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Filters in .NET
Filters Make Code 128 In .NET Using Barcode creation for VS .NET Control to generate, create Code 128C image in .NET applications. Decoding Code 128 In VS .NET Using Barcode reader for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications. Characterizing the plot of H ( j ) for a lter is so important that we revisit it here Let s consider some examples Bar Code Creation In Visual Studio .NET Using Barcode creator for Visual Studio .NET Control to generate, create barcode image in Visual Studio .NET applications. Barcode Recognizer In .NET Framework Using Barcode reader for VS .NET Control to read, scan read, scan image in .NET applications. Bode Plots and Butterworth Filters
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So we say that this transfer function has a simple zero at s We can see that this is not a bandpass lter because as s 0, H (s) 1/4 Now let s j Then H ( j ) = Hence H ( j ) = 16 2 1 + 2 = 2 + 16)2 2 ( ( + 16) 2 + 16 1 1 = 4 + j 4 + j 4 j 4 j = 4 j 2 + 16 Let s plot this The plot is shown in Fig 156, from which it s clear that this circuit will function as a lowpass lter Butterworth Filters
A Butterworth lowpass lter has a transfer function of the form H ( ) = 1+ 1 c
(1510) Bode Plots and Butterworth Filters
The number of inductors and capacitors in the circuit used to construct the lter is given by n We call nthe order of the Butterworth lter By increasing n, we can get closer to an ideal lowpass lter with a sharp cutoff This is because lim 2 (1511) when > c Then lim H ( ) = lim 1 1+ c
1 1+ Hence, high frequencies are not passed through the lter On the other hand, when < c lim 2
Therefore lim H ( ) = lim 1 1+ c
1 = 1 1+0 So for frequencies below the cutoff, the transmission is perfect, as if the lter was described by a transfer function given by a unit step with cutoff at c A Butterworth lter of order 1 has a transfer function given by H (s) = Since H ( j ) = 1 1 + 2 1 s+1 Here we are implicitly setting the cutoff frequency to 1
Circuit Analysis Demysti ed
H( ) (rad/sec) Fig 157 A plot of a rstorder Butterworth lter
Let s verify the improved performance of a Butterworth lter as n gets larger We start with the rstorder lter, showing a plot in Fig 157 This lter is far from the ideal case; the cutoff drops gradually rather than sharply Now let s let n = 4 In this case the transfer function is given by H ( ) = 1 1 + 8 The characteristics of the lter are signi cantly better already The response of the lter drops quickly over a small frequency range, and in practice it might be enough for many purposes A plot of a fourthorder Butterworth lter is shown in Fig 158 As n gets even larger, the behavior of the lter quickly approaches the ideal case In Fig 159, we show a plot for a twentiethorder Butterworth lter This is an essentially ideal lowpass lter, with a transfer function that rapidly drops to zero at the cutoff frequency c = 1 It is often preferred to have a lowpass lter with a cutoff of a desired sharpness That is, we may specify how rapidly the magnitude of the transfer function drops off This can be speci ed by choosing the order n of the Butterworth lter To determine the order of a Butterworth lter we begin by considering the limit of the transfer function at high frequency Looking at (1510) and letting c = 1 without loss of generality, notice that as gets large we can ignore the 1 in the H( ) Bode Plots and Butterworth Filters
(rad/sec) Fig 158 A plot of a fourthorder Butterworth lter
H( ) (rad/sec) Fig 159 A twentiethorder Butterworth lter is essentially an ideal lowpass lter
Circuit Analysis Demysti ed
denominator Considering the gain function, then we have lim 20 log10 1 2n = 20n log10
Remember that a decade is de ned by using (154), so this tells us that a Butterworth lter has a dropoff or attenuation of 20n dB/decade Or the attenuation can be described as 6n dB/octave (1513) (1512) EXAMPLE 156 A lowpass lter is to be designed with the following characteristic There must be an attenuation of 390 dB at the frequency given by = 20 c Find the required order for the circuit and write down the magnitude of the transfer function SOLUTION Since = 20 c this tells us that we are two decades past the critical frequency With an attenuation of 20n dB/decade, the order of our circuit must satisfy 40n 390 That is n 975 The order of a Butterworth lter is an integer, so we choose the smallest integer satisfying this inequality, n = 10 The transfer function is given by 1 H ( ) = 1 + 20 EXAMPLE 157 A lowpass lter is to be designed with the following characteristic There must be an attenuation of 80 dB at the frequency given by = 6 c Find the required order for the circuit and write down the magnitude of the transfer function

