vb.net barcode reader code Figure 616 High-pass lter in Software

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Figure 616 High-pass lter
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Thus, the frequency response of the lter is: Vo j CR (j ) = Vi 1 + j CR which can be expressed in magnitude-and-phase form by H (j ) = = or H (j ) = |H |ej H with H (j ) = CR 1 + ( CR)2
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(625)
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Vo j CR = (j ) = Vi 1 + j CR CR 1 + ( CR)2 ej (90
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1 + ( CR)2 ej arctan( CR/1)
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H (j ) = 90 arctan( CR) You can verify by inspection that the amplitude response of the high-pass lter will be zero at = 0 and will asymptotically approach 1 as approaches in nity, while the phase shift is 90 at = 0 and tends to zero for increasing Amplitudeand-phase response curves for the high-pass lter are shown in Figure 617 These plots have been normalized to have the lter cutoff frequency 0 = 1 rad/s Note that, once again, it is possible to de ne a cutoff frequency at 0 = 1/RC in the same way as was done for the low-pass lter
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1 Phase, degrees 10 1 10 0 101 102 103 Radian frequency (logarithmic scale) 104 08 Amplitude 06 04 02 0 10 2 80 60 40 20 0 10 2 10 1 10 0 101 102 103 Radian frequency (logarithmic scale) 104
Figure 617 Frequency response of a high-pass lter
Part I
Circuits
EXAMPLE 65 Frequency Response of RC High-Pass Filter
Problem
Compute the response of the RC lter in the circuit of Figure 616 Evaluate the response of the lter at = 2 100 and 2 10,000 rad/s
Solution
Known Quantities: R = 200
; C = 0199 F
Find: The frequency response, HV (j ) Assumptions: None Analysis: We rst recognize that the cutoff frequency of the high-pass lter is 0 = 1/RC = 2 4,000 rad/s Next, we write the frequency response as in equation 625: Vo j CR 0 HV (j ) = = arctan (j ) = Vi 1 + j CR 2 0 2 1+ 0
We can now evaluate the response at the two frequencies: HV ( = 2 100) = 100 4000 1+ 100 4000
100 arctan 2 4000
= 0025 1546
HV ( = 2 10,000) =
10,000 4000 1+ 10,000 4000
10,000 arctan 2 4000
= 0929 038
The frequency response plots are shown in Figure 618
1 Phase, degrees 08 Amplitude 06 04 02 0 101 10 2 103 104 105 106 Radian frequency (logarithmic scale) 107 80 60 40 20 0 101 10 2 103 104 105 106 Radian frequency (logarithmic scale) 107
Figure 618 Response of high-pass lter of Example 65
6
Frequency Response and System Concepts
Comments: The effect of this high-pass lter is to preserve the amplitude of the input signal at frequencies substantially greater than 0 , while signals at frequencies below 0 would be strongly attenuated With 0 = 2 4,000 (ie, 4,000 Hz), this lter would pass only the treble range of the audio frequency spectrum
Band-Pass Filters Building on the principles developed in the preceding sections, we can also construct a circuit that acts as a band-pass lter, passing mainly those frequencies within a certain frequency range The analysis of a simple second-order band-pass lter (ie, a lter with two energy-storage elements) can be conducted by analogy with the preceding discussions of the low-pass and high-pass lters Consider the circuit shown in Figure 619, and the related frequency response function for the lter H (j ) = Noting that Vo (j ) = Vi (j ) R R + 1/j C + j L Vo (j ) Vi
RLC band-pass filter The circuit preserves frequencies within a band C L + + Vi _ R Vo _
Figure 619 RLC band-pass lter
j CR = Vi (j ) 1 + j CR + (j )2 LC we may write the frequency response of the lter as Vo j CR (j ) = Vi 1 + j CR + (j )2 LC Equation 629 can often be factored into the following form: Vo j A (j ) = Vi (j / 1 + 1)(j / 2 + 1)
(628)
(629)
(630)
where 1 and 2 are the two frequencies that determine the pass-band (or bandwidth) of the lter that is, the frequency range over which the lter passes the input signal and A is a constant that results from the factoring An immediate observation we can make is that if the signal frequency, , is zero, the response of the lter is equal to zero, since at = 0 the impedance of the capacitor, 1/j C, becomes in nite Thus, the capacitor acts as an open circuit, and the output voltage equals zero Further, we note that the lter output in response to an input signal at sinusoidal frequency approaching in nity is again equal to zero This result can be veri ed by considering that as approaches in nity, the impedance of the inductor becomes in nite, that is, an open circuit Thus, the lter cannot pass signals at very high frequencies In an intermediate band of frequencies, the band-pass lter circuit will provide a variable attenuation of the input signal, dependent on the frequency of the excitation This may be veri ed by taking a closer look at
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