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equation 630: H (j ) = = 1+ = 1+ 1 1 Vo j A (j ) = Vi (j / 1 + 1)(j / 2 + 1) A ej 90
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ej arctan( / 1 ) ej arctan( / 2 ) ej [90
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(631)
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Equation 631 is of the form H (j ) = |H |ej H , with |H (j )| = 1+ and H (j ) = 90 arctan 1 arctan 2 1 A
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(632) 1+ 2
The magnitude and phase plots for the frequency response of the band-pass lter of Figure 619 are shown in Figure 620 These plots have been normalized to have the lter pass-band centered at the frequency = 1 rad/s The frequency response plots of Figure 620 suggest that, in some sense, the band-pass lter acts as a combination of a high-pass and a low-pass lter As illustrated in the previous cases, it should be evident that one can adjust the lter response as desired simply by selecting appropriate values for L, C, and R The expression for the frequency response of a second-order band-pass lter (equation 629) can also be rearranged to illustrate two important features of this circuit: the quality factor, Q, and the resonant frequency, 0 Let 1 0 = LC Then we can write CR = 0 CR =Q 0 0 0 and Q = 0 CR = R 0 L (633)
and rearrange equation 629 as follows: Vo (j ) = Vi jQ j 0
+ jQ +1 0
(634)
In equation 634, the resonant frequency, 0 , corresponds to the center frequency of the lter, while Q, the quality factor, indicates the sharpness of the resonance,
6
Frequency Response and System Concepts
Band-pass filter amplitude response 1 08 Amplitude 06 04 02 0 _ 10 3
10 1 100 101 Radian frequency (logarithmic scale) Band-pass filter phase response
50 Phase, degrees
0 _50
_3 _2
10 1 100 101 Radian frequency (logarithmic scale)
Figure 620 Frequency response of RLC band-pass lter
that is, how narrow or wide the shape of the pass-band of the lter is The width of the pass-band is also referred to as the bandwidth, and it can easily be shown that the bandwidth of the lter is given by the expression B= 0 Q (635)
Thus, a high-Q lter has a narrow bandwidth, while a low-Q lter has a large bandwidth and is therefore less selective The quality factor of a lter provides an immediate indication of the nature of the lter The following examples illustrate the signi cance of these parameters in the response of various RLC lters
EXAMPLE 66 Frequency Response of Band-Pass Filter
Problem
Compute the frequency response of the band-pass lter of Figure 619 for two sets of component values
Multisim
Solution
Known Quantities:
(a) R = 1 k ; C = 10 F; L = 5 mH (b) R = 10 ; C = 10 F; L = 5 mH
Find: The frequency response, HV (j )
Part I
Circuits
Assumptions: None Analysis: We write the frequency response of the band-pass lter as in equation 629:
HV (j ) = =
Vo j CR (j ) = Vi 1 + j CR + (j )2 LC CR 1 2 LC
+ ( CR)2
CR arctan 2 1 2 LC
We can now evaluate the response for two different values of the series resistance The frequency response plots for case a (large series resistance) are shown in Figure 621 Those for case b (small series resistance) are shown in Figure 622 Let us calculate some quantities for each case Since L and C are the same in both cases, the resonant frequency of the two circuits will be the same: 1 0 = = 447 103 rad/s LC On the other hand, the quality factor, Q, will be substantially different: Qa = 0 CR 045 Qb = 0 CR 45 case a case b
From these values of Q we can calculate the approximate bandwidth of the two lters: 0 10,000 rad/s case a Ba = Qa 0 100 rad/s case b Bb = Qb The frequency response plots in Figures 621 and 622 con rm these observations
Broad-band filter amplitude response 1 08 Amplitude 06 04 02 0 101 102 103 104 105 Radian frequency (logarithmic scale) Broad-band filter phase response 106 107
50 Phase, degrees
0 _50
103 104 105 Radian frequency (logarithmic scale)
Figure 621 Frequency response of broad-band band-pass lter of Example 66
6
Frequency Response and System Concepts
Narrow-band filter amplitude response 1 08 Amplitude 06 04 02 0 101 102 103 104 105 Radian frequency (logarithmic scale) Narrow-band filter phase response 106 107
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