopband ripple δ 2 , and the transition band controlled by ω s . The following formulas determine the moduli k and k 1 from the passband ripple δ 1 or its dB equavilent a, and the stopband ripple δ 2 or its dB attenuation equivalent b:

ϵ = 2 δ 1 - δ 1 2 1 - 2 δ 1 - δ 1 2 = 10 a / 10 - 1
k 1 = ϵ 1 / δ 2 2 - 1 = ϵ 10 b / 10 - 1
k 1 ' = 1 - k 1 2
k = ω p / ω s k ' = 1 - k 2

The order N is the smallest integer satisfying

N K K 1 ' K ' K 1

This integer order N will not in general exactly satisfy [link] , i.e., will not satisfy [link] with equality. Either k or k 1 must to recalculated to satisfy [link] and [link] . The various possibilities for this are developed below.

Methods for meeting specifications

Fixed order, passband ripple, and transition band

Given N from [link] and the specifications δ 1 , ω p , and ω s , the parameters ϵ and k are found from [link] and (refcc50). From k , the complete elliptic integrals K and K' are calculated [link] . From [link] , the ratio K / K ' determines the ratio K 1 ' / K 1 . Using numerical methods from [link] , k 1 is calculated. This gives the desired δ 1 , ω p , and ω s and minimizes the stopband ripple δ 2 (or maximizes the stopband attenuation b ).

Using these parameters, the zeros are calculated from (refcc31) and the poles from (refcc39). Note the zero locations do not depend on ϵ or k 1 , but only on N and ω s . This makes the tradeoff between stop and passband occur in (refcc48) and only affectsthe calculation of n u 0 in (refcc38)

This approach which minimizes the stopband ripple is used in the IIR filter design program in the appendix of this book.

Fixed order, stopband rejection, and transition band

Given N from [link] and the specifications δ 2 , ω p , and ω s , the parameter k is found from (refcc50). From k, the complete ellipticintegrals K and K' are calculated [link] . From [link] , the ratio K/K' determines the ratio K 1 ' / K 1 . Using numerical methods from [link] , k 1 is calculated. From k 1 and δ 2 , ϵ and δ 1 are found from

ϵ = k 1 1 / δ 2 2 - 1

and

δ 1 = 1 - 1 1 + ϵ 2

This set of parameters gives the desired ω p , ω s , and stopband ripple and minimizes the passband ripple. The zero and polelocations are found as above.

Fixed order, stopband, and passband ripple

Given N from [link] and the specifications δ 1 , δ 2 , and either ω p or ω s , the parameters ϵ and k 1 are found from [link] and (refcc48). From k 1 , the complete elliptic integrals K 1 and K 1 ' are calculated [link] . From [link] , the ratio K 1 / K 1 ' determines the ratio K ' / K . Using numerical methods from [link] , k is calculated. This gives the desired passband and stopband ripple andminimizes the transition-band width. The pole and zero locations are found as above.

An approximation

In many filter design programs, after the order N is found from [link] , the design proceeds using the original e, k , and k 1 , even though they do not satisfy [link] . The resulting design has the desired transition band, but both pass and stopband rippleare smaller than specified. This avoids the calculation of the modulus k or k 1 from a ratio of complete elliptic integrals as was necessary in all three cases above, but produces results thatare difficult to exactly predict.

Design of a third-order elliptic-function filter

A lowpass elliptic-function filter is desired with a maximum passband ripple of δ 1 = 0 . 1 or a = 0 . 91515 dB, a maximum stopband ripple of δ 2 = 0 . 1 or b = 20 dB rejection, and a normalized stopband edge of ω s = 1 . 3 radians per second. The first step is to determine the order of the filter.

From ω s , the modulus k is calculated and then the complimentary modulus using the relations in (refcc50). Specialnumerical algorithms illustrated in Program 8 are then used to find the complete elliptic integrals K and K ' [link] .

k = 1 / 1 . 3 = 0 . 769231 , k ' = 1 - k 2 = 0 . 638971
K = 1 . 940714 , K ' = 1 . 783308

From δ 1 , ϵ is calculated using [link] , and from ϵ and δ 2 , k 1 is calculated from (refcc48). k 1 ' , K 1 , and K 1 ' are then calculated.

