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Characterizing the plot of |H ( j )| for a lter is so important that we revisit it here Let s consider some examples
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Bode Plots and Butterworth Filters
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Fig 15-5 A Bode plot of the transfer function of Example 15-4
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EXAMPLE 15-5 A lter has a transfer function given by H (s) = 1 s+4
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Plot its magnitude versus frequency What type of lter does this represent SOLUTION First, looking at the function notice that the function has a simple pole at s = 4 since 1 = s 4 s + 4 lim Also notice that 1 =0 s s + 4 lim
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0 1 5
0 0 5
(rad/sec)
Fig 15-6 A plot of the magnitude of H (s) =
1 s+4
So we say that this transfer function has a simple zero at s We can see that this is not a band-pass 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 15-6, from which it s clear that this circuit will function as a low-pass lter
Butterworth Filters
A Butterworth low-pass 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 low-pass 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
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H( )
(rad/sec)
Fig 15-7 A plot of a rst-order Butterworth lter
Let s verify the improved performance of a Butterworth lter as n gets larger We start with the rst-order lter, showing a plot in Fig 15-7 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 fourth-order Butterworth lter is shown in Fig 15-8 As n gets even larger, the behavior of the lter quickly approaches the ideal case In Fig 15-9, we show a plot for a twentieth-order Butterworth lter This is an essentially ideal low-pass lter, with a transfer function that rapidly drops to zero at the cutoff frequency c = 1 It is often preferred to have a low-pass 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 15-8 A plot of a fourth-order Butterworth lter
H( )
(rad/sec)
Fig 15-9 A twentieth-order Butterworth lter is essentially an ideal low-pass 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 15-6 A low-pass 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 15-7 A low-pass 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
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