 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
barcode lib ssrs and the selectivity factor. in Software
and the selectivity factor. Recognizing Code128 In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128B Drawer In None Using Barcode generation for Software Control to generate, create Code 128 Code Set A image in Software applications. Thus, we have
Recognizing Code 128C In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. Draw USS Code 128 In Visual C# Using Barcode printer for .NET Control to generate, create Code 128 Code Set B image in Visual Studio .NET applications. log d = 5.71 log k
Draw Code128 In .NET Framework Using Barcode drawer for ASP.NET Control to generate, create Code128 image in ASP.NET applications. Encode Code 128B In .NET Framework Using Barcode creation for Visual Studio .NET Control to generate, create Code128 image in VS .NET applications. which, when rounded up, gives N = 6. For the 3dB cutoff frequency of the Butterworth filter, we will select 52, so that Code 128A Maker In VB.NET Using Barcode printer for Visual Studio .NET Control to generate, create USS Code 128 image in .NET applications. Generating EAN 128 In None Using Barcode encoder for Software Control to generate, create EAN128 image in Software applications. that is, so that the Butterworth filter satisfies the passband specifications exactly (this will provide for some allowance for aliasing in the stopband). With UPCA Creation In None Using Barcode generator for Software Control to generate, create UPCA image in Software applications. Code 3 Of 9 Maker In None Using Barcode creator for Software Control to generate, create Code 39 Full ASCII image in Software applications. we have
ANSI/AIM Code 128 Generator In None Using Barcode generation for Software Control to generate, create Code128 image in Software applications. Bar Code Maker In None Using Barcode generation for Software Control to generate, create bar code image in Software applications. which gives
ISBN Generator In None Using Barcode maker for Software Control to generate, create ISBN  10 image in Software applications. Barcode Encoder In .NET Framework Using Barcode creation for ASP.NET Control to generate, create bar code image in ASP.NET applications. Therefore, the magnitude of the frequency response squared is
Recognize Barcode In Visual Studio .NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET applications. GTIN  12 Printer In None Using Barcode printer for Font Control to generate, create GS1  12 image in Font applications. I ~HU(J')~' = I + c n / o , 7 0 9 ~
Reading Bar Code In Visual Basic .NET Using Barcode Control SDK for VS .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. European Article Number 13 Maker In ObjectiveC Using Barcode generation for iPad Control to generate, create GS1  13 image in iPad applications. and the 12 poles of Ha(s)Ha(s) = lie on a circle of radius R, = 0.7090, at angles
Scanning Barcode In Visual C# Using Barcode scanner for .NET framework Control to read, scan read, scan image in VS .NET applications. Code39 Scanner In C# Using Barcode recognizer for VS .NET Control to read, scan read, scan image in VS .NET applications. + (~/jS2(.)'~
FILTER DESIGN as illustrated in the following figure: [CHAP. 9
Thus, the poles of the Butterworth filter are the three complex conjugate pole pairs of H,(s)H,(s) lefthalf splane: that are in the
Therefore, with
forming secondorder polynomials from each conjugate pole pair, we have
The next steps, which are algebraically very tedious, are lo perform a partial fraction expansion of H,,(s), perform the transformation
and then recombine the terms. The result is
H (z) = 0.0105z~' 0.0168z~ 0 . 0 0 4 2 ~ + 0.0001z~ ~ 0.0007~ 3.3431~I 5 . 0 1 5 0 ~  ~ 4.215322 + 2 . 0 7 0 3 ~  0.5593zr5 O.0646zr6 ~ The magnitude of the frequency response in decibels is plotted in the following figure.
CHAP. 91
FILTER DESIGN
at and , As a final check on the design, evaluating H ( e J U ) w = 0 . 2 ~ w = 0 . 3 ~we find that
Therefore, the filter exceeds the passband specifications and comes close to meeting the stopband specifications. If this filter is unacceptable, the design could be modified by decreasing Q, to improve the stopband performance. Repeat Prob. 9.39 using the bilinear transformation.
Using the bilinear transformation to design a Butterworth filter according to the specifications given in Prob. 9.39, we first use the transformation to map the passband and stopband frequencies of the digital filter to the cutoff frequencies of the analog filter. With Ts = 2, we have = tan(0. I n ) == 0.3249 R, = tan 2 = t a n ( 0 . 1 5 ~L 0.5095 ) As we found in Prob. 9.39, the required filter order IS N = 6. For the 3dB cutoff frequency of the analog Butterworth filter, we may choose any frequency in the range 0.3667 5 R,. 5 0.39 10 If we select Q,. =0.3667, the passband specifications will be met exactly, and the stopband specifications will be exceeded. Conversely. if we set R, = 0.3910. the stopband specifications will be met exactly, and the passband specifications exceeded. Picking a frequency between the two extremes will produce an improvement in both bands. Because the stopband deviation is twice that of the deviation in the passband. we will set 0,.= 0.3667 in order to improve the stopband performance. From Table 94, we find the coel'ticients in the system function of a sixthorder normalized (52, = I) Butterworth filter to be To obtain a Butterworth filter with a cutofl' frequency R, = 0.3666, we perform the lowpasstolowpass transformation r FILTER DESIGN This gives Hob) = [CHAP. 9
+ 1.4 165s5+ 1 .0033sJ + 0.4505s3 + 0. 1349s2+ 0.0256s + 0.0024
(0.3666)' Finally, we apply the bilinear transformation I  z' I 2' which yields the digital filter H (z) = 0 . 0 0 9 0 ~  ~ 0.0120~' 0 . 0 0 9 0 ~  ~ 0 . 0 0 3 6 ~  ~ 0.0006~' 0.0006 0.0036z' I  3 . 2 9 4 2 ~ ' 4.8985zr2  4.0857zr3 1 . 9 9 3 2 ~  ~0 . 5 3 5 3 ~ + 0 . 0 6 1 5 ~  ~ ~ At the passband cutoff frequency, o, = 0.217, the magnitude of the frequency response is
and at the stopband cutoff frequency, o,= 0.317, the magnitude of the frequency response is
Therefore, this filter meets the given specifications.
Use the bilinear transformation to design a firstorder lowpass Butterworth filter that has a 3dB cutoff frequency w,. = 0 . 2 ~ . If a digital lowpass filter is to have a 3dB cutoff frequency at w,. = 0.217, the analog Butterworth filter should have a 3dB cutoff frequency Q. = tan($) = tan(0. I n ) = 0.3249 For a firstorder Butterworth filter.
Therefore, the system function is H0(s) = s Q,. With Q, = 0.3249, applying the bilinear transformation
we have
H (z) = 0.3249 0.3249(1 + ZI) 0.2452(1 + zI) 1  z' 1  0.5095~I (1  2I) O.3249(1 zI) 0.3249 1 z' A secondorder continuoustime filter has a system function
H,(s) = + sa sb
where a < 0 and b
0 are real.
( a ) Determine the locations of the poles and zeros of H ( z ) if the filter is designed using the bilinear transformation with T, = 2. (b) Repeat part (a)for the impulse invariance technique, again with T, = 2.

