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barcode lib ssrs Thus, the system function of the seventhorder Butterworth filter is in Software
Thus, the system function of the seventhorder Butterworth filter is Code 128C Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Encoding Code 128 Code Set C In None Using Barcode creator for Software Control to generate, create Code 128 Code Set C image in Software applications. Let apand 52, be the desired passband and stopband cutoff frequencies of an analog lowpass filter, and let 6, and 6, be the passband and stopband ripples. Show that the order of the Butterworth filter required to meet these s~ecifications is log d N Z log k with the 3dB cutoff frequency a, being any value within the range Scan Code 128A In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Print Code 128A In Visual C#.NET Using Barcode creator for Visual Studio .NET Control to generate, create Code 128 image in VS .NET applications. The squared magnitude of the frequency response of the Butterworth filter is
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Draw Code39 In None Using Barcode drawer for Software Control to generate, create Code39 image in Software applications. Printing Bar Code In None Using Barcode encoder for Software Control to generate, create bar code image in Software applications. Because I H,(jQ)l is monotonically decreasing, the maximum error in the passband and stopband occurs at the band edges, Q, and a,, respectively. Therefore, we want UPCA Supplement 2 Maker In None Using Barcode creation for Software Control to generate, create Universal Product Code version A image in Software applications. GTIN  13 Maker In None Using Barcode maker for Software Control to generate, create UPC  13 image in Software applications. and From the first equation, we have
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("n) we have
(note that the inequality is reversed because log k < 0). Because the right side of this equation will not generally be an integer, the order N is taken to be the smallest integer larger than (logd)/(log k ) . Finally, once the order N is fixed, it follows from Eqs. (9.20) and (9.21) that a,, may be any value in the range Suppose that w e would like to design an analog Chebyshev lowpass filter s o that
1  Jp 5
IHa(jQ)l I 1
I H a ( j Q ) l 5 J,T
IQI 1 Qp Qs I IQI
Find an expression for the required filter order, N , as a function of
a a,, tip,and , For a Chebyshev filter, the magnitude of the frequency response squared is
where is an Nthorder Chebyshev polynomial. Over the passband, In1 < Q,, the magnitude of the frequency response oscillates between 1 and ( l c2)'I2. Therefore, the ripple amplitude, 6,. is 6, = 1 +c ~ )  ' / ~
CHAP. 91
or At the stopband frequency R , we have
FILTER DESIGN
= (1  a,)' I I H ~ ( ~ Q= ~ ) I ~ . I c~T,:(R~I
which we want to be less than or equal to 8:: Therefore, )), Because (Q,/R,) > I, then TN(R,/i2,) = cosh(N ~ o s h  ~ ( ~ , / R , and we have
or which is the desired expression.
cosh'(I I d ) C O S ~  I (QJR,)  cosh'( l / d ) COS~~(I,X) If Ha(s) is a thirdorder type I Chebyshev lowpass filter with a cutoff frequency
find Ha(s)H,(s). = I and
= 0. I , The magnitude of the frequency response squared for an Nthorder type 1 Chebyshev filter is
where TN(x)is an Nthorder Chebyshev polynomial that is defined recursively as follows, with To(x) = 1 and Tl(x) = x. Therefore, to find the thirdorder Chebyshev polynomial, we first find T2(.u) as follows, T2(x)= 2xTl(x)  To(x) = 2x2  I and then we have T3(x) = 2xT2(x)  Tl(x) = 4x3  2x  x = x(4x2  3) Thus, the denominator polynomial in I Hu(jR)12is and we have
we make the substitution R = s / j in 1H,(jC2)l2as follows: 398 9.28 FILTER DESIGN
[CHAP. 9
Show that the bilinear transformation maps the jQaxis in the splane onto the unit circle, lzl = 1, and maps the lefthalf splane. Re(s) < 0 insidc the unit circle, l z ( < 1. To investigate the characteristics of the bilinear transformation, let z = reJ"' and s = a transformation may then be written as + JR.
The bilinear
='(T , Therefore, 3r sin w r2 l I +r2+2rcosw +'1+r2+2rcosw 2
=T, I
+ rr' + 2r cosw
r2  I
and Note that if r < I, then o < 0, and if r 1, then 0 z 0. Consequently, the lefthalf splane is mapped inside the unit circle. and the righthalf splane is mapped outside the unit circle. If 1, = I, then a = 0, and Thus, the JRaxis is mapped onto the unit circle. Using trigonometric identities, thismay be written in the equivalent Let H&) be an allpole filter with no zeros in the finite splane, If H,,(s) is mapped into a digital filter using the bilinear transformation, will H ( z ) be an allpole filter With T, = 2, the bilinear transformation is
:I
+z' and the system function of the digital filter is
This may be written in the more conventional form as follows, where
CHAP. 9 1
FILTER DESIGN
and Therefore, H ( z ) has p poles (inside the unit circle if Re(sd) < 0) and p zeros at z =  I . Note that rhese zeros come from the p zeros in HJs) at s = m. which are mapped to z =  I by the bilinear transformation. Thus. H ( : ) will nor be an allpole filter. Shown in the figure below is the magnitude of the frequency response of a lowpass tilter that was designed by mapping a type I analog Chebyshev filter into a discretelime filter using the bilinear transformation. Find the filter order (i.e., the number of poles and zeros in H(::)). The magnitudesquared response of a type I analog Chebyshev filter is
where is an Nthorder Chebyshev polynomial. Over the passband, i R,, the magnitude of the frequency response oscillates between I and ( 1 c ' )  ' / ~ . AS the frequency varies from I2 = O to R = R,,, H = N cos'(R/R,,) varies from 8 = N 7r/2 to 0 = 0. Therefore. reaches its maximum or oscillates between zero and 1 N $ 1 times over the interval [O, Q,] [i.e., T,~(R/Q,,) I times]. The bilinear transformation is a onetoone mapping of the jRaxis onto the unit minimum value N I times belween I and I/([ 6') over the interval 10, w , ) , where circle. Therefore. I H ( ~ J ' " ) ~ alternate N will ~