ϵ = 0 . 4843221 as for the Chebyshev example.
k 1 = 0 . 0486762 , k 1 ' = 0 . 9988146
K 1 = 1 . 571727 , K 1 ' = 4 . 4108715

The order is obtained from [link] by calculating

K K ' K ' K 1 = 3 . 0541

This is close enough to 3 to set N = 3 . Rather than recalculate k and k 1 , the already calculated values are used as discussed in the design method D in this section. The zeros are found from(refcc31) using only N and k from above.

ω z = ± 1 k s n ( 2 K / N , k ) = ± 1 . 430207

To find the pole locations requires the calculation of ν 0 from (refcc38) which is somewhat complicated. It is carried out using the algorithms in Program 8 in the appendix.

ν 0 = K N K 1 s c - 1 ( 1 / ϵ , k 1 ' ) = 0 . 6059485

From this value of ν 0 , and k and N above, the elliptic functions in (refcc40) are calculated to give

s n ' = . 557986 , c n ' = 0 . 829850 , d n ' = 0 . 934281

which, for the single real pole corresponding to i = 0 in (refcc39), gives

s p = 0 . 672393

For the complex conjugate pair of poles corresponding to i = 2 , the other elliptic functions in (refcc40) are

s n = 0 . 908959 , c n = 0 . 416886 , d n = 0 . 714927

which gives from (refcc39) for the poles

s p = 0 . 164126 ± j 1 . 009942

The complete transfer function is

F ( s ) = s 2 + 2 . 045492 ( s + 0 . 672393 ) ( s 2 + 0 . 328252 s + 1 . 046920 )

This design should be compared to the Chebyshev and inverse- Chebyshev designs.

Questions & Answers

a diagram of an adult mosquito
mubarak Reply
what are white blood cells
Mlungisi Reply
white blood cell is part of the immune system. that help fight the infection.
MG
what about tissue celss
Mlungisi
Cells with a similar function, form a tissue. For example the nervous tissue is composed by cells:neurons and glia cells. Muscle tissue, is composed by different cells.
Anastasiya
I need further explanation coz celewi anything guys,,,
Calvin Reply
hey guys
Isala
on what?
Anastasiya
is air homogenous or hetrogenous
damiane Reply
homogenous
Kevin
why saying homogenous?
Isala
explain if oxygen is necessary for photosynthesis
Allice Reply
explain if oxygen is necessary for photosynthesis
Allice Reply
Yes, the plant does need oxygen. The plant uses oxygen, water, light, and produced food. The plant use process called photosynthesis.
MG
By using the energy of sunlight, plants convert carbon dioxide and water into carbohydrates and oxygen by photosynthesis. This happens during the day and sunlight is needed.
NOBLE
no. it s a product of the process
Anastasiya
yet still is it needed?
NOBLE
no. The reaction is: 6CO2+6H20+ solar energy =C6H12O6(glucose)+602. The plant requires Carbon dioxyde, light, and water Only, and produces glucose and oxygen( which is a waste).
Anastasiya
what was the question
NOBLE Reply
joining
Godfrey
the specific one
NOBLE
the study of non and living organism is called.
Godfrey
Is call biology
Alohan
yeah
NOBLE
yes
Usher
what Is ecology
Musonda Reply
what is a cell
Emmanuel Reply
A cell is a basic structure and functional unit of life
Ndongya
what is biolgy
Hawwi Reply
is the study of living and non living organisms
Ahmed
may u draw the female organ
MARTIN Reply
i dont understand
Asal
:/
Asal
me too
DAVID
anabolism and catabolism
Sani Reply
Anabolism refers to the process in methabolism in which complex molecules are formed "built" and requires energy to happen. Catabolism is the opposite process: complex molecules are deconstructed releasing energy, such as during glicolysis.
Anastasiya
Explain briefly independent assortment gene .
Otu Reply
hi
Amargo
hi I'm Anatalia
Joy
what do you mean by pituitary gland
Digambar
draw and lable the cell
Ameh Reply
why homones are destroyed soon after completing their role
Nyirenda Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Digital signal processing and digital filter design (draft). OpenStax CNX. Nov 17, 2012 Download for free at http://cnx.org/content/col10598/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Digital signal processing and digital filter design (draft)' conversation and receive update notifications?

Ask