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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

can someone help me with some logarithmic and exponential equations.
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sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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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
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